derived category methods in commutative algebralchriste/download/dcmca.pdf · practitioner of...

Lars Winther ChristensenHans-Bjørn FoxbyHenrik Holm

Derived Category Methodsin Commutative Algebra

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Preface

Homological algebra originated in late 19th century topology. Homological studiesof algebraic objects, such as rings and modules, only got under way in the middle ofthe 20th century—Cartan and Eilenberg’s classic text “Homological Algebra” servesas a historic marker. The utility of homological methods in commutative algebra wasfirmly established in the mid 1950s through the proofs of Krull’s conjectures on reg-ular local rings, which were achieved, independently, by Auslander and Buchsbaumand by Serre.

The technically more involved methods of derived categories came into com-mutative algebra through the work of Grothendieck and his school. This happenedonly some ten years after the initial successes of classic homological algebra; anearly, and still central, text is Hartshorne’s exposition “Residues and Duality” ofGrothendieck’s 1963–64 seminar at Harvard. Applications of derived categories incommutative algebra have grown steadily, albeit slowly, since the mid 1960s. Onereason for the modest pace has, quite likely, been the absence of a coherent intro-duction to the topic. There are several excellent textbooks from which one can learnabout classic homological algebra and its applications, but to become an efficientpractitioner of derived category methods in commutative algebra one must be well-versed in a train of research articles and lecture notes, including unpublished ones.

With this book, we aim to provide an accessible and coherent introduction to de-rived category methods—in the past known as hyperhomological algebra—and theirapplications in commutative algebra. We want to make the case that these methods,compared to those of classic homological algebra, provide broader and stronger re-sults with cleaner proofs—this to an extent that outweighs the effort it requires tomaster them. Moreover, there are important results in commutative algebra whosenatural habitat is the derived category. The Local Duality Theorem, for example, hasan elegant formulation in the derived category, but only for a limited class of ringsdoes it have a satisfactory one within classic homological algebra.

The book is intended to double as a graduate textbook and a work of reference.It is organized into three parts “Foundations”, “Tools”, and “Applications”. In thefirst part, we introduce the fundamental homological machinery and construct thederived category over a ring. The second part continues with a systematic studyof functors and invariants of utility in ring theory. In the third part we assemble

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xii Preface

textbook applications of derived category methods in commutative ring theory intoa short treatise on homological commutative algebra.

The organization of the book serves several purposes. Readers familiar with de-rived categories may skip “Foundations”. The tools are kept separate from theirapplications in order to develop them in higher generality; this should make “Foun-dations” and “Tools” useful, not only to commutative algebraists, but also to readersfrom neighboring fields. The third part “Applications” is intended to be a high-levelprimer to homological aspects of commutative algebra.

We have learned the material in this book from our teachers, our collaborators, andother colleagues—we hope that they will find that we have done it justice.

The forerunners of this book are two collections of lecture notes by H.-B.F.,that were used at the University of Copenhagen from 1982. Indeed, L.W.C. andH.H. were first introduced to the subject through these notes, and through the notes“Differential Graded Homological Algebra" by Avramov, H.-B.F., and Halperin.

Preliminary versions of this book have since 2006 been used in graduate classesand lectures at Beijing Normal University, Texas Tech University, and at the Uni-versity of Nebraska-Lincoln. We thank the many students who gave their commentson these early versions of the manuscript.

Great strides were made on the book during sojourns at Banff International ResearchStation, Centre Internationale de Rencontres Mathématiques (Luminy), and Mathe-matisches Forschungsinstitut Oberwolfach, as well as during L.W.C.’s visits to theUniversity of Copenhagen and H.H.’s visits to Texas Tech University; the supportand hospitality of these institutions is greatly appreciated.

At various stages of the writing, the authors have been supported by grants fromthe Carlsberg Foundation, the Danish Confederation of Professional Associations(AC), the Danish Council for Independent Research, the (U.S.) National SecurityAgency, and the Simons Foundation.

Lubbock and Copenhagen, L.W.C.August 2017 H.-B.F.

H.H.


Contents

Introduction xviiTheme and Goal xviiContents and Organization xviiiExposition xxReader’s Guide xxiTeacher’s Guide xxii

Conventions and Notation xxiii

Part I Foundations

1 Modules 31.1 Prerequisites 3

Five Lemma; Snake Lemma; Hom; tensor product; linear category; linear functor; biproduct;product; coproduct; direct sum; bimodule; split exact sequence; (half/left/right) exact functor;faithful functor.

1.2 Standard Isomorphisms 14Unitor; counitor; commutativity; associativity; swap; adjunction.

1.3 Exact Functors and Classes of Modules 18Basis; free module; unique extension property; finite generation of Hom and tensor product;projective module; injective module; lifting properties; semi-simple ring; Baer’s criterion;finitely presented module; flat module; von Neumann regular ring.

1.4 Evaluation Homomorphisms 32Biduality; tensor evaluation; homomorphism evaluation.

2 Complexes 392.1 Definitions and Examples 39

Graded module; graded homomorphism; complex; chain map; Five Lemma; Snake Lemma;(degreewise) split exact sequence; \-functor.

2.2 Homology 52Shift; homology; connecting morphism; homotopy.

2.3 Homomorphisms 62Hom complex; chain map; homotopy; Hom functor; exactness.

2.4 Tensor Products 68Tensor product complex; chain map, homotopy; tensor product functor; exactness.

2.5 Boundedness and Finiteness 74Degreewise finite generation/presentation; boundedness (above/below); supremum; infimum;amplitude; boundedness of Hom complex; boundedness of tensor product complex; hard/softtruncation; graded basis; graded-free module.

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xiv Contents

3 Categorical Constructions 833.1 Products and Coproducts 83

Coproduct; product; universal properties; direct sum; functor that preserves (co)products.

3.2 Colimits 93Direct system; colimit; universal property; functor that preserves colimits; pushout; filteredcolimit; telescope.

3.3 Limits 107Inverse system; limit; universal property; functor that preserves limits; pullback; tower;Mittag-Leffler Condition.

4 Equivalences and Isomorphisms 1254.1 Mapping Cone 125

Mapping cone (sequence); homotopy; Σ-functor.

4.2 Quasi-isomorphisms 133Quasi-isomorphism; functor that preserves quasi-isomorphisms; semi-simple module.

4.3 Homotopy Equivalences 138Homotopy equivalence; functor that preserves homotopy; contractible complex; mappingcylinder (sequence).

4.4 Standard Isomorphisms for Complexes 147Unitor; counitor; commutativity; associativity; swap; adjunction.

4.5 Evaluation Morphisms for Complexes 154Biduality; tensor evaluation; homomorphism evaluation.

5 Resolutions 1655.1 Semi-freeness 165

Semi-basis; semi-free complex; semi-free resolution; free resolution of module.

5.2 Semi-projectivity 170Graded-projective module; complex of projective modules; semi-projective complex;semi-projective resolution; lifting property; projective resolution of module.

5.3 Semi-injectivity 178Character complex; graded-injective module, complex of injective modules; semi-injectivecomplex; semi-injective resolution; lifting property; injective resolution of module.

5.4 Semi-flatness 190Graded-flat module; complex of flat modules; semi-flat complex; perfect ring.

6 The Derived Category 1996.1 Construction of the Homotopy Category K 199

Homotopy category; quasi-isomorphism; product; coproduct; homotopy invariant functor;universal property; functor that preserves homotopy.

6.2 Triangulation of K 207Strict triangle; distinguished triangle; quasi-triangulated functor; universal property; homology.

6.3 Resolutions 218Unique lifting property; functoriality of resolution.

6.4 Construction of the Derived Category D 225Fraction; derived category; product; coproduct; universal property.

6.5 Triangulation of D 241Distinguished triangle; triangulated functor; universal property; distinguished triangle fromshort exact sequence; homology; Five Lemma, Horseshoe Lemma.

Part II Tools


Contents xv

7 Derived Functors 2557.1 Induced Functors on the Homotopy Category 255

Functor that preserves homotopy; Hom functor; tensor product functor; unitor; counitor;commutativity; associativity; swap; adjunction; biduality; tensor evaluation; homomorphismevaluation.

7.2 Induced Functors on the Derived Category 260Functor that preserves quasi-isomorphisms; universal property of left derived functor; universalproperty of right derived functor.

7.3 Right Derived Hom Functor 267RHom; Ext; exact Ext sequence; complexes of bimodules.

7.4 Left Derived Tensor Product Functor 275⊗L; Tor; exact Tor sequence; complexes of bimodules.

7.5 Standard Isomorphisms in the Derived Category 281Unitor; counitor; commutativity; associativity; swap; adjunction.

7.6 Boundedness and Finiteness 288Bounded derived category; preservation of boundedness and finiteness by RHom and ⊗L.

8 Homological Dimensions 2938.1 Projective Dimension 293

Vanishing of Ext; semi-projective replacement; projective dimension; Schanuel’s lemma;projective dimension over Noetherian ring; ∼ over semi-perfect ring; ∼ over perfect ring; ∼ ofmodule.

8.2 Injective Dimension 301Vanishing of Ext; semi-injective replacement; injective dimension; Schanuel’s lemma; minimalsemi-injective resolution; injective dimension over Artinian ring; ∼ of module; coproduct ofinjective modules.

8.3 Flat Dimension 308Vanishing of Tor; semi-flat replacement, flat dimension; ∼ vs. injective dimension; Schanuel’slemma; flat dimension over Noetherian ring; ∼ over perfect ring; ∼ of module; product of flatmodules.

8.4 Evaluation Morphisms in the Derived Category 318Biduality; tensor evaluation; homomorphism evaluation.

8.5 Global and Finitistic Dimensions 322(Weak) global dimension, ∼ of Noetherian ring, finitistic projective/injective/flat dimension, ∼of perfect ring, ∼ of Noetherian ring.

9 Dualizing Complexes 3299.1 Grothendieck Duality 329

Homothety formation; dualizing complex; Grothendieck Duality.

9.2 Morita Equivalence 337Invertible complex; (co)unit of adjunction; Morita Equivalence; inverse complex.

9.3 Foxby Equivalence 342Auslander Category; Bass Category; Foxby Equivalence; projective dimension of flat modules;parametrization of dualizing complexes; invertible complex.

10 Gorenstein Homological Dimensions 35110.1 Gorenstein Projective Dimension 351

Totally acyclic complex of (finitely generated) projective modules; Gorenstein projectivemodule; Gorenstein projective dimension; ∼ is refinement of projective dimension; ∼ overNoetherian ring; ∼ of module.

10.2 Gorenstein Injective Dimension 367Totally acyclic complex of injective modules; Gorenstein injective module; Gorensteininjective dimension; ∼ is refinement of injective dimension; ∼ of module.


xvi Contents

10.3 Gorenstein Flat Dimension 378Totally acyclic complex of flat modules; Gorenstein flat module; Gorenstein flat dimension; ∼vs. Gorenstein injective dimension; ∼ is refinement of flat dimension; ∼ vs. Gorensteinprojective dimension; ∼ over Noetherian ring; ∼ of module.

10.4 Gorenstein Dimensions and Foxby Equivalence 390Gorenstein projective/flat dimension and the Auslander category; Gorenstein injectivedimension and the Bass category; derived reflexive complex; Iwanaga-Gorenstein ring.

11 Derived Torsion and Completion 40111.1 Torsion and Completion 401

a-torsion functor; a-torsion subcomplex; a-adic completion functor; torsion(-free) module; flatmodule.

11.2 Right Derived Torsion Functor 407RΓa; local cohomology; derived a-torsion complex; Koszul complex; Cech complex; weaklyproregular sequence.

11.3 Left Derived Completion Functor 415LΛa; local homology; derived a-adically complete complex; semi-free resolution of Cechcomplex.

11.4 Greenlees–May Equivalence 422Idempotence of RΓa and LΛa; Greenlees–May Equivalence.

Appendices

A Vanishing and Boundedness of Functors 429Vanishing of functor on module category; acyclicity of Hom complex; acyclicity of tensor productcomplex; way-out functor.

B Minimality 439Minimal complex; injective envelope; minimal semi-injective resolution; Nakayama’s lemma;projective cover; minimal semi-projective resolution; semi-perfect ring; perfect ring.

C Structure of Semi-flat Complexes 461Degreewise finitely presented complex; Govorov and Lazard’s theorem; categorically flat complex.

D Triangulated Categories 475Axioms for triangulated category; triangulated functor; triangulated subcategory; Five Lemma.

E Projective Dimension of Flat Modules 485Jensen’s theorem; Homotopy Lemma; pure exact sequence; ℵ-generated module; countably related flatmodule; continuous chain of modules; transfinite extension of projective modules; flat preenvelopes.

F Structure of Injective Modules 501Hom-vanishing Lemma; faithfully injective module; structure of injective module over Noetherian ring;injective precover; indecomposable injective module; endomorphism of ∼.

Bibliography 509

Glossary 513

List of Symbols 519

Index 523


Introduction

The year 1956 saw the first major applications of homological algebra in commuta-tive ring theory. In seminal papers by Auslander and Buchsbaum [2] and Serre [76],homological algebra was used to characterize regular local rings in a way thatopened to proofs of two conjectures by Krull [48, 49]. This breakthrough right awayestablished homological algebra as a powerful research tool in commutative ringtheory and, in the words of Kaplansky [46], “marked a turning point of the sub-ject.” It coincided with the appearance of the magnum opus “Homological Algebra”[16], by Cartan and Eilenberg, which made the tools of classic homological algebrabroadly available. The present book deals with the more advanced methods of de-rived categories and their applications in commutative algebra; applications to otherareas of mathematics are beyond the scope of the book.

Theme and Goal

Under the heading “Hyperhomology”, a predecessor to the framework of derivedcategories was briefly treated by Cartan and Eilenberg in the final chapter of [16].Yet, it was the work of Grothendieck—in particular, Hartshorne’s notes “Residuesand Duality” [35] published in 1966—that truly brought derived category meth-ods into commutative algebra. In the 1970s, works by Iversen [39] and Roberts[72] emphasized the utility of these methods, and since then their importance hasgrown steadily.

THE THEME OF THIS BOOK, as indicated in the Preface, is that derived categorymethods have significant advantages over those of classic homological algebra,which they extend. To facilitate a discussion of their similarities and differences,we sketch elements of the two pieces of machinery in the algebraic context.

Classic homological algebra is used to study modules through the behavior—notably vanishing—of derived functors on modules. Let F : M(R)→M(R) be anadditive functor on the category of modules over a ring R. The value of the nth leftderived functor Ln F on an R-module M is computed follows:

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xviii Contents and Organization

(1) Choose any free resolution L• of M(2) Apply the functor F to this resolution to get a complex F(L•) of R-modules(3) By definition, Ln F(M) is the homology module Hn(F(L•))

This procedure determines Ln F(M) up to isomorphism in M(R), and a similar onefor homomorphisms establishes Ln F as a functor. The purpose of step (3) is to returnan object in M(R); this is not only aesthetically pleasing but also useful, as it allowsthe procedure to be iterated. However, the honest output is the complex F(L•).

Derived category methods are used to study R-modules and complexes of suchthrough the behavior—notably boundedness—of derived functors on complexes.Let F: C(R)→ C(R) be an additive functor on the category of complexes of R-modules. The value of the left derived functor LF on a complex M• of R-modules iscomputed as follows:

(1) Choose any semi-free resolution L• of the complex M•

(2) By definition, LF(M•) is the complex F(L•) of R-modulesThis procedure only determines LF(M•) up to homology isomorphism in C(R).Thus, the homology H(LF(M•)) is a well-defined object in C(R), but LF(M•) itself isnot. This is overcome by passage to the derived category D(R), which has the sameobjects as C(R) but more morphisms—enough to make homology isomorphisms in-vertible. In this category, LF(M•) is unique up to isomorphism, and LF is a functor.The motivation for focusing on the object LF(M•) over H(LF(M•)) is evident: theformer contains more information than the latter as one cannot, in general, retrievea complex from its homology.

A module M can be viewed as a complex, and in that perspective H(LF(M)) isnothing but the assembly of all the modules Ln F(M). The theme of the book, ina nutshell, is that working with the functors LF on D(R), as opposed to Ln F onM(R), yields broader and stronger results, even for modules. An early example ofthe case in point is Roberts’ “No Holes Theorem” [72] in local algebra, which gavenew insight into the structure of injective resolutions of modules. Another examplecomes from algebraic geometry. Grothendieck’s localization problem for flat localhomomorphisms was stated in EGA [34], but a solution was only obtained 30 yearslater, when derived category methods were applied to study a wider class of ringhomomorphisms [6].

THE GOAL OF THIS BOOK is to expound the utility of derived category methods inring theory. For commutative rings, the goal is further to collect textbook applica-tions of these methods into a compact, yet comprehensive, introduction to homolo-gical aspects of commutative algebra. The book is intended to serve as a graduatetextbook as well as a work of reference for professional mathematicians.

Contents and Organization

The facets of the goal, as described above, are reflected in the organization of thebook. First we build the framework, that is, the derived category over a ring. Within


Contents and Organization xix

this framework, we develop a set of efficacious tools, such as derived functors, in-despensable isomorphisms, categorical equivalences, and homological invariants.Finally, we apply these tools to prove classic and modern results in commutativealgebra. The body of the book hence consists of three parts.

PART I “FOUNDATIONS” presents the primary machinery of homological algebra.Chapter 1 is focused on the category of modules over a ring. For convenience, itsfirst section recounts a few basic notions and results that we expect the reader tobe familiar with, such as bimodule structures and diagram lemmas. Chapters 2–4treat the category of complexes of modules; the treatment includes categorical con-structions, important functors, and special types of morphisms. Resolutions play akey role in homological algebra; they are the topic of Chap. 5. The derived category,which provides the framework for the rest of the book, is constructed in Chap. 6.

PART II “TOOLS” is devoted to homological invariants of complexes and studiesof special complexes, morphisms, and functors in the derived category. Chapter 7covers the derived Hom and tensor product functors; together with the standard iso-morphisms that link them, these functors are key players in the rest of the book.Homological dimensions are developed in Chaps. 8 and 10, and they are tied in tothe so-called evaluation morphisms. Together with the standard isomorphisms fromChap. 7, the evaluation morphisms play a central role in our approach to the mate-rial in Part III. Another powerful technical tool, dualizing complexes, is presentedin Chap. 9, where also Grothendieck Duality, Morita Equivalence, and Foxby Equi-valence are treated. Local homology and cohomology over commutative rings isdeveloped in Chap. 11 via Koszul and Cech complexes.

PART III “APPLICATIONS” focuses on the homological theory of commutativeNoetherian rings. It opens with a “Brief for Commutative Ring Theorists” (Chap.??),which recapitulates central results from Part II in the simpler form they take in thesetting of commutative Noetherian rings. In the remaining chapters, we refer to thestatements in Chap. ??; as such, Part III is essentially self-contained.

Chapter ?? extends standard notions from commutative algebra, such as supportand Krull dimension, to objects in the derived category. With these notions in place,the tools from Part II are applied to prove time-honored and recent theorems on com-mutative rings and their modules. Chapter ?? treats numerical invariants of modulesover local rings, and Grothendieck’s Local Duality Theorem is proved in Chap. ??.Chapters ?? and ?? focus on regular rings and on Gorenstein rings, respectively.Intersection theorems and their applications to the study of Cohen–Macaulay ringsare the topic of Chap. ??. The final two chapters, ??–??, cover the theory of localring homomorphisms and modules over such.

THE CHOICE OF TOPICS is detailed in the synopses that open each section; they arealso embedded in the table of contents.

A few topics that one might expect to find in this book, spectral sequences forexample, are absent. The simplistic reason is that we manage without them. Forexample, standard arguments based on collapsing spectral sequences coming from


xx Exposition

double complexes are naturally replaced by isomorphisms in the derived category.More to the point, our rationale is that, within the scope of this book, we see no wayto improve on existing expositions of these topics, and since we do not need them,we have decided to avoid them all together.

At the same time, the book covers two topics that could be considered non-standard: Gorenstein homological dimensions and local ring homomorphisms. Themotivation for including each of these topics is two-fold. Significant progress hasbeen made within the last two decades and derived category methods have been cru-cial for the successes of this research. The literature is ponderous to penetrate, andthe existing surveys [18, 19] and [3, 5] are not up to date or present no proofs.

The derived category over a ring is a particular example of a triangulated cate-gory. Other such categories, stable module categories for example, play significantroles in contemporary homological algebra, but they are not treated in this book.

Exposition

We have striven to make the book coherent and self-contained. Assuming somebackground knowledge—specified in the Reader’s Guide below—we provide proofsof all results in the book with two exceptions. We use Cohen’s structure theorem forcomplete local rings [20]; its proof is beyond the scope of this book but can befound in [58]. We also use the New Intersection Theorem; it was proved by Peskineand Szpiro [68] for equicharacteristic rings, and by Roberts [73] for rings of mixedcharacteristic. Robert’s detailed proof makes a book of its own [74].

Constructions and results that are required to keep the book self-contained, butmight otherwise disrupt the flow of the material, have been relegated to appendices.

To keep the text accessible to a wide audience, we have resisted temptations toincrease the level of generality beyond what is justified by the goal of the book, evenwhen it would come at low or no cost. For example, in the construction of the ho-motopy category over a ring, one could easily replace the module category with anyother additive category. Similarly, parts of the material on standard isomorphismsand evaluation morphisms could be developed in the more general context of aclosed monoidal category. In the same vein, we emphasize explicit constructionsover axiomatic approaches. For example, we use resolutions to establish certainproperties of the derived category over a ring, though they could be deduced fromformal properties of triangulated categories.

THE MATERIAL in this textbook is not new, and the fact that references are sparsein the main text should not be interpreted as the authors’ subtle claim for credit. Infact, every significant statement in the text is either folklore or can be traced to thepapers and books listed in the Bibliography. Among the texts that have inspired thecontents, we emphasize two that have also influenced the exposition: “Homologi-cal Algebra” [16] by Cartan and Eilenberg and “Differential Graded HomologicalAlgebra” [7] by Avramov, Foxby, and Halperin.


Reader’s Guide xxi

Reader’s Guide

This book is written for researchers and advanced graduate students, so the readershould possess the mathematical maturity of a doctoral student of algebra.

THE PREREQUISITES include familiarity with basic notions from set theory, cate-gory theory, ring theory and, for Part III, commutative algebra. Notice, though, thatwe assume no prerequisites in homological algebra.

Commutative algebraists who want to dive into Part III will find an extract ofessential results from Part II in Chap. ??.

TEXT ELEMENTS that are typeset in small font are auxiliary. They provide com-mentary and perspective to the main text, but they can be read or skipped at will, asthe main text does not depend on them.

THE CHAPTERS of the book depend on each other as depicted below.

Part I 1

...

6

Part II 7

8

11 9

10

Part III 12

13

14

15 16 18 19 17

20

THE CONVENTIONS we employ are explained right after this introduction.

A GLOSSARY is provided towards the end of the book. It lists terms that are usedbut not defined in the text along with their definitions and textbook references. Werefer to Lam’s books “A First Course in Noncommutative Rings” and “Lectureson Modules and Rings” [50, 51] for notions in ring theory, to “Categories for theWorking Mathematician” [54] by MacLane for notions in category theory, to and toMatsumura’s “Commutative Ring Theory” [58] for notions in commutative algebra.

A LIST OF SYMBOLS follows the glossary; it includes most symbols used—andcertainly all those defined—in the book.


xxii Teacher’s Guide

EXERCISES are included at the end of each section. In addition to the purpose thatexercises ordinarily serve in a textbook, they supplement the text in three ways.• Examples in the text are kept as simple as reasonably possible; exercises are

used to develop more interesting ones.• To some results there are parallel or dual versions that are not needed in the

exposition; these are relegated to exercises.• Exercises are used to explore directions beyond the scope of the book; some of

these come with references to the literature to facilitate further study.We emphasize that the main text does not depend on the exercises.

Teacher’s Guide

The nature of the subject leaves a teacher of homological algebra with hard choicesto make about the amount of detail to present in lectures. An over- or underempha-sis on technical details can easily create an irksome feeling that the topic belongsto accounting or religion rather than mathematics. The favored compromise is totreat some constructions and arguments in full detail and leave it to the students toreassure themselves that the balance of the material is also solid. We facilitate thisstyle by supplying a fair and consistent amount of detail; certainly more than onewould attempt to include in lectures. With some restraint, we follow the traditionfor recycling arguments by saying that a proof is “parallel” or “dual” to one that hasalready been presented. We apply this technique when consecutive statements havesimilar proofs, but we avoid the wholesale application that would dispense with thetheory of injective modules in one sentence, “dual to the projective case.”

For pedagogical reasons, we have included several exercises that ask the studentsto perform elementary verifications and computations that are omitted from the text.These exercises are easily recognized as they open with a reference like (Cf. 1.1.1).The details examined in these exercises are ones that a professional mathematiciancan fill in on the fly, so we stand by the claim made right above: the main text doesnot depend on the exercises. Another category of exercises explore simple versionsof constructions and arguments that are treated later in the text; they are not markedin any particular way, but they should be easily recognizable to the teacher.

Here are a few suggestions for how to use the book as a graduate textbook.

A FIRST COURSE IN HOMOLOGICAL ALGEBRA could consist of Chaps. 1–7 andparts of Chap. 8. A sequel course with emphasis on commutative algebra could bebased on Chaps. 8–9, ??–?? and, possibly, the first few sections of Chaps. ??–??.

A COURSE ON DERIVED CATEGORY METHODS IN COMMUTATIVE ALGEBRA foran audience with prior knowledge of classic homological algebra could be based onChaps. 2–7 and 11–?? and, even, selected sections of Chaps. ??–??.

A TOPICS COURSE on Gorenstein dimensions can be built around Chaps. 9–10 and??; one on local ring homomorphisms could pivot on Chaps. ??–??.


Conventions and Notation

Throughout the book, the symbols Q, R, S, and T denote non-zero associative unitalrings. They are assumed to be algebras over a common commutative unital ring k,and homomorphisms between them are tacitly assumed to be k-algebra homomor-phisms. The generic choice of k is the ring Z of integers, but in concrete settingsother choices may be useful. For example, in studies of algebras over a field k, thenatural choice is k= k, and in studies of Artin A-algebras, the commutative Artinianring A is the natural candidate for k. In a different direction, the center of R is apossible choice for k, when one studies ring homomorphisms R→ S. Finally, k= Ris the default choice in studies of a commutative ring R; that is our choice in Part III.

IDEALS in a ring are subsets that are both left ideals and right ideals. Similarly, aring is called Artinian/Noetherian/perfect etc. if it is both left and right Artinian/Noetherian/perfect etc. In Part III, the rings Q, R, S, and T are assumed to be com-mutative and such left/right distinctions conveniently disappear.

MODULES are assumed to be unitary, and by convention the ring acts on the left.That is, an R-module is a left R-module. Right R-modules are, consequently, con-sidered to be modules (i.e. left modules) over Ro, the opposite ring of R. Thus, a leftideal in R is a submodule of the R-module R, while a right ideal in R is a submoduleof the Ro-module R. When needed, R is adorned with a subscript to show whichmodule structure is used; that is, RR is the R-module R and RR is the Ro-module R.

FUNCTORS are by convention covariant. Replacing one category by its opposite,a contravariant functor F: U→ V is hence considered to be a (covariant) functorUop→ V or U→ Vop. For a natural transformation τ : F→ G of functors U→ V wewrite τop : Gop→ Fop for the natural transformation of opposite functors Uop→Vop.

THE NOTATION is either standard or explained in the text, starting here:( 0) 0 sufficiently (small) large

injective map surjective map

(⊂) ⊆ (proper) subset\ difference of sets⊎

disjoint union of sets∼= isomorphism

C complex numbersN (N0) natural numbers (and 0)

Q rational numbersR real numbersZ integers

Id identity functorcard cardinality

xxiii

Part I

Foundations


If I could only understand the beautiful consequencefollowing from the concise proposition ∂2 = 0

Henri Cartan, 1980(translated from Latin)

THE OBJECTIVE of the six chapters that lay ahead is to construct the derived cat-egory of the module category over a ring. In order to not loose track of the overallaim—applications of homological algebra in (commutative) ring theory—we sus-pend work on the construction every now and then to tie up connections to ringtheory as they come within reach. Such connections include homological character-izations of principal ideal domains, semi-simple rings, von Neumann regular rings,perfect rings, and semi-perfect rings—and in exercises, hereditary rings. Most ofthese excursions into ring theory are short, but the homological characterizations of(semi-)perfect rings, including Bass’ Theorem P, are intertwined with the construc-tions of minimal resolutions, which is the topic of the rather lengthy Appn. B.

HOMOLOGICAL ALGEBRA has a reputation for being (tediously) technical, per se.It is somewhat befitting, and we shall make no attempt to hide or trivialize the the-ory’s nature. Rather, we expound it as we adopt the point of view that the fabric ofhomological algebra is technical constructions and statements about them. With thisunapologetic style we aspire to put a transparent hood on the homological machine.


Chapter 1Modules

The collection of all R-modules and all homomorphisms of R-modules forms anAbelian category with set indexed products and coproducts (also, but not here,known as direct sums); it is denoted M(R). We take this for granted, and the firstsection of this chapter sums up the key properties of this category that we will buildon throughout the book.

Although the applications of homological algebra that we ultimately aim for areto modules, it is advantageous to work with the wider notion of complexes. Modulesare simple complexes, or complexes are modules with extra structure, depending onthe point of view. In any event, there is a decision to make about when to leave mod-ules and pass to complexes. We make this transition as soon as we have establisheda baseline of homological module theory that allows us to interpret results aboutcomplexes in the realm of modules. That baseline is established in this first chapter,and the transition to complexes takes place in Chap. 2.

1.1 Prerequisites

SYNOPSIS. Five Lemma; Snake Lemma; Hom; tensor product; linear category; linear functor;biproduct; product; coproduct; direct sum; bimodule; split exact sequence; (half/left/right) exactfunctor; faithful functor.

The primary purpose of this section is to remind the reader of some basic materialand, in that process, to introduce the accompanying symbols and nomenclature. Thematerial in this section is used throughout the book and usually without reference.

1.1.1 Definition. A sequence of R-modules is a, possibly infinite, diagram in M(R),

(1.1.1.1) · · · −→M0 α0−−→M1 α1

−−→M2 α2−−→ ·· · ;

it is called exact if one has Imαn−1 = Kerαn for all n. Notice that (1.1.1.1) is exactif and only if every sequence 0→ Imαn−1 →Mn → Imαn → 0 is exact. An exactsequence of the form 0→M′→M→M′′→ 0 is called a short exact sequence.

3

4 1 Modules

Two sequences αn : Mn→Mn+1n∈Z and βn : Nn→ Nn+1n∈Z of R-modulesare called isomorphic if there exists a family of isomorphisms ϕnn∈Z such thatthe diagram

Mn αn//

ϕn

Mn+1

ϕn+1

Nn βn// Nn+1

is commutative for every n ∈ Z.

Diagram lemmas are workhorses of homological algebra. Two of the most fre-quently used are known as the Five Lemma and the Snake Lemma.

1.1.2 Five Lemma. Let

M1 //

ϕ1

M2 //

ϕ2

M3 //

ϕ3

M4 //

ϕ4

M5

ϕ5

N1 // N2 // N3 // N4 // N5

be a commutative diagram in M(R) with exact rows. The following assertions hold.(a) If ϕ1 is surjective, and ϕ2 and ϕ4 are injective, then ϕ3 is injective.(b) If ϕ5 is injective, and ϕ2 and ϕ4 are surjective, then ϕ3 is surjective.(c) If ϕ1, ϕ2, ϕ4, and ϕ5 are isomorphisms, then ϕ3 is an isomorphism.

1.1.3 Construction. LetM′ α′

//

ϕ′

M α//

ϕ

M′′ //

ϕ′′

0

0 // N′β′// N

β// N′′

be a commutative diagram in M(R) with exact rows. Given an element x′′ ∈ Kerϕ′′

choose, by surjectivity of α, a preimage x in M. Set y = ϕ(x) and note that y be-longs to Kerβ, by commutativity of the right-hand square. By exactness at N choosey′ ∈ N′ with β′(y′) = y. It is straightforward to verify that the element [y′]Imϕ′ inCokerϕ′ does not depend on the choices of x and y′, so this procedure defines a mapδ : Kerϕ′′→ Cokerϕ′.

1.1.4 Snake Lemma. The map δ : Kerϕ′′→ Cokerϕ′ defined in 1.1.3 is a homomor-phism of R-modules, called the connecting homomorphism, and there is an exactsequence in M(R),

Kerϕ′ α′−−→ Kerϕ α−−→ Kerϕ′′ δ−−→ Cokerϕ′β′−−→ Cokerϕ

β−−→ Cokerϕ′′ .

Moreover, if α′ is injective then so is the restriction α′ : Kerϕ′→ Kerϕ, and if β issurjective, then so is the induced homomorphism β : Cokerϕ→ Cokerϕ′′.

For historical reasons, the morphisms in M(R) are called homomorphisms andthe hom-sets are written Hom.


1.1 Prerequisites 5

1.1.5 Homomorphisms. From the hom-sets in M(R) one can construct a functor

HomR(–,–) : M(R)op×M(R) −→ M(k) .

For homomorphisms α : M′→M and β : N→ N′ of R-modules, the functor acts asfollows,

HomR(α,β) : HomR(M,N) −→ HomR(M′,N′) is given by ϑ 7−→ βϑα .

1.1.6 Example. Let a be a left ideal in R and let M be an R-module. There is anisomorphism of k-modules HomR(R/a,M)→ (0 :M a) given by α 7→ α([1]a).

1.1.7 Tensor Products. The tensor product of modules yields a functor

–⊗R – : M(Ro)×M(R) −→ M(k) .

For a homomorphism α : M→M′ of Ro-modules and a homomorphism β : N→ N′

of R-modules, the functor acts as follows,

α⊗R β : M⊗R N −→ M′⊗R N′ is given by m⊗n 7−→ α(m)⊗β(n) .

1.1.8 Example. Let b be a right ideal in R and let M be an R-module. There is anisomorphism of k-modules R/b⊗R M→M/bM given by [r]b⊗m 7→ [rm]bM .

1.1.9 Example. Let R be commutative and let U ⊆ R be a multiplicative subset. Forthe localization there is an isomorphism of U−1R-modules (U−1R)⊗R M→U−1Mgiven by r/u⊗m 7→ rm/u.

LINEARITY

Because R is assumed to be a k-algebra, the module category M(R) is k-linear inthe following sense.

1.1.10 Definition. A category U is called k-linear if it satisfies the next conditions.(1) For every pair of objects M and N in U, the hom-set U(M,N) is a k-module,

and composition of morphisms U(M,N)×U(L,M)→ U(L,N) is k-bilinear.(2) There is a zero object, 0, in U. That is, for each object M in U there is a unique

morphism M→ 0 and a unique morphism 0→M.(3) For every pair of objects M and N in U there is a biproduct, M⊕N, in U. That

is, given M and N there is a diagram in U,

MεM// M⊕N

$Moo

$N// N ,

εNoo

such that $MεM = 1M , $NεN = 1N , and εM$M +εN$N = 1M⊕N hold. Here1X denotes the identity morphism of the object X .

A category that satisfies (1) is called k-prelinear. It is evident that the oppositecategory of a k-(pre)linear category is k-(pre)linear.


6 1 Modules

REMARK. The module category M(R) has additional structure; indeed, it is Abelian, and so isthe category C(R) of R-complexes, which is the topic of the next chapter. However, the homotopycategory K(R) and the derived category D(R), which are constructed in Chap. 6, are not Abelianbut triangulated. All four categories are k-linear, hence the focus on that notion.

1.1.11. Let U be a k-linear category. The biproduct ⊕ is both a product and a co-product, and it is elementary to verify that it is associative. For a finite set U anda family Muu∈U of objects in U the notation

⊕u∈U Mu is, therefore, unambigu-

ous. It is convenient to identify a morphism α :⊕n

v=1 Mv→⊕m

u=1 Nu by an m× nmatrix,

M1⊕...⊕Mn

α11 . . . α1n.... . .

...αm1 . . . αmn

//

N1⊕...⊕Nm ,

with αuv =$NuαεMv

, where εMvand $Nu

are the morphisms from 1.1.10.

The homomorphism functor 1.1.5 and the tensor product functor 1.1.7 are bothk-multilinear in the following sense.

1.1.12 Definition. A functor F: U→ V between k-prelinear categories is called k-linear if it satisfies the following conditions.

(1) F(α+β) = F(α)+F(β) for all parallel morphisms α and β in U.(2) F(xα) = xF(α) for all morphisms α in U and all x ∈ k.Let U1, . . . ,Un and V be k-prelinear categories. A functor

F : U1×·· ·×Un −→ V

is called k-multilinear if it is k-linear in each variable.

1.1.13. There is a unique ring homomorphism Z→ k; therefore, every k-linear cat-egory/functor is Z-linear in a canonical way. A Z-linear category/functor is alsocalled additive. A Z-prelinear category is also called preadditive.

1.1.14. Let M and N be objects in an additive category U. The zero morphism fromM to N is the composite of the unique morphisms M→ 0 and 0→N; this morphismis denoted by 0.

Let F: U→ V be an additive functor between additive categories. It takes zeromorphisms in U to zero morphisms in V. In particular, F takes zero objects in U tozero objects in V.

PRODUCTS, COPRODUCTS, AND DIRECT SUMS OF MODULES

1.1.15. For a family Nuu∈U in M(R), the product∏

u∈U Nu in M(R) is the Carte-sian product of the underlying sets, with the R-module structure given by coordinate-wise operations, together with the coordinatewise projections $u :

∏u∈U Nu Nu.


1.1 Prerequisites 7

Indeed, given a family of homomorphisms αu : M→ Nuu∈U , the map defined bym 7→ (αu(m))u∈U is the unique homomorphism that makes the diagram∏

u∈UNu

$u

M

??

βu// Nu

commutative for every u∈U . For a family αu : Mu→ Nuu∈U of homomorphisms,the product

∏u∈U α

u :∏

u∈U Mu→∏

u∈U Nu is given by (mu)u∈U 7→ (αu(mu))u∈U .If one has Mu = M for every u ∈U , then the product

∏u∈U Mu is denoted MU

and called the U-fold product of M.

REMARK. Other names for the product are categorical product and direct product.

1.1.16. For a family Muu∈U in M(R), the coproduct∐

u∈U Mu is the submodule

(mu)u∈U ∈∏

u∈UMu | mu = 0 for all but finitely many u ∈U

of the product, together with the coordinatewise injections εu : Mu∐

u∈U Mu. Ev-ery element of

∐u∈U Mu has the form ∑u∈U ε

u(mu) for a unique family (mu)u∈Uwith mu = 0 for all but finitely many u ∈ U . Given a family of homomorphismsαu : Mu→ Nu∈U , the map defined by ∑u∈U ε

u(mu) 7→ ∑u∈U αu(mu) is the unique

homomorphism that makes the following diagram commutative

Mu

αu

//εu//∐

u∈UMu

||

N

for every u∈U . For a family αu : Mu→ Nuu∈U of homomorphisms the coproduct∐u∈U α

u :∐

u∈U Mu→∐

u∈U Nu is given by ∑u∈U εu(mu) 7→ ∑u∈U α

u(mu).If one has Mu = M for every u∈U , then the coproduct

∐u∈U Mu is denoted M(U)

and called the U-fold coproduct of M.

REMARK. Other names for the coproduct are categorical sum and direct sum; we reserve the latterfor the (iterated) biproduct; see below.

1.1.17. If U is a set with two elements (or, more generally, a finite set) then theproduct and the coproduct of a family Muu∈U in M(R) coincide, and this module∏

u∈U Mu =∐

u∈U Mu is the (iterated) biproduct M =⊕

u∈U Mu of the family. Werefer to M as the direct sum of the family Muu∈U , and each module Mu is calleda direct summand of M.

1.1.18. Let Muu∈U be a family of R-modules. Notice that for every finite subsetU ′ of U there are canonical isomorphisms

∏u∈U Mu ∼=

(⊕u∈U ′M

u)⊕∏

u∈U\U ′Mu

and∐

u∈U Mu ∼=(⊕

u∈U ′Mu)⊕∐

u∈U\U ′Mu.


8 1 Modules

1.1.19 Definition. Let M be an R-module and let Muu∈U be a family of submod-ules of M. There is a surjective homomorphism,∐

u∈UMu −→ ∑

u∈UMu given by ∑

u∈Uεu(mu) 7−→ ∑

u∈Umu ;

if it is an isomorphism, then the family Muu∈U is called independent.

REMARK. Often, but not here, an independent family of submodules is said to form an internaldirect sum. We have, though, reserved direct sum to mean a finite (co)product.

1.1.20. Let M be an R-module. A family Muu∈U of submodules of M is indepen-dent if and only if one has (∑u∈V Mu)∩Mw = 0 for every finite subset V of U andevery element w ∈U \V . Notice that if Muu∈U is independent with Mu 6= 0 forinfinitely many u ∈U , then M contains an infinite ascending chain of submodules.

BIMODULES

An Abelian group M that is both an R-module and an S-module is called an R–S-bimodule if the two module structures are compatible, i.e. one has s(rm) = r(sm) forall r∈R, s∈ S, and m∈M. A homomorphism of R–S-bimodules is a homomorphismof Abelian groups that is both R- and S-linear.

The convention to identify right modules with left modules over the opposite ringis convenient for abstract considerations. However, in concrete computations withelements in, say, an R–So-bimodule it often adds clarity to write the S-action on theright; the bimodule condition then reads (rm)s = r(ms).

An R–Ro-bimodule is called symmetric if one has rm = mr for all r ∈ R andm ∈ M. If R is commutative, then every R-module has a canonical structure of asymmetric R–Ro-bimodule. When no other convention is specified, modules over acommutative ring R are considered to be symmetric bimodules, and the categoriesM(R) and M(Ro) are identified.

As R and S are k-algebras, every R–So-bimodule is a k–ko-bimodule. We assumethat this structure is symmetric; i.e. we only consider k-symmetric R–So-bimodules.

1.1.21. By the conventions above, every R-module is an R–ko-bimodule, and everySo-module is a k–So-bimodule.

A ring homomorphism R→ S induces R–So- and S–Ro-bimodule structures onevery S–So-bimodule M; in particular on S.

REMARK. Neither of the two R–Ro-bimodule structures induced on a ring R by an endomorphismϕ : R→ R is symmetric, unless ϕ is the identity—not even if R is commutative.

1.1.22 Example. For integers m,n > 1 let Mm×n(R) denote the R–Ro-bimodule ofm×n-matrices with entries from R. Set M = M1×n(R) and N = Mn×1(R), and let Qdenote the ring Mn×n(R). Now M is an R–Qo-bimodule and N is a Q–Ro-bimodule.

1.1.23. There is an equivalence of k-linear Abelian categories,


1.1 Prerequisites 9

R–So-bimodules and their homomorphismsF//M(R⊗k So) .

Goo

The functor F assigns to an R–So-bimodule M the R⊗k So-module with actiongiven by (r ⊗ s)m = rms. Conversely, G assigns to an R⊗k So-module M theR–So-bimodule with R-action given by rm = (r⊗ 1)m and right S-action given byms = (1⊗ s)m. Homomorphisms are treated analogously.

1.1.24 Definition. Let M(R–So) denote the category M(R⊗k So).

Per 1.1.23 the category M(R–So) is naturally identified with the category ofR–So-bimodules and their homomorphisms.

1.1.25. The ring R⊗k Ro is called the enveloping algebra of R and denoted Re. Per1.1.23 the category of R–Ro-bimodules is equivalent to the category of Re-modules.The set r⊗ 1− 1⊗ r | r ∈ R generates an ideal C in Re, and as in 1.1.23 it isstraightforward to verify that the full subcategory of M(R⊗k Ro) whose objects aresymmetric R–Ro-bimodules is equivalent to the category of modules over the ringRe = Re/C. If R is commutative, then the rings R and Re are isomorphic.

The set HomR(M,N) of R-linear maps between R-modules is a k-module but ithas, in general, no built-in R-module structure. The reason is, so to say, that the R-structures on M and N are taken up by the R-linearity of the maps. Similarly, theR-balancedness takes up the Ro- and R-structures on the factors in a tensor productM⊗R N and leaves only a k-module. If R is commutative, then one can take k= R.At work here is the tacit assumption that a module over a commutative ring R isa symmetric R–Ro-bimodule. Also in a non-commutative setting, access to richermodule structures on hom-sets and tensor products is via bimodules.

1.1.26. If M is an R–Qo-bimodule and N is an R–So-bimodule, then the k-moduleHomR(M,N) has a Q–So-bimodule structure given by

(qϕ)(m) = ϕ(mq) and (ϕs)(m) = (ϕ(m))s .

Moreover, if α : M→M′ is a homomorphism of R–Qo-bimodules, and β : N→ N′

is a homomorphism of R–So-bimodules, then HomR(α,β), as defined in 1.1.5, is ahomomorphism of Q–So-bimodules. Thus, there is an induced k-bilinear functor,

HomR(–,–) : M(R–Qo)op×M(R–So) −→ M(Q–So) .

1.1.27 Example. Let R→ S be a ring homomorphism, and consider S as an R–So-bimodule; see 1.1.21. Now HomR(S,–) is a functor from M(R) to M(S).

1.1.28 Example. Set Q = Mn×n(R) and M = M1×n(R) as in 1.1.22. An applica-tion of 1.1.26 with S = k yields a functor HomR(M,–) : M(R)→M(Q). Anotherapplication of 1.1.26, this time with S = Mn(R) = Q, shows that HomR(M,–) is afunctor from M(R–Qo) to M(Q–Qo). In particular, HomR(M,M) has the structureof a Q–Qo-bimodule.


10 1 Modules

1.1.29. If M is a Q–Ro-bimodule and N is an R–So-bimodule, then the k-moduleM⊗R N has a Q–So-bimodule structure given by

q(m⊗n) = (qm)⊗n and (m⊗n)s = m⊗ (ns) .

Moreover, if α : M→M′ is a homomorphism of Q–Ro-bimodules, and β : N→ N′ isa homomorphism of R–So-bimodules, then α⊗β, as defined in 1.1.7, is a homomor-phism of Q–So-bimodules. Thus, there is an induced k-bilinear functor,

–⊗R – : M(Q–Ro)×M(R–So) −→ M(Q–So) .

1.1.30 Example. Let R→ S be a ring homomorphism, and consider S as an S–Ro-bimodule; see 1.1.21. Now S⊗R – is a functor from M(R) to M(S).

1.1.31 Example. Set Q = Mn×n(R), M = M1×n(R), and N = Mn×1(R) as in 1.1.22.An application of 1.1.29 with S = k yields a functor N⊗R –: M(R)→M(Q). An-other application, this time with S = Mn×n(R) = Q, shows that N⊗R – is a functorfrom M(R–Qo) to M(Q–Qo). In particular, N⊗R M is a Q–Qo-bimodule.

EXACTNESS

Though we are not concerned with abstract Abelian categories, the language is use-ful for the following reason. The convention that every functor is covariant forcesone to consider, for example, HomR(–,–) as a functor on the product categoryM(R)op×M(R) which is Abelian but not a module category.

REMARK. The class of all Abelian categories is closed under products and opposites of categories,that is, if U and V are Abelian categories, then so are U×V and Uop. The class of “modulecategories”—i.e. categories that are equivalent to M(R) for some ring R—is also closed underproducts; indeed M(R)×M(S) is equivalent to M(R×S); see E 1.1.21. However, M(R)op is nota module category; see E 1.1.23. Yet, by Freyd’s theorem [29], as strengthened by Mitchell [59],every Abelian category is, in fact, a full subcategory of a module category.

1.1.32 Split Exact Sequences. An exact sequence 0−→M′ α′−→M α−→M′′ −→ 0 inM(R) is called split if it satisfies the following equivalent conditions.

(i) There exist homomorphisms % : M→M′ and σ : M′′→M such that one has

%α′ = 1M′ , α′%+σα = 1M , and ασ = 1M′′ .

(ii) There exists a homomorphism % : M→M′ such that %α′ = 1M′ .(iii) There exists a homomorphism σ : M′′→M such that ασ= 1M′′ .(iv) The sequence is isomorphic to 0 −→M′ ε−→M′⊕M′′ $−→M′′ −→ 0, where ε

and $ are the injection and the projection, respectively.

Moreover, if 0 −→ M′ α′−→ M α−→ M′′ −→ 0 is split exact, then also the sequence0−→M′′ σ−→M

%−→M′ −→ 0, where % and σ are as in part (i), is split exact.

REMARK. There exist non-split short exact sequences of Z-modules 0→ M′ → M → M′′ → 0with M ∼= M′⊕M′′, via some isomorphism. By a result of Miyata [60] this phenomenon can notoccur for finitely generated modules over a commutative Noetherian ring.


1.1 Prerequisites 11

The definitions 1.1.1 and 1.1.32 of (split) exact sequences in M(R) make sensein any Abelian category. A (split) exact sequence in M(R)op is just a (split) exactsequence in M(R) with the arrows reversed.

1.1.33. Let F: U→ V be an additive functor between Abelian categories. For everysplit exact sequence 0 −→M′ −→M −→M′′ −→ 0 in U the induced sequence in V,0−→ F(M′)−→ F(M)−→ F(M′′)−→ 0, is split exact.

1.1.34 Example. The assignments M 7→ k⊕M and α 7→ 1k⊕α define a functorF: M(k)→M(k), which is not additive, as F(0) is non-zero; cf. 1.1.14.

A frequently used consequence of 1.1.33 is that additive functors preservebiproducts and hence direct sums; cf. 1.1.10 and 1.1.17. In particular, there is acanonical isomorphism F(⊕u∈U Mu) ∼= ⊕u∈U F(Mu) for an additive functor F and afinite family of objects Muu∈U .

1.1.35 Half Exactness. A functor F: U→ V between Abelian categories is calledhalf exact if for every short exact sequence 0 −→ M′ −→ M −→ M′′ −→ 0 in U, thesequence F(M′)−→ F(M)−→ F(M′′) in V is exact. A half exact functor is additive.

REMARK. It is not too hard to construct an additive functor that is not half exact. An example ofsuch a functor comes up naturally in 11.1.20. See also E 2.2.1.

The Hom functor 1.1.5 is left exact in the following sense.

1.1.36 Left Exactness. A functor F: U→ V between Abelian categories is calledleft exact if it satisfies the following equivalent conditions.

(i) For every short exact sequence 0−→M′−→M−→M′′−→ 0 in U, the sequence0−→ F(M′)−→ F(M)−→ F(M′′) in V is exact.

(ii) For every (left) exact sequence 0 −→ M′ −→ M −→ M′′ in U, the sequence0−→ F(M′)−→ F(M)−→ F(M′′) in V is exact.

REMARK. If G: M(R)op→M(S) is an additive left exact functor that preserves products, thenthere is an R–So-bimodule N and a natural isomorphism G ∼= HomR(–,N). If F : M(R)→M(Z)is an additive left exact functor that preserves filtered limits, then there exists an R-module M anda natural isomorphism F∼= HomR(M,–); see Watts [85].

The tensor product functor 1.1.7 is right exact in the following sense.

1.1.37 Right Exactness. A functor F: U→ V between Abelian categories is calledright exact if it satisfies the following equivalent conditions.

(i) For every short exact sequence 0−→M′−→M−→M′′−→ 0 in U, the sequenceF(M′)−→ F(M)−→ F(M′′)−→ 0 in V is exact.

(ii) For every (right) exact sequence M′ −→ M −→ M′′ −→ 0 in U, the sequenceF(M′)−→ F(M)−→ F(M′′)−→ 0 in V is exact.

REMARK. If F : M(R)→M(S) is an additive right exact functor that preserves coproducts, thenthere exists an S–Ro-bimodule M and a natural isomorphism F∼= M⊗R –; see Watts [85].


12 1 Modules

Given a homomorphism of rings ϕ : R→ S, the restriction of scalars functorM(S)→M(R) assigns to each S-module M the R-module with the action inducedby ϕ. This functor is exact in the following sense.

1.1.38 Exactness. A functor F: U→ V between Abelian categories is called exactif it satisfies the following equivalent conditions.

(i) For every short exact sequence 0−→M′−→M−→M′′−→ 0 in U, the sequence0−→ F(M′)−→ F(M)−→ F(M′′)−→ 0 in V is exact.

(ii) F preserves exactness of sequences.(iii) F is left exact and right exact.

REMARK. It is not too hard to construct a half functor that is neither left nor right exact. See 2.2.22and E 8.2.2 for examples of such a functors.

FAITHFULNESS

An additive functor F: U→ V between additive categories is faithful if for everymorphism α in U one has F(α) = 0 in V only if α = 0 in U. In that case it followsthat for objects M in U one has F(M)∼= 0 in V only if M ∼= 0 in U.

1.1.39 Faithful Exactness. An additive functor F: U→ V between Abelian cate-gories is called faithfully exact if it satisfies the following equivalent conditions.

(i) F is exact and faithful.(ii) F is exact and for every M in U one has F(M)∼= 0 in V only if M ∼= 0 in U.

Faithful functors have convenient cancellation properties.

1.1.40. Let F: U→ V be an additive functor between Abelian categories. If F isfaithfully exact and M′→M→M′′ is a sequence in U, then exactness of the inducedsequence F(M′)→ F(M)→ F(M′′) in V implies exactness of M′→M→M′′ in U.

Let G: V→W be a faithfully exact functor between Abelian categories. Thecomposite GF: U→W is then (faithfully) exact if and only if F is (faithfully) exact.

EXERCISES

E 1.1.1 Let ϕ : M→ N be a homomorphism of R-modules. Show that if ϕ is bijective, then theinverse map N→M is also a homomorphism of R-modules; that is, ϕ is an isomorphism.

E 1.1.2 Show that a morphism in M(R) is a monomorphism if and only if it is injective, andthat it is an epimorphism if and only if it is surjective.

E 1.1.3 Show that the category M(R) is Abelian.E 1.1.4 Show that in the category of unital rings the embedding ZQ is both a monomorphism

and an epimorphism though not an isomorphism.E 1.1.5 Consider a commutative diagram in M(R),

M′α′// M

α//

ϕ

M′′

ϕ′′

0 // N′β′// N

β// N′′ .


1.1 Prerequisites 13

Assuming that the lower row is exact and αα′ = 0 holds, show that there is a uniquehomomorphism ϕ′ : M′→ N′ such that β′ϕ′ = ϕα′ holds.

E 1.1.6 Consider a commutative diagram in M(R),

M′α′//

ψ′

Mα//

ψ

M′′ // 0

N′β′// N

β// N′′ .

Assuming that the upper row is exact and ββ′ = 0 holds, show that there is a uniquehomomorphism ψ′′ : M′′→ N′′ such that ψ′′α= βψ holds.

E 1.1.7 LetM′

α′//

ϕ′

Mα//

ϕ

M′′

ϕ′′

N′β′// N

β// N′′

be a commutative diagram in M(R) with exact rows. (a) Show that if α is surjective, thenthe sequence Cokerϕ′ → Cokerϕ→ Cokerϕ′′ is exact. (b) Show that if β′ is injective,then the sequence Kerϕ′→ Kerϕ→ Kerϕ′′ is exact.

E 1.1.8 Determine the inverse of the map HomR(R/a,M)→ (0 :M a) given in 1.1.6.E 1.1.9 Determine the inverse of the map R/b⊗R M→M/bM given in 1.1.8.E 1.1.10 Let Muu∈U be a family of R-modules and set M =

∐u∈U Mu. Show that the family

εu(Mu)u∈U of submodules of M is independent.E 1.1.11 Let αu : Mu→ Nuu∈U be a family of homomorphisms in M(R).

(a) Show that∏

u∈U αu, as defined in 1.1.15, is the unique homomorphism α thatmakes the following diagram commutative for every u ∈U .∏

u∈U Mu

α//∏

u∈U Nu

Mu αu// Nu

(b) Show that∐

u∈U αu, as defined in 1.1.16, is the unique homomorphism α thatmakes the following diagram commutative for every u ∈U .

Mu

αu

// //∐

u∈U Mu

α

Nu // //∐

u∈U Nu

E 1.1.12 Show that R is an R–Ro-bimodule, but only a symmetric one if R is commutative.E 1.1.13 (Cf. 1.1.25) Show that if R is commutative, then the rings R and Re are isomorphic.E 1.1.14 Show that the enveloping algebra (Ro)e is the opposite ring of Re and isomorphic to Re.

Deduce that the quotient Re defined in 1.1.25 is isomorphic to the opposite ring (Re)o.E 1.1.15 Show that a non-zero R-module is semi-simple if and only if it is isomorphic to a co-

product of simple R-modules.E 1.1.16 For Q and M as in 1.1.28 decide if HomR(M,M) and Q are isomorphic Q–Qo-bimodules.E 1.1.17 For Q, M, and N as in 1.1.31 decide if N⊗R M and Q are isomorphic Q–Qo-bimodules.E 1.1.18 Let M be an R-module. (a) Show that S = HomR(M,M) is a k-algebra with multiplica-

tion given by composition of homomorphisms. (b) Show that M is an S-module.E 1.1.19 Show that the endomorphism algebra HomR(RR,RR), see E 1.1.18, is isomorphic to Ro.E 1.1.20 In the algebra M = M2×2(R), consider the subset

C =

(x −yy x

)∣∣∣∣ x,y ∈ R

.


14 1 Modules

(a) Show that C is a subring isomorphic to C. (b) Show that the C-actions afforded bythe canonical ring homomorphism C→M make M a non-symmetric C–Co-bimodule.

E 1.1.21 Consider the functors F: M(R)×M(S)→M(R×S) given by F(M,N) = M×N andG: M(R×S)→M(R)×M(S) given by G(X) = ((1,0)X ,(0,1)X). Show that F and Gyield an equivalence of categories.

E 1.1.22 Let U be a set with cardU >maxcardR,ℵ0. Show that R(U) has cardinality strictlyless than RU and conclude that there exists no epimorphism R(U)→ RU in M(R).

E 1.1.23 Consider the category V=M(R)op. Coproducts in M(R) yields products in V, denoted∏V, and products in M(R) yield coproducts in V, denoted

∐V. (a) Show that for every

family Xuu∈U of objects in V there exists an epimorphism∐V

u∈U Xu→∏V

u∈U Xu.(b) Use E 1.1.22 to show that unless R is the zero ring, there exists no ring S such thatV is equivalent to M(S).

E 1.1.24 Show that the sequence of Z-modules 0−→ Z α−→ Z⊕ (Z/2Z)Nβ−→ (Z/2Z)N −→ 0,

with α(m) = (2m,0,0, . . .) and β(n,x1,x2, . . .) = ([n]2Z,x1,x2, . . .), is exact but not split.E 1.1.25 Show that the Z-submodule Z(N) of ZN is not a direct summand.E 1.1.26 (a) Show that a half exact functor between Abelian categories is additive. (b) For every

Z-module M set F(M) = 2M, and for every homomorphism α : M→ N of Z-moduleslet F(α) : 2M→ 2N be the restriction of α. Show that F : M(Z)→M(Z) is an additivefunctor but not half exact.

E 1.1.27 Show that a functor F: U→V between Abelian categories is exact if and only if theinduced sequence F(M′)→ F(M)→ F(M′′) in V is exact for every exact sequence M′→M→M′′ in U.

E 1.1.28 Show that a functor F: U→V between Abelian categories is right exact if and only ifthe opposite functor Fop : Uop→Vop is left exact.

E 1.1.29 Let F: U→V and G: V→W be functors between Abelian categories. (a) Show thatif both functors are right exact, then GF is right exact. (b) Show that if both functors areleft exact, then GF is left exact. (c) Show that if one functor is left exact and the other isright exact, then GF need not be half exact.

E 1.1.30 Let M be an R–Ro-bimodule. (a) Show that M⊗kM has two compatible structures asa module over Re, an “inner” and an “outer”. (b) Show that the assignments M 7→HomRe (R,M⊗kM) and α 7→ HomRe (R,α⊗k α) define an endofunctor on M(Re) thatis not additive.

1.2 Standard Isomorphisms

SYNOPSIS. Unitor; counitor; commutativity; associativity; swap; adjunction.

Under suitable assumptions about bimodule structures, it makes sense to considercomposites like Hom(X⊗M,N) and Hom(M,Hom(X ,N)), and they turn out tobe isomorphic. At work here is adjunction, one of several natural isomorphisms ofcomposites of Hom and tensor product functors, which are the focus of this section.

We start by recalling that both functors R⊗R – and HomR(R,–) are naturallyisomorphic to the identity functor on M(R).

UNITOR AND COUNITOR

1.2.1. For every R-module M there are isomorphisms of R-modules,


1.2 Standard Isomorphisms 15

µMR : R⊗R M −→ M given by µM

R (r⊗m) = rm and(1.2.1.1)

εMR : M −→ HomR(R,M) given by εM

R (m)(r) = rm ,(1.2.1.2)

called the unitor and counitor; they are natural in M. Moreover, if M is an R–So-bimodule, then these maps are isomorphisms of R–So-bimodules.

The tensor product behaves as one would expect of a “product”. Being additive,the tensor product distributes over direct sums, and the first two standard isomor-phisms below show that it is commutative and associative.

COMMUTATIVITY

1.2.2 Proposition. Let M be an Ro-module and N be an R-module. The commuta-tivity map,

υMN : M⊗R N −→ N⊗Ro M given by υMN(m⊗n) = n⊗m ,

is an isomorphism of k-modules, and it is natural in M and N. Moreover, if M is inM(Q–Ro) and N is in M(R–So), then υMN is an isomorphism in M(Q–So).

PROOF. For every Ro-module M and every R-module N, the map π from M×N toN⊗Ro M given by (m,n) 7→ n⊗m is k-bilinear and middle R-linear. Indeed, one has

π(m,n+n′) = (n+n′)⊗m = n⊗m+n′⊗m = π(m,n)+π(m,n′)

π(m+m′,n) = n⊗ (m+m′) = n⊗m+n⊗m′ = π(m,n)+π(m′,n)

π(mr,n) = n⊗mr = rn⊗m = π(m,rn)

π(mx,n) = π(m,xn) = xn⊗m = x(n⊗m) = xπ(m,n)

for all m∈M, n∈N, r ∈ R, and x∈ k. Thus, υMN is a homomorphism of k-modules.Let α : M→M′ be a homomorphism of Ro-modules and let β : N→ N′ be a ho-

momorphism of R-modules. The diagram

M⊗R N υMN//

α⊗β

N⊗Ro M

β⊗α

M′⊗R N′ υM′N′// N′⊗Ro M′

is commutative, as one has

(υM′N′ (α⊗R β))(m⊗n) = υM′N′(α(m)⊗β(n))= β(n)⊗α(m)

= (β⊗Ro α)(n⊗m)

= ((β⊗Ro α)υMN)(m⊗n)

for all m ∈ M and all n ∈ N. Thus, υ is a natural transformation of functors fromM(Ro)×M(R) to M(k).


16 1 Modules

For M in M(Ro) and N in M(R) the map from N⊗Ro M to M⊗R N given byn⊗m 7→ m⊗n is an inverse of υMN , so υMN is an isomorphism of k-modules.

If M is a Q–Ro-bimodule and N is an R–So-bimodule, then M⊗R N and N⊗Ro Mare Q–So-bimodules. The computation

υMN(q(m⊗n)s) = υMN(qm⊗ns) = ns⊗qm = q(n⊗m)s = q(υMN(m⊗n))s ,

which holds for all q ∈ Q, s ∈ S, m ∈ M, and n ∈ N, shows that the isomorphismυMN is Q- and So-linear.

ASSOCIATIVITY

1.2.3 Proposition. Let M be an Ro-module, X be an R–So-bimodule, and N be anS-module. The associativity map,

ωMXN : (M⊗R X)⊗S N −→ M⊗R (X⊗S N) ,

given byωMXN((m⊗ x)⊗n) = m⊗ (x⊗n)

is an isomorphism of k-modules, and it is natural in M, X , and N. Moreover, if M isin M(Q–Ro) and N is in M(S–T o), then ωMXN is an isomorphism in M(Q–T o).

PROOF. Proceeding as in the proof of 1.2.2, it is straightforward to verify that ωis a natural transformation of functors from M(Ro)×M(R–So)×M(S) to M(k).Further, for modules M, X , and N as in the statement, the assignment m⊗ (x⊗n) 7→(m⊗ x)⊗ n defines a map M⊗R (X⊗S N) → (M⊗R X)⊗S N; it is an inverse ofωMXN which, therefore, is an isomorphism of k-modules.

If M is in M(Q–Ro) and N is in M(S–T o), then the modules (M⊗R X)⊗S N andM⊗R (X⊗S N) are in M(Q–T o), and a computation similar to the one performed inthe proof of 1.2.2 shows that ωMXN is Q- and T o-linear.

SWAP

For modules M, N, and X the bilinear maps M×N → X are in one-to-one corre-spondence with elements in Hom(M,Hom(N,X)) and Hom(N,Hom(M,X)). Swapcompares these two hom-sets directly.

1.2.4 Proposition. Let M be an R-module, X be an R–So-bimodule, and N be anSo-module. The swap map,

ζMXN : HomR(M,HomSo(N,X)) −→ HomSo(N,HomR(M,X)) ,

given byζMXN(ψ)(n)(m) = ψ(m)(n)

is an isomorphism of k-modules, and it is natural in M, X , and N. Moreover, if M isin M(R–Qo) and N is in M(T –So), then ζMXN is an isomorphism in M(Q–T o).


1.2 Standard Isomorphisms 17

PROOF. It is straightforward to verify that ζ is a natural transformation of functorsfrom M(R)op×M(R–So)×M(So)op to M(k). Further, for modules M, X , and N asin the statement, it is immediate that the swap map ζNXM is an inverse of ζMXN .

If M is in M(R–Qo) and N is in M(T –So), then HomR(M,HomSo(N,X)) andHomSo(N,HomR(M,X)) are Q–T o-bimodules. The computation

ζMXN(qψt)(n)(m) = (qψt)(m)(n)

= ψ(mq)(tn)

= ζMXN(ψ)(tn)(mq)

= (q(ζMXN(ψ))t)(n)(m) ,

which holds for all q ∈ Q, t ∈ T , ψ ∈ HomR(M,HomSo(N,X)), m ∈M, and n ∈ N,shows that the isomorphism ζMXN is Q- and T o-linear.

ADJUNCTION

The next isomorphism expresses that Hom and tensor product are adjoint functors.

1.2.5 Proposition. Let M be an R-module, X be an R–So-bimodule, and N be anS-module. The adjunction map,

ρMXN : HomR(X⊗S N,M) −→ HomS(N,HomR(X ,M)) ,

given byρMXN(ψ)(n)(x) = ψ(x⊗n)

is an isomorphism of k-modules, and it is natural in M, X , and N. Moreover, if M isin M(R–T o) and N is in M(S–Qo), then ρMXN is an isomorphism in M(Q–T o).

PROOF. It is straightforward to verify that ρ is a natural transformation of functorsfrom M(R)×M(R–So)op×M(S)op to M(k). Further, for modules M, X , and N asin the statement, it is elementary to verify that the map

κ : HomS(N,HomR(X ,M)) −→ HomR(X⊗S N,M)

given by κ(ϕ)(x⊗n) = ϕ(n)(x) is an inverse of ρMXN .If M is in M(R–T o) and N is in M(S–Qo), then HomR(X⊗S N,M) is a Q–T o-

bimodule and so is HomS(N,HomR(X ,M)). The computation

ρMXN(qψt)(n)(x) = (qψt)(x⊗n)

= (ψ(x⊗nq))t

= (ρMXN(ψ)(nq)(x))t

= ((q(ρMXN(ψ)))(n)(x))t

= (q(ρMXN(ψ))t)(n)(x) ,

which holds for all q ∈ Q, t ∈ T , ψ ∈ HomR(X⊗S N,M), n ∈ N, and x ∈ X , showsthat the isomorphism ρMXN is Q- and T o-linear.


18 1 Modules

EXERCISES

E 1.2.1 Determine the inverse maps of the unitor (1.2.1.1) and the counitor (1.2.1.2).

E 1.2.2 Let α be an m×n matrix with entries in R. (a) Show that Rm ·α−→ Rn, i.e. the map given byright multiplication by α, is a homomorphism of R-modules. (b) Let M be an R-moduleand show that HomR(·α,M) is a homomorphism of k-modules naturally identified withMn α·−→Mm.

E 1.2.3 Let k be a field and set (–)∗ = Homk(–,k). Let L be a k-vector space with basis euu∈U .For each u ∈U let e∗u : L→ k be the functional given by e∗u(ev) = δuv. (a) Show that theassignment eu 7→ e∗u defines a homomorphism ε : L→ L∗ of k-vector spaces. (b) Show thatε is an isomorphism if L has finite rank. (c) Assume that L has rank at least 2; show thatthere is an automorphism α : L→ L such that the following diagram is not commutative,

Lε//

α

L∗

Lε// L∗ .

α∗OO

E 1.2.4 Let k be a field and set (–)∗ =Homk(–,k). Let M be a k-vector space. (a) For m∈M, showthat the map εm : M∗→ k given by ϕ 7→ϕ(m) is an element in M∗∗=(M∗)∗. (b) Show thatthe map δM : M→M∗∗ given by m 7→ εm is k-linear. (c) Show that δ : IdM(k)→ (–)∗∗ isa natural transformation of functors from M(k) to M(k).

E 1.2.5 Let M be an Ro-module, X be an R–So-bimodule, and N be an S-module. Show that(M⊗R X)⊗S N is an Rc–Sc-bimodule, where Rc is the center of R.

E 1.2.6 Let M be a k-module. Show that the functor M⊗k – is left adjoint for Homk(M,–)E 1.2.7 Let M be an R-module, X be an R–So-bimodule, and N be an S-module. Without using

1.2.2–1.2.4, show that there is a natural isomorphism of k-modulesHomR(N⊗So X ,M)−→ HomS(N,HomR(X ,M)) .

E 1.2.8 Show that under suitable assumptions one can derive swap 1.2.4 from adjunction 1.2.5.E 1.2.9 Show that the restriction of scalars functor U: M(R)→M(k) has a left adjoint and a

right adjoint. Hint: Use U∼= RR⊗R – and U∼= HomR(RR,–).

1.3 Exact Functors and Classes of Modules

SYNOPSIS. Basis; free module; unique extension property; finite generation of Hom and tensorproduct; projective module; injective module; lifting properties; semi-simple ring; Baer’s criterion;finitely presented module; flat module; von Neumann regular ring.

We start by recalling the language of generators of modules and ideals.

1.3.1. Let M be an R-module, and let X = xuu∈U be a subset of M. It is simple toverify that the subset

R〈X〉 = ∑u∈U

ruxu | ru ∈ R and ru = 0 for all but finitely many u ∈U

of M is a submodule; it is called the submodule of M generated by X . For a finite setX = x1, . . . ,xn we often write R〈x1, . . . ,xn〉 instead of R〈X〉. By convention, R〈∅〉is the zero module. If one has R〈X〉= M, then X is called a set of generators for M.If M has a finite set of generators, then M is called finitely generated.


1.3 Exact Functors and Classes of Modules 19

If M is generated by one element, then M is called cyclic. A cyclic R-modulegenerated by an element x is isomorphic to R/a for the left ideal a= (0 :R x) in R.

1.3.2. Left ideals and right ideals in R generated by elements x1, . . . ,xn are de-noted R(x1, . . . ,xn) and (x1, . . . ,xn)R, respectively. The abridged notations Rx andxR are used for principal left and right ideals. If R is commutative, then the idealR(x1, . . . ,xn) = (x1, . . . ,xn)R is written (x1, . . . ,xn), and a principal ideal may bewritten using any of the notations Rx, xR, and (x).

REMARK. Though left and right ideals in R are submodules of the R-module R and the Ro-moduleR, respectively, it would be awkward to insist on applying the notation from 1.3.1 to ideals. Indeed,a principal right ideal would be written Ro〈x〉; even worse, writing k〈x1,x2〉 for the ideal in kgenerated by elements x1 and x2 would conflict with the standard notation for a free k-algebra.

FREE MODULES

The gateway to projective objects, and also to injective objects, in module categoriesis free modules.

1.3.3 Definition. Let L be an R-module and let E = euu∈U be a set of generatorsfor L. Every element in L can then be expressed on the form ∑u∈U rueu; if this ex-pression is unique, then E is called a basis for L, and L is called free. By convention,the zero module is free with the empty set as basis.

For a set E, not a priori assumed to be a subset of a module, the free R-modulewith basis E is denoted R〈E〉.

1.3.4 Example. If R is a division ring, then every R-module is free; indeed, everymodule over a division ring has a basis.

The Z-module M = Z/2Z is not free; indeed, a set of generators for M mustinclude [1]2Z, and one has 0[1]2Z = 2[1]2Z .

Bases of free modules have the following unique extension property.

1.3.5. Let L be a free R-module with basis E = euu∈U and let M be an R-module.For every map α : E→M there is a unique R-module homomorphism, α : L→M,such that the diagram

E

α

// // L

α

M

is commutative; the homomorphism is given by α(∑u∈U rueu) = ∑u∈U ruα(eu).

REMARK. The unique extension property characterizes bases; see E 1.3.3.

1.3.6. Let L be a free R-module with basis euu∈U and let fuu∈U be the standardbasis for R(U). There is an isomorphism of R-modules, L→ R(U), given by eu 7→ fu.


20 1 Modules

Important families of rings—commutative rings, left Noetherian rings, and localrings included—have the invariant basis number property (IBN).

1.3.7 Definition. A free module is said to have finite rank if it is finitely generated;that is, it has a finite basis. A free module that is not finitely generated is said to haveinfinite rank. For a finitely generated free module L over a ring R that has IBN, therank of L, written rankR L, is the number of elements in a basis for L.

REMARK. If U is infinite, then one has R(U) ∼= R(V ) only if the sets U and V have the samecardinality, even if R does not have IBN; see [50, cor. (1.2)].

1.3.8. Let Luu∈U be a family of free R-modules with bases Euu∈U . The co-product

∐u∈U Lu is then a free R-module with basis

⋃u∈U ε

u(Eu), where εu is theinjection Lu

∐u∈U Lu. Notice that if U is a finite set, each module Lu is finitely

generated, and R has IBN, then one has rankR(⊕

u∈U Lu) = ∑u∈U rankR Lu.

1.3.9. Let L and L′ be free k-modules with bases euu∈U and fvv∈V . It is elemen-tary to verify that the k-module L⊗k L′ is free with basis eu⊗ fvu∈U, v∈V . Thus, ifL and L′ are finitely generated, then one has rankk(L⊗k L′) = (rankkL)(rankkL′).

1.3.10 Theorem. If R is a principal left ideal domain, then every submodule of afree R-module is free.

PROOF. Let L be a free R-module with basis euu∈U , and let M be a submodule ofL. Choose a well-ordering 6 on U . For u ∈U define submodules of L as follows:

L<u = R〈ev | v < u〉 and L6u = R〈ev | v6 u〉 .

Let u ∈U be given. Every element l in L6u has a unique decomposition l = l′+ reuwith l′ ∈ L<u and r ∈ R, so there is a split exact sequence of R-modules,

0−→ L<u −→ L6u $u−−→ Reu −→ 0 ,

where $u is given by l′+ reu 7→ reu. Let ϕu be the restriction of $u to M ∩ L6u;one has Kerϕu = M ∩Ker$u = M ∩L<u. The image of ϕu is a submodule of Reuand, hence, isomorphic to a left ideal in R. It follows from the assumption on R thatImϕu is cyclic and free, so choose xu ∈ R with Imϕu = Rxueu and note that if xu isnon-zero, then xueu is a basis for Imϕu. Now there is a short exact sequence

0−→M∩L<u −→M∩L6u ϕu−−→ Rxueu −→ 0 ,

and it is split. Indeed, if xu 6= 0 choose an element fu in M∩L6u with ϕu( fu) = xueu,and if xu = 0 set fu = 0. The assignment xueu 7→ fu then defines a right inversehomomorphism to ϕu; cf. 1.3.5. Thus, one has M ∩ L6u = (M ∩ L<u)⊕R fu; thatis, every element m in M ∩ L6u has a unique decomposition m = m′ + r fu withm′ ∈M∩L<u and r ∈ R.

Set U ′ = u ∈U | xu 6= 0; we will show that fuu∈U ′ is a basis for M. To seethat every linear combination of the elements fu is unique, suppose that one has arelation r1 fu1 + · · ·+ rn fun = 0 with ri ∈ R and ui ∈ U ′. We may assume that theelements ui are ordered u1 < · · · < un, and thus consider the relation in M∩L6un .



Applying ϕun to the relation one gets rnxuneun = 0. As un is in U ′, the singletonxun eun is a basis for Imϕu, whence one has rn = 0. Continuing in this manner, onegets rn = · · ·= r1 = 0, as desired. If F = R〈 fu | u ∈U ′〉 were a proper submodule ofM, then there would be a least u in U ′ such that M∩L6u contains an element m notin F . This element has a unique decomposition m = m′+r fu with m′ ∈M∩L<u andr ∈ R. The element m′ is in M∩L6v for some v < u and hence in F by minimalityof u. However, r fu is also in F , so one has m = m′+ r fu ∈ F ; a contradiction.

REMARK. A commutative ring is a principal ideal domain if (and only if) every submodule of afree module is free; see ??.

The content of the next result is often phrased as: the category of R-modules hasenough free modules.

1.3.11 Lemma. For every R-module M there is a surjective homomorphism L→Mof R-modules where L is free. Moreover, if M is generated by n elements, then onecan choose L such that it has a basis with n elements.

PROOF. Choose a set G of generators for M and let E = eg | g ∈G be an abstractset. Consider the free R-module L = R〈E〉 and define by 1.3.5 a homomorphismπ : L→M by π(∑g∈G rgeg) = ∑g∈G rgg; it is surjective by the assumption on G.

FINITE GENERATION OF HOM AND TENSOR PRODUCT

Under suitable assumptions on the ring, the Hom and tensor product functors restrictto the subcategories of finitely generated modules.

1.3.12 Proposition. Let S be right Noetherian. Let M be an R-module and X be anR–So-bimodule. If M is finitely generated and X is finitely generated over So, thenthe So-module HomR(M,X) is finitely generated.

PROOF. Choose by 1.3.11 a surjective homomorphism of R-modules L→M, whereL is free and finitely generated; say, L∼= Rn as R-modules. Apply the left exact func-tor HomR(–,X) to get an injective homomorphism HomR(M,X)→ HomR(L,X).The counitor (1.2.1.2) and additivity of the Hom functor yield an isomorphism ofSo-modules, HomR(L,X)∼= Xn. Thus, HomR(M,X) is a submodule of a finitely gen-erated So-module and hence finitely generated, as S is right Noetherian.

REMARK. For M and X as in 1.3.12 the S-module HomR(X ,M) need not be finitely generated, noteven if R and S are Noetherian; see E 1.3.31.

1.3.13 Proposition. Let N be an S-module and let X be an R–So-bimodule. If N isfinitely generated and X is finitely generated over R, then the R-module X⊗S N isfinitely generated.

PROOF. Choose by 1.3.11 a surjective homomorphism of S-modules L→ N, whereL is free and finitely generated; say, L ∼= Sn as S-modules. Apply the right ex-act functor X⊗S – to get a surjective homomorphism X⊗S L→ X⊗S N. The uni-tor (1.2.1.1) and additivity of the tensor product yield an isomorphism of R-modules,


22 1 Modules

X⊗S L ∼= Xn. Thus, X⊗S N is a homomorphic image of a finitely generated R-module and hence finitely generated.

PROJECTIVE MODULES

For an R-module M, the functors HomR(M,–), HomR(–,M), and –⊗R M are, ingeneral, not exact. Modules that make one or more of these functors exact are ofparticular interest and play a pivotal role in homological algebra.

1.3.14 Definition. An R-module P is called projective if the functor HomR(P,–)from M(R) to M(k) is exact.

Part (ii) below captures the lifting property of projective modules, which amountsto the definition of projective objects in a general category.

1.3.15 Proposition. For an R-module P, the following conditions are equivalent.(i) P is projective.

(ii) For every homomorphism α : P→ N and every surjective homomorphismβ : M→ N, there exists a homomorphism γ : P→M such that the diagram

Pγ

~~

α

Mβ// // N

in M(R) is commutative; that is, there is an equality α= βγ.(iii) Every exact sequence 0→M′→M→ P→ 0 of R-modules is split.(iv) P is a direct summand of a free R-module.

PROOF. Conditions (i) and (ii) are equivalent, as the Hom functor is left exact.(ii)=⇒ (iii): Let β denote the homomorphism M P. by (ii) there exists a ho-

momorphism γ : P→M such that 1P = βγ holds, whence the sequence is split.(iii)=⇒ (iv): Choose by 1.3.11 a surjective homomorphism β : L→ P, where L is

free. The associated exact sequence 0→Kerβ→ L→ P→ 0 is split, so P is a directsummand of L.

(iv)=⇒ (ii) As the Hom functor is additive, it is sufficient to prove that every freeR-module has the lifting property. Let β : M N and α : L→ N be homomorphismsof R-modules. Assume that L is free with basis E = eu | u ∈U. For every u inU , choose by surjectivity of β a preimage mu ∈M of α(eu). The map E→M givenby eu 7→ mu extends by the unique extension property 1.3.5 to a homomorphismγ : L→M with α= βγ.

1.3.16 Corollary. Every free R-module is projective.

A homomorphism PM with P projective is called a projective precover of M.It follows from 1.3.11 and 1.3.16 that every R-module has a projective precover.



1.3.17 Corollary. Let R be a principal left ideal domain. An R-module is projectiveif and only if it is free.

PROOF. Combine 1.3.10 and 1.3.15.

REMARK. That every projective module is free is also true over a local ring (Kaplansky [45]) andover a polynomial algebra k[x1, . . . ,xn], where k is a field (Bass [11], Quillen [69], and Suslin [82]).The result for the polynomial algebra resolved Serre’s problem on projective modules; Lam hasgiven a thorough account of its history in [52].

1.3.18 Example. In a product ring R×S, the ideal a= R×0 is a projective module;if S is non-zero then a is not free.

1.3.19 Example. In the commutative ring Z[√−5] = a+ b

√−5 | a,b ∈ Z, the

subset a= a+b√−5 | a = b mod 2 is an ideal. The elements 1±

√−5 in a have

no common factor, so a is not a principal ideal, whence it is not free. Yet, the mapZ[√−5]⊕Z[

√−5]→ a⊕a given by (r,s) 7→ (2r+(1+

√−5)s,2s+(1−

√−5)r)

is an isomorphism of Z[√−5]-modules, so a is a direct summand of a free module

and hence projective.

REMARK. A classic example of a non-free projective can be found, for example, in [24, 19.17].Let R be the coordinate ring of the real 2-sphere, R = R[x,y,z]/(x2 +y2 + z2−1), and consider thehomomorphism ϕ : R→ R3 given by r 7→ (rx,ry,rz) where, by abuse of notation, x, y, and z nowdenote the residue classes of the indeterminates. The sequence 0−→ R ϕ−→ R3 −→ Cokerϕ−→ 0 issplit exact, so the module Cokerϕ is projective; one can, however, show that it is not free.

1.3.20 Proposition. Let Puu∈U be a family of R-modules. The coproduct∐

u∈U Pu

is projective if and only if each module Pu is projective.

PROOF. If the coproduct∐

u∈U Pu is projective, then by 1.3.15 it is a direct summandof a free module, and hence so is each module Pu. Conversely, if each module Pu

is projective and hence a direct summand of a free module Lu, then the coproduct∐u∈U Pu is a direct summand of the free module

∐u∈U Lu; cf. 1.3.8.

REMARK. A product of projective modules need not be projective; see E 1.3.15.

INJECTIVE MODULES

Injective modules are categorically dual to projective modules.

1.3.21 Definition. An R-module I is called injective if the functor HomR(–, I) fromM(R)op to M(k) is exact. If the functor HomR(–, I) is faithfully exact, then I iscalled faithfully injective.

1.3.22. An R-module I is injective if and only if it has the following lifting property,which amounts to the definition of an injective object in a general category. Given ahomomorphism α : K→ I and an injective homomorphism β : K→M, there existsa homomorphism γ : M→ I such that the diagram


24 1 Modules

K //β//

α

M

γ~~

I

in M(R) is commutative; that is, there is an equality γβ= α.

REMARK. The epimorphisms in the category M(R) are exactly the surjective homomorphisms;see E 1.1.2. In view of this and 1.3.15, an R-module P is projective if and only if the functorHomR(P,–) : M(R)→M(k) takes epimorphisms to epimorphisms.

The epimorphisms in the category M(R)op correspond to monomorphisms—which by E 1.1.2are the injective homomorphisms—in M(R). In view of this and 1.3.22, an R-module I is injectiveif and only if the functor HomR(–, I) : M(R)op→M(k) takes epimorphisms to epimorphisms.

1.3.23 Proposition. Let Iuu∈U be a family of R-modules. The product∏

u∈U Iu isinjective if and only if each module Iu is injective.

PROOF. Let β : K→M be an injective homomorphism of R-modules, let Iuu∈Ube a family of R-modules, and set I =

∏u∈U Iu. Assume first that each module Iu

is injective. Let a homomorphism α : K→ I be given. For each u ∈ U set αu =$uα, where $u is the projection I Iu. By assumption there exist homomorphismsγu : M→ Iu, such that γuβ= αu holds for every u ∈U . The unique homomorphismγ : M→ I with γu =$uγ now satisfies γβ= α, so I is injective.

Assume now that I is injective, fix an element u ∈U , and let εu denote the injec-tion Iu I. Given a homomorphism α : K→ Iu, one has a homomorphism εuα fromK to I. By injectivity of I there is a homomorphism γ : M→ I such that γβ = εuαholds and, therefore, one has ($uγ)β= α.

REMARK. A coproduct of injective modules need not be injective; see E 1.3.21.

1.3.24 Theorem. The following conditions are equivalent.(i) R is semi-simple.

(ii) Every short exact sequence of R-modules is split.(iii) Every R-module is projective.(iv) Every R-module is injective.

PROOF. Recall that R being semi-simple means that every submodule M′ of an R-module M is a direct summand of that module; conditions (i) and (ii) are, therefore,equivalent. It is evident that (ii) implies (iii) and (iv). The converse implicationsfollow from 1.3.15 and 1.3.22.

REMARK. The Artin–Wedderburn Theorem asserts that a ring is semi-simple if and only if it isisomorphic to a product Mn1×n1 (D1)×·· ·×Mnk×nk (Dk) where D1, . . . ,Dk are division rings; see[51, (3.5)]. In particular, a commutative ring is semi-simple if and only if it is a finite product offields. An arbitrary product of fields is a von Neumann regular ring; see 1.3.40.

1.3.25 Example. Every module over a division ring is injective, as division ringsare semi-simple.

The next result is known as Baer’s criterion.



1.3.26 Theorem. Let I be an R-module, let a be a left ideal in R, and let ι : a Rbe the embedding. The module I is injective if and only if for every R-module ho-momorphism ϕ : a→ I there exists a homomorphism ϕ′ : R→ I with ϕ′ι= ϕ.

PROOF. The “only if” part of the statement follows from 1.3.22. To prove “if”, letα : K→ I and β : K→M be homomorphisms of R-modules and assume that β isinjective. Denote by G the set of all homomorphisms γ′ : M′→ I with Imβ ⊆ M′

and γ′β = α. Since αβ−1 : Imβ→ I belongs to G, this set is non-empty. By declar-ing (γ′ : M′→ I) 6 (γ′′ : M′′→ I) if M′ ⊆ M′′ and γ′′|M′ = γ′, the set G becomesinductively ordered. Hence Zorn’s lemma guarantees the existence of a maximal el-ement γ′′ : M′′→ I. To prove the lemma, we show the equality M′′ = M. Assume,towards a contradiction, that M′′ is a proper submodule of M and choose an elementm ∈ M \M′′. The set a= (M′′ :R m) is a left ideal of R. The map ϕ : a→ I givenby ϕ(r) = γ′′(rm) is an R-module homomorphism, so by assumption it has a liftϕ′ : R→ I. It follows from the definition of a that the map γ′ : M′′+Rm→ I givenby γ′(m′′+ rm) = γ′′(m′′)+ϕ′(r) is well-defined. It is evidently a homomorphismwhose restriction to M′′ is γ′′, so one has γ′β= γ′′β= α. Hence γ′ belongs to G andsatisfies γ′ > γ′′, which contradicts the maximality of γ′′.

REMARK. It appears that Baer’s criterion has no real counterpart for projective modules; perhapsE 1.4.7 comes as close as one can get.

1.3.27. Let R be a domain. Recall that an R-module M is called divisible if rM = Mholds for all r 6= 0 in R. Every injective R-module I is divisible. Indeed, right-multiplication by r 6= 0 yields an injective homomorphism R r−→ R of R-modules,so the induced homomorphism HomR(R, I)→ HomR(R, I) of k-modules is surjec-tive, whence one has rI = I; cf. (1.2.1.2). The converse statement in 1.3.28 below,however, hinges crucially on the principal ideal hypothesis, as illustrated in 1.3.29.

1.3.28 Proposition. Let R be a principal left ideal domain. An R-module is injectiveif and only if it is divisible. Moreover, every quotient of an injective R-module isinjective.

PROOF. Once the first claim is proved, the second one follows, as the divisibilityproperty is inherited by quotient modules. Every injective R-module is divisible by1.3.27. Let I be a divisible R-module. Let a be a left ideal in R; by assumption thereexists an element x ∈ R with Rx = a. A homomorphism of R-modules α : a→ I isdetermined by the value α(x) = i, and it can be lifted to a homomorphism R→ Ias there exists an element i′ ∈ I with xi′ = i. Thus, it follows from Baer’s crite-rion 1.3.26 that I is injective.

REMARK. The field of fractions of an integral domain is evidently divisible and, in fact, injective.For principal ideal domains this follows from 1.3.28; see E 11.1.5 for the general case.

1.3.29 Example. Let k be a field and consider the polynomial ring R = k[x,y];its field of fractions is the field Q = k(x,y) of rational functions. Evidently, Q and,therefore, Q/R are divisible R-modules. However, Q/R is not injective. Indeed, con-sider map R→ Q/R given by


26 1 Modules

f 7−→[

f (x,0)xy

]R.

While it is additive, it is not R-linear. However, its restriction to the ideal M ofpolynomials with zero constant term (called the irrelevant maximal ideal) is R-linear.This homomorphism M→ Q/R does not extend to a homomorphism ϕ : R→ Q/R.If it did, then ϕ(1), which has the form [gh−1]R for some g and h in R, would satisfyyϕ(1) = ϕ(y) = [0]R and xϕ(1) = ϕ(x) = [y−1]R. The first equation shows that therewould be a k ∈ R with ygh−1 = k and, therefore, ϕ(1) = [gh−1]R = [ky−1]R. Tosatisfy the second equation there would be an l in R with xky−1 = y−1 + l, that is,xk = 1+ ly, which is absurd.

FAITHFUL INJECTIVITY

It follows from 1.3.28 that Q and Q/Z are injective Z-modules; one can say evenmore about the latter module.

1.3.30 Proposition. The Z-module Q/Z is faithfully injective.

PROOF. Let G be a non-zero Z-module, and choose an element g 6= 0 in G. Definea homomorphism ξ from the cyclic submodule Z〈g〉 to Q/Z as follows. If g is tor-sion set ξ(g) = [ 1

n ]Z, where n be the least positive integer with ng = 0. If g is nottorsion, set ξ(g) = [ 1

2 ]Z. By the lifting property of injective modules 1.3.22, there isa homomorphism G→ Q/Z that restricts to ξ on Z〈g〉, whence HomZ(G,Q/Z) isnon-zero.

1.3.31. Let E be a k-module. By adjunction 1.2.5 and the unitor (1.2.1.1) there arenatural isomorphisms of functors from M(R) to M(k),

HomR(–,Homk(R,E)) ∼= Homk(R⊗R –,E) ∼= Homk(–,E) .

Thus, if the k-module E is (faithfully) injective, then the R-module Homk(R,E) is(faithfully) injective.

1.3.32 Definition. The faithfully injective R-module ER =HomZ(RR,Q/Z) is calledthe character module of R. The abbreviated notation E is used for Ek.

FINITELY PRESENTED MODULES

1.3.33. Let M be an R-module. By 1.3.11 there exist free R-modules L and L′ suchthat there is an exact sequence

(1.3.33.1) L′ −→ L−→M −→ 0 .

1.3.34 Definition. Let M be an R-module. The exact sequence (1.3.33.1) is called afree presentation of M. If M has a free presentation with L and L′ finitely generated,then M is called finitely presented.



Every finitely presented module is finitely generated; the converse holds over leftNoetherian rings; in fact, it characterizes left Noetherian rings.

1.3.35 Lemma. Let M be a finitely presented R-module. The next assertions hold.(a) For every surjective homomorphism π : L→M with L finitely generated and

free, the submodule Kerπ is finitely generated.(b) For every finitely generated submodule N of M, the quotient module M/N is

finitely presented.

PROOF. As M is finitely presented, there exists a surjective homomorphism of R-modules π′ : L′→M with L′ finitely generated free and Kerπ′ finitely generated.

(a): Denote by X the kernel of the homomorphism L⊕L′→M that maps (l, l′) toπ(l)−π′(l′). As π and π′ are surjective, the canonical homomorphisms X → L andX → L′ that map (l, l′) to l and l′, respectively, are surjective; notice that they havekernels 0⊕Kerπ′ and Kerπ⊕0. It follows from 1.3.15 that there are isomorphismsKerπ′⊕L ∼= X ∼= Kerπ⊕L′. Thus, Kerπ is isomorphic to a direct summand of thefinitely generated module Kerπ′⊕L and hence finitely generated.

(b): There is a commutative diagram with exact rows,

0 // 0 //

L′ =//

π′

L′ //

κπ′

0

0 // N // M κ// M/N // 0 ,

where κ is the canonical surjection. By the Snake Lemma 1.1.4 there is an exactsequence 0→ Kerπ′→ Ker(κπ′)→ N→ 0. As Kerπ′ and N are finitely generated,it follows that also Ker(κπ′) is finitely generated, whence M/N is finitely presented.

FLAT MODULES

1.3.36 Definition. An R-module F is called flat if the functor –⊗R F from M(Ro)to M(k) is exact. If –⊗R F is faithfully exact, then F is called faithfully flat.

1.3.37 Example. Let R be commutative and let U ⊆ R be a multiplicative subset.As localization at U is an exact functor, the isomorphism in 1.1.9 shows that theR-module U−1R is flat. In particular, if R is an integral domain, then its field offractions is flat as an R-module.

1.3.38 Example. It is elementary to verify that every non-zero free module is faith-fully flat. Hence, by additivity of the tensor product, every projective module is flat;see 1.3.15. The Z-module Q is flat by 1.3.37, but it is not faithfully flat, and in viewof 1.3.17 it is elementary to show that it is not projective either.

1.3.39 Lemma. Let M be an R-module and K be a submodule of M. If the quotientmodule M/K is flat, then one has bK = bM∩K for every right ideal b in R.


28 1 Modules

PROOF. Let ι be the embedding b R. For every module X set µXb = µX

R (ι⊗R X),where µX

R is the unitor (1.2.1.1). By assumption, the homomorphism ι⊗R M/K isinjective, hence so is µM/K

b . There is a commutative diagram with exact rows

b⊗R K //

µKb

b⊗R M //

µMb

b⊗R M/K //

µM/Kb

0

0 // K // M // M/K // 0

where the upper row is obtained by application of b⊗R – to the lower row. One hasImµK

b = bK and ImµMb = bM, so it follows from the Snake Lemma 1.1.4 that the

canonical homomorphism K/bK→M/bM is injective. This yields the containmentK∩bM ⊆ bK; the opposite containment is trivial, so equality holds.

Our first application of 1.3.39 is to von Neumann regular rings.

1.3.40 Example. Let Ruu∈U be a family of von Neumann regular rings, for ex-ample fields. Let x = (xu)u∈U be an element in the product ring R =×u∈U Ru andset r = (ru)u∈U , where ru satisfies xu = xuruxu in Ru. One then has x = xrx in R, soR is a von Neumann regular ring.

1.3.41 Proposition. If every R-module is flat, then R is von Neumann regular.

PROOF. Let x be an element in R and set M = R, K = Rx, and b= xR; then one hasx ∈ bM∩K. If every R-module is flat, then 1.3.39 yields x ∈ bK = (xR)(Rx), so R isvon Neumann regular.

The von Neumann regular rings are, in fact, precisely the rings over which everymodule is flat; this is proved in 8.5.8.

REMARK. Von Neumann regular rings are also known as absolutely flat rings.

The next theorem is an easy consequence of a theorem due to Govorov andLazard, see C.19. Here we present a direct proof.

1.3.42 Theorem. For an R-module F , the following conditions are equivalent.(i) F is a finitely presented flat R-module.

(ii) F is a finitely generated projective R-module.(iii) F is a direct summand of a finitely generated free R-module.

PROOF. (i)=⇒ (ii): Let L′ → L→ F → 0 be a presentation with L and L′ finitelygenerated free R-modules. It follows that F and the kernel K of the surjection L→ Fare finitely generated, so one has a short exact sequence 0→ K → L→ F → 0 offinitely generated R-modules. Showing that F is projective is by 1.3.15 tantamountto showing that this sequence is split. To this end we construct a homomorphism% : L→ K with %|K = 1K .

Let e1, . . . ,em be a basis for L and let k1, . . . ,kn be a set of generators forthe submodule K. Let k ∈ K be given; we start by constructing a homomorphism%k : L→ K with %k(k) = k. Write k in terms of the basis: k = r1e1 + · · ·+ rmem. Let



b be the right ideal in R generated by r1, . . . ,rm, then one has k ∈ K∩bL, whence kis in bK by 1.3.39. It follows that there are elements bi ∈ b and x ji ∈ R such that thefollowing equalities hold,

(?) k =n∑

i=1biki =

n∑

i=1

( m∑j=1

r jx ji)ki =

m∑j=1

r j( n

∑i=1

x jiki).

Define %k by e j 7→ ∑ni=1 x jiki, then %k(k) = %k

(∑

mj=1 r je j

)= k holds by (?).

To construct a homomorphism % : L→ K whose restriction to K is the identity, itsuffices to construct % with %(ki) = ki for all the generators k1, . . . ,kn. Proceed byinduction on n; the construction above settles the base case n = 1. For n > 1 thereexists by the same construction a homomorphism %kn that fixes kn. For i < n setk′i = ki−%kn(ki). By the induction hypothesis, there is a homomorphism %′ : L→ Kwith %′(k′i) = k′i for i < n. Now, set %= %′−%′%kn +%kn , then one has

%(kn) = %′(kn)−%′(kn)+ kn = kn,

and for i < n one has

%(ki) = %′(ki)−%′%kn(ki)+%kn(ki) = %′(k′i)+%kn(ki) = k′i +%kn(ki) = ki .

(ii)=⇒ (iii): By 1.3.11 there is a short exact sequence 0→ K → L→ F → 0,where L is a finitely generated free R-module, and by 1.3.15 the sequence is split.

(iii)=⇒ (i): There is a split exact sequence 0→ K→ L→ F → 0 of R-modules,where L is finitely generated and free. The module K is also a homomorphic imageof L, in particular it is finitely generated, so by 1.3.11 there is a finitely generatedfree R-module L′ and a surjective homomorphism L′ → K. Thus, one has a finitepresentation L′→ L→ F → 0, and F is flat by 1.3.38.

A characterization in non-homological terms of flat modules over principal idealdomains—akin to 1.3.17 and 1.3.28—is given in 11.1.25.

FLAT–INJECTIVE DUALITY

Projective and injective objects are categorically dual. In module categories there isanother important duality: one between flat and injective modules which is rootedin the adjointness of Hom and tensor product.

1.3.43 Proposition. For an R-module F , the following conditions are equivalent.(i) F is flat.

(ii) For every right ideal b in R, the homomorphism ι⊗R F , induced by the em-bedding ι : b R, is injective.

(iii) The Ro-module Homk(F,E) is injective for every injective k-module E.(iv) The Ro-module Homk(F,E) is injective for a faithfully injective k-module E.

Moreover, if E is a faithfully injective k module, then the Ro-module Homk(F,E) isfaithfully injective if and only if F is a faithfully flat R-module.


30 1 Modules

PROOF. Condition (iv) clearly follows from (iii), and it follows from the definition1.3.36 that (i) implies (ii).

(ii)=⇒ (iii): Apply the exact functors –⊗R F followed by (–)∨ = Homk(–,E) tothe short exact sequence 0→ b→ R→ R/b→ 0. This yields the upper row in thefollowing commutative diagram

0 // (R/b⊗R F)∨ //

∼=

(R⊗R F)∨ //

∼=

(b⊗R F)∨ //

∼=

0

0 // HomRo(R/b,F∨) // HomRo(R,F∨) // HomRo(b,F∨) // 0 .

The vertical isomorphisms follow from commutativity 1.2.2 and adjunction 1.2.5.By commutativity of the diagram, the lower row is an exact sequence. Thus, theRo-module F∨ = Homk(F,E) is injective by Baer’s criterion 1.3.26.

(iv)=⇒ (i): By adjunction 1.2.5 and commutativity 1.2.2 there is a natural iso-morphism of functors from M(R) to M(k),

(?) HomRo(–,Homk(F,E)) ∼= Homk(–⊗R F ,E) .

By assumption, the left-hand functor is exact and Homk(–,E) is faithfully exact; itfollows that –⊗R F is exact, whence F is flat.

Assuming that E is faithfully injective, it also follows from (?) that the functorHomRo(–,Homk(F,E)) is faithfully exact if and only if –⊗R F is so.

In Sect. 5.3 we shall make extensive use of the following special case.

1.3.44 Corollary. For every non-zero free Ro-module L, the R-module Homk(L,E)is faithfully injective.

PROOF. Immediate from 1.3.38 and 1.3.43.

EXERCISES

E 1.3.1 Show that Q is not a finitely generated (f.g.) Z-module and that R is not a f.g. Q-module.E 1.3.2 Let L be an R-module and let E = euu∈U be a set of generators for L. Show that

every element in L can be expressed uniquely on the form ∑u∈U rueu if and only if someelement in L can be expressed uniquely on that form.

E 1.3.3 Let L be an R-module and let E be a subset of L. Show that if every map from E to anR-module M extends uniquely to a homomorphism L→ M of R-modules, then E is abasis for L; in particular, L is free.

E 1.3.4 Let k be a field and let M be a k-vector space of infinite rank. Show that the endomor-phism ring Homk(M,M) does not have IBN.

E 1.3.5 Show that every division ring has IBN.E 1.3.6 Let L and L′ be finitely generated free k-modules. Show that the k-module Homk(L,L′)

is free and find its rank as a function of the ranks of L and L′.E 1.3.7 Denote by S the category of sets. Show that the functor S→M(R) given by U 7→ R(U)

is a left adjoint for the forgetful functor M(R)→ S.



E 1.3.8 Let L be a free module with basis euu∈U and let K be a submodule of L generated byelements kvv∈V . (a) Show that if V is a finite set, then there is a finite subset U ′ of Usuch that K is contained in L′ = R〈euu∈U ′ 〉. (b) Show that if V is infinite, then thereis a subset U ′ of U with cardU ′ 6 cardV such that K is contained in L′ = R〈euu∈U ′ 〉.

E 1.3.9 Let M be an Ro-module generated by elements x1, . . . ,xm, and let N be an R-modulegenerated by y1, . . . ,yn. Assume that R is generated as a k-module by r1, . . . ,rl . Showthat the elements in the set xi⊗ rhy j | 1 6 h 6 l, 1 6 i 6 m, 1 6 j 6 n generate thek-module M⊗R N.

E 1.3.10 Let L be a free k-module with basis e1, . . . ,en. Let l1, . . . , ln be elements in L andwrite l j = ∑

nj=1 ai je j . Show that l1, . . . , ln is a basis for L if and only if the matrix

(ai j)16i, j6n is invertible.E 1.3.11 Let P be a projective R-module. Show that there exists a free R-module L with P⊕L∼= L.

This is known as Eilenberg’s swindle.E 1.3.12 Show that a direct summand of a projective/injective/flat module is projective/injective/

flat.E 1.3.13 Dualize the proof of 1.3.23 to show that a coproduct of projective modules is projective.

This provides an alternative proof of 1.3.20.E 1.3.14 (Cf. 1.3.38) Show that every non-zero free module is faithfully flat, and show that Q as

Z-module is flat but not projective.E 1.3.15 Show that ZN is not a projective Z-module.E 1.3.16 Show that Z is not a homomorphic image of an injective Z-module.E 1.3.17 Let R be left hereditary. Show that a submodule of a projective R-module is projective.E 1.3.18 Show that if P is a free/projective R-module and P′ is a free/projective k-module, then

the R-module P⊗k P′ is free/projective.E 1.3.19 Let F be an R-module and let F ′ be a flat k-module. Show: (a) If F is flat, then the

R-module F⊗k F ′ is flat. (b) If F ′ is faithfully flat, then the R-module F⊗k F ′ is (faith-fully) flat if and only if F is (faithfully) flat.

E 1.3.20 Let 0→K→ L→M→ 0 be an exact sequence of R-modules with L free. If K is finitelygenerated, then M is said to be finitely related. Show that a finitely related flat module isprojective. (A countably related flat module is almost projective; see E.8.)

E 1.3.21 In the ring R =×n∈NQ consider the ideals ai = (qn)n∈N ∈ R | qn = 0 for all n 6= ifor every i∈N. Show that each ai is an injective R-module but that

∐i∈N a

i is not. Showalso that the R-module R is injective.

E 1.3.22 Show that if I is an injective R-module and P is a projective k-module, then the R-moduleHomk(P, I) is injective.

E 1.3.23 (a) Show that HomZ(I,Z) = 0 holds for every injective Z-module I. (b) Consider the fullsubcategory of M(Z) whose objects are the injective Z-modules. Show that in this cate-gory the canonical homomorphism Q Q/Z is a monomorphism and an epimorphismbut not an isomorphism. Hint: 1.3.28.

E 1.3.24 Let R→ S be a ring homomorphism. Show that if P is a free/projective R-module, thenthe S-module S⊗R P is free/projective.

E 1.3.25 Let R→ S be a ring homomorphism. Show that, with the induced structure, S is faithfullyflat as an R-module if and only if R→ S is injective and S/R is flat as an R-module.

E 1.3.26 Let R→ S be a ring homomorphism. Show that if F is a (faithfully) flat R-module, thenthe S-module S⊗R F is (faithfully) flat.

E 1.3.27 Let R→ S be a ring homomorphism. Show that if I is a (faithfully) injective R-module,then the S-module HomR(S, I) is (faithfully) injective.

E 1.3.28 Let M be an R-module and E be a faithfully injective R-module. Show that for everym 6= 0 in M there exists a homomorphism ϕ ∈ HomR(M,E) with ϕ(m) 6= 0.

E 1.3.29 Give a proof of 1.3.31 that does not use adjunction 1.2.5. Hint: Let γ : M→ E be a non-zero homomorphism of k-modules. For m ∈ M consider the map γm : R→ E defined


32 1 Modules

by γm(r) = γ(rm). Show that m 7→ γm is a non-zero homomorphism of R-modules. Toprove injectivity, turn a map to Homk(R,E) into a map to E by evaluation at 1.

E 1.3.30 There exists a family Auu∈U of infinite subsets of N such that cardU = 2ℵ0 andAu ∩Av is finite for all u 6= v in U ; see for example Sierpinski [78, IV. §14, thm. 1]. Letk be a field and use this fact to show that a basis for kN has cardinality at least 2ℵ0 .

E 1.3.31 Let k be a field and set R = k[x]. Show that Homk(R,k) is not a finitely generated R-module. Hint: E 1.3.30.

1.4 Evaluation Homomorphisms

SYNOPSIS. Biduality; tensor evaluation; homomorphism evaluation.

With projective, injective, and flat modules now available, we continue the compar-isons, started in Sect. 1.2, of composites of Hom and tensor product functors.

1.4.1 Definition. Let L be a free R-module with basis euu∈U . For each u ∈U lete∗u ∈ HomR(L,R) be given by ev 7→ δuv, where δ denotes the Kronecker delta.

If euu∈U is a basis for a vector space L of finite rank, then the family of func-tionals e∗uu∈U is a basis for the dual space L∗, and it is referred to as the dual basis.

BIDUALITY

For a vector space M of finite rank there is a canonical isomorphism from M to itsdouble dual space M∗∗; it is given by evaluation: A basis element ev is mapped tothe functional given by e∗u 7→ e∗u(ev) = δuv. This isomorphism is an instance of thebiduality map that we consider next.

1.4.2 Lemma. Let X be an R–So-bimodule. For an R-module M, the biduality map,

δMX : M −→ HomSo(HomR(M,X),X)

given byδM

X (m)(ψ) = ψ(m) ,

is a homomorphism of R-modules, and it is natural in M. Moreover, if M is inM(R–T o), then δM

X is a homomorphism in M(R–T o).

PROOF. It is straightforward to verify that δ is a natural transformation of endofunc-tors on M(R); see the proof of 1.2.2.

If M is an R–T o-bimodule, then HomSo(HomR(M,X),X) is an R–T o-bimoduleas well. For t ∈ T , m ∈M, and ψ ∈ HomR(M,X) one has

δMX (mt)(ψ) = ψ(mt) = (tψ)(m) = δM

X (m)(tψ) = (δMX (m)t)(ψ) ;

that is, the homomorphism δMX is T o-linear.


1.4 Evaluation Homomorphisms 33

1.4.3 Proposition. Let P be a finitely generated projective R-module. The Ro-module HomR(P,R) is finitely generated and projective, and biduality

δPR : P −→ HomRo(HomR(P,R),R)

is an isomorphism.

PROOF. By 1.3.42 the module P is a direct summand of a finitely generated freeR-module L. Let euu∈U be a basis for L. The Ro-module L∗ = HomR(L,R) isfree with basis e∗uu∈U ; see 1.4.1. Indeed, for every element ϕ in L∗ one has ϕ =

∑u∈U e∗uϕ(eu). The Hom functor is additive, so the Ro-module HomR(P,R) is a directsummand of L∗, whence it is finitely generated and projective.

Again because the Hom functor is additive, it is sufficient to show that δLR is an

isomorphism for a cyclic free module L = R〈e〉. Let e∗ be the functional L→ Rgiven by e∗(e) = 1; every element ψ in HomR(L,R) has the form e∗a with a =ψ(e). As one has δL

R(e)(e∗) = e∗(e) = 1 the homomorphism δL

R is injective. Let ϑbe a homomorphism in HomRo(HomR(L,R),R); for every ψ ∈ HomR(L,R) one hasϑ(ψ) = ϑ(e∗ψ(e)) = ϑ(e∗)ψ(e) = ϑ(e∗)δL

R(e)(ψ), and hence ϑ = bδLR(e) = δL

R(be)with b = ϑ(e∗), so δL

R is surjective as well.

TENSOR EVALUATION

For a vector space M of finite rank there is an isomorphism M∗⊗M→ Hom(M,M);it assigns to a basis element e∗u⊗ev the linear map given by ew 7→ e∗u(ew)ev = δuwev.It is a special case of a map known as tensor evaluation.

1.4.4 Lemma. Let M be an R-module, X be an R–So-bimodule, and N be an S-module. The tensor evaluation map,

θMXN : HomR(M,X)⊗S N −→ HomR(M,X⊗S N)

given byθMXN(ψ⊗n)(m) = ψ(m)⊗n ,

is a homomorphism of k-modules, and it is natural in M, X , and N. Moreover, if Mis in M(R–Qo) and N is in M(S–T o), then θMXN is a homomorphism in M(Q–T o).

PROOF. It is straightforward to verify that θ is a natural transformation of functorsfrom M(R)op×M(R–So)×M(S) to M(k); see the proof of 1.2.2.

If M is in M(R–Qo) and N is in M(S–T o), then HomR(M,X)⊗S N is a Q–T o-bimodule and so is HomR(M,X⊗S N). The computation

θMXN(q(ψ⊗n)t)(m) = θMXN(qψ⊗nt)(m)

= (qψ)(m)⊗nt

= ψ(mq)⊗nt

= (ψ(mq)⊗n)t

= (θMXN(ψ⊗n)(mq))t

= (q(θMXN(ψ⊗n))t)(m) ,


34 1 Modules

which holds for all q ∈ Q, t ∈ T , ψ ∈ HomR(M,X), m ∈M, and n ∈ N, shows thatthe homomorphism θMXN is Q- and T o-linear.

1.4.5 Example. Set R = S = k= Z. For the Z-modules M = Z/2Z= N and X = Z,the homomorphism θMXN maps from 0 to Z/2Z, so it is not an isomorphism.

1.4.6 Proposition. Let M be an R-module, X be an R–So-bimodule, and N be anS-module. Tensor evaluation 1.4.4,

θMXN : HomR(M,X)⊗S N −→ HomR(M,X⊗S N) ,

is an isomorphism under any one of the following conditions.(a) M or N is finitely generated and projective.(b) M is projective and N is finitely presented.(c) M is finitely presented and N is flat.

PROOF. (a): For every R–So-bimodule X and S-module N the counitor (1.2.1.2)yields a commutative diagram,

HomR(R,X)⊗S N θRXN// HomR(R,X⊗S N)

X⊗S N ,

∼=εX⊗N

dd∼=εX⊗N

::

which shows that θRXN is an isomorphism. A similar diagram involving the uni-tor (1.2.1.1) shows that θMXS is an isomorphism for every R-module M and everyR–So-bimodule X . By additivity of the involved functors, it now follows that θMXN

is an isomorphism if M or N is finitely generated and projective.(b): Choose a presentation of N by finitely generated free S-modules

(?) L′ −→ L−→ N −→ 0 .

Consider the following diagram, which is commutative as θ is natural by 1.4.4.

HomR(M,X)⊗S L′ //

∼= θMXL′

HomR(M,X)⊗S L //

∼= θMXL

HomR(M,X)⊗S N //

θMXN

0

HomR(M,X⊗S L′) // HomR(M,X⊗S L) // HomR(M,X⊗S N) // 0 .

Either row in this diagram is exact. Indeed, they are obtained by applying the rightexact functors HomR(M,X)⊗S – and HomR(M,X⊗S –) to (?). Right exactness ofthe latter functor hinges on the assumption that M is projective. The maps θMXL′

and θMXL are isomorphisms by part (a), and it follows from the Five Lemma thatθMXN is an isomorphism.

(c): Choose a presentation L′ → L → M → 0 of M by finitely generated freeR-modules. As in the proof of (b), we get the following commutative diagram withexact rows, since the functors HomR(–,X)⊗S N and HomR(–,X⊗S N) are left exact.Left exactness of the former functor hinges on the assumption that N is flat.



0 // HomR(M,X)⊗S N //

θMXN

HomR(L,X)⊗S N //

∼= θLXN

HomR(L′,X)⊗S N

∼= θL′XN

0 // HomR(M,X⊗S N) // HomR(L,X⊗S N) // HomR(L′,X⊗S N) .

The maps θLXN and θL′XN are isomorphisms by part (a), and it follows from the FiveLemma that θMXN is an isomorphism.

HOMOMORPHISM EVALUATION

For a vector space M of finite rank there is an isomorphism M⊗M→Hom(M∗,M);it assigns to a basis element ev⊗ew the linear map given by e∗u 7→ e∗u(ew)ev = δuwev.It is a special case of a map known as homomorphism evaluation.

1.4.7 Lemma. Let M be an R-module, X be an R–So-bimodule, and N be an So-module. The homomorphism evaluation map

ηMXN : N⊗S HomR(X ,M) −→ HomR(HomSo(N,X),M)

given byηMXN(n⊗ψ)(ϑ) = ψϑ(n)

is a homomorphism of k-modules, and it is natural in M, X , and N. Moreover, if Mis in M(R–T o) and N is in M(Q–So), then ηMXN is a homomorphism in M(Q–T o).

PROOF. It is straightforward to verify that η is a natural transformation of functorsfrom M(R)×M(R–So)op×M(So) to M(k). See the proof of 1.2.2.

If M is in M(R–T o) and N is in M(Q–So), then N⊗S HomR(X ,M) is a Q–T o-bimodule, and so is HomR(HomSo(N,X),M). The computation

ηMXN(q(n⊗ψ)t)(ϑ) = ηMXN(qn⊗ψt)(ϑ)

= (ψt)ϑ(qn)

= (ψϑ(qn))t

= (ψ(ϑq)(n))t

= (ηMXN(n⊗ψ)(ϑq))t

= (q(ηMXN(n⊗ψ))t)(ϑ) ,

which holds for all q ∈ Q, t ∈ T , ψ ∈ HomR(X ,M), n ∈ N, and ϑ ∈ HomSo(N,X),shows that the homomorphism ηMXN is Q- and T o-linear.

1.4.8 Example. Set R = S = k= Z. For the Z-modules M = Z/2Z= N and X = Z,the homomorphism ηMXN maps from Z/2Z to 0, so it is not an isomorphism.

1.4.9 Proposition. Let M be an R-module, X be an R–So-bimodule, and N be anSo-module. Homomorphism evaluation 1.4.7,

ηMXN : N⊗S HomR(X ,M) −→ HomR(HomSo(N,X),M) ,

is an isomorphism under either one of the following conditions.


36 1 Modules

(a) N is finitely generated and projective.(b) N is finitely presented and M is injective.

PROOF. (a): Using the unitor (1.2.1.1) and counitor (1.2.1.2) it is elementary toverify that ηRXN is an isomorphism for all R–So-bimodules X and all So-modules N.The claim then follows by additivity of the involved functors.

(b): Choose a presentation of N by finitely generated free So-modules

(?) L′ −→ L−→ N −→ 0 .

Consider the following diagram, which is commutative as η is natural by 1.4.7.

L′⊗S HomR(X ,M) //

∼=ηMXL′

L⊗S HomR(X ,M) //

∼=ηMXL

N⊗S HomR(X ,M) //

ηMXN

0

HomR(HomSo(L′,X),M) // HomR(HomSo(L,X),M) // HomR(HomSo(N,X),M) // 0 .

Either row in this diagram is exact. Indeed, they are obtained by applying the rightexact functors –⊗S HomR(X ,M) and HomR(HomSo(–,X),M) to (?). Right exact-ness of the latter functor hinges on the assumption that M is injective. The mapsηMXL′ and ηMXL are isomorphisms by part (a), and it follows from the Five Lemmathat ηMXN is an isomorphism.

EXERCISES

E 1.4.1 Let L be a finitely generated free R-module with basis euu∈U ; show that the function-als e∗u form a basis for the Ro-module HomR(L,R).

E 1.4.2 Let P be a finitely generated projective R-module; show that HomR(P,R) is a projectiveRo-module.

E 1.4.3 Let R be right Noetherian and let M be a finitely generated R-module. Show that theRo-module HomR(M,R) is finitely generated.

E 1.4.4 Show that biduality δMER is injective for every R-module M.

E 1.4.5 Show that biduality 1.4.2 need not be injective nor surjective.E 1.4.6 Show that for every R-module M there is an injective homomorphism M→ I where I is

injective.E 1.4.7 Show that an R-module P is projective if and only if every homomorphism I N with

I injective induces a surjective homomorphism HomR(P, I)→ HomR(P,N).E 1.4.8 Show that R is left hereditary if and only if every quotient of an injective R-module is

injective.E 1.4.9 Show that an R-module is injective if and only if it is a direct summand of a module

Homk(L,E) where L is a free Ro-module.E 1.4.10 For an R-module I, show that the following conditions are equivalent. (i) I is injective.

(ii) For every left ideal a in R, the homomorphism HomR(ι, I), induced by the embeddingι : a R, is surjective. (iii) Every exact sequence 0→ I→M→M′′→ 0 is split.

E 1.4.11 Let k be a field and let Vii∈Z and W j j∈Z be families of finite dimensional k-vector spaces. Show that the vector spaces

∐i∈ZVi and

∏j∈ZW j are isomorphic if and

only if the sets I = i ∈ Z | Vi 6= 0 and J = j ∈ Z |W j 6= 0 are finite and one has∑i∈I dimkVi = ∑ j∈J dimkW j . Hint: E 1.3.30.



E 1.4.12 Let k be a field, and let M and X 6= 0 and N be k-vector spaces. Show that tensor evalu-ation θMXN : Homk(M,X)⊗kN→ Homk(M,X⊗kN) is an isomorphism if and only ifM or N has finite rank. Hint: E 1.4.11.

E 1.4.13 Let k be a field, and let M and X 6= 0 and N 6= 0 be k-vector spaces. Show that homo-morphism evaluation ηMXN : M⊗kHomk(X ,N)→ Homk(Homk(M,X),N) is an iso-morphism if and only if M has finite rank. Hint: E 1.4.11.

E 1.4.14 Use homomorphism evaluation 1.4.7 to show that every finitely presented flat module isprojective.


Chapter 2Complexes

We now shift focus to complexes of modules. They are the objects in the derivedcategory, and our first step towards that category is to study the Abelian category ofcomplexes and their morphisms.

2.1 Definitions and Examples

SYNOPSIS. Graded module; graded homomorphism; complex; chain map; Five Lemma; SnakeLemma; (degreewise) split exact sequence; \-functor.

Many modules carry an intrinsic structure: a grading. By imposing an additionalstructure, a square zero endomap that respects the grading, one arrives at the notionof a complex. The zero map respects any grading, so one can always consider agraded module as a complex. After a short opening discussion of graded modules,we move on to complexes, which abound in mathematics. We illustrate the conceptwith examples from algebra, geometry, and topology.

GRADED MODULES

2.1.1 Definition. Let U be a set. An R-module M is called U-graded if there exists afamily Muu∈U of submodules of M such that M =

∐u∈U Mu. A Z-graded module

is simply called a graded module.

2.1.2 Example. Considered as an k-module, the ring of polynomials M = k[x] isgraded with Mv = 0 for v < 0 and Mv = m ∈M | m is a monomial of degree v forv> 0, with the convention that 0 is a monomial of every degree.

2.1.3 Definition. Let M =∐

v∈ZMv be a graded R-module. The submodule Mv iscalled the module in degree v. An element m in M that belongs to a submodule Mvis said to be homogeneous of degree v. Thus, the zero element is homogeneous of

39

40 2 Complexes

every degree. A homogeneous element m 6= 0 is homogeneous of exactly one degree,which is called the degree of m and denoted |m|; that is

|m|= v ⇐⇒ m ∈Mv .

All subsequent formulas that involve the degree |m| of an arbitrary homogeneouselement m will be valid no matter what value one assigns to |0|.

2.1.4. Let M =∐

v∈ZMv be a graded R-module and let N be an R-module. A homo-morphism α from M to N is identified with the family (αv)v∈Z in

∏v∈ZHomR(Mv,N),

given by αv = αεv where εv is the injection MvM.

2.1.5 Definition. Let M and N be graded R-modules and let p ∈ Z. A homomor-phism α : M→ N with α(Mv) ⊆ Nv+p for all v ∈ Z is called a (graded) homomor-phism of degree p. Under the identification α = (αv)v∈Z in 2.1.4 such homomor-phisms are in bijective correspondence with elements in

∏v∈ZHomR(Mv,Nv+p),

which is denoted by HomR(M,N)p. Write HomR(M,N) for the graded k-modulethat has the module HomR(M,N)p in degree p; it is called the graded Hom.

2.1.6 Example. Set M = k[x] as in 2.1.2. The derivative ddx yields a homomorphism

M→M of degree −1.

For graded R-modules M and N the symbol HomR(M,N) has two possible inter-pretations; namely the k-module of all R-linear maps from M to N, and the gradedk-module defined in 2.1.5. The two interpretations do, in general, not agree, butwhenever we consider graded modules, we have the second in mind.

2.1.7. Let L, M, and N be graded R-modules, and let p and q be integers. There is ak-bilinear composition rule for graded homomorphisms:

HomR(M,N)p×HomR(L,M)q −→ HomR(L,N)p+q .

For α= (αv)v∈Z and β= (βv)v∈Z it is given by (α,β) 7→ αβ= (αv+qβv)v∈Z . Noticethat the collection of homomorphisms of degree 0 is closed under composition.

2.1.8 Definition. A homomorphism of degree 0 is called a morphism of graded R-modules. The category of graded R-modules and their morphisms is denoted Mgr(R).

2.1.9 Example. Let M be a graded R-module and r be an element of R. The maprM : M→M given by m 7→ rm for m ∈M is called a homothety. It is a morphism inMgr(k), and if r is central, e.g. if r comes from k, then rM is a morphism in Mgr(R).

Notice that the notation rM for a homothety is in line with the notation 1M for theidentity morphism.

2.1.10 Definition. Let M be a graded R-module. A graded submodule of M is agraded R-module K that is a submodule of M as an R-module such that the embed-ding K→M is a morphism of graded R-modules.


2.1 Definitions and Examples 41

Given a graded submodule K of a graded R-module M, the graded quotient is thegraded module M/K =

∐v∈ZMv/Kv, and the canonical surjection M → M/K is a

morphism of graded R-modules.

2.1.11 Example. Let α : M→ N be a graded homomorphism of graded R-modules.The kernel Kerα= m∈M | α(m) = 0 is a graded submodule of M, and the imageImα = α(m) | m ∈ M is a graded submodule of N. The cokernel is the gradedquotient module Cokerα= N/Imα.

2.1.12. With the notions of kernels and cokernels from 2.1.11, the category Mgr(R)is k-linear and Abelian. The biproduct M⊕N of M =

∐v∈ZMv and N =

∐v∈ZNv

in Mgr(R) is the graded module∐

v∈Z(Mv⊕Nv); one refers to M⊕N as a gradeddirect sum and to M and N as graded direct summands.

2.1.13. There are exact k-linear functors between Abelian categories

M(R) //Mgr(R) .oo

The functor from M(R) to Mgr(R) is full and faithful; it equips an R-module Mwith the trivial grading M0 = M and Mv = 0 for v 6= 0, and it acts analogously onhomomorphisms. The functor from Mgr(R) to M(R) forgets the grading.

At the level of symbols, applications of these functors is suppressed. However,when we write e.g. “as an R-module” about a graded R-module, it means that weapply the forgetful functor.

REMARK. The category Mgr(R) also has products and coproducts (and limits and colimits); theywill all be treated within the context of complexes; see Chap. 3.

2.1.14 Definition. Let M be a graded Ro-module and let N be a graded R-module.Denote by M⊗R N the graded k-module with

(M⊗R N)p =∐v∈Z

Mv⊗R Np−v ;

it is called the graded tensor product of M and N. Notice that if m and n are ho-mogeneous elements in M and N, then m⊗n is homogeneous in M⊗R N of degree|m⊗n|= |m|+ |n|.

The graded tensor product has the expected universal property.

2.1.15 Proposition. Let M be a graded Ro-module, let N be a graded R-module, letX be a graded k-module, and let p be an integer. For every family of k-bilinear andmiddle R-linear maps

Φv :⊎i∈Z

Mi×Nv−i −→ Xv+pv∈Z

there is a unique degree p homomorphism of graded k-modules

ϕ : M⊗R N −→ X

with ϕ(m⊗n) =Φv(m,n) for all elements m ∈M and n ∈ N with |m⊗n|= v.


42 2 Complexes

PROOF. It is evident that such a morphism is unique if it exists. For existence, it issufficient to verify that for each integer v there is a homomorphism of k-modulesϕv : (M⊗R N)v =

∐i∈ZMi⊗R Nv−i→ Xv+p with ϕv(m⊗n) =Φv(m,n). Fix v; for

each i ∈ Z it follows from the universal property of tensor products that there isa homomorphism ϕi

v : Mi⊗R Nv−i→ Xv+p with ϕiv(m⊗n) =Φv(m,n). Now the uni-

versal property of coproducts yields the desired homomorphism ϕv.

2.1.16 Definition. Let Mgr(R–So) denote the category Mgr(R⊗k So).

2.1.17. Graded R–So-bimodules and graded homomorphisms of such are defined asin 2.1.3 and 2.1.5. The category Mgr(R–So) is naturally identified with the categorywhose objects are graded R–So-bimodules, and whose morphisms are R- and So-linear homomorphisms of degree 0; cf. 1.1.23.

In particular, Mgr(R–Ro) is identified with the category of R–Ro-bimodules, andthe category Mgr(Re) is naturally identified with the full subcategory of Mgr(R–Ro)whose objects are symmetric R–Ro-bimodules; cf. 1.1.25.

2.1.18 Addenddum to 2.1.5 and 2.1.14. If M is a graded R–Qo-bimodule and Nis a graded R–So-bimodule, then it follows from 1.1.26 that the graded k-moduleHomR(M,N) has a graded Q–So-bimodule structure.

If M is a graded Q–Ro-bimodule and N is a graded R–So-bimodule, then thegraded k-module M⊗R N has a graded Q–So-bimodule structure; cf. 1.1.29.

2.1.19 Addendum to 2.1.15. If M is in Mgr(Q–Ro) and N is in Mgr(R–So), thenthe tensor product M⊗R N is a graded Q–So-bimodule; see 2.1.18. If X is also inMgr(Q–So) and Φv :

⊎i∈ZMi×Nv−i→ Xv+pv∈Z is a family of Q- and So-linear

maps that are middle R-linear, then ϕ : M⊗R N→ X is a morphism in Mgr(Q–So).

It follows from 2.1.7 that if M is a graded R-module, then HomR(M,M) has agraded k-algebra structure with multiplication given by composition of homomor-phisms. The tensor powers of a module can also be assembled into a graded algebra.

2.1.20 Example. Let M be a k-module. Consider the graded k-module Tk(M) with

Tkp(M) = 0 for p < 0 , Tk0(M) = k , and Tkp(M) = M⊗p for p > 0 ;

where M⊗p = M⊗k · · ·⊗kM is the p-fold tensor product of M. The module Tkp(M)

is called the pth tensor power of M. With multiplication given by concatenation ofelementary tensors, Tk(M) is a graded k-algebra called the tensor algebra of M. Tobe precise, the product of elements x1⊗·· ·⊗xp in Tkp(M) and y1⊗·· ·⊗yq in Tkq(M)

is the element x1⊗·· ·⊗ xp⊗ y1⊗·· ·⊗ yq in Tkp+q(M).Denote by I the ideal in Tk(M) generated by the set x⊗ x | x ∈ M of homo-

geneous elements of degree 2; the graded quotient algebra∧k(M) = Tk(M)/I is

called the exterior algebra of M, and the module∧k

p(M) in degree p is called thepth exterior power of M. One writes x1 ∧ ·· · ∧ xp for the coset [x1⊗ ·· · ⊗ xp]I in∧k(M) and it is called the wedge product of the elements x1, . . . ,xn. As elements



of the form x⊗ y+ y⊗ x = (x+ y)⊗ (x+ y)− x⊗ x− y⊗ y belong to I, one hasx1∧·· ·∧ xp =−x1∧·· ·∧ xi+1∧ xi∧·· ·∧ xp.

We return to the exterior algebra in 2.1.24.

COMPLEXES

2.1.21 Definition. An R-complex, also called a complex of R-modules, is a gradedR-module M equipped with a homomorphism ∂M : M→M of degree −1, called thedifferential, that satisfies ∂M∂M = 0. It can be visualized as follows:

· · · −→Mv+1∂M

v+1−−→Mv∂M

v−−→Mv−1 −→ ·· · .

The module Mv is the module in degree v; the homomorphism ∂Mv : Mv→Mv−1 is

the vth differential, and one has ∂Mv ∂

Mv+1 = 0 for all v ∈ Z.

Given an R-complex M, the underlying graded R-module is denoted M\.

2.1.22 Example. Over the ring Z/4Z consider the graded module with Z/4Z ineach degree. Endowed with the degree −1 homomorphism that in each degree ismultiplication by 2, it is a Z/4Z-complex,

· · · −→ Z/4Z 2−−→ Z/4Z 2−−→ Z/4Z−→ ·· · ,

called the Dold complex.

REMARK. Another word for complex is differential graded module. In the notation for complexesintroduced in 2.1.21, degrees are written as subscripts and descend in the direction of the arrows;this is called homological notation. In cohomological notation, degrees are written as superscriptsand ascend in the direction of the arrows; in this notation, a complex M is visualized as follows:

· · · −→Mv−1 ∂v−1M−−→Mv ∂v

M−−→Mv+1 −→ ·· · .In the literature, there is a strong tradition for employing homological (cohomological) notation forcomplexes that are bounded below (above) in the sense of 2.5.2; such complexes are often referredto as chain (cochain) complexes. Switching between homological and cohomological notation, itis standard to set Mv = M−v. We have no proclivity for complexes bounded on either side, but wesettle on homological notation and only deviate from it in a few examples, such as 2.1.26 below.

2.1.23 Definition. Let M be a complex, and let u>w be integers. The complex M issaid to be concentrated in degrees u, . . . ,w if one has Mv = 0 for all v /∈ w, . . . ,u;it is then visualized like this:

0−→Mu −→Mu−1 −→ ·· · −→Mw+1 −→Mw −→ 0 .

2.1.24 Example. Let M be a k-module and let ε : M→ k be a homomorphism. Re-call from 2.1.20 that Tk(M) and

∧k(M) are the tensor algebra and the exterior alge-bra of M. Consider the degree −1 homomorphism δ : Tk(M)→

∧k(M) given by

δ(x1⊗·· ·⊗ xp) =p∑

i=1(−1)i+1ε(xi)x1∧·· ·∧ xi−1∧ xi+1∧·· ·∧ xp .


44 2 Complexes

Every element in the ideal I ⊆ Tk(M) from 2.1.20 is a k-linear combination ofelements of the form y1⊗·· ·⊗yp⊗x⊗x⊗ z1⊗·· ·⊗ zq; since δ is zero on such ele-ments, it factors through the exterior algebra, yielding a degree −1 homomorphism∂ :∧k(M)→

∧k(M) given by

∂(x1∧·· ·∧ xp) =p∑

i=1(−1)i+1ε(xi)x1∧·· ·∧ xi−1∧ xi+1∧·· ·∧ xp .

This homomorphism is square zero; indeed, one has

∂∂(x1∧·· ·∧ xp)

=p∑

i=1(−1)i+1ε(xi)∂(x1∧·· ·∧ xi−1∧ xi+1∧·· ·∧ xp)

= ∑j<i

(−1)(i+1)( j+1)ε(xi)ε(x j)x1∧·· ·∧ x j−1∧ x j+1∧·· ·∧ xi−1∧ xi+1∧·· ·∧ xp

+ ∑i<k

(−1)(i+1)kε(xi)ε(xk)x1∧·· ·∧ xi−1∧ xi+1∧·· ·∧ xk−1∧ xk+1∧·· ·∧ xp

= 0 .

The k-complex with underlying graded module∧k(M) and differential ∂ is called

the Koszul complex over ε and denoted Kk(ε). For a sequence x1, . . . ,xn in k, onewrites Kk(x1, . . . ,xn) for the Koszul complex over the canonical homomorphismfrom the free module k〈e1, . . . ,en〉 to the ideal (x1, . . . ,xn). Notice that in this impor-tant special case, the differential is given by

∂Kk(x1,...,xn)(eh1 ∧·· ·∧ ehp) =p∑

i=1(−1)i+1xieh1 ∧·· ·∧ ehi−1 ∧ ehi+1 ∧·· ·∧ ehp .

It is standard to set KRv (x1, . . . ,xn) = KR(x1, . . . ,xn)v.

2.1.25 Example. Consider for each integer n> 0 the standard n-simplex,

∆n = (t0, . . . , tn) ∈ Rn+1 |

n∑j=0

t j = 1 and t0, . . . , tn > 0 .

For n> 1 and 06 i6 n the i th face map εni : ∆n−1→ ∆n is given by

(t0, . . . , tn−1) 7−→ (t0, . . . , ti−1,0, ti, . . . , tn−1) .

Let X be a topological space. Denote by C(∆n,X) the set of all continuous mapsfrom ∆n to X , and consider the free Abelian group Sn(X) = Z〈C(∆n,X)〉 on this set.The elements of Sn(X) are called singular n-chains. For n> 1, the map

C(∆n,X)−→ Sn−1(X) given by σ 7−→n∑

i=0(−1)iσεn

i

extends uniquely by 1.3.5 to a group homomorphism ∂Xn : Sn(X)→ Sn−1(X). Set

∂X0 = 0, then one has ∂X

0 ∂X1 = 0; for n> 1 and a basis element σ in Sn+1(X) one has



∂Xn ∂

Xn+1(σ) = ∂X

n (n+1∑

i=0(−1)iσεn+1

i )

=n∑j=0

(−1) j(n+1∑

i=0(−1)iσεn+1

i )εnj

= ∑j<i

(−1)i+ jσεn+1i εn

j + ∑i6 j


j .

For 06 j < i6 n+1 one has εn+1i εn

j = εn+1j εn

i−1 and, therefore,

∑j<i


j = ∑j<i

(−1)i+ jσεn+1j εn

i−1 = ∑j6i

(−1)i+ j+1σεn+1j εn

i .

In combination, these displays yield ∂Xn ∂

Xn+1(σ) = 0, and thus one has a Z-complex,

S(X) = · · · −→ S2(X)∂X

2−−→ S1(X)∂X

1−−→ S0(X)−→ 0 .

This complex is called the singular chain complex of the space X .Let A be an Abelian group. Application of the functors –⊗Z A and HomZ(–,A)

to the complex S(X) in each degree yields complexes called the singular chain andsingular cochain complex of X with coefficients in A.

2.1.26 Example. Let M be a smooth d-dimensional real manifold. For a point xon M, denote by C∞

x (M) the germ of smooth functions at x; it is an R-algebra. Alinear map υ : C∞

x (M)→ R with υ( f g) = υ( f )g(x)+ f (x)υ(g) for all f ,g ∈ C∞x (M)

is called a derivation at x. The set Mx of all derivations is a d-dimensional vectorspace, called the tangent space at x. Given a chart ϕ= (ϕ1, . . . ,ϕd) : U → Rd , whereU is a neighborhood of x, one can form a basis ∂

∂ϕ1

∣∣x, . . . ,

∂∂ϕd

∣∣x for Mx; for f in

C∞x (M) the derivation ∂

∂ϕi

∣∣x is defined by∂∂ϕi

∣∣x( f ) = Di( f ϕ−1)

∣∣ϕ(x) ,

where Di( ·)∣∣ϕ(x) is the i th derivative at the point ϕ(x) ∈ Rd . The dual space M∗x is

called the cotangent space at x, and the dual basis is denoted by (dϕ1)x, . . . ,(dϕd)x.More generally, for a smooth function f : U → R its differential at x is the functional(d f )x ∈M∗x given by (d f )x(υ) = υ( f ) for υ ∈Mx.

Fix 0 6 k 6 d and denote by∧R

k (M∗x ) the kth exterior power of M∗x , which is a

vector space of dimension q =(d

k

). One can show that the family

∧Rk (M

∗x )x∈M

can be assembled to a smooth vector bundle Ek on M of rank q; the total space isEk =

⊎x∈M

∧Rk (M

∗x ) and the bundle projection π : EkM maps

∧Rk (M

∗x ) to x.

A differential form of degree k is a smooth section of π : EkM, i.e. a smoothmapω : M→ Ek with πω= 1M . The vector space of all such maps is denotedΩk(M).

As one has∧R

0 (M∗x ) = R, the bundle E0 = M×R is trivial. Thus, an element in

Ω0(M) is nothing but a smooth map of the form (1M, f ) : M→M×R. Hence onenaturally identifies Ω0(M) with the set C∞(M) of smooth functions M→ R.

As one has∧R

1 (M∗x ) = M∗x , an element in Ω1(M) is a smooth map ω : M→ E1

such that ωx = ω(x) belongs to M∗x for every x ∈ M. In particular, ω = d f is adifferential form of degree 1 for every f in C∞(M). Thus, the differential yields a


46 2 Complexes

map Ω0(M)→ Ω1(M). This map d0 = d is part of the so-called de Rham complex,which traditionally is written in cohomological notation,

Ω(M) = 0−→Ω0(M)

d0−−→Ω

1(M)d1−−→Ω

2(M)d2−−→ ·· · .

The differential on Ω(M) is called the exterior derivative. To define it, notice thatthe wedge product ∧ :

∧Ri (M

∗x )×

∧Rj (M

∗x )→

∧Ri+ j(M

∗x ) on the exterior algebra in-

duces a pairing ∧ :Ωi(M)×Ω j(M)→Ωi+ j(M), defined by (ω∧ψ)x =ωx∧ψx. Theexterior derivative is defined recursively by the formula

di+ j(ω∧ψ) = di(ω)∧ψ+(−1)iω∧d j(ψ)

for ω ∈Ωi(M) and ψ ∈Ω j(M).

2.1.27 Definition. A homomorphism of complexes is a graded homomorphism ofthe underlying graded modules; see 2.1.5. Let M and N be R-complexes, a chainmap M→ N is homomorphism α : M→ N that satisfies

∂Nα = (−1)|α|α∂M .

This means that every square in the next diagram commutes up to the sign (−1)|α|.

· · · // Mv+1

(−1)|α|

∂Mv+1

//

αv+1

Mv

(−1)|α|

∂Mv

//

αv

Mv−1 //

αv−1

· · ·

· · · // Nv+1+|α|∂N

v+1+|α|

// Nv+|α|∂N

v+|α|

// Nv−1+|α| // · · · ,

A chain map of degree 0 is called a morphism of R-complexes.

REMARK. The sign above follows Koszul’s sign convention: a sign (−1)mn is introduced whenelements of degree m and n are interchanged; in this case, (−1)|α||∂| = (−1)−|α| = (−1)|α|.

2.1.28 Example. Let M be an R-complex. The differential ∂M is a chain map ofdegree −1. For a central element x ∈ R the homothety xM is a morphism.

2.1.29 Example. Let (x1, . . . ,xn) ⊆ (x1, . . . ,xn,y1, . . . ,ym) be ideals in k. Withthe notation from 2.1.24 it is elementary to verify that the map Kk(x1, . . . ,xn)→Kk(x1, . . . ,xn,y1, . . . ,ym) given by x1 ∧ ·· · ∧ xp 7→ x1 ∧ ·· · ∧ xp is a morphism of k-complexes.

2.1.30. It is straightforward to verify that a composite of chain maps is a chain map.In particular, a composite of morphisms is a morphism; cf. 2.1.7.

2.1.31 Definition. The category R-complexes and their morphisms is denoted C(R).

2.1.32 Definition. Let M be an R-complex. A graded submodule K of M\, withthe property that ∂M(K) is contained in K, is a complex when endowed with thedifferential ∂M|K ; it is called a subcomplex of M. In this case, the embedding KMis a morphism of R-complexes.



Given a subcomplex K of M, the differential ∂M induces a differential on thegraded module M\/K\. The resulting complex M/K is called a quotient complex.Note that the canonical map MM/K is a morphism of R-complexes.

2.1.33 Example. Let M be a smooth real manifold. Recall from 2.1.25 that the mod-ules in the singular chain complex S(M) are Sn(M) = Z〈C(∆n,M)〉 for n> 0 andSn(M) = 0 for n < 0. It is immediate from the definitions that the graded submoduleS∞(M) with S∞

n (M) = Z〈C∞(∆n,M)〉 for n> 0 is a subcomplex of S(M).With the notation from 2.1.25 and 2.1.26 there exists a morphism of R-complexes

Ω(M)→HomZ(S∞(M),R) that maps a differential form ω of degree n to the grouphomomorphism S∞

n (M)→ R given by σ 7→∫σω=

∫∆n σ∗ω.

2.1.34. Let α : M→ N be a chain map of R-complexes. It is simple to verify thatKerα, Imα, and Cokerα are R-complexes, see 2.1.11, and that the category C(R) isk-linear and Abelian. The biproduct of complexes M and N is the graded direct sumM\⊕N\ from 2.1.12 endowed with the differential ∂M⊕∂N .

2.1.35. There are exact k-linear functors between Abelian categories

Mgr(R)//C(R) .

(–)\oo

The functor from Mgr(R) to C(R) is full and faithful; it equips a graded moduleM with the trivial differential ∂M = 0, and it is the identity on morphisms. Thefunctor (–)\ from C(R) to Mgr(R) forgets the differential on R-complexes, and it isthe identity on morphisms. At the level of symbols, applications of these functors isoften suppressed. When we write e.g. “as an R-complex” about a graded R-module,it means that we apply the functor Mgr(R)→ C(R).

In particular, graded R-modules are isomorphic if and only if they are isomorphicas R-complexes. Further, a morphism M→ N of R-complexes is an isomorphism ifand only if it is an isomorphism of the underlying graded modules.

The module category M(R) is a full subcategory of C(R) via the full and faithfulfunctors M(R)→Mgr(R)→ C(R); see 2.1.13.

2.1.36 Definition. Let C(R–So) denote the category C(R⊗k So).

2.1.37. Complexes of R–So-bimodules as well as homomorphisms and chain mapsof such are defined as in 2.1.21 and 2.1.27. The category C(R–So) is naturally iden-tified with the category whose objects are complexes of R–So-bimodules (with R-and So-linear differentials), and whose morphisms are R- and So-linear chain mapsof degree 0; cf. 2.1.17.

In particular, C(R–Ro) is identified with the category of complexes of R–Ro-bimodules, and the category C(Re) is naturally identified with the full subcategoryof C(R–Ro) whose objects are complexes of symmetric R–Ro-bimodules.


48 2 Complexes

DIAGRAM LEMMAS

In the next several paragraphs, in 2.1.38–2.1.44 to be precise, the category of com-plexes could be replaced by any Abelian category, and in that sense they repeatmaterial from Sect. 1.1. We include them, nevertheless, for ease of reference, andwe provide proofs, because the material is central.

2.1.38 Definition. A sequence of R-complexes is a, possibly infinite, diagram,

(2.1.38.1) · · · −→M0 α0−−→M1 α1

−−→M2 α2−−→ ·· · ,

in C(R); it is called exact if one has Imαn−1 = Kerαn for all n. Notice that (2.1.38.1)is exact if and only if every sequence 0→ Imαn−1→Mn→ Imαn→ 0 is exact. Anexact sequence 0→M′→M→M′′→ 0 is called a short exact sequence.

Two sequences αn : Mn→Mn+1n∈Z and βn : Nn→ Nn+1n∈Z of R-complex-es are called isomorphic if there exists a family of isomorphisms ϕnn∈Z such thatthe diagram

Mn αn//

ϕn

Mn+1

ϕn+1

Nn βn// Nn+1

is commutative for every n ∈ Z.

2.1.39. Exactness of a sequence in C(R) does not depend on the differentials, so itcan be detected degreewise. That is, (2.1.38.1) is exact if and only the sequence

· · · −→M0v

α0v−−→M1

vα1

v−−→M2v

α2v−−→ ·· ·

in M(R) is exact for every v ∈ Z.

The next result is known as the Five Lemma.

2.1.40 Lemma. Let

M1 //

ϕ1

M2 //

ϕ2

M3 //

ϕ3

M4 //

ϕ4

M5

ϕ5

N1 // N2 // N3 // N4 // N5

be a commutative diagram in C(R) with exact rows. The following assertions hold.(a) If ϕ1 is surjective, and ϕ2 and ϕ4 are injective, then ϕ3 is injective.(b) If ϕ5 is injective, and ϕ2 and ϕ4 are surjective, then ϕ3 is surjective.(c) If ϕ1, ϕ2, ϕ4, and ϕ5 are isomorphisms, then ϕ3 is an isomorphism.

PROOF. (a): The assumptions imply for each v ∈ Z that the homomorphism ϕ1v is

surjective, and the homomorphisms ϕ2v and ϕ4

v are injective. Thus, the Five Lemmafor modules 1.1.2 applied to the degree v part of the given diagram yields that ϕ3

v isinjective. Since this holds for every v ∈ Z, it follows that ϕ3 is injective.

The proofs of parts (b) and (c) are similar.



2.1.41 Construction. LetM′ α′

//

ϕ′

M α//

ϕ

M′′ //

ϕ′′

0

0 // N′β′// N

β// N′′

be a commutative diagram in C(R) with exact rows. In each degree v it yields acommutative diagram in M(R), to which 1.1.3 applies to give a homomorphismδv : Kerϕ′′v → Cokerϕ′v. Denote by δ : (Kerϕ′′)\→ (Cokerϕ′)\ the morphism (δv)v∈Zof graded R-modules.

The next result is known as the Snake Lemma, and the morphism δ is known asthe connecting morphism.

2.1.42 Lemma. The morphism δ of graded R-modules, defined in 2.1.41, is a mor-phism of R-complexes, and there is an exact sequence in C(R),

Kerϕ′ α′−−→ Kerϕ α−−→ Kerϕ′′ δ−−→ Cokerϕ′β′−−→ Cokerϕ

β−−→ Cokerϕ′′ .

Moreover, if α′ is injective then so is the restricted morphism α′ : Kerϕ′→ Kerϕ,and if β is surjective, then so is the induced morphism β : Cokerϕ→ Cokerϕ′′.

PROOF. By the Snake Lemma for modules 1.1.4 there is an exact sequence

(Kerϕ′)vα′v−→ (Kerϕ)v

αv−→ (Kerϕ′′)vδv−→ (Cokerϕ′)v

β′v−→ (Cokerϕ)vβv−→ (Cokerϕ′′)v

for every v ∈ Z. The restricted maps α′ : Kerϕ′→ Kerϕ and α : Kerϕ→ Kerϕ′′

are morphisms because Kerϕ′, Kerϕ, and Kerϕ′′ are subcomplexes. Similarly,β′ = (β′v)v∈Z : Cokerϕ′→ Cokerϕ and β= (βv)v∈Z : Cokerϕ→ Cokerϕ′′ are mor-phisms. It remains to show that δ is a morphism. For a homogeneous elementm′′ ∈Kerϕ′′ one has δ(m′′) = [n′]Imϕ′ for a homogeneous element n′ ∈ N′ that satis-fies β′(n′) = ϕ(m) for some m ∈M with α(m) = m′′; see 1.1.3. Now one has

∂Cokerϕ′(δ(m′′)) = ∂Cokerϕ′([n′]Imϕ′) = [∂N′(n′)]Imϕ′ .

The element ∂N′(n′) satisfies β′(∂N′(n′))= ∂N(β′(n′))= ∂N(ϕ(m))=ϕ(∂M(m)), andone has α(∂M(m)) = ∂M′′(α(m)) = ∂M′′(m′′). Thus, by the definition of δ one hasδ(∂M′′(m′′)) = [∂N′(n′)]Imϕ′ = ∂Cokerϕ′(δ(m′′)).

SPLIT SEQUENCES

2.1.43 Definition. An exact sequence 0−→M′ α′−→M α−→M′′ −→ 0 in C(R) is calledsplit if there exist morphisms % : M→M′ and σ : M′′→M such that one has

%α′ = 1M′ , α′%+σα = 1M , and ασ = 1M′′ .

2.1.44 Proposition. Let 0−→M′ α′−→M α−→M′′ −→ 0 be an exact sequence in C(R).The following conditions are equivalent.


50 2 Complexes

(i) The sequence is split.(ii) There exists a morphism % : M→M′ such that %α′ = 1M′ .

(iii) There exists a morphism σ : M′′→M such that ασ= 1M′′ .(iv) The sequence is isomorphic to 0 −→M′ ε−→M′⊕M′′ $−→M′′ −→ 0, where ε

and $ are the injection and the projection, respectively.

Moreover, if 0 −→ M′ α′−→ M α−→ M′′ −→ 0 is split exact, then also the sequence0−→M′′ σ−→M

%−→M′ −→ 0, where % and σ are as in 2.1.43, is split exact.

PROOF. Conditions (ii) and (iii) follow from (i). To see that (ii) implies (iv), let% : M→M′ be a morphism with %α′ = 1M′ . There is then a commutative diagram,

0 // M′ α′//

1M′

M α//

ϕ

M′′ //

1M′′

0

0 // M′ εM′// M′⊕M′′ $M′′

// M′′ // 0 ,

in C(R), where ϕ is given by m 7→ (%(m),α(m)); it follows from the Five Lemma2.1.40 that ϕ is an isomorphism. A parallel argument shows that (iii) implies (iv).

Given a commutative diagram,

0 // M′ α′//

ϕ′∼=

M α//

ϕ∼=

M′′ //

ϕ′′∼=

0

0 // M′ εM′// M′⊕M′′ $M′′

// M′′ // 0 ,

set % = (ϕ′)−1$M′ϕ and σ = ϕ−1εM′′ϕ′′. Now one has %α′ = 1M′ , and ασ = 1M′′ ,and α′%+σα= α′(ϕ′)−1$M′ϕ+ϕ−1εM′′ϕ′′α= ϕ−1(εM′$M′+εM′′$M′′)ϕ= 1M , so(iv) implies (i).

Finally, assume that 0−→M′ α′−→M α−→M′′ −→ 0 is split exact, and let % and σbe as in 2.1.43. The sequence 0−→M′′ σ−→M

%−→M′ −→ 0 is exact. Indeed % isinjective and σ is surjective; moreover, one has %σ = %σασ = %(1M −α′%)σ = 0,and for m ∈ M with %(m) = 0 one gets m = (α′%+σα)(m) = σα(m). Thus thesequence is split exact by 2.1.43.

2.1.45. A functor F: M(R)→M(S) extends to a functor C(R)→ C(S) that is alsodenoted F and acts as follows: For a morphism α : M→ N in C(R) and v ∈ Z one has

F(M)v = F(Mv) , ∂F(M)v = F(∂M

v ) , and F(α)v = F(αv) .

A functor G: M(R)op→M(S) extends to a functor C(R)op→ C(S) that is also de-noted G and acts as follows: For a morphism α : M→ N in C(R)op and v ∈ Z one has

G(N)v = G(N−v) , ∂G(N)v = G(∂N

−v+1) , and G(α)v = G(α−v) .

If the original functor is k-linear then so is the extended functor, and if the origi-nal functor is (left/right) exact, then so is the extended functor.



2.1.46 Definition. A short exact sequence 0→M′→M→M′′→ 0 in C(R) is calleddegreewise split if the exact sequence 0→M′ \→M\→M′′ \→ 0 is split.

Only exact functors C(R)→ C(S) preserve short exact sequences of complexes.It follows, however, from the next lemma that every functor F that satisfies a naturalcondition—the differential on a complex M has no influence on the graded moduleunderlying F(M)—preserves degreewise split exact sequences.

2.1.47 Definition. A functor F: C(R)→ C(S) is called a \-functor if it is additive and(–)\ F and (–)\ F(–)\ are naturally isomorphic functors from C(R) to Mgr(S).

A functor G: C(R)op→ C(S) is called a \-functor if it is additive and (–)\ G and(–)\ G((–)\)op are naturally isomorphic functors from Uop to Mgr(S).

2.1.48 Example. Every functor on complexes that is extended from an additivefunctor on modules, as described in 2.1.45, is per construction a \-functor.

Further examples of \-functors are given in 2.3.12 and 2.4.11. A few functorsthat are not \-functors can be found in 2.2.17.

2.1.49 Lemma. Let F: C(R)→ C(S) be a \-functor. For every degreewise split exactsequence of R-complexes, 0→ M′ → M → M′′ → 0, the induced sequence of S-complexes, 0→ F(M′)→ F(M)→ F(M′′)→ 0, is degreewise split exact.

PROOF. It follows from the assumptions on F that the sequence

(?) 0−→ (F(M′))\ −→ (F(M))\ −→ (F(M′′))\ −→ 0

is isomorphic to

(‡) 0−→ (F(M′ \))\ −→ (F(M\))\ −→ (F(M′′ \))\ −→ 0

in Mgr(S). The sequence (‡) arises from application of the additive functor (–)\ Fto the split exact sequence 0→M′\→M\→M′′\→ 0. Therefore, (‡) is split exact,and hence so is the isomorphic sequence (?).

2.1.50 Lemma. Let G: C(R)op→ C(S) be a \-functor. For every degreewise splitexact sequence of R-complexes, 0→M′→M→M′′→ 0, the induced sequence ofS-complexes, 0→ G(M′′)→ G(M)→ G(M′)→ 0, is degreewise split exact.

PROOF. Similar to the proof of 2.1.49.

EXERCISES

E 2.1.1 Let M =∐

v∈ZMv be a graded R-module and let K be a submodule of M. Show that thefollowing conditions are equivalent. (i) K is a graded submodule of M. (ii) K is generatedby homogeneous elements. (iii) The equality K =

∐v∈ZK∩Mv holds.

E 2.1.2 Let n be an integer. (a) Show that there is a full and faithful functor M(R)→Mgr(R)that equips an R-module M with the grading Mn = M and Mv = 0 for n 6= v. (b) Showthat there is a functor Mgr(R)→M(R) that forgets the modules Mv for v 6= n.

E 2.1.3 Let ϕ : M→ N be a graded homomorphism. Show that if ϕ is a bijective map, then itsinverse is a graded homomorphism of degree −|ϕ|.


52 2 Complexes

E 2.1.4 Show that the assignment M 7→ Tk(M) yields a functor from M(k) to the category ofk-algebras and that it is left adjoint to the forgetful functor.

E 2.1.5 Show that the Koszul complexes KZ(2,3) and KZ(4,5) are not isomorphic.E 2.1.6 Let x,y be elements in k; show that the complexes Kk(x,y) and Kk(y,x) are isomorphic.E 2.1.7 Let f : X → Y be a continuous map between topological spaces; that is, a morphism in

the category T of topological spaces. (a) Show that the assignment σ 7→ fσ defines amorphism of singular chain complexes S(X)→ S(Y ). (b) Show that there is a functorT→C(Z) that assigns to a topological space X its singular chain complex S(X).

E 2.1.8 Show that the category C(R) is isomorphic to Mgr(R[x]/(x2)).E 2.1.9 Show that a morphism in C(R) is a monomorphism if and only if it is injective, and

show that it is an epimorphism if and only if it is surjective.E 2.1.10 Show that a surjective morphism of complexes is surjective on boundaries and show that

a morphism that is injective on cycles is injective.E 2.1.11 Show that a morphism of complexes is surjective if it is so on boundaries and on cycles.E 2.1.12 (Cf. 2.1.9) Let x be a central element in R and let M be a graded R-module. Show that

the homothety xM : M→M is a morphism of graded R-modules.E 2.1.13 (Cf. 2.1.30) Let β : L→M and α : M→ N be chain maps of R-complexes; show that the

composite αβ : L→ N is a chain map of R-complexes of degree |α|+ |β|.E 2.1.14 (Cf. 2.1.34) Let α : M→ N be a chain map of R-complexes. Show that Kerα is a sub-

complex of M and that Imα is a subcomplex of N.E 2.1.15 (Cf. 2.1.34) Verify that C(R) is an Abelian category.E 2.1.16 (Cf. 2.1.35) Let α be a morphism of R-complexes. Show that α is an isomorphism of

R-complexes if and only if it is an isomorphism of the underlying graded R-modules.E 2.1.17 Show that every split exact sequence in C(R) is degreewise split. Is the converse true?E 2.1.18 Give a proof of 2.1.50.

2.2 Homology

SYNOPSIS. Shift; homology; connecting morphism; homotopy.

The requirement that the differential ∂ on a complex has to be square zero yieldsan inclusion Im∂⊆ Ker∂; the homology of the complex is the quotient Ker∂/Im∂.The implications of (non-)vanishing of homology depend on the underlying com-plex, but it is often illustrative to interpret homology as a measure of deviation froma paragon or obstructions to an aim. For instance, the homology of the de Rhamcomplex Ω(M) from 2.1.26 detects closed differential forms on the manifold M thatare not exact. The homology of the singular chain complex S(X) from 2.1.25 tellswhether it is possible to contract the space X to a single point, and much more. Thehomology of the Koszul complex Kk(x1, . . . ,xn) from 2.1.24 shows, among otherthings, if one or more of the elements x1, . . . ,xn are non-zerodivisors.

SHIFT

Before we proceed with the precise definition of homology, we formalize the intu-itively natural process of renumbering the modules in a complex.


2.2 Homology 53

2.2.1 Definition. Let M be an R-complex and let s be an integer. The s-fold shift ofM is the complex Σs M given by

(Σs M)v = Mv−s and ∂Σs M

v = (−1)s∂Mv−s

for all v∈ Z. For a homomorphism α= (αv)v∈Z : M→ N of R-complexes, the s-foldshift Σsα : Σs M→ Σs N is the homomorphism with (Σsα)v = αv−s for all v ∈ Z.

REMARK. Other words for shift are suspension and translation.

2.2.2 Example. Every functor on complexes that is extended from a functor onmodules, as described in 2.1.45, commutes with shift.

2.2.3. For an R-complex M one has ∂Σs M = (−1)sΣs ∂M . For a homomorphism of

R-complexes α one has |Σsα| = |α|, and for composable homomorphisms α andβ one has (Σs β)(Σsα) = Σs (βα). It follows that if α : M→ N is a chain map ofR-complexes then so is Σsα; indeed, there are equalities

∂Σs N(Σsα) = (−1)s(Σs ∂N)(Σsα)

= (−1)sΣ

s (∂Nα)

= (−1)|α|(−1)sΣ

s (α∂M)

= (−1)|Σs α|(Σsα)∂Σ

s M .

In particular, Σs takes morphisms to morphisms, and it follows that Σs is an exact k-linear automorphism on C(R) with inverse Σ−s. Evidently, Σs is the s-fold compositeof the functor Σ= Σ1.

By means of shift, every chain map can be identified with a morphism.

2.2.4. Let M be an R-complex and s be an integer. There is a canonical chain mapςM

s : M→ Σs M of degree s that maps a homogeneous element m in M to the corre-sponding element of degree |m|+ s in Σs M. The map is invertible, and (ςM

s )−1 isthe chain map ςΣ

s M−s : Σs M→M. Notice that for every homomorphism α : M→ N

of complexes the following diagram is commutative,

(2.2.4.1)M

ςMs//

α

Σs M

Σs α

ςΣs M−s// M

α

NςN

s// Σs N

ςΣs N−s// N .

These degree shifting maps are occasionally suppressed. In particular, a homomor-phism (chain map) β : M→ N can be identified with the homomorphism (chain map)βςΣ

s M−s : Σs M→ N of degree |β|− s or with ςN

s β : M→ Σs N of degree |β|+ s.

HOMOLOGY

2.2.5. Let M be an R-complex. The differential ∂M is a chain map, so it follows from2.1.34 that Ker∂M and Im∂M are subcomplexes of M. As ∂M∂M = 0 holds, there is


54 2 Complexes

an inclusion Im∂M ⊆ Ker∂M , and the induced differential on either subcomplex is0. Likewise, the differential on the quotient complex Coker∂M is 0.

2.2.6 Definition. Let M be an R-complex; set

Z(M) = Ker∂M , B(M) = Im∂M , and C(M) = Coker∂M .

Notice that C(M) is the quotient complex M/B(M). Elements in the subcomplexZ(M) are called cycles, and elements in B(M) are called boundaries. The module ofhomogeneous cycles in M of degree v, that is, the module in degree v in the complexZ(M), is written Zv(M). Similarly the modules in the complexes B(M) and C(M)are written Bv(M) and Cv(M). As these complexes have trivial differentials, theycould alternately be defined by specifying their modules as follows,

Zv(M) = Ker∂Mv , Bv(M) = Im∂M

v+1 , and Cv(M) = Coker∂Mv+1 .

The quotientH(M) = Z(M)/B(M)

is called the homology complex of M. The module in degree v is written Hv(M) andcalled the vth homology module of M. If one has H(M) = 0, then M is called acyclic.

Notice that a complex is acyclic (see 2.2.6) as an object in C(R) if and only if itis exact (see 1.1.1) when regarded as a sequence in M(R).

REMARK. Other words for acyclic are exact and homologically trivial. For a complex writtenin cohomological notation M = · · · → Mv−1 → Mv → Mv+1 → ··· it is standard to write thehomology complex H(M) in cohomological notation as well. The module in cohomological degreev of H(M) is denoted Hv(M) and called the vth cohomology module of M.

The terms “boundary” and “cycle” are borrowed from specific homology theo-ries, where this terminology comes natural. Briefly: paths π and π′ in R2 are deemedequivalent if they differ by a boundary; that is, the concatenation of π and −π′ isthe boundary of a region. The reason is that by Stokes’ theorem the path integral ofan exact differential form is then the same along π and π′. In a topological spacethat has a structure of a simplicial complex—a higher triangulation; cf. 2.1.25—it iscompelling to use the name cycle for a potential boundary. For a rigid, but still brief,explanation see the first section of Rotman’s book [75]. Weibel’s historical survey[86] has plenty of illuminating details.

2.2.7 Example. The Dold complex in 2.1.22 is acyclic.

2.2.8 Example. The Koszul homology of a sequence x1, . . . ,xn in k is the homologyof the Koszul complex Kk(x1, . . . ,xn) from 2.1.24. It is elementary to verify that thehomology in degree 0 is H0(Kk(x1, . . . ,xn)) = k/(x1, . . . ,xn).

For a single element x the Koszul complex Kk(x) has the form

(2.2.8.1) 0−→ k〈e〉 ∂−−→ k−→ 0 ,

where ∂(e) = x. Hence one has H1(Kk(x)) ∼= (0 :k x). In particular, this homologymodule is zero if and only if x is not a zerodivisor in k.


2.2 Homology 55

The Koszul complex K = Kk(x1,x2) has the form

(2.2.8.2) 0−→ k〈e1∧ e2〉∂K

2−−→ k〈e1,e2〉∂K

1−−→ k−→ 0 ,

where ∂K1 (k1e1 + k2e2) = k1x1 + k2x2 and ∂K

2 (e1 ∧ e2) = x1e2− x2e1. The cycles indegree 2 are elements ke1 ∧ e2 with kx1 = 0 = kx2; hence there is an isomorphismH2(K) ∼= (0 :k x1)∩ (0 :k x2) = (0 :k (x1,x2)). The cycles in degree 1 embody therelations between x1 and x2 while the boundaries embody the simple relation x1x2−x2x1 = 0. Notice that H2(K) is zero if x1 or x2 is not a zerodivisor in k. The homologymodule H1(K) is zero if and only if [x2](x1) is not a zerodivisor in the quotient ringk/(x1) and [x1](x2) is not a zerodivisor in k/(x2).

The complex K = Kk(x1, . . . ,xn) is concentrated in degrees n, . . . ,1,0. The ho-mology module Hn(K) is isomorphic to the ideal of common annihilators of theelements x1, . . . ,xn. The module H1(K) is the module generated by all relations be-tween the generators modulo the one generated by the simple ones xix j− x jxi = 0.In intermediate degrees n > i > 1, the module Hi(K) is generated by “higher” re-lations between the generators, and the boundaries that are factored out are simple“higher” relations usually referred to as Koszul relations.

2.2.9 Example. The singular homology H∗(X ;A) of a topological space X withcoefficients in an Abelian group A is the homology of the singular chain complexS(X)⊗Z A from 2.1.25. Singular homology with coefficients in Z is written H∗(X).The singular homology group Hn(X) detects “n-dimensional holes” in X .

For every space X , the Abelian group H0(X) is free, and its rank (possibly an in-finite cardinal) is the number of path components of X . The singular chain complexof a one-point space pt is

S(pt) = · · · −→ Z〈σ2〉∂

pt2−−→ Z〈σ1〉

∂pt1−−→ Z〈σ0〉 −→ 0 .

where σn for each n > 0 denotes the unique (continuous) map ∆n → pt from thestandard n-simplex to a point. The definition 2.1.25 yields

∂ptn = 0 for n odd and ∂pt

n (σn) = σn−1 for n even ,

and thus one has H0(pt)∼= Z and Hn(pt) = 0 for all n 6= 0. Every contractible spacehas the same singular homology as pt; see 4.3.5. For m > 0 the singular homologyof the m-sphere Sm is H0(Sm)∼= Z∼= Hm(Sm) and Hn(Sm) = 0 for all n /∈ 0,m.

2.2.10 Example. The de Rham cohomology HdR(M) of a smooth real manifold Mis the cohomology of the de Rham complex Ω(M) from 2.1.26; it detects closeddifferential forms on M that are not exact. A closed 0-form is constant on everyconnected component, so the rank of the real vector space H0

dR(M) is the number ofconnected components of M. The cohomology group H1

dR(M) is zero if and only ifevery closed differential 1-form is the differential of a smooth function f : M→ R.

The de Rham complex Ω(S1) of the 1-sphere is concentrated in degrees 1 and 0;in particular, the cohomology modules Hn

dR(S1) are zero for n /∈ 0,1. As S1 is

connected, one has H0dR(M) ∼= R. Every 1-form is closed (d1 = 0). Despite the ap-


56 2 Complexes

pearance, dθ, the differential of the polar coordinate “function” θ is not exact, soH1

dR(M) is non-zero and, in fact, isomorphic to R.The de Rham cohomology Hn

dR(Bn) of the open unit ball in Rn is zero for n 6= 0.

FUNCTORIALITY

2.2.11. Let M be an R-complex. The subcomplexes and (sub-)quotient complexesintroduced in 2.2.6 are linked by the following exact sequences in C(R).

0−→ Z(M)−→M −→ ΣB(M)−→ 0 ,(2.2.11.1)0−→ B(M)−→M −→ C(M)−→ 0 ,(2.2.11.2)

0−→ B(M)−→ Z(M)−→ H(M)−→ 0 , and(2.2.11.3)0−→ H(M)−→ C(M)−→ ΣB(M)−→ 0 .(2.2.11.4)

2.2.12. Let α : M→ N be a morphism of R-complexes. It follows from the equalityα∂M = ∂Nα that α restricts to a morphism of cycle complexes Z(M)→ Z(N) andfurther to a morphism of boundary complexes B(M)→ B(N). In particular, α in-duces a morphism of cokernel complexes, α : C(M)→ C(N). It is straightforwardto verify that Z(–), B(–), and C(–) are k-linear endofunctors on C(R). For brevity,we continue to write α for C(α) and α for Z(α) = α|Z(M) and for B(α) = α|B(M).

2.2.13 Lemma. Let 0→M′→M→M′′→ 0 be an exact sequence of R-complexes.(a) For every v ∈ Z there is an exact sequence

0→ Zv(M′)→ Zv(M)→ Zv(M′′)→ Hv−1(M′)→ Hv−1(M)→ Hv−1(M′′) .

In particular, the functors Zv(–) and Z(–) are left exact.(b) For every v ∈ Z there is an exact sequence

Hv+1(M′)→ Hv+1(M)→ Hv+1(M′′)→ Cv(M′)→ Cv(M)→ Cv(M′′)→ 0 .

In particular, the functors Cv(–) and C(–) are right exact.

PROOF. For every v ∈ Z the Snake Lemma 1.1.4 applies to the diagram,

0 // M′v //

∂M′v

Mv //

∂Mv

M′′v //

∂M′′v

0

0 // M′v−1// Mv−1 // M′′v−1

// 0 ,

and yields an exact sequence

0→ Zv(M′)→ Zv(M)→ Zv(M′′)→ Cv−1(M′)→ Cv−1(M)→ Cv−1(M′′)→ 0 .

Thus, every functor Zv(–) is left exact, and it follows that Z(–), which is computeddegreewise, is left exact. Similarly, each functor Cv(–) and hence C(–) is right exact.

To finish part (a) apply the Snake Lemma 1.1.4 to the diagram,


2.2 Homology 57

0 // M′v //

∂M′v

Mv //

∂Mv

M′′v //

∂M′′v

0

0 // Zv−1(M′) // Zv−1(M) // Zv−1(M′′) .

To finish part (b) apply the Snake Lemma 1.1.4 to the diagram,

Cv+1(M′) //

∂M′v+1

Cv+1(M) //

∂Mv+1

Cv+1(M′′) //

∂M′′v+1

0

0 // M′v // Mv // M′′v // 0 .

REMARK. The functor B(–) is not half exact, but it preserves surjective morphisms; see E 2.1.10.

2.2.14. Let α : M→ N be a morphism of R-complexes. It follows from 2.2.12 thatα induces morphism of R-complexes

H(α) : H(M) −→ H(N) ,

which is given by the assignment [z]B(M) 7→ [α(z)]B(N) for z ∈ Z(M).The equivalence class [z]B(M) is called the homology class of z. Hereafter we drop

the subscript on homology classes and write [z] for the homology class of a cycle z.By the definition of H(α) there is a commutative diagram,

0 // B(M) //

α

Z(M) //

α

H(M) //

H(α)

0

0 // B(N) // Z(N) // H(N) // 0 .

(2.2.14.1)

2.2.15 Theorem. Homology H is a k-linear endofunctor on C(R). Moreover, thereis an equality HΣ= ΣH of endofunctors on C(R).

PROOF. It is straightforward to verify that H: C(R)→ C(R) is a k-linear functor. Forthe proof of the second claim, let M be an R-complex. The equality ∂ΣM = −Σ∂M

and exactness of the shift functor yields

H(ΣM) = Ker∂ΣM/Im∂ΣM

= Ker(Σ∂M)/Im(Σ∂M)

= (ΣKer∂M)/(Σ Im∂M)

= Σ(Ker∂M/Im∂M)

= ΣH(M) .

2.2.16. For a complex Z with zero differential one has H(Z) = Z. In particular,H(H(M)) = H(M) holds for every complex.


58 2 Complexes

2.2.17 Example. The functors B,C,H,Z: C(R)→ C(R), see 2.2.12 and 2.2.15, arek-linear but not \-functors. Indeed, with D = 0→ R =−→ R→ 0, concentrated in de-grees 1 and 0, the exact sequence η= 0→ R→D→ ΣR→ 0 in C(R) is degreewisesplit, but none of the sequences B(η), C(η), H(η), and Z(η) are even exact. Theassertion now follows from 2.1.49.

An exact functor C(R) → C(S) need not commute with homology. Far from,actually, and most of Chap. 5 is devoted to a study of exact functors that, at least,preserve acyclicity of complexes. Functors that are extended from exact functors onmodules do, however, commute with homology.

2.2.18. A functor F: C(R)→ C(S) that per 2.1.45 is extended from an exact functorM(R)→M(S) commutes with homology. Indeed, it is straightforward to verify thatfor every v ∈ Z one has

Zv(F(M)) ∼= F(Zv(M)) , Bv(F(M)) ∼= F(Bv(M)) ,

Cv(F(M)) ∼= F(Cv(M)) , and Hv(F(M)) ∼= F(Hv(M)) ;

similarly H(F(α)) = F(H(α)) for every morphism α in C(R).Analogously, for a functor G: C(R)op→ C(S) that is extended from an exact

functor M(R)op→M(S) one has

Zv(G(M)) ∼= G(C−v(M)) , Bv(G(M)) ∼= G(B−v−1(M)) ,

Cv(G(M)) ∼= G(Z−v(M)) , and Hv(G(M)) ∼= G(H−v(M)) ;

similarly H(G(α)) = G(H(α)) for every morphism α in C(R).

CONNECTING MORPHISM IN HOMOLOGY

The homology functor is only half exact, but the Snake Lemma facilitates a closercomparison of the homology of complexes in a short exact sequence.

2.2.19 Construction. Consider a short exact sequence of R-complexes,

0−→M′ α′−−→M α−−→M′′ −→ 0 .

It induces a commutative diagram in C(R) with exact rows,

C(M′) α′//

ςM′1 ∂M′

C(M)α//

ςM1 ∂

M

C(M′′) //

ςM′′1 ∂M′′

0

0 // ΣZ(M′) Σα′// ΣZ(M)

Σα// ΣZ(M′′) ,

and the Snake Lemma 2.1.42 yields an exact sequence

(2.2.19.1) H(M′)H(α′)−−−→ H(M)

H(α)−−−→ H(M′′) ð−−→ ΣH(M′)ΣH(α′)−−−−→ ΣH(M) ,

where the connecting morphism in homology, ð, maps a homology class [z′′] withz′′ = α(m) to [z′] with (Σα′)(z′) = ςM

1 ∂M(m); cf. 2.1.41 and 1.1.3.


2.2 Homology 59

The complexes in the exact sequence (2.2.19.1) have zero differentials, and it isoften written as an exact sequence in M(R):

· · · −→ Hv(M′)Hv(α

′)−−−−→ Hv(M)Hv(α)−−−→ Hv(M′′)

ðv−−→ Hv−1(M′)−→ ·· · .

A useful consequence is a convenient two-out-of-three criterion for acyclicity.

2.2.20. It follows from (2.2.19.1) that if two of the complexes M′, M, and M′′ in ashort exact sequence 0→M′→M→M′′→ 0 are acyclic, then so is the third.

REMARK. The acyclicity criterion above can be considered the degenerate case of E 2.5.5.

The connecting morphism is natural in the following sense.

2.2.21 Proposition. For every commutative diagram of R-complexes

0 // M′ α′//

ϕ′

M α//

ϕ

M′′ //

ϕ′′

0

0 // N′β′// N

β// N′′ // 0

with exact rows, there is a commutative diagram

H(M′)H(α′)

//

H(ϕ′)

H(M)H(α)//

H(ϕ)

H(M′′) ð//

H(ϕ′′)

ΣH(M′)ΣH(α′)

//

ΣH(ϕ′)

ΣH(M)

ΣH(ϕ)

H(N′)H(β′)

// H(N)H(β)// H(N′′) δ

// ΣH(N′)ΣH(α′)

// ΣH(N)

with exact rows. Here ð and δ are the connecting morphisms from 2.2.19.

PROOF. In view of 2.2.15 and 2.2.19 it remains to prove that the square

(?)

H(M′′) ð//

H(ϕ′′)

ΣH(M′)

ΣH(ϕ′)

H(N′′) δ// ΣH(N′)

is commutative. To this end, let [z′′] be an element in H(M′′) = Ker ∂M′′ . By thedefinition of the connecting morphism ð, see 2.2.19, one has ð([z′′]) = [z′], wherez′ ∈ ΣZ(M′) satisfies (Σα′)(z′) = ςM

1 ∂M([x]B(M)) for some element x ∈ M with

α([x]B(M)) = [α(x)]B(M′′) = [z′′]. Thus, one has (ΣH(ϕ′)ð)([z′′]) = [(Σϕ′)(z′)]. Thecycle (Σϕ′)(z′) satisfies

(Σβ′)(Σϕ′)(z′) = (Σϕ)(Σα′)(z′) = (Σϕ)ςM1 ∂

M([x]B(M)) = ςN1 ∂

N([ϕ(x)]B(N)) ,

where the last equality follows from (2.2.4.1). The element [ϕ(x)]B(N) satisfies

β([ϕ(x)]B(N)) = [βϕ(x)]B(N′′) = [ϕ′′α(x)]B(N′′) = H(ϕ′′)([α(x)]) = H(ϕ′′)([z′′]) .

By the definition of δ one now has (δH(ϕ′′))([z′′]) = [(Σϕ′)(z′)]. That is, the square(?) is commutative.


60 2 Complexes

It is evident from 2.2.21 that homology is a half exact functor, as the next exampleshows it is neither left nor right exact.

2.2.22 Example. Let 0→N′→N→N′′→ 0 be an exact sequence of R-complexes.If N is acyclic but N′,N′′ are not, then the sequence 0→ H(N′)→ 0→ H(N′′)→ 0is neither exact at H(N′) nor at H(N′′).

HOMOTOPY

Like many other notions in homological algebra, homotopy originates in topology.Homotopy works behind the scenes: it is preserved by most functors of interest, butit has zero homological footprint.

2.2.23 Definition. A chain map of R-complexes α : M→ N is called null-homotopicif there exists a homomorphism σ : M\→ N\ of degree |α|+1,

· · · // Mv+1∂M

v+1//

αv+1

Mv

σvyy

∂Mv

//

αv

Mv−1

σv−1yy

//

αv−1

· · ·

· · · // Nv+1+|α|∂N

v+1+|α|

// Nv+|α|∂N

v+|α|

// Nv−1+|α| // · · · ,

such that the equality α= ∂Nσ+(−1)|α|σ∂M holds.Two chain maps of R-complexes α,α′ : M→ N of the same degree are called

homotopic, in symbols α ∼ α′, if α− α′ is null-homotopic. A homomorphism σwith α−α′ = ∂Nσ+(−1)|α|σ∂M is called a homotopy from α to α′.

2.2.24 Example. Let M be an R-module; the identity morphism for the complex0 −→M =−→M −→ 0 is null-homotopic. If the complex is concentrated in degrees,say, 1 and 0, then the required homotopy is σ with σ0 = 1M and σv = 0 for v 6= 0.Complexes with this property are treated towards the end of Sect. 4.2.

The next proposition shows that homotopy yields a congruence relation in C(R),i.e. an equivalence relation that is compatible with k-linearity and composition.

2.2.25 Proposition. Let L, M and N be R-complexes. For every integer n, homotopy‘∼’ is an equivalence relation on the set of chain maps M→ N of degree n. Further,

(a) Let α,α′,β,β′ : M→ N be chain maps of the same degree. If α∼α′ and β∼ β′,then one has α+ xβ∼ α′+ xβ′ for every x ∈ k.

(b) Let β,β′ : L→M and α,α′ : M→ N be chain maps. If α∼ α′ and β∼ β′, thenone has αβ∼ α′β′.

PROOF. (a): Let σ be a homotopy from α to α′ and τ be a homotopy from β to β′. Itis immediate that σ+ xτ is a homotopy from α+ xβ to α′+ xβ′.

In now follows that ’∼’ is transitive: For chain maps γ ∼ γ′ and γ′ ∼ γ′′ one hasγ−γ′′ = (γ−γ′)+(γ′−γ′′)∼ 0+0 = 0 and hence γ∼ γ′′. As it is evident from thedefinition, 2.2.23, that ∼ is reflexive and symmetric, it is an equivalence relation.


2.2 Homology 61

(b): Letσ be a homotopy from α to α′ and τ be a homotopy from β to β′. Using theidentity αβ−α′β′ = α(β−β′)+(α−α′)β′ it is straightforward to verify that degree|α|+ |β|+1 homomorphism (−1)|α|ατ+σβ′ is a homotopy from αβ to α′β′.

2.2.26. Let α : M→ N be a morphism of R-complexes. If α is null-homotopic, thenthere is an inclusion α(Z(M)) ⊆ B(N), so the induced morphism H(α) is 0. Ashomology is an additive functor, one has H(α) = H(α′) for morphisms α∼ α′.

2.2.27 Example. Consider an R-complex

M = 0−→M2 −→M1 −→M0 −→ 0 .

One has H(1M) = 0 if and only if M is exact as a sequence in M(R), while 1M isnull-homotopic if and only if M is split exact as a sequence in M(R).

2.2.28. A diagram in C(R),M

ϕ

α// M′

ϕ′

Nβ// N′ ,

is called commutative up to homotopy if one has ϕ′α ∼ βϕ. Notice that, in view of2.2.26, the induced diagram in homology is then commutative.

EXERCISES

E 2.2.1 (Cf. 2.2.12) Show that B(–) is k-linear functor that is not half exact.E 2.2.2 (Cf. 2.2.15) Show that homology H: C(R)→C(R) is a k-linear half exact functor.E 2.2.3 (Cf. 2.2.18) Let F: M(R)→M(S) be a functor and consider the extended functor

C(R) → C(S); see 2.1.45. (a) Show that if the functor F is left exact, then one hasZv(F(M)) ∼= F(Zv(M)) for every v ∈ Z. (b) Show that if F is right exact, then one hasCv(F(M)) ∼= F(Cv(M)) for every v ∈ Z. (c) Conclude that if F is exact, then one hasBv(F(M))∼= F(Bv(M)) and Hv(F(M))∼= F(Hv(M)) for every v ∈ Z.

E 2.2.4 (Cf. 2.2.18) Let G: M(R)op→M(S) be a functor and consider the extended functorC(R)op → C(S); see 2.1.45. (a) Show that if the functor G is left exact, then one hasZv(G(M))∼= G(C−v(M)) for every v ∈ Z. (b) Show that if G is right exact, then one hasCv(G(M)) ∼= G(Z−v(M)) for every v ∈ Z. (c) Conclude that if G is exact, then one hasBv(G(M))∼= G(B−v−(M)) and Hv(G(M))∼= G(H−v(M)) for every v ∈ Z.

E 2.2.5 Let 0→M1→M2→M3→M4→ 0 be an exact sequence of R-complexes. (a) Showthat if three of the complexes Mu are exact, then so is the fourth. (b) Generalize.

E 2.2.6 Let M be an R-complex. Show that for every v ∈ Z with Hv(M) = 0 there is an isomor-phism Cv(M)∼= Bv−1(M).

E 2.2.7 Assume that R is semi-simple. Show that for every R-complex M there are morphismsα : H(M)→M and β : M→ H(M) such that H(α) and H(β) are isomorphisms.

E 2.2.8 For n > 0 show that singular homology Hn(S(–)) is a functor from the category T oftopological spaces to M(Z). Hint: E 2.1.2 and E 2.1.7.

E 2.2.9 Let f and g be elements in the ring Z[x] of polynomials with integer coefficients, andlet K denote the Koszul complex KZ(x)( f ,g). (a) Show that if H0(K) is zero, then K isacyclic. (b) Show that if H1(K) is zero, then H2(K) is zero as well.


62 2 Complexes

E 2.2.10 Let f ,g : X → Y be continuous maps between topological spaces, and assume that fand g are homotopic in the topological sense; that is, there exists a continuous mapH : X× [0,1]→ Y such that H(x,0) = f (x) and H(x,1) = g(x) for all x ∈ X . Show thatthe induced morphisms S( f ),S(g) : S(X)→ S(Y ) in C(Z) are homotopic in the senseof 2.2.23. Is the converse true?

E 2.2.11 Let M be an R-complex. Show that ∂M is a null-homotopic chain map.E 2.2.12 Show that a homotopy between homotopic chain maps may not be unique.E 2.2.13 Let x1 and x2 be elements in k and set K = Kk(x1,x2). Show that the homothety xK is

null-homotopic for every x ∈ (x1,x2).E 2.2.14 Let M be a k-complex. Show that x ∈ k | xM ∼ 0 is an ideal in k and notice that it

coincides with (0 :k M) if M is a module.E 2.2.15 Show that the identity morphism on the Koszul complex KZ(2,3) is null-homotopic.E 2.2.16 Show that the inclusion functor Mgr(R)→C(R) has a left adjoint C: C(R)→Mgr(R).E 2.2.17 Show that the inclusion functor Mgr(R)→C(R) has a right adjoint Z: C(R)→Mgr(R).

2.3 Homomorphisms

SYNOPSIS. Hom complex; chain map; homotopy; Hom functor; exactness.

Complexes M and N are graded modules with differentials, and those differentialsinduce a differential on the graded module HomR(M,N), which then becomes theHom complex. Recall from 2.1.27 that the homogeneous elements in HomR(M,N)are homomorphisms M→ N.

2.3.1 Definition. For R-complexes M and N, the complex HomR(M,N) is the k-complex with underlying graded module

HomR(M,N)\ = HomR(M\,N\)

and differential given by

∂HomR(M,N)(α) = ∂Nα− (−1)|α|α∂M

for a homogeneous element α.

2.3.2. It is elementary to verify that ∂HomR(M,N) is square zero. Indeed, one has

∂HomR(M,N)∂HomR(M,N)(α) = ∂HomR(M,N)(∂Nα− (−1)|α|α∂M)

= ∂N(∂Nα− (−1)|α|α∂M)

− (−1)|α|−1(∂Nα− (−1)|α|α∂M)∂M

= 0 .

The next result interprets boundaries and cycles in Hom complexes in familiarterms and provides, thus, a retrospective motivation for the definition of the differ-ential on Hom complexes.

2.3.3 Proposition. Let M and N be R-complexes.


2.3 Homomorphisms 63

(a) A homomorphism M→ N is a cycle in the complex HomR(M,N) if and onlyif it is a chain map.

(b) A homomorphism M→ N is a boundary in the complex HomR(M,N) if andonly if it is a null-homotopic chain map.

PROOF. The assertions are immediate from the definition of the differential on theHom complex; see 2.3.1.

2.3.4. Let Lβ−→M α−→ N be homomorphisms of R-complexes. The differentials on

the Hom complexes interact with the composition rule from 2.1.7 as follows,

∂HomR(L,N)(αβ) = ∂Nαβ− (−1)|αβ|αβ∂L

=(∂Nα− (−1)|α|α∂M)β+(−1)|α|α

(∂Mβ− (−1)|β|β∂L)

= ∂HomR(M,N)(α)β+(−1)|α|α∂HomR(L,M)(β) .

Assume that α and β are chain maps. It follows from the identity above and 2.3.3that αβ is a chain map; this recovers 2.1.30.

REMARK. A differential graded (for short, DG) k-algebra is a k-complex A endowed with a gradedk-algebra structure, such that the Leibniz rule ∂A(ab) = ∂A(a)b+(−1)|a|a∂A(b) holds for all ho-mogeneous elements a,b ∈ A. In other words: the differential ∂A is a derivation on A. For anR-complex M, it follows from 2.1.7 and 2.3.4 that the complex HomR(M,M) is a DG k-algebra.Moreover, the Koszul complex from 2.1.24 is a DG k-algebra; so is the de Rham complex underthe wedge product, and the singular cochain complex HomZ(S(X),k) from 2.1.25 endowed withthe so-called cup product is a DG algebra as well. An algebra structure on a complex A induces analgebra structure, notably a product, in homology; this yields another tool for investigating H(A).

FUNCTORIALITY

We now set out to prove that Hom is a functor. Our efforts culminate in Theo-rem 2.3.10; the intermediate paragraphs establish the necessary definitions and ver-ify the technical requirements. We start by defining how Hom acts on homomor-phisms, in particular, on morphisms.

2.3.5 Definition. Let α : M′→M and β : N→ N′ be homomorphisms of R-com-plexes. Denote by HomR(α,β) the degree |α|+ |β| homomorphism of k-complexes

HomR(M,N)−→ HomR(M′,N′) given by ϑ 7−→ (−1)|α|(|β|+|ϑ|)βϑα .

Furthermore, set HomR(α,N) = HomR(α,1N) and HomR(M,β) = HomR(1M,β).

2.3.6. It is straightforward to verify that the assignment (α,β) 7→ HomR(α,β), forhomomorphisms α and β of R-complexes, is k-bilinear. It is also immediate fromthe definition that one has

HomR(1M,1N) = 1HomR(M,N) .

For homomorphisms M′′ α−→M α′−→M′ and N′β′−→ N

β−→ N′′ of R-complexesthere is an equality


64 2 Complexes

HomR(α′α,ββ′) = (−1)|α

′|(|α|+|β|) HomR(α,β)HomR(α′,β′) .

Indeed, for every homomorphism ϑ : M′→ N′ one has

HomR(α′α,ββ′)(ϑ) = (−1)|α

′α|(|ββ′|+|ϑ|)(ββ′)ϑ(α′α)

= (−1)|α′α|(|ββ′|+|ϑ|)(−1)|α|(|β|+|β

′ϑα′|) HomR(α,β)(β′ϑα′)

= (−1)|α′|(|α|+|β|)(−1)|α

′|(|β′|+|ϑ|) HomR(α,β)(β′ϑα′)

= (−1)|α′|(|α|+|β|) HomR(α,β)HomR(α

′,β′)(ϑ) .

In particular, there are equalities,

HomR(α′α,N) = (−1)|α

′||α|HomR(α,N)HomR(α′,N) ,

HomR(M,ββ′) = HomR(M,β)HomR(M,β′) ,

and, furthermore, a commutative diagram of homomorphisms of k-complexes,

HomR(M,N)Hom(M,β)

//

Hom(α,β)

))

(−1)|α||β|Hom(α,N)

HomR(M,N′′)

Hom(α,N′′)

HomR(M′′,N)Hom(M′′,β)

// HomR(M′′,N′′) .

2.3.7 Lemma. Let α : M′→M and β : N→ N′ be homomorphisms of R-complexes.With H = Homk(HomR(M,N),HomR(M′,N′)) there is an equality

∂H(HomR(α,β)) = HomR(∂HomR(M′,M)(α),β)+(−1)|α|HomR(α,∂

HomR(N,N′)(β)) .

PROOF. Let ϑ : M→ N be a homomorphism. The definitions yield(∂H(HomR(α,β))

)(ϑ)

=(∂HomR(M′,N′) HomR(α,β)

)(ϑ)

− (−1)|α|+|β|(

HomR(α,β)∂HomR(M,N)

)(ϑ)

= ∂HomR(M′,N′)((−1)|α|(|β|+|ϑ|)βϑα)

− (−1)|α|+|β|HomR(α,β)(∂Nϑ− (−1)|ϑ|ϑ∂M)

= (−1)|α|(|β|+|ϑ|)(∂N′βϑα− (−1)|β|+|ϑ|+|α|βϑα∂M′)

− (−1)|α|+|β|(−1)|α|(|β|+|ϑ|−1)β(∂Nϑ− (−1)|ϑ|ϑ∂M)α .

Further, one has(HomR(∂

HomR(M′,M)(α),β)+(−1)|α|HomR(α,∂HomR(N,N′)(β))

)(ϑ)

= (−1)(|α|−1)(|β|+|ϑ|)βϑ∂HomR(M′,M)(α)

+(−1)|α|(−1)|α|(|β|−1+|ϑ|)∂HomR(N,N′)(β)ϑα

= (−1)(|α|−1)(|β|+|ϑ|)βϑ(∂Mα− (−1)|α|α∂M′)

+(−1)|α|(−1)|α|(|β|−1+|ϑ|)(∂N′β− (−1)|β|β∂N)ϑα .



Compare the two expressions above to see that they are identical.

Now we focus on chain maps.

2.3.8 Proposition. Let α and β be chain maps of R-complexes. The homomorphismHomR(α,β) is a chain map of degree |α|+ |β|, and if α or β is null-homotopic, thenHomR(α,β) is null-homotopic.

PROOF. Let α : M′→M and β : N→ N′ be chain maps; by 2.3.3 they are cycles inthe complexes HomR(M′,M) and HomR(N,N′). That is, one has ∂HomR(M′,M)(α) = 0and ∂HomR(N,N′)(β) = 0. With H =Homk(HomR(M,N),HomR(M′,N′)), 2.3.7 yields∂H(HomR(α,β)) = 0, so HomR(α,β) is a chain map by 2.3.3.

If α is null-homotopic, then one has α= ∂HomR(M′,M)(ϑ) for a ϑ in HomR(M′,M);see 2.3.3. Now 2.3.7 yields HomR(α,β) = ∂H(HomR(ϑ,β)), whence HomR(α,β) isnull-homotopic. Similarly, if one has β= ∂HomR(N,N′)(ϑ) for some ϑ in HomR(N,N′),then 2.3.7 yields HomR(α,β) = ∂H(HomR(α,(−1)|α|ϑ)).

2.3.9 Corollary. Let α,α′ : M′→M and β,β′ : N→ N′ be chain maps of R-complex-es. If α∼ α′ and β∼ β′, then one has HomR(α,β)∼ HomR(α

′,β′).

PROOF. By bilinearity one has

HomR(α,β)−HomR(α′,β′) = HomR(α−α′,β)+HomR(α

′,β−β′) ,

and by 2.3.8 that is a sum of null-homotopic chain maps; now invoke 2.2.25.

The functor described in the next theorem is called the Hom functor. The lastequality requires a preparatory remark. In the k-linear category C(R), each hom-setC(R)(M,N) is a k-module. Homotopy ’∼’ is by 2.2.25 a congruence relation onC(R)(M,N), so the set C(R)(M,N)/∼ of equivalence classes is a k-module underinduced addition, [α]∼+[β]∼ = [α+β]∼, and k-multiplication x[α]∼ = [xα]∼.

2.3.10 Theorem. The functions

HomR(–,–) : C(R)op×C(R) −→ C(k) ,

defined on objects in 2.3.1 and on morphisms in 2.3.5, constitute a k-bilinear andleft exact functor.

For R-complexes M and N, the complex HomR(M,N) and the module C(R)(M,N)are related by the following identities of k-modules,

Z0(HomR(M,N)) = C(R)(M,N) andH0(HomR(M,N)) = C(R)(M,N)/∼ .

PROOF. It follows from 2.3.1 and 2.3.8 that HomR(–,–) takes objects and mor-phisms in C(R)op×C(R) to objects and morphisms in C(k). The functoriality andthe k-bilinearity are established in 2.3.6. To prove left exactness, let

0−→ (M′,N′)(α′,β′)−−−−→ (M,N)

(α,β)−−−→ (M′′,N′′)−→ 0

be an exact sequence in C(R)op×C(R). It is sufficient to verify that the sequence


66 2 Complexes

(?) 0−→ HomR(M′,N′)vHom(α′,β′)v−−−−−−−→ HomR(M,N)v

Hom(α,β)v−−−−−−→ HomR(M′′,N′′)v

in M(k) is exact for every v ∈ Z. By 2.1.5 and 2.3.5 such a sequence is a product ofexact sequences; indeed HomR(α

′,β′)v is the product∏i∈Z

HomR(α′i,β′i+v) :

∏i∈Z

HomR(M′i ,N′i+v) −→

∏i∈Z

HomR(Mi,Ni+v) ,

and HomR(α,β)v has a similar form; it follows that (?) is exact.Finally, the equalities of k-modules follow from 2.3.3.

REMARK. The proof above uses the fact, immediate from 1.1.15, that a product of exact sequencesin M(R) is exact. This property does not hold in every Abelian category; a counterexample is thecategory of sheaves of Abelian groups on a suitable topological space; see [33].

2.3.11 Addendum. If M is in C(R–Qo) and N is in C(R–So), then it follows from2.1.37 and 2.1.18 that HomR(M,N)\ is a graded Q–So-bimodule, and it is elementaryto verify that the differential ∂HomR(M,N) is Q- and So-linear. That is, HomR(M,N) isan object in C(Q–So). For morphisms α in C(R–Qo) and β in C(R–So) it is straight-forward to verify that HomR(α,β) is a morphism in C(Q–So). Thus, there is aninduced k-bilinear functor,

HomR(–,–) : C(R–Qo)op×C(R–So) −→ C(Q–So) .

2.3.12. Let M be an R-complex. It follows from 2.3.1 and 2.3.5 that HomR(M,–)is a \-functor. For every degreewise split exact sequence 0→ N′ → N → N′′ → 0in C(R), the sequence 0→ HomR(M,N′)→ HomR(M,N)→ HomR(M,N′′)→ 0 isdegreewise split exact by 2.1.49.

Similarly, for every R-complex N the functor HomR(–,N) is a \-functor, whenceit preserves degreewise split exact sequences by 2.1.50.

HOM AND SHIFT

The shift functor allows one to consider chain maps as morphisms; for this reason,and for deeper reasons that will become apparent in Chap. 6, we record how shiftinteracts with Hom.

2.3.13 Proposition. Let M and N be R-complexes and let s be an integer. The com-posite HomR(ς

Σs M−s ,N)ς

Σ−s Hom(M,N)s is an isomorphism of k-complexes,

Σ−s HomR(M,N)

∼=−−→ HomR(Σs M,N) ,

and it is natural in M and N.

PROOF. Recall from 2.2.4 that ςΣ−s Hom(M,N)

s is an invertible chain map of degrees and that ςΣ

s M−s is a degree −s invertible chain map with inverse ςM

s . It followsfrom 2.3.6 and 2.3.8 that HomR(ς

Σs M−s ,N) is a degree −s invertible chain map with

inverse (−1)s HomR(ςMs ,N). Thus, the composite HomR(ς

Σs M−s ,N)ς

Σ−s Hom(M,N)s is

an invertible chain map of degree 0, i.e. an isomorphism in the category C(k).



To prove that this isomorphism is natural in M and N, let α : M′→M andβ : N→ N′ be morphisms of R-complexes. It suffices to show that the followingdiagram of homomorphisms of k-complexes is commutative,

Σ−s HomR(M,N)ςΣ−s Hom(M,N)s

//

Σ−s Hom(α,β)

HomR(M,N)Hom(ςΣ

s M−s ,N)

//

Hom(α,β)

HomR(Σs M,N)

Hom(Σs α,β)

Σ−s HomR(M′,N′)ςΣ−s Hom(M′,N′)s

// HomR(M′,N′)HomR(ς

Σs M′−s ,N′)

// HomR(Σs M′,N′) .

The left-hand square is commutative by (2.2.4.1). The equality αςΣs M′−s = ςΣ

s M−s (Σsα)

from (2.2.4.1) conspires with 2.3.6 to give

HomR(ςΣs M′−s ,N′)HomR(α,β) = HomR(ας

Σs M′−s ,β)

= HomR(ςΣs M−s (Σsα),β)

= HomR(Σsα,β)HomR(ς

Σs M−s ,N) .

This shows that the right-hand square is commutative.

2.3.14. If one suppresses the degree changing maps, then the isomorphism in 2.3.13is given by the assignment ϑ 7→ (−1)s|ϑ|ϑ, where |ϑ| is the degree of ϑ as an elementof HomR(M,N); see 2.3.5. Similarly, the isomorphism in 2.3.15 below is the identityupon suppression of the degree changing chain maps.

2.3.15 Proposition. Let M and N be R-complexes, and let s be an integer. Thecomposite ςHom(M,N)

s HomR(M,ςΣs N−s ) is an isomorphism of k-complexes,

HomR(M,Σs N)∼=−−→ Σ

s HomR(M,N) ,


PROOF. Parallel to the proof of 2.3.13.

EXACTNESS

2.3.16 Proposition. Let M be an R-complex and 0→ N′ → N → N′′ → 0 be asequence in C(R). If 0→ HomR(Mv,N′i )→ HomR(Mv,Ni)→ HomR(Mv,N′′i )→ 0is an exact sequence of modules for all v, i ∈ Z, then the sequence of complexes,0→ HomR(M,N′)→ HomR(M,N)→ HomR(M,N′′)→ 0, is exact.

PROOF. By 2.1.39 exactness of a sequence of complexes can be verified degree-wise. Fix p ∈ Z; it follows from 2.1.5 and 2.3.5 that the sequence of modules,0→ HomR(M,N′)p→ HomR(M,N)p→ HomR(M,N′′)p→ 0, is the product of thesequences 0→ HomR(Mv,N′v+p)→ HomR(Mv,Nv+p)→ HomR(Mv,N′′v+p)→ 0 forall v ∈ Z. By assumption all these sequences are exact, hence so is the product.

2.3.17 Example. Let P be a complex of projective R-modules. For every exactsequence 0→ N′ → N → N′′ → 0 of R-complexes the sequence of k-complexes,0→ HomR(P,N′)→ HomR(P,N)→ HomR(P,N′′)→ 0, is exact.


68 2 Complexes

2.3.18 Proposition. Let N be an R-complex and 0→ M′ → M → M′′ → 0 be asequence in C(R). If 0→ HomR(M′′v ,Ni)→ HomR(Mv,Ni)→ HomR(M′v,Ni)→ 0is an exact sequence of modules for all v, i ∈ Z, then the sequence of complexes,0→ HomR(M′′,N)→ HomR(M,N)→ HomR(M′,N)→ 0, is exact.


2.3.19 Example. Let I be a complex of injective R-modules. For every exact se-quence 0→ M′ → M → M′′ → 0 of R-complexes the sequence of k-complexes,0→ HomR(M′′, I)→ HomR(M, I)→ HomR(M′, I)→ 0, is exact.

EXERCISES

E 2.3.1 Apply 2.3.4 to prove that if β : L→M and α : M→ N are chain maps and one of them isnull-homotopic, then so is the composite αβ. From there recover 2.2.25(b).

E 2.3.2 Generalize the identity in 2.3.4 as follows. If M1 α1−→ M2 α2

−→ ·· · −→ Mn αn−→ Mn+1 are

homomorphisms of R-complexes, then one has

∂HomR(M1,Mn+1)(αn · · ·α2α1)

= ∑ni=1(−1)|α

i+1|+···+|αn|αn · · ·αi+1∂HomR(Mi,Mi+1)(αi)αi−1 · · ·α1 .

E 2.3.3 Give a proof of part (b) in 2.3.9.

E 2.3.4 For R-complexes M and N consider the homomorphisms ∂HomR(M,N), HomR(∂M ,N), and

HomR(M,∂N) of degree −1 from the complex HomR(M,N) to itself. Verify that

∂HomR(M,N) = HomR(M,∂N)−HomR(∂M ,N) .

E 2.3.5 (Cf. 2.3.11) Let α be a homomorphism of complexes of R–Qo-bimodules and let β be ahomomorphism of complexes of R–So-bimodules. Show that HomR(α,β) is a homomor-phism of complexes of Q–So-bimodules.

E 2.3.6 Give a proof of 2.3.15.E 2.3.7 Let M be an R-module and consider the functor HomR(M,–) : M(R)→M(k). Show

that the extended functor C(R) → C(k) described in 2.1.45 agrees with the functorHomR(M,–) defined in this section. Is the parallel statement true for HomR(–,M)?

E 2.3.8 Let η = 0→ M′ → M→ M′′ → 0 be an exact sequence of R-complexes. Show that thefollowing conditions are equivalent. (i) η is degreewise split. (ii) HomR(η,N) is exact forevery R-module/complex N. (iii) HomR(N,η) is exact for every R-module/complex N.

2.4 Tensor Products

SYNOPSIS. Tensor product complex; chain map, homotopy; tensor product functor; exactness.

The graded tensor product of two complexes can be endowed with a differentialconstructed from the differentials of the factors.

2.4.1 Definition. Let M be an Ro-complex and let N be an R-complex. The tensorproduct complex M⊗R N is the k-complex with underlying graded module

(M⊗R N)\ = M\⊗R N\


2.4 Tensor Products 69

and differential given by

∂M⊗RN(m⊗n) = ∂M(m)⊗n+(−1)|m|m⊗∂N(n)

for homogeneous elements m and n.

2.4.2. It is elementary to verify that ∂M⊗RN is square zero. Indeed, one has

∂M⊗RN∂M⊗RN(m⊗n) = ∂M⊗RN(∂M(m)⊗n+(−1)|m|m⊗∂N(n))

= ∂M∂M(m)⊗n+(−1)|m|−1∂M(m)⊗∂N(n)

+(−1)|m|(∂M(m)⊗∂N(n)+(−1)|m|m⊗∂N∂N(n))

= 0 .

2.4.3 Example. Let x1 and x2 be elements in k. The tensor product of Koszul com-plexes Kk(x1)⊗kKk(x2) is isomorphic to Kk(x1,x2). To see this, let KR(x1) andKR(x2) be generated by free k-modules k〈e1〉 and k〈e2〉; that is,

KR(xi) = 0−→ k〈ei〉∂−−→ k−→ 0 with ∂(ei) = xi ,

cf. (2.2.8.1). The tensor product Kk(x1)⊗kKk(x2) then has the form

0−→ k〈e1〉⊗k k〈e2〉∂2−−→ (k〈e1〉⊗k k)⊕ (k⊗R k〈e2〉)

∂1−−→ k⊗k k−→ 0with

∂2(e1⊗ e2) = (−e1⊗ x2,x1⊗ e2) and ∂1(e1⊗ k2,k1⊗ e2) = x1⊗ k2 + k1⊗ x2 .

Similarly, consider Kk(x1,x2) to be generated by the free module k〈 f1, f2〉. Com-paring to (2.2.8.2) it is now elementary to verify that the assignments

e1⊗ e2 7−→ f1∧ f2 , (e1⊗ k2,k1⊗ e2) 7−→ k2 f1 + k1 f2 , and k1⊗ k2 7−→ k1k2

define an isomorphism Kk(x1)⊗kKk(x2)−→ Kk(x1,x2).

FUNCTORIALITY

The next several paragraphs lead up to Theorem 2.4.9, which shows that the tensorproduct yields a functor.

2.4.4 Definition. Let α : M→M′ be a homomorphism of Ro-complexes and letβ : N→ N′ a homomorphism of R-complexes. Denote by α⊗R β the degree |α|+ |β|homomorphism of k-complexes

M⊗R N −→M′⊗R N′ given by m⊗n 7−→ (−1)|β||m|α(m)⊗β(n) .

Furthermore, set α⊗R N = α⊗R 1N and M⊗R β= 1M⊗R β.

2.4.5. It is straightforward to verify that the assignment (α,β) 7→ α⊗R β, for ho-momorphisms α and β of complexes, is k-bilinear. It is also immediate from thedefinition that one has


70 2 Complexes

1M⊗R 1N = 1M⊗RN .

For homomorphisms M′ α′−→ M α−→ M′′ of Ro-complexes and homomorphisms

N′β′−→ N

β−→ N′′ of R-complexes there is an equality

(αα′)⊗R (ββ′) = (−1)|α

′||β|(α⊗R β)(α′⊗R β

′) .

Indeed, for homogeneous elements m′ in M′ and n′ in N′ one has

((αα′)⊗R (ββ′))(m′⊗n′) = (−1)|ββ

′||m′|(αα′)(m′)⊗ (ββ′)(n′)

= (−1)|ββ′||m′|(−1)|β||α

′(m′)|(α⊗R β)(α′(m′)⊗β′(n′))

= (−1)|α′||β|(−1)|β

′||m′|(α⊗R β)(α′(m′)⊗β′(n′))

= (−1)|α′||β|(α⊗R β)((α

′⊗R β′)(m′⊗n′)) .

In particular, there are equalities,

(αα′)⊗R N = (α⊗R N)(α′⊗R N) ,

M⊗R (ββ′) = (M⊗R β)(M⊗R β

′) ,

and, furthermore, a commutative diagram of homomorphisms of k-complexes,

M⊗R NM⊗β

//

α⊗β

((

(−1)|α||β|α⊗N

M⊗R N′′

α⊗N′′

M′′⊗R NM′′⊗β

// M′′⊗R N′′ .

2.4.6 Lemma. Let α : M→M′ be a homomorphism of Ro-complexes and β : N→N′

be a homomorphisms of R-complexes. With H = Homk(M⊗R N,M′⊗R N′) one has

∂H(α⊗R β) = ∂HomRo(M,M′)(α)⊗R β+(−1)|α|α⊗R ∂HomR(N,N′)(β) .

PROOF. Let m and n be homogeneous elements in M and N. There are equalities,(∂H(α⊗R β)

)(m⊗n)

=(∂M′⊗RN′(α⊗R β)− (−1)|α|+|β|(α⊗R β)∂

M⊗RN)(m⊗n)

= (−1)|β||m|∂M′⊗RN′(α(m)⊗β(n))

− (−1)|α|+|β|(α⊗R β)(∂M(m)⊗n+(−1)|m|m⊗∂N(n))

= (−1)|β||m|(∂M′α(m)⊗β(n)+(−1)|α|+|m|α(m)⊗∂N′β(n))

− (−1)|α|+|β|((−1)|β|(|m|−1)α∂M(m)⊗β(n)

+(−1)|m|(−1)|β||m|α(m)⊗β∂N(n)).

Further, one has(∂HomRo(M,M′)(α)⊗R β+(−1)|α|α⊗R ∂

HomR(N,N′)(β))(m⊗n)



= (−1)|β||m|(∂HomRo(M,M′)(α))(m)⊗β(n)

+(−1)|α|(−1)(|β|−1)|m|α(m)⊗ (∂HomR(N,N′)(β))(n)

= (−1)|β||m|(∂M′α− (−1)|α|α∂M)(m)⊗β(n)

+(−1)|α|(−1)(|β|−1)|m|α(m)⊗ (∂N′β− (−1)|β|β∂N)(n) .

Compare the two expressions above to see that they are identical.

2.4.7 Proposition. Let α : M→M′ be a chain map of Ro-complexes and β : N→ N′

be a chain map of R-complexes. The homomorphism α⊗R β is a chain map of degree|α|+ |β|, and if α or β is null-homotopic, then α⊗R β is null-homotopic.

PROOF. By 2.3.3 the chain maps α and β are cycles in the complexes HomRo(M,M′)and HomR(N,N′). That is, one has ∂HomRo(M,M′)(α) = 0 and ∂HomR(N,N′)(β) = 0. De-note by H the complex Homk(M⊗R N,M′⊗R N′), then 2.4.6 yields ∂H(α⊗R β) = 0,so α⊗R β is a chain map by 2.3.3.

If α is null-homotopic, then α = ∂HomRo(M,M′)(ϑ) for some ϑ ∈ HomRo(M,M′);see 2.3.3. Now 2.4.6 yields α⊗R β= ∂H(ϑ⊗R β), whence α⊗R β is null-homotopic.Similarly, β= ∂HomR(N,N′)(ϑ) implies α⊗R β= ∂H((−1)|α|α⊗R ϑ).

2.4.8 Corollary. Let α,α′ : M→M′ be chain maps of Ro-complexes and β,β′ : N→ N′

be chain maps of R-complexes. If α∼ α′ and β∼ β′, then one has α⊗R β∼ α′⊗R β′.

PROOF. By bilinearity one has

α⊗R β−α′⊗R β′ = (α−α′)⊗R β+α′⊗R (β−β′) ,

and by 2.4.7 that is a sum of null-homotopic chain maps; now invoke 2.2.25.

The functor described in the next theorem is called the tensor product functor.

2.4.9 Theorem. The functions

–⊗R – : C(Ro)×C(R) −→ C(k) ,

defined on objects in 2.4.1 and on morphisms in 2.4.4, constitute a k-bilinear andright exact functor.

PROOF. It follows from 2.4.1 and 2.4.7 that –⊗R – takes objects and morphismsin C(Ro)× C(R) to objects and morphisms in C(k). The functoriality and the k-bilinearity are established in 2.4.5. To prove right exactness, let

0−→ (M′,N′)(α′,β′)−−−−→ (M,N)

(α,β)−−−→ (M′′,N′′)−→ 0

be an exact sequence in C(Ro)×C(R). It is sufficient to verify that the sequence

(?) (M′⊗R N′)v(α′⊗β′)v−−−−−→ (M⊗R N)v

(α⊗β)v−−−−→ (M′′⊗R N′′)v −→ 0

in M(k) is exact for every v ∈ Z. By 2.1.14 and 2.4.4 such a sequence is a coproductof exact sequences; indeed (α′⊗R β

′)v is the coproduct


72 2 Complexes∐i∈Z

α′i⊗R β′v−i :

∐i∈Z

M′i ⊗R N′v−i −→∐i∈Z

Mi⊗R Nv−i ,

and (α⊗R β)v has a similar form; it follows that (?) is exact.

2.4.10 Addendum. If M is in C(Q–Ro) and N is in C(R–So), then it follows from2.1.37 and 2.1.18 that (M⊗R N)\ is a graded Q–So-bimodule, and it is elementaryto verify that ∂M⊗RN is Q- and So-linear. That is, M⊗R N is an object in C(Q–So).For morphisms α in C(Q–Ro) and β in C(R–So) it is straightforward to verify thatα⊗R β is a morphism in C(Q–So). Thus, there is an induced k-bilinear functor

–⊗R – : C(Q–Ro)×C(R–So) −→ C(Q–So) .

2.4.11. Let M be an Ro-complex. It follows from 2.4.1 and 2.4.4 that M⊗R – is a \-functor. For every degreewise split exact sequence 0→ N′→ N→ N′′→ 0 in C(R),the sequence 0→M⊗R N′→M⊗R N→M⊗R N′′→ 0 is degreewise split exact by2.1.49.

Similarly, for every R-complex N the functor –⊗R N is a \-functor, whence itpreserves degreewise split exact sequences.

TENSOR PRODUCT AND SHIFT

Recall from 2.2.4 that for every complex X and every integer s there is an invertiblechain map ςX

s : X → Σs X . If one suppresses these maps, then the isomorphism in2.4.12 below is given by the assignment m⊗n 7→ (−1)s|m|m⊗n.

2.4.12 Proposition. Let M be an Ro-complex and let N be an R-complex. For everys ∈ Z the composite ςM⊗N

s (M⊗R ςΣs N−s ) is an isomorphism of k-complexes,

M⊗R Σs N

∼=−−→ Σs (M⊗R N) ,


PROOF. As recalled above, ςM⊗Ns is an invertible chain map of degree s, and ςΣ

s N−s

is a degree −s invertible chain map with inverse ςNs ; see 2.2.4. It follows from 2.4.5

and 2.4.7 that M⊗R ςΣs N−s is a degree −s invertible chain map with inverse M⊗R ς

Ns .

Thus, the composite ςM⊗Ns (M⊗R ς

Σs N−s ) is an invertible chain map of degree 0, i.e.

an isomorphism in the category C(k).To prove that this isomorphism is natural in M and N, let α : M→M′ and

β : N→ N′ be morphisms of complexes. It suffices to show that the following di-agram of homomorphisms of k-complexes is commutative,

M⊗R Σs N

α⊗Σs β

M⊗ςΣs N−s

// M⊗R NςM⊗N

s//

α⊗β

Σs (M⊗R N)

Σs (α⊗β)

M′⊗R Σs N′

M′⊗ςΣs N′−s// M′⊗R N′

ςM′⊗N′s

// Σs (M′⊗R N′) .

The equality βςΣs N−s = ςΣ

s N′−s Σs β from (2.2.4.1) conspires with 2.4.5 to give



(α⊗R β)(M⊗R ςΣs N−s ) = α⊗R (βς

Σs N−s )

= α⊗R (ςΣs N′−s Σ

s β)

= (M′⊗R ςΣs N′−s )(α⊗R Σ

s β) .

This shows that the left-hand square is commutative. The right-hand square is com-mutative by (2.2.4.1).

If one suppresses the degree changing chain maps ς defined in 2.2.4, then theisomorphism in 2.4.13 below is an equality.

2.4.13 Proposition. Let M be an Ro-complex and let N be an R-complex. For everys ∈ Z the composite ςM⊗N

s (ςΣs M−s ⊗R N) is an isomorphism of k-complexes,

Σs M⊗R N

∼=−−→ Σs (M⊗R N) ,



EXACTNESS

2.4.14 Proposition. Let M be an Ro-complex and 0→ N′ → N → N′′ → 0 be asequence in C(R). If 0→Mv⊗R N′i →Mv⊗R Ni→Mv⊗R N′′i → 0 is an exact se-quence of modules for all v, i∈ Z, then 0→M⊗R N′→M⊗R N→M⊗R N′′→ 0 isan exact sequence of complexes.

PROOF. By 2.1.39 exactness of a sequence of complexes can be verified degree-wise. Fix p ∈ Z; it follows from 2.1.14 and 2.4.4 that the sequence of mod-ules, 0→ (M⊗R N′)p→ (M⊗R N)p→ (M⊗R N′′)p→ 0, is the coproduct of thesequences 0→Mv⊗R N′v−p→Mv⊗R Nv−p→Mv⊗R N′′v−p→ 0 for all v ∈ Z. By as-sumption each of these sequences is exact, hence so is the coproduct.

2.4.15 Proposition. Let N be an R-complex and 0→ M′ → M → M′′ → 0 be asequence in C(Ro). If 0→M′v⊗R Ni→Mv⊗R Ni→M′′v ⊗R Ni→ 0 is an exact se-quence of modules for all v, i∈ Z, then 0→M′⊗R N→M⊗R N→M′′⊗R N→ 0 isan exact sequence of complexes.


2.4.16 Example. Let F be a complex of flat R-modules. For every exact sequence0→ M′ → M → M′′ → 0 of Ro-complexes the induced sequence of k-complexes,0→M′⊗R F →M⊗R F →M′′⊗R F → 0, is exact.

EXERCISES

E 2.4.1 Let M be an Ro-complex and N be an R-complex. Consider the degree −1 homomor-phisms ∂M⊗RN , ∂M⊗R N, and M⊗R ∂

N from M⊗R N to itself. Verify the identity

∂M⊗RN = ∂M⊗R N +M⊗R ∂N .


74 2 Complexes

E 2.4.2 Give a proof of part (b) in 2.4.8.E 2.4.3 Show that the isomorphism in 4.4.7 is an isomorphism of (differential) graded k-algebras.E 2.4.4 (Cf. 2.4.10) Let α be a homomorphism of complexes of Q–Ro-bimodules and let β be a

homomorphism of complexes of R–So-bimodules. Show that α⊗R β is a homomorphismof complexes of Q–So-bimodules.

E 2.4.5 Let x1 and x2 be central elements in R and consider the complexes K1 = 0→ Rx1−→ R→ 0

and K2 = 0→ Rx2−→ R→ 0. Construct the complex K1⊗R K2.

E 2.4.6 Let M be an Ro-module and consider the functor M⊗R –: M(R)→M(k). Show thatthe extended functor C(R)→ C(k) described in 2.1.45 agrees with the functor M⊗R –defined in this section. Is the similar statement true for an R-module N and –⊗R N?

2.5 Boundedness and Finiteness

SYNOPSIS. Degreewise finite generation/presentation; boundedness (above/below); supremum;infimum; amplitude; boundedness of Hom complex; boundedness of tensor product complex;hard/soft truncation; graded basis; graded-free module.

In this section, we examine conditions on complexes that naturally extend finitegeneration and finite presentation of modules.

2.5.1 Definition. An R-complex M is called degreewise finitely generated if the R-module Mv is finitely generated for every v ∈ Z. Similarly, M is called degreewisefinitely presented if the R-module Mv is finitely presented for every v ∈ Z.

REMARK. A complex may be degreewise finitely generated without the underlying graded modulebeing finitely generated; see E 2.5.1.

2.5.2 Definition. An R-complex M is called bounded above if Mv = 0 holds forv 0, bounded below if Mv = 0 holds for v 0, and bounded if it is boundedabove and below.

REMARK. Other words for bounded above/below are left/right bounded.

2.5.3. Suppressing the full and faithful functors M(R) −→Mgr(R) −→ C(R), see2.1.13 and 2.1.34, an R-module M is considered as an R-complex 0→M→ 0 con-centrated in degree 0. Conversely, an R-complex M that is concentrated in degree 0is identified with the module M0; this amounts to suppressing the forgetful functorsC(R)−→Mgr(R)−→M(R).

More generally, a diagram in M(R),

M0 α0−−→ ·· · −→Mp−1 αp−1

−−−→Mp ,

such that αnαn−1 = 0 holds for all n in 1, . . . , p−1, can be considered, for everyu ∈ Z, as a complex concentrated in degrees p+u, . . . ,u.

REMARK. In the literature, a complex 0→M→ 0 with the module M in degree zero is sometimesreferred to as a stalk complex.


2.5 Boundedness and Finiteness 75

The supremum and infimum of a complex, which will be defined next, captureits homological position; the amplitude captures its homological size.

2.5.4 Definition. Let M be a complex. The supremum and infimum of M are definedas follows,

supM = supv ∈ Z | Hv(M) 6= 0 and infM = infv ∈ Z | Hv(M) 6= 0 ,

with the convention that one sets supM = −∞ and infM = ∞ if M is acyclic. Theamplitude of M is the difference

ampM = supM− infM ,

with the convention that one sets ampM =−∞ if M is acyclic.

2.5.5. For every complex M and every integer s one evidently has

supΣs M = supM+ s , infΣs M = infM+ s , and ampΣs M = ampM .

For every complex M one has H(M\) = M\ and, therefore,

supM\ = supv ∈ Z |Mv 6= 0 and infM\ = infv ∈ Z |Mv 6= 0 .

Thus, a complex M 6= 0 is bounded above if and only if supM\ is not ∞; and M isbounded below if and only if infM\ is not −∞.

2.5.6. We record a few consequences of 2.2.18. Let M be an R-complex.(a) For every projective R-module P the next inequalities hold,

infHomR(P,M) > infM and supHomR(P,M) 6 supM ;

if P is non-zero and free, then equalities hold.(b) For every injective R-module E the next inequalities hold,

infHomR(M,E) > −supM and supHomR(M,E) 6 − infM ;

if E is faithfully injective, then equalities hold. In particular, if E faithfully injective,then HomR(M,E) is acyclic if and only if M is acyclic.

(c) For every flat Ro-module F the next inequalities hold,

inf(F⊗R M) > infM and sup(F⊗R M) 6 supM ;

if F is faithfully flat, then equalities hold. In particular, if F faithfully flat, thenF⊗R M is acyclic if and only if M is acyclic.

HOM COMPLEXES

2.5.7 Lemma. Let M and N be R-complexes with Mv = 0 for all v < 0 and Nv = 0for all v > 0. There are isomorphisms of k-modules,

H0(HomR(M,N)) ∼= Z0(HomR(M,N)) ∼= HomR(H0(M),H0(N)) .


76 2 Complexes

PROOF. It follows by the assumptions that the only degree 1 homomorphism fromM to N is the zero map. In particular, the only null-homotopic morphism M→ N isthe zero map. In view of this fact and 2.3.10 one sees that the canonical map

Z0(HomR(M,N)) = C(R)(M,N) C(R)(M,N)/∼ = H0(HomR(M,N))

is an isomorphism. This proves the first isomorphism in 2.5.7.To prove the second isomorphism, note that the assumptions on M and N yield

equalities H0(M) = C0(M) and H0(N) = Z0(N), so that one has

HomR(H0(M),H0(N)) = HomR(C0(M),Z0(N)) .

Let π0 : M0 C0(M) be the canonical map, and let ι0 : Z0(N) N0 be the embed-ding. There is an isomorphism of k-modules,

ϕ : HomR(C0(M),Z0(N))∼=−−→ C(R)(M,N) ,

defined by assigning to a homomorphism α : C0(M)→ Z0(N) of modules the mor-phism of complexes given by:

· · · // M1

∂M1// M0

ι0απ0

// 0

// · · ·

· · · // 0 // N0∂N

0// N−1 // · · · .

The inverse of ϕ is defined as follows. Let β= (βv)v∈Z : M→ N be a morphism. Thehomomorphism β0 : M0→ N0 satisfies β0∂

M1 = 0 and ∂N

0 β0 = 0. Hence there is aunique homomorphism β : C0(M)→ Z0(N) that makes the diagram

M0

β0

π0// // C0(M)

β

N0 Z0(N)ooι0oo

commutative. Now, set ϕ−1(β) = β.

2.5.8. Let M and N be R-complexes. Suppose there exist integers w and u such thatone has Mv = 0 for all v < u and Nv = 0 for all v > w. For each v ∈ Z the moduleHomR(M,N)v is then a direct sum

HomR(M,N)v =∏i∈Z

HomR(Mi,Ni+v) =w−v⊕i=u

HomR(Mi,Ni+v) .

If one has w− v < u, then HomR(M,N)v = 0 holds.

2.5.9 Proposition. Let M and N be R-complexes. If M is bounded below and N isbounded above, then the complex HomR(M,N) is bounded above. More precisely,if one has Mv = 0 for all v< u and Nv = 0 for all v>w, then the next assertions hold.

(a) HomR(M,N)v = 0 for v > w−u.



(b) HomR(M,N)w−u = HomR(Mu,Nw).(c) Hw−u(HomR(M,N))∼= HomR(Hu(M),Hw(N)).

PROOF. Parts (a) and (b) are immediate from 2.5.8.(c): By 2.2.15, 2.3.13, and 2.3.15 there is an isomorphism

Hw−u(HomR(M,N)) = H0(Σu−w HomR(M,N)) ∼= H0(HomR(Σ

−u M,Σ−w N)) .

The complexes Σ−u M and Σ−w N are concentrated in non-negative and non-positivedegrees, respectively, so by 2.5.7 there is an isomorphism

H0(HomR(Σ−u M,Σ−w N)) ∼= HomR(H0(Σ

−u M),H0(Σ−w N))

= HomR(Hu(M),Hw(N)) .

2.5.10 Proposition. Let S be right Noetherian. Let M be a bounded below R-complex and let X a bounded above complex of R–So-bimodules. If M is degree-wise finitely generated and X is degreewise finitely generated over So, then the So-complex HomR(M,X) is bounded above and degreewise finitely generated.

PROOF. For every v ∈ Z and i ∈ Z the So-module HomR(Mi,Xi+v) is finitely gener-ated; see 1.3.12. By assumption there exist integers w and u such that Mv = 0 for allv < u and Xv = 0 for all v > w. It follows from 2.5.8 that the module HomR(M,X)vis finitely generated for every v, and by 2.5.9 it is zero for v > w−u.

TENSOR PRODUCT COMPLEXES

2.5.11 Lemma. Let M be an Ro-complex and let N be an R-complex with Mv = 0and Nv = 0 for all v < 0. There are isomorphisms of k-modules

H0(M⊗R N) ∼= C0(M⊗R N) ∼= H0(M)⊗R H0(N) .

PROOF. By definition of the tensor product complex, see 2.4.1, and by the assump-tions on M and N, it follows that the sequence

(M⊗R N)1∂

M⊗RN1−−−−→ (M⊗R N)0

∂M⊗RN0−−−−→ (M⊗R N)−1

equals

(?)(M1⊗R N0)⊕

(M0⊗R N1)

(∂M1 ⊗RN0 M0⊗R∂

N1 )−−−−−−−−−−−−−→M0⊗R N0 −→ 0 .

The first isomorphism in 2.5.11 follows as ∂M⊗RN0 is 0.

To prove the second isomorphism, notice that one has H0(M) = C0(M) andH0(N) = C0(N), as ∂M

0 = 0 and ∂N0 = 0. Thus, we must establish an isomorphism of

k-modules, C0(M⊗R N)∼= C0(M)⊗R C0(N). Consider the homomorphism,

ϕ : M0⊗R N0 −→ C0(M)⊗R C0(N) ,

defined by m⊗n 7→ [m]B0(M)⊗ [n]B0(N). From (?) one gets B0(M⊗R N)⊆Kerϕ, andhence ϕ induces a homomorphism,


78 2 Complexes

ϕ : C0(M⊗R N) −→ C0(M)⊗R C0(N) ,

given by [m⊗ n]B0(M⊗RN) 7→ [m]B0(M)⊗ [n]B0(N). To see that ϕ is an isomorphism,note that there is a well-defined map,

ψ : C0(M)×C0(N) −→ C0(M0⊗R N0) ,

given by ([m]B0(M), [n]B0(N)) 7→ [m⊗n]B0(M⊗RN). Indeed, if m−m′ is in B0(M) andn−n′ is in B0(N), then one has m−m′ = ∂M

1 (m0) and n−n′ = ∂N1 (n0) for suitable

m0 ∈M0 and n0 ∈ N0, and consequently,

m⊗n−m′⊗n′ = (∂M1 ⊗R N0)(m0⊗n)+(M0⊗R ∂

N1 )(m

′⊗n0) ∈ B0(M⊗R N) .

It is straightforward to verify that ψ is k-bilinear and middle R-linear, so by theuniversal property of tensor products it induces a homomorphism of k-modules,

ψ : C0(M)⊗R C0(N) −→ C0(M0⊗R N0) ,

given by [m]B0(M)⊗ [n]B0(N) 7→ [m⊗n]B0(M⊗RN). Clearly, ψ is an inverse of ϕ.

2.5.12. Let M be an Ro-complex and let N be an R-complex. Suppose there existintegers u and w such that Mv = 0 for v < u and Nv = 0 for v < w. For each v ∈ Zthe module (M⊗R N)v is then a direct sum

(M⊗R N)v =∐i∈Z

Mi⊗R Nv−i =v−w⊕i=u

Mi⊗R Nv−i .

If one has v−w < u, then (M⊗R N)v = 0 holds.

2.5.13 Proposition. Let M be an Ro-complex and let N be an R-complex. If M andN are bounded below, then the complex M⊗R N is a bounded below. More precisely,if one has Mv = 0 for v < u and Nv = 0 for v < w, then the next assertions hold.

(a) (M⊗R N)v = 0 for v < u+w.(b) (M⊗R N)u+w = Mu⊗R Nw.(c) Hu+w(M⊗R N)∼= Hu(M)⊗R Hw(N).

PROOF. Parts (a) and (b) are immediate from 2.5.12.(c): By 2.2.15, 2.4.12, and 2.4.13 there is an isomorphism

Hu+w(M⊗R N) = H0(Σ−u−w (M⊗R N)) ∼= H0((Σ

−u M)⊗R (Σ−w N)) .

The complexes Σ−u M and Σ−w N are concentrated in non-negative degrees, so by2.5.11 there is an isomorphism

H0((Σ−u M)⊗R (Σ

−w N)) ∼= H0(Σ−u M)⊗R H0(Σ

−w N)

= Hu(M)⊗R Hw(N) .

2.5.14 Proposition. Let N be a bounded below S-complex and let X be a boundedbelow complex of R–So-bimodules. If N is degreewise finitely generated and X isdegreewise finitely generated over R, then the R-complex X⊗S N is bounded belowand degreewise finitely generated.



PROOF. For every v ∈ Z and i ∈ Z the R-module Xi⊗S Nv−i is finitely generated;see 1.3.13. By assumption there exist integers u and w such that Xv = 0 for v < uand Nv = 0 for v < w. It follows from 2.5.12 that the module (X⊗S N)v is finitelygenerated for every v, and by 2.5.13 it is zero for v < u+w.

TRUNCATIONS

To handle unbounded complexes it is at times convenient to cut them into boundedpieces. The instruments for such procedures are known as truncations.

2.5.15 Definition. Let M be an R-complex and n be an integer. The hard truncationabove of M at n is the complex Mďn defined by (Mďn)v = 0 for v > n and ∂Mďn

v = ∂Mv

for v6 n. It can be visualized as follows,

Mďn = 0−→Mn∂M

n−−→Mn−1∂M

n−1−−→Mn−2 −→ ·· · .

Similarly, the hard truncation below of M at n is the complex Měn defined by(Měn)v = 0 for v < n and ∂Měn

v = ∂Mv for v > n. It can be visualized as follows,

Měn = · · · −→Mn+2∂M

n+2−−→Mn+1∂M

n+1−−→Mn −→ 0 .

Let α : M→ N be a morphism of R-complexes and n be an integer. The symbolαďn denotes the induced morphism Mďn→ Nďn, and αěn is defined similarly.

2.5.16. For every n ∈ Z, hard truncation above (–)ďn and hard truncation below(–)ěn are evidently k-linear \-functors from C(R) to C(R).

2.5.17. Let M be an R-complex. For every n∈Z, the truncation Mďn is a subcomplexof M, and the quotient complex M/Mďn is the truncation Měn+1. In particular, thereis a degreewise split exact sequence of R-complexes

0−→Mďn −→M −→Měn+1 −→ 0 .

2.5.18 Definition. Let M be an R-complex and n be an integer. The soft truncationabove of M at n is the complex MĎn defined by

(MĎn)v =

0 for v > nCn(M) for v = n

and ∂MĎnv =

∂M

n for v = n∂M

v for v < n

where ∂Mn : Cn(M)→Mn−1 is the homomorphism induced by ∂M

n . The complex canbe visualized as follows,

MĎn = 0−→ Cn(M)∂M

n−−→Mn−1∂M

n−1−−→Mn−2 −→ ·· · .

Similarly, the soft truncation below of M at n is the complex MĚn defined by:

(MĚn)v =

Zn(M) for v = n0 for v < n

and ∂MĚnv = ∂M

v for v > n .


80 2 Complexes

The complex can be visualized as follows,

MĚn = · · · −→Mn+2∂M

n+2−−→Mn+1∂M

n+1−−→ Zn(M)−→ 0 .

Let α : M→ N be a morphism of R-complexes and n be an integer. The symbolαĎn denotes the induced morphism MĎn→ NĎn, and αĚn is defined similarly.

2.5.19. It is immediate from the definitions that soft truncation above (–)Ďn and softtruncation below (–)Ěn are k-linear endofunctors on C(R) for every n ∈ Z.

2.5.20. Let M be an R-complex. For every n∈Z, the truncation MĚn is a subcomplexof M, and the embedding

(2.5.20.1) τMĚn : MĚn M

induces an isomorphism in homology in degree n and above. That is, Hv(τMĚn) is an

isomorphism for v> n. Similarly, MĎn is a quotient complex and the canonical map

(2.5.20.2) τMĎn : M MĎn

induces an isomorphism in homology in degree n and below. That is, Hv(τMĎn) is an

isomorphism for v6 n.For n > supM and for n < infM it follows from 2.2.11 that there is an exact

sequence of R-complexes

(2.5.20.3) 0−→MĚn+1 −→M −→MĎn −→ 0 .

GRADED-FREE MODULES

2.5.21 Definition. Let L be a graded R-module. A set E = euu∈U of generatorsfor L, a basis for L in particular, is called graded if each element eu is homogeneous.The module L is called graded-free if has a graded basis.

For a graded set E, not a priori assumed to be a subset of a module, the graded-free R-module with graded basis E is denoted R〈E〉.

2.5.22 Proposition. A graded R-module L is graded-free if and only if the R-moduleLv is free for every v ∈ Z.

PROOF. If each module Lv is free with basis Ev, then E =⋃

v∈ZEv is a graded basisfor L. For the converse, let E be a graded basis for L and fix v. Every element in Lvis a unique linear combination of elements in E. Only elements of degree v occurwith non-zero coefficients, so the elements of degree v in E form a basis for Lv.

2.5.23. Let M be a graded R-module. In view of 2.5.22 it follows from 1.3.11 thatthere is a surjective morphism L→M of graded R-modules with L graded-free. If Mis degreewise finitely generated, then L can be chosen degreewise finitely generated.

To realize a complex as the image of a complex of free modules, one has to takethe differentials into account. To this end, the next construction is key.



2.5.24 Construction. For an R-module F and an integer v, let Dv(F) denote thedisc complex 0−→ F =−→ F −→ 0 concentrated in degrees v and v−1.

Let M be an R-complex. For every homomorphism ϕ : F →Mv of R-modules,there is a morphism of R-complexes, Dv(F)→M, given by the diagram

0 // F

ϕ

1F// F

∂Mv ϕ

// 0

· · · // Mv+1∂M

v+1// Mv

∂Mv// Mv−1

∂Mv−1// Mv−2 // · · · .

Choose for every v ∈ Z a surjective homomorphism ϕv : Fv→Mv with Fv free, anddo it such that Fv has a basis with n> 0 elements if Mv is generated by n elements;cf. 1.3.11. Consider the complex L given by

Lv =Fv

⊕Fv+1

and ∂Lv =

(0 0

1Fv0

),

and the map π : L→M given by πv = (ϕv ∂Mv+1ϕ

v+1); it is straightforward to verifythat π is a morphism of complexes.

REMARK. Presumably, disc complexes are called so because they are simple examples of con-tractible complexes, see 4.3.23, just as topological discs are simple examples of contractible spaces.

2.5.25 Lemma. Let M be an R-complex. The complex L and morphism π : L→Mconstructed in 2.5.24 have the following properties.

(a) The morphism π is surjective.(b) The complex L is acyclic, and the graded module L\ is graded-free.(c) If M is degreewise finitely generated, then L is degreewise finitely generated.(d) If M is bounded (above/below), then L is bounded (above/below).

PROOF. As the morphisms ϕv are surjective, so is π; this proves part (a).It is evident from the definition of ∂L that L is acyclic. Each module Lv is a

direct sum of free modules, Fv and Fv+1, and hence free. Thus L\ is graded-freeby 2.5.22; this proves (b). Moreover, if Mv is finitely generated (zero), then Fv isfinitely generated (zero), and parts (c) and (d) follow.

2.5.26 Proposition. For an R-complex M, the following conditions are equivalent.(i) M is degreewise finitely presented.

(ii) There exists an R-complex L of finitely generated free modules and a surjec-tive morphism L→M whose kernel is degreewise finitely generated.

(iii) There exist R-complexes L and L′ of finitely generated free modules and anexact sequence of R-complexes L′→ L→M→ 0.

Moreover, if M is bounded (above/below) and degreewise finitely presented, thenthe complexes L and L′ in (iii) can be chosen to be bounded (above/below).

PROOF. The implication (iii)=⇒ (i) is trivial.


82 2 Complexes

(i)=⇒ (ii): By 2.5.25 there exists a surjective morphism π : L→M where L isa complex of finitely generated free R-modules and, moreover, if M is bounded(above/below), then so is L. Since each Mv is finitely presented, it follows from1.3.35 that Kerπ is degreewise finitely generated.

(ii)=⇒ (iii): By assumption there exists a short exact sequence of R-complexes0→ K→ L→M→ 0 where L is a complex of finitely generated free modules, andK is degreewise finitely generated. By 2.5.25 there is a surjective morphism L′→ Kwhere L′ is a complex of finitely generated free modules. Moreover, if M and henceL and K are bounded (above/below), then so is L′. The composite L′ K L nowyields the left-hand morphism in an exact sequence L′→ L→M→ 0.

EXERCISES

E 2.5.1 Let M be an R-complex. Show that if M\ is finitely generated in Mgr(R), then M isdegreewise finitely generated. Show that the converse is not true.

E 2.5.2 Show that the isomorphism classes of degreewise finitely presented complexes form aset.

E 2.5.3 Let M be a bounded above R-complex and N be a bounded below R-complex. Establisha result about HomR(M,N) akin to 2.5.9.

E 2.5.4 Let M be a bounded above Ro-complex and N be a bounded above R-complex. Establisha result about M⊗R N akin to 2.5.13.

E 2.5.5 Let 0→M′→M→M′′→ 0 be an exact sequence in C(R).(a) Show that with s′ = supM′, s = supM, and s′′ = supM′′ there are inequalities

s′ 6 maxs,s′′−1 , s 6 maxs′,s′′ , and s′′ 6 maxs′+1,s .(b) Show that with u′ = infM′, u = infM, and u′′ = infM′′ there are inequalities

u′ > minu,u′′−1 , u > minu′,u′′ , and u′′ > minu′+1,u .E 2.5.6 (Cf. 2.5.16) Show that hard truncations yield exact k-linear endofunctors (–)ďn and

(–)ěn on C(R).E 2.5.7 (Cf. 2.5.17) Let M be an R-complex and n be an integer. Show that the canonical se-

quence 0→Mďn→M→Měn+1→ 0 is degreewise split exact and decide if it is splitexact.

E 2.5.8 For n ∈ Z define full subcategories of C(R) by specifying their objects as follows,C6n(R) = M ∈C(R) |Mv = 0 for all v > n and

C>n(R) = M ∈C(R) |Mv = 0 for all v < n .Show that the functors (–)ďn and (–)ěn are right and left adjoints for the inclusionfunctors C6n(R)→C(R) and C>n(R)→C(R), respectively.

E 2.5.9 (Cf. 2.5.19) Show that soft truncation above and below yield right and left exact, respec-tively, endofunctors (–)Ďn and (–)Ěn on C(R).

E 2.5.10 Show that soft truncations are not \-functors.E 2.5.11 Let M be an R-complex. Show that for every n ∈ Z with Hn(M) = 0 there is an exact

sequence 0→MĚn→M→MĎn→ 0.E 2.5.12 (Cf. 2.5.20) Let M be an R-complex and n be an integer. Show that the canonical maps

τMĎn : M→MĎn and τM

Ěn : MĚn→M are morphisms of R-complexes.E 2.5.13 Give an example of a graded R-module that is free but not graded-free.E 2.5.14 Let L be a graded R-module and let E = euu∈U be a subset of L consisting of homo-

geneous elements. Show that L is graded-free with graded basis E if and only if it hasthe following graded extension property. Given a graded R-module M and a graded mapα : E→M there exists a unique graded homomorphism α : L→M with α|E = α.


Chapter 3Categorical Constructions

Products and coproducts are categorical devices defined by universal properties. Wesay that a category has products (coproducts) if all set-indexed, also known as small,products (coproducts) exist in the category. In the first section of this chapter, it isestablished that the category of R-complexes has products and coproducts. As thiscategory is Abelian, it then follows from general principles that it has small limitsand colimits. In Sects. 3.2 and 3.3 we give a detailed treatment of (co)limits overpreordered sets, which is sufficient for our purposes.

3.1 Products and Coproducts

SYNOPSIS. Coproduct; product; universal properties; direct sum; functor that preserves (co)products.

In the category of complexes, like in the category of modules, products and co-products are palpable objects. After describing their construction, we consider theirinteractions with the functors from Chap. 2. Homology and shift preserve productsand coproducts alike. The Hom functor preserves products, while the tensor prod-uct functor preserves coproducts. Under assumptions of boundedness and finiteness,Hom also preserves coproducts and the tensor product preserves products.

COPRODUCTS

3.1.1 Construction. Let Muu∈U be a family of R-complexes. One defines a com-plex

∐u∈U Mu by setting(∐

u∈U

Mu)v =

∐u∈U

Muv and ∂

∐u∈U Mu

v =∐u∈U

∂Mu

v ,

where the right-hand side of either equality is given by the coproduct in M(R). Forevery u∈U the injections εu

v : Muv

∐u∈U Mu

v in M(R) yield an injective morphism

83

84 3 Categorical Constructions

(3.1.1.1) εu : Mu ∐u∈U

Mu

of R-complexes given by εu(muv) = εu

v(muv) on homogeneous elements mu

v ∈Muv .

It is straightforward to verify that every element in Mu has the form ∑u∈U εu(mu)

for a unique family of elements mu ∈Mu with mu = 0 for all but finitely many u∈U .

3.1.2 Proposition. For a family Muu∈U in C(R), the complex∐

u∈U Mu togetherwith the family of morphisms εuu∈U , constructed in 3.1.1, is the coproduct.

For every family αu : Mu→ Nu∈U of morphisms in C(R), the unique morphismthat makes the diagram

Mu // εu//

αu

∐u∈U

Mu

||

N

commutative for every u ∈U is given by ∑u∈U εu(mu) 7→ ∑u∈U α

u(mu).

PROOF. The assignment ∑u∈U εu(mu) 7→∑u∈U α

u(mu) yields by 3.1.1 a well-definedmap α :

∐u∈U Mu→ N, and it is straightforward to verify that it is a morphism of

R-complexes. The equality αu = αεu holds for all u∈U by definition of α. Any mor-phism α′ :

∐u∈U Mu→ N with αu = α′εu for all u ∈U satisfies α′(∑u∈U ε

u(mu)) =

∑u∈U α′εu(mu) = ∑u∈U α

u(mu), and hence α′ = α.

3.1.3. It follows from 3.1.1 and 3.1.2 that the full subcategory Mgr(R) of C(R) isclosed under coproducts in C(R). In particular, the category Mgr(R) has coproducts.

3.1.4. Let αu : Mu→ Nuu∈U be a family of morphisms of R-complexes. It followsfrom 3.1.2 that the coproduct morphism

∐u∈U α

u :∐

u∈U Mu→∐

u∈U Nu is given by∑u∈U ε

u(mu) 7→ ∑u∈U εuαu(mu).

Above, and throughout this section, we use the same symbol εu for the injectionsin different coproducts.

3.1.5. Let αu : Mu→ Nuu∈U and βu : Nu→ Xuu∈U be families of morphisms ofR-complexes. It follows from 3.1.4 that the sequence∐

u∈U

Mu∐

u∈U αu

−−−−−→∐u∈U

Nu∐

u∈U βu

−−−−−→∐u∈U

Xu

is exact if and only if the sequence Mu αu−→ Nu βu

−→ Xu is exact for every u ∈U .

3.1.6 Proposition. Let αu : Mu→ Nu∈U be a family of morphisms in C(R). Ifeach αu is null-homotopic, then so is the canonical morphism α :

∐u∈U Mu→ N.

In particular, a coproduct of null-homotopic morphisms is null-homotopic.

PROOF. By assumption there are degree 1 homomorphisms τu : Mu→ N such thatαu = ∂Nτu + τu∂Mu

holds for every u ∈U . Consider each homomorphism τu as amorphism Mu\→ Σ−1 N\ of graded R-modules; see 2.2.4. Set M =

∐u∈U Mu; as M\


3.1 Products and Coproducts 85

together with the injections εu : Mu\M\u∈U is the coproduct of Mu\u∈U inMgr(R), there is a morphism τ : M\→ Σ−1 N\ with τεu = τu for all u ∈U . Viewingτ as a degree 1 homomorphism M→ N, one has

αεu = αu = ∂Nτu +τu∂Mu= ∂Nτεu +τεu∂Mu

= (∂Nτ+τ∂M)εu

for every u ∈U ; here the third equality holds by definition of τ, and the last equalityholds as εu is a morphism. Since ∂Nτ+ τ∂M is a morphism of R-complexes, it fol-lows from the universal property of coproducts that one has α= ∂Nτ+τ∂M . That is,α is null-homotopic.

If αu : Mu→ Nuu∈U is a family of null-homotopic morphisms, then the com-posites Mu→ Nu

∐u∈U Nu are null-homotopic by 2.2.25. It follows that the ca-

nonical morphism∐

u∈U αu :∐

u∈U Mu→∐

u∈U Nu is null-homotopic; cf. 3.1.4.

FUNCTORS THAT PRESERVE COPRODUCTS

3.1.7. Let Muu∈U be a family of R-complexes and let s be an integer. There is anequality of complexes

∐u∈U Σ

s Mu = Σs∐u∈U Mu. Moreover, if αu : Mu→ Nuu∈U

is a family of morphisms in C(R), then one has∐

u∈U Σsαu = Σs∐

u∈U αu.

3.1.8. Let Muu∈U be a family of R-complexes. The differential ∂∐

u∈U Mu, con-

sidered as a morphism∐

u∈U Mu → Σ∐

u∈U Mu, is the coproduct of the family ofmorphisms ∂Mu

: Mu→ ΣMuu∈U . Hence, there are equalities

Z(∐

u∈U

Mu) = ∐u∈U

Z(Mu) and B(∐

u∈U

Mu) = ∐u∈U

B(Mu)

of subcomplexes of∐

u∈U Mu; cf. (2.2.11.1) and 3.1.5.

3.1.9. Let F: C(R)→ C(S) be a functor and let Muu∈U be a family of R-complexes.The canonical morphism∐

u∈U

F(Mu)−→ F(∐u∈U

Mu) is given by ∑u∈U

εu(xu) 7−→ ∑u∈U

F(εu)(xu) .

Recall that F is said to preserve coproducts if this morphism is an isomorphism forevery family Muu∈U of R-complexes.

3.1.10 Proposition. The homology functor H: C(R)→ C(R) preserves coproducts.That is, for every family Muu∈U R-complexes, the canonical morphism

(3.1.10.1)∐u∈U

H(Mu)−→ H(∐

u∈U

Mu) ,given by ∑u∈U ε

u(hu) 7→ ∑u∈U H(εu)(hu), is an isomorphism.

PROOF. If we write the element hu ∈ H(Mu) as hu = [zu] with zu ∈ Z(Mu), thenthe canonical morphism is given by ∑u∈U ε

u([zu]) 7→ ∑u∈U [εu(zu)] = [∑u∈U ε

u(zu)],which is an isomorphism by 3.1.8.



3.1.11 Proposition. For every R-complex N the functor –⊗R N : C(Ro)→ C(k) pre-serves coproducts. That is, for every family Muu∈U Ro-complexes, the canonicalmorphism

(3.1.11.1)∐u∈U

(Mu⊗R N)−→(∐

u∈U

Mu)⊗R N ,

given by ∑u∈U εu(tu) 7→ ∑u∈U (ε

u⊗R N)(tu), is an isomorphism.

PROOF. To show that (3.1.11.1) is an isomorphism, it suffices to construct an inverseat the level of graded modules. The map

⊎i∈Z(

∐u∈U Mu)i×Nv−i→

∐u∈U(Mu⊗R N)v

defined by (∑u∈U εu(mu),n) 7→ ∑u∈U ε

u(mu⊗ n) is k-bilinear and middle R-linear.By the universal property 2.1.15 there is a unique morphism of graded k-modules((∐

u∈U Mu)⊗R N)\→∐

u∈U (Mu⊗R N)\ that maps the element (∑u∈U εu(mu))⊗ n

to ∑u∈U εu(mu⊗n). This gives the desired inverse of the canonical morphism.

3.1.12 Proposition. For every Ro-complex M the functor M⊗R – : C(R)→ C(k) pre-serves coproducts. That is, for every family Nuu∈U R-complexes, the canonicalmorphism

(3.1.12.1)∐u∈U

(M⊗R Nu)−→M⊗R∐u∈U

Nu ,

given by ∑u∈U εu(tu) 7→ ∑u∈U (M⊗R ε

u)(tu), is an isomorphism.


REMARK. For an Ro-complex M, the functor M⊗R –: C(R)→C(k) has a right adjoint, namelythe functor Homk(M,–); see E 4.4.3. From this fact alone, it follows that M⊗R – preserves co-products; see E 3.1.6.

PRODUCTS

3.1.13 Construction. Let Nuu∈U be a family of R-complexes. One defines a com-plex

∏u∈U Nu by setting(∏

u∈U

Nu)v =

∏u∈U

Nuv and ∂

∏u∈U Nu

v =∏u∈U

∂Nu

v ,

where the right-hand side of either equality is given by the product in M(R). For ev-ery u ∈U the projections $u

v :∏

u∈U Nuv Nu

v in M(R) yield a surjective morphism

(3.1.13.1) $u :∏u∈U

Nu Nu

of R-complexes given by $u((nuv)u∈U ) = $u

v((nuv)u∈U ) = nu

v on homogeneous ele-ments (nu

v)u∈U ∈ (∏

u∈U Nu)v =∏

u∈U Nuv .

It is straightforward to verify that every element n ∈∏

u∈U Nu has the form n =(nu)u∈U = ($u(n))u∈U , where nu =$u(n) belongs to Nu.



3.1.14 Proposition. For a family Nuu∈U in C(R), the complex∏

u∈U Nu togetherwith the family of morphisms $uu∈U , constructed in 3.1.13, is the product.

For every family αu : M→ Nuu∈U of morphisms in C(R), the unique morphismthat makes the diagram ∏

u∈UNu

$u

M

??

αu// Nu

commutative for every u ∈U is given by m 7→ (αu(m))u∈U .

PROOF. The assignment m 7→ (αu(m))u∈U yields by 3.1.13 a map α : M→∏

u∈U Nu,and it is straightforward to verify that it is a morphism of R-complexes. The equalityαu =$uα holds for all u ∈U by definition of α. Any morphism α′ : M→

∏u∈U Nu

with αu = $uα′ for all u ∈U satisfies α′(m) = ($uα′(m))u∈U = (αu(m))u∈U , andhence α′ = α.

3.1.15. It follows from 3.1.13 and 3.1.14 that the full subcategory Mgr(R) of C(R)is closed under products in C(R). In particular, the category Mgr(R) has products.

3.1.16. Let αu : Mu→ Nuu∈U be a family of morphisms of R-complexes. It fol-lows from 3.1.14 that the product morphism

∏u∈U α

u :∏

u∈U Mu→∏

u∈U Nu isgiven by (mu)u∈U 7→ (αu(mu))u∈U .

3.1.17. Let αu : Mu→ Nuu∈U and βu : Nu→ Xuu∈U be families of morphisms ofR-complexes. It follows from 3.1.16 that the sequence∏

u∈U

Mu∏

u∈U αu

−−−−−→∏u∈U

Nu∏

u∈U βu

−−−−−→∏u∈U

Xu

is exact if and only if the sequence Mu αu−→ Nu βu

−→ Xu is exact for every u ∈U .

3.1.18 Proposition. Let αu : M→ Nuu∈U be a family of morphisms in C(R). Ifeach αu is null-homotopic, then so is the canonical morphism α : M→

∏u∈U Nu.

In particular, a product of null-homotopic morphisms is null-homotopic.

PROOF. By assumption there are degree 1 homomorphisms τu : M→ Nu such thatαu = ∂Nu

τu + τu∂M holds for every u ∈U . Consider each homomorphism τu as amorphism M\→ Σ−1 Nu\ of graded R-modules; see 2.2.4. Set N =

∏u∈U Nu; as N\

together with the projections $u : N\ Nu\u∈U is the product of Nu\u∈U inMgr(R), there is a morphism τ : M\→ Σ−1 N\ with $uτ= τu for all u ∈U . Viewingτ as a degree 1 homomorphism M→ N, one has

$uα = αu = ∂Nuτu +τu∂M = ∂Nu

$uτ+$uτ∂M = $u(∂Nτ+τ∂M)

for every u ∈U ; here the third equality holds by definition of τ, and the last equalityholds as $u is a morphism. Since ∂Nτ+ τ∂M is a morphism of R-complexes, itfollows from the universal property of products that one has α = ∂Nτ+ τ∂M . Thatis, α is null-homotopic.



If αu : Mu→ Nuu∈U is a family of null-homotopic morphisms, then the com-posites

∏u∈U MuMu→Nu are null-homotopic by 2.2.25. It follows that the cano-

nical morphism∏

u∈U αu :∏

u∈U Mu→∏

u∈U Nu is null-homotopic; cf. 3.1.16.

FUNCTORS THAT PRESERVE PRODUCTS

3.1.19. Let Nuu∈U be a family of R-complexes and let s be an integer. There is anequality of complexes Σs∏

u∈U Nu =∏

u∈U Σs Nu. Moreover, if βu : Nu→Muu∈U

is a family of morphisms in C(R), then one has Σs∏u∈U β

u =∏

u∈U Σs βu.

3.1.20. Let Nuu∈U be a family of R-complexes. The differential ∂∏

u∈U Nu, con-

sidered as a morphism∏

u∈U Nu→ Σ∏

u∈U Nu, is the product of the family of mor-phisms ∂Nu

: Nu→ ΣNuu∈U . It follows that there are equalities

Z(∏

u∈U

Nu) = ∏u∈U

Z(Nu) and B(∏

u∈U

Nu) = ∏u∈U

B(Nu)

of subcomplexes of∏

u∈U Nu; cf. (2.2.11.1) and 3.1.17.

3.1.21. Let F: C(R)→ C(S) be a functor and let Nuu∈U be a family of R-com-plexes. The canonical morphism

F(∏u∈U

Nu)−→∏u∈U

F(Nu) is given by x 7−→ (F($u)(x))u∈U .

Recall that F is said to preserve products if this morphism is an isomorphism forevery family Nuu∈U of R-complexes.

3.1.22 Proposition. The homology functor H: C(R)→ C(R) preserves products.That is, for every family Nuu∈U of R-complexes, the canonical morphism

(3.1.22.1) H(∏

u∈U

Nu)−→ ∏u∈U

H(Nu) ,

given by h 7→ (H($u)(h))u∈U , is an isomorphism.

PROOF. By 3.1.20 every element h ∈ H(∏

u∈U Nu) has the form h = [(zu)u∈U ] withzu ∈Z(Nu), and the canonical morphism is therefore given by [(zu)u∈U ] 7→ ([zu])u∈U ,which is evidently bijective.

If M is an object in a k-linear category U, then the functor U(M,–) : U→M(k)preserves products. In particular, for every M in C(R), the functor C(R)(M,–) =Z0(HomR(M,–)) from C(R) to M(k) preserves products; cf. 2.3.10. The next resultis stronger, it shows that the functor HomR(M,–) preserves products.

3.1.23 Proposition. For every R-complex M the functor HomR(M,–) : C(R)→ C(k)preserves products. That is, for every family Nuu∈U of R-complexes, the canonicalmorphism

(3.1.23.1) HomR(M,∏u∈U

Nu)−→ ∏u∈U

HomR(M,Nu) ,



given by ϑ 7→ (HomR(M,$u)(ϑ))u∈U = ($uϑ)u∈U , is an isomorphism.

PROOF. Assign to an element (ϑu)u∈U in∏

u∈U HomR(M,Nu) the homomorphismϑ in HomR(M,

∏u∈U Nu) that maps an element m ∈M to (ϑu(m))u∈U in

∏u∈U Nu.

This assignment defines an inverse to (3.1.23.1).

REMARK. For an R-complex M, the functor HomR(M,–) : C(R)→C(k) has a left adjoint, namelythe functor M⊗k –; see E 4.4.3. From this fact alone, it follows that HomR(M,–) preserves prod-ucts; see E 3.1.13.

3.1.24. Coproducts in C(R) correspond to products in C(R)op. I.e. let Muu∈U be afamily of R-complexes whose coproduct in C(R) is M with injections εu : Mu→M.The R-complex M is also an object in C(R)op and the morphisms εu in C(R) cor-respond to morphisms εu : M→Mu in C(R)op. As the family εu : Mu→Mu∈U inC(R) has the universal property of a coproduct in C(R), the family εu : M→Muu∈Uin C(R)op has the universal property of a product in C(R)op (and vice versa).

Thus, if F : C(R)op→ C(S) is a functor and Muu∈U is a family of R-complexes,then the canonical morphism from F of the product of Muu∈U in C(R)op to theproduct of F(Mu)u∈U in C(R) takes the form

F(∐u∈U

Mu)−→∏u∈U

F(Mu) given by x 7−→ (F(εu)(x))u∈U ,

where∐

u∈U Mu is the coproduct in C(R) with injections εu : Mu→∐

u∈U Mu.Recall that F : C(R)op→ C(S) is said to preserve products if this morphism is an

isomorphism for every family Muu∈U of R-complexes.

The next result and 3.1.23 show that the functor HomR(–,–) preserves products.

3.1.25 Proposition. For every R-complex N the functor HomR(–,N) : C(R)op→ C(k)preserves products. That is, for every family Muu∈U of R-complexes, the canoni-cal morphism

(3.1.25.1) HomR(∐

u∈U

Mu,N)−→

∏u∈U

HomR(Mu,N) ,

given by ϑ 7→ (HomR(εu,N)(ϑ))u∈U = (ϑεu)u∈U , is an isomorphism.

PROOF. Assign to an element (ϑu)u∈U in∏

u∈U HomR(Mu,N) the homomorphismin HomR(

∐u∈U Mu,N) given by ∑u∈U ε

u(mu) 7→ ∑u∈U ϑu(mu). This defines an in-

verse to (3.1.25.1).

BOUNDEDNESS AND FINITENESS

3.1.26. For every family Muu∈U of R-complexes, the coproduct∐

u∈U Mu is asubcomplex of the product

∏u∈U Mu. If U is a finite set, then the product and co-

product coincide. Per 1.1.11 the biproduct notation⊕

u∈U Mu is used for this com-plex, which is called the direct sum of the family Muu∈U in C(R); each complexMu is called a direct summand.



3.1.27 Example. There is no isomorphism of Z-modules,

Q⊗Z∏n∈NZ/2nZ−→

∏n∈N

(Q⊗Z Z/2nZ) .

Indeed, one has∏

n∈Z(Q⊗Z Z/2nZ) = 0. The family Z Z/2nZn∈N of canonicalhomomorphisms induces by the universal property of products an injective homo-morphism Z→

∏n∈NZ/2nZ. As the Z-module Q is flat, there is an injective homo-

morphism Q→ Q⊗Z∏

n∈NZ/2nZ, in particular, Q⊗Z∏

n∈NZ/2nZ is non-zero.

Under suitable finiteness conditions on M the functor M⊗R – preserves products.

3.1.28 Proposition. If M is a bounded and degreewise finitely presented Ro-complex, then the functor M⊗R – : C(R)→ C(k) preserves products. That is, forevery family Nuu∈U of R-complexes, the canonical morphism

(3.1.28.1) M⊗R∏u∈U

Nu −→∏u∈U

(M⊗R Nu) ,

given by t 7→ ((M⊗R$u)(t))u∈U , is an isomorphism.

PROOF. Assume that M is bounded and degreewise finitely presented. It followsfrom 2.5.26 that there is an exact sequence L′ → L→ M → 0 where L and L′ arebounded complexes of finitely generated free Ro-modules. Consider the followingcommutative diagram, whose upper row is obtained by application of the functor–⊗R

∏u∈U Nu to this sequence,

L′⊗R∏

u∈UNu //

κL′

L⊗R∏

u∈UNu //

κL

M⊗R∏

u∈UNu //

κM

0

∏u∈U

(L′⊗R Nu) //∏

u∈U(L⊗R Nu) //

∏u∈U

(M⊗R Nu) // 0 .

The rows are exact by right exactness of tensor products, 2.4.9, and exactness ofproducts, 3.1.17. The canonical morphism (3.1.28.1) yields the vertical morphisms.For v ∈ Z and n > 1 additivity of the tensor product and the unitor (1.2.1.1) yieldRn⊗R (

∏u∈U Nu)v ∼= (

∏u∈U Nu

v )n ∼=

∏u∈U (Rn⊗R Nu

v ); this accounts for the first iso-morphism below. The module (L′⊗R

∏u∈U Nu)v is a direct sum,

supL′⊕i=infL′

L′i⊗R (∏

u∈UNu)v−i ∼=

∏u∈U

supL′⊕i=infL′

(L′i⊗R Nuv−i)∼=

∏u∈U

(L′⊗R Nu)v ,

and this composite isomorphism is κL′ in degree v. Similarly, κL is an isomorphism,so it follows from the Five Lemma 2.1.40 that κM is an isomorphism.

REMARK. A module M is finitely presented if and only if (3.1.28.1) is an isomorphism for everyfamily Nuu∈U ; see E 3.1.16. There exists aQ-module M such that M⊗QQN and (M⊗QQ)N areisomorphic whilst the canonical map M⊗QQN→ (M⊗QQ)N is not an isomorphism; see E 3.1.14.

3.1.29 Proposition. If N is a bounded and degreewise finitely presented R-complex,then the functor –⊗R N : C(Ro)→ C(k) preserves products. That is, for every familyMuu∈U of Ro-complexes, the canonical morphism



(3.1.29.1)(∏

u∈U

Mu)⊗R N −→∏u∈U

(Mu⊗R N) ,

given by t 7→ (($u⊗R N)(t))u∈U , is an isomorphism.


3.1.30 Example. The canonical homomorphism of R-modules,

(3.1.30.1) HomR(R(N),R)(N) −→ HomR(R(N),R(N)) ,

given by (ϑn)n∈N 7→ ∑n∈N εnϑn, is not surjective. Indeed, the identity homomor-

phism on R(N) is not in the image of (3.1.30.1).

Under suitable conditions on M, the functor HomR(M,–) preserves coproducts.

3.1.31 Proposition. If M is a bounded and degreewise finitely generated R-complex,then the functor HomR(M,–) : C(R)→ C(k) preserves coproducts. That is, for everyfamily Nuu∈U of R-complexes, the canonical morphism

(3.1.31.1)∐u∈U

HomR(M,Nu)−→ HomR(M,∐u∈U

Nu) ,given by ∑u∈U ε

u(ϑu) 7→ ∑u∈U HomR(M,εu)(ϑu) = ∑u∈U εuϑu, is an isomorphism.

PROOF. Assume that M is bounded and degreewise finitely generated. The moduleM\ is finitely generated, so a homomorphism ϑ : M→

∐u∈U Nu factors through a

subcomplex⊕

u∈U ′ Nu, where U ′ is a finite subset of U . Denote by$u :

⊕u∈U ′ N

u Nu

the projection for u ∈U ′. Assign to a homomorphism ϑ in HomR(M,∐

u∈U Nu) theelement ∑u∈U ε

u(ϑu) in∐

u∈U HomR(M,Nu) where ϑu =$uϑ for u∈U ′ and ϑu = 0for u /∈U ′. This defines an inverse to (3.1.31.1).

REMARK. Rentschler [70] says that a module M is of type Σ if the functor Hom(M,–) preservescoproducts. More recent names used in the literature for such modules are small and dually slender.A dually slender module need not be finitely generated, however, if R is left Noetherian, then everydually slender R-module is finitely generated. See E 3.1.18–E 3.1.21.

EXERCISES

E 3.1.1 Let εu : Mu→Cu∈U be a family of morphisms in C(R) with the property thatfor every family αu : Mu→ Nu∈U of morphisms in C(R) there is a unique mor-phism α : C→ N with αεu = αu for all u ∈ U . Show that there is an isomorphismϕ :∐

u∈U Mu→C with ϕεu = εu for all u ∈ U . Conclude that the universal propertydetermines the coproduct uniquely up to isomorphism.

E 3.1.2 Let α :∐

u∈U Mu→ N be the morphism induced by a family αu : Mu→ Nu∈U ofmorphisms in C(R). Show that α is surjective if one has

⋃u∈U Imαu = N.

E 3.1.3 (Cf. 3.1.3) Show that the coproduct in C(R) of a family of graded R-modules is a gradedR-module. Conclude, in particular, that the category Mgr(R) has coproducts.

E 3.1.4 Let U be a set. Show that U-indexed families of R-complexes form an Abelian categoryand that the product and coproduct are exact functors from this category to C(R).



E 3.1.5 Let τ : F→ G be a natural transformation of functors C(R)→ C(S) and let Muu∈Ube a family of R-complexes. Show that there is a commutative diagram in C(S),∐

u∈U F(Mu)

∐τMu

//∐

u∈U G(Mu)

F(∐

u∈U Mu)τ∐

Mu// G(

∐u∈U Mu) ,

where the vertical morphisms are the canonical ones; see 3.1.9.E 3.1.6 Show that every functor F: C(R)→C(S) that has a right adjoint preserves coproducts.

E 3.1.7 Show that for every complex N there are isomorphisms∐

v∈ZΣv Nv ∼= N\ ∼=

∏v∈ZΣ

v Nv.E 3.1.8 Let Muu∈U be a family of R-complexes. Show that there are equalities

sup(∐

u∈U Mu) = supu∈UsupMu= sup(∏

u∈U Mu) and

inf(∐

u∈U Mu) = infu∈UinfMu = inf(∏

u∈U Mu) .E 3.1.9 Let $u : P→ Nuu∈U be a family of morphisms in C(R) with the property that

for every family αu : M→ Nuu∈U of morphisms in C(R) there is a unique mor-phism α : M→ P with $uα = αu for all u ∈ U . Show that there is an isomorphismϕ : P→

∏u∈U Nu with $uϕ = $u for all u ∈U . Conclude that the universal property

determines the product uniquely up to isomorphism.E 3.1.10 Let α : M→

∏u∈U Nu be the morphism induced by a family αu : M→ Nuu∈U of mor-

phisms in C(R). Show that α is injective if one has⋂

u∈U Kerαu = 0.E 3.1.11 (Cf. 3.1.15) Show that the product in C(R) of a family of graded R-modules is a graded

R-module. Conclude, in particular, that the category Mgr(R) has products.E 3.1.12 Let τ : F→ G be a natural transformation of functors C(R)→ C(S) and let Muu∈U

be a family of R-complexes. Show that there is a commutative diagram in C(S),

F(∏

u∈U Mu)

τ∏

Mu// G(

∏u∈U Mu)

∏u∈U F(Mu)

∏τMu

//∏

u∈U G(Mu) ,

where the vertical morphisms are the canonical ones; see 3.1.21.E 3.1.13 Show that every functor F: C(R)→C(S) that has a left adjoint preserves products.E 3.1.14 Let k be a field. Show that the canonical homomorphism kN⊗k kN→ (kN⊗k k)N is not

an isomorphism. Show that kN⊗k kN and (kN⊗k k)N are isomorphic.E 3.1.15 Let M be an Ro-module. Show that the next conditions are equivalent. (i) M is finitely

generated. (ii) The morphism (3.1.28.1) is surjective for every family Nuu∈U . (iii) Themorphism (3.1.28.1) is surjective for every family Nuu∈U with Nu = R for all u ∈U .

E 3.1.16 Let M be an Ro-module. Show that the next conditions are equivalent. (i) M is finitelypresented. (ii) The morphism (3.1.28.1) is bijective for every family Nuu∈U . (iii) Themorphism (3.1.28.1) is bijective for every family Nuu∈U with Nu = R for all u ∈U .

E 3.1.17 Show that the canonical morphism (3.1.31.1) is always injective.E 3.1.18 Let M be an R-module. Show that HomR(M,–) preserves coproducts if and only if for

every ascending chain M1 ⊆M2 ⊆ ·· · of submodules of M with M =⋃

n∈NMn one hasM = Mn for some n ∈ N.

E 3.1.19 Let M be an R-module. Show that if HomR(M,–) preserves coproducts and M is count-ably generated, then M is finitely generated.

E 3.1.20 Give an example of an R-module M such that M is not finitely generated and HomR(M,–)preserves coproducts. Hint: See e.g. [36] or [70].

E 3.1.21 Assume that R is left Noetherian and let M be an R-module. Show that HomR(M,–)preserves coproducts if and only if M is finitely generated.


3.2 Colimits 93

E 3.1.22 Let V be a category. Show that products in V correspond to coproducts in the oppositecategory Vop and that coproducts in V correspond to products in Vop.

3.2 Colimits

SYNOPSIS. Direct system; colimit; universal property; functor that preserves colimits; pushout;filtered colimit; telescope.

A colimit is a quotient of a coproduct. Structurally, our treatment of colimits fol-lows the pattern established in Sect. 3.1. Recall that a set endowed with a reflexiveand transitive binary relation ‘6’ is called preordered. Partially ordered sets are, inparticular, preordered.

3.2.1 Definition. Let (U,6) be a preordered set. A U-direct system in C(R) is afamily µvu : Mu→Mvu6v of morphisms in C(R) with the following properties.

(1) µuu = 1Mufor all u ∈U .

(2) µwvµvu = µwu for all u6 v6 w in U .Any mention of a U-direct system µvu : Mu→Mvu6v includes the tacit assump-tion that (U,6) is a preordered set. A Z-direct system is called a direct system.

3.2.2 Construction. Let µvu : Mu→Mvu6v be a U-direct system in C(R). Wedescribe the quotient of the coproduct

∐u∈U Mu by the subcomplex generated by

the set of elements εu(mu)− εvµvu(mu) | mu ∈ Mu, u 6 v as the cokernel of amorphism between coproducts in C(R).

Set ∇(U) = (u,v) ∈U ×U | u 6 v and set M(u,v) = Mu for all (u,v) ∈ ∇(U).The assignment

(m(u,v))(u,v)∈∇(U) 7−→ ∑(u,v)∈∇(U)

εu(m(u,v))−εvµvu(m(u,v)) ,

where ε is the injection (3.1.1.1), defines a morphism of R-complexes

∆µ :∐

(u,v)∈∇(U)

M(u,v) −→∐u∈U

Mu .

The cokernel of ∆µ is denoted colimu∈U Mu.Notice that for every u ∈U the composite of the injection εu with the canonical

map onto colimu∈U Mu is a morphism of R-complexes,

(3.2.2.1) µu : Mu −→ colimu∈U

Mu ,

and one has µu = µvµvu for all u 6 v in U . Every element in colimu∈U Mu has theform ∑u∈U µ

u(mu) for some element ∑u∈U εu(mu) in

∐u∈U Mu, and one has

(3.2.2.2) ∂colimu∈U Mu(∑

u∈Uµu(mu)

)= ∑

u∈Uµu(∂Mu

(mu)) .

REMARK. Though the complex colimu∈U Mu depends on the morphisms µvu : Mu→Mv, it isstandard to use this symbol that suppresses the morphisms.



The next definition is justified by 3.2.4; it shows that the complex colimu∈U Mu

and the canonical morphisms µu have the universal property that defines a colimit.In any category, this property determines the colimit uniquely up to isomorphism.

3.2.3 Definition. For a U-direct system µvu : Mu→Mvu6v in C(R) the complexcolimu∈U Mu together with the canonical morphisms µuu∈U , constructed in 3.2.2,is called the colimit of µvu : Mu→Mvu6v in C(R).

REMARK. Other names for the colimit defined above are direct limit, inductive limit, and injectivelimit; other symbols used for this gadget are lim−→ and inj lim.

3.2.4 Lemma. Let µvu : Mu→Mvu6v be a U-direct system in C(R). For everyfamily of morphisms αu : Mu→ Nu∈U in C(R) with αu = αvµvu for all u6 v, thereis a unique morphism α that makes the next diagram commutative for all u6 v,

Mv

µv &&

αv

##

colimu∈U

Mu α// N .

Mu

µu 88

µvu

OO

αu

;;

The morphism α is given by ∑u∈U µu(mu) 7→ ∑u∈U α

u(mu).

PROOF. Let αu : Mu→ Nu∈U be a family of morphisms with αu = αvµvu for allu6 v. The equalities αu =αvµvu ensure that the morphism

∐u∈U Mu→N from 3.1.2

factors through colimu∈U Mu to yield a morphism α with the stipulated definition.It is evident from the definition that α satisfies αu = αµu for all u ∈U . Moreover,

for any morphism α′ : colimu∈U Mu→ N that satisfies αu = α′µu for all u ∈U , onehas α′(∑u∈U µ

u(mu)) = ∑u∈U α′µu(mu) = ∑u∈U α

u(mu), hence α′ = α.

3.2.5. For µvu : Mu→Mvu6v and αu : Mu→ Nu∈U as in 3.2.4, notice that themorphism α : colimu∈U Mu→ N is surjective if one has

⋃u∈U Imαu = N.

3.2.6. It follows readily from 3.2.2 and 3.2.4 that the full subcategories M(R) andMgr(R) of C(R) are closed under colimits. In particular, with the notation from3.2.4, one has colimu∈U (Mu)\ = (colimu∈U MU )\ and for every n ∈ Z the familyµvu

n : Mun →Mv

nu6v of homomorphisms of modules is a direct system with colimit(colimu∈U Mu)n.

3.2.7 Example. Let Muu∈U be a family of R-complexes. Endowed with the dis-crete order, U is a preordered set, and µuu = 1Muu∈U is a U-direct system withcolimu∈U Mu =

∐u∈U Mu and µu = εu for all u ∈U . Thus, a coproduct is a colimit.

3.2.8 Definition. Let µvu : Mu→Mvu6v and νvu : Nu→ Nvu6v be U-direct sys-tems in C(R). A family of morphisms αu : Mu→ Nuu∈U in C(R) that satisfyνvuαu = αvµvu for all u 6 v in U is called a morphism of U-direct systems. Sucha morphism is called injective (surjective) if each map αu is injective (surjective).


3.2 Colimits 95

Given a morphism αu : Mu→ Nuu∈U of U-direct systems, it follows from theuniversal property of colimits that the map given by ∑u∈U µ

u(mu) 7→∑u∈U νuαu(mu)

is the unique morphism that makes the next diagram commutative for all u6 v,

Mv

µv &&

αv// Nv

νv

xx

colimu∈U

Mu // colimu∈U

Nu

Mu

µu 88

µvu

OO

αu// Nu .

νvff

νvu

OO

This morphism is called the colimit of αu : Mu→ Nuu∈U and denoted colimu∈U αu.

3.2.9. Let µvu : Mu→Mvu6v be a U-direct system in C(R) and let s be an inte-ger. It follows from 2.2.3 and 3.1.7 that Σs µvu : Σs Mu→ Σs Mvu6v is a U-directsystem with colimu∈U Σ

s Mu = Σs colimu∈U Mu. Moreover, if αu : Mu→ Nuu∈U isa morphism of U-direct systems, then one has colimu∈U Σ

sαu = Σs colimu∈U αu.

3.2.10 Example. Let µvu : Mu→Mvu6v be a U-direct system in C(R). As themaps µvu are morphisms in C(R), the family ∂Mu

: Mu→ ΣMuu∈U is a morphismof U-direct systems. From the definitions one has colimu∈U ∂

Mu= ∂colimu∈U Mu

.

The next statement sums up as: colimits are right exact. While general colimitsare not exact—see 3.2.24 for an example—we shall see in 3.2.32 that colimits overfiltered sets are exact.

3.2.11 Lemma. Let αu : Mu→ Nuu∈U and βu : Nu→ Xuu∈U be morphisms ofU-direct systems in C(R). If the sequence

Mu αu−−→ Nu βu

−−→ Xu −→ 0

is exact for every u ∈U , then the next sequence is exact,

colimu∈U

Mu colimu∈U αu

−−−−−−−→ colimu∈U

Nu colimu∈U βu


Xu −→ 0 .

PROOF. Let ∇(U) be as in 3.2.2, and for all (u,v)∈∇(U) set α(u,v) =αu and β(u,v) =βu . By 3.1.5 and 3.2.8 there is a commutative diagram with exact columns and exactupper and middle rows,

∐(u,v)∈∇(U)

M(u,v)∐

(u,v)∈∇(U)α(u,v)

//

∆

∐(u,v)∈∇(U)

N(u,v)∐

(u,v)∈∇(U) β(u,v)

//

∆

∐(u,v)∈∇(U)

X (u,v) //

∆

0

∐u∈U

Mu

∐u∈U αu

//∐

u∈UNu

∐u∈U βu

//∐

u∈UXu

// 0

colimu∈U

Mu colimu∈U αu// colim

u∈UNu colimu∈U βu

// colimu∈U

Xu // 0 ,



where the vertical morphisms ∆ are defined in 3.2.2 An elementary diagram chaseshows that the lower row in the diagram is exact.

FUNCTORS THAT PRESERVES COLIMITS

3.2.12. Let F: C(R)→ C(S) be a functor and let µvu : Mu→Mvu6v be a U-directsystem in C(R). It is easy to see that the morphisms F(µvu) : F(Mu)→ F(Mv)u6vform a U-direct system in C(S). For every u ∈ U let λu be the canonical mor-phism F(Mu)→ colimu∈U F(Mu). For every u∈U , the canonical morphism (3.2.2.1)induces a morphism F(µu) : F(Mu)→ F(colimu∈U Mu). These morphisms satisfyF(µu) = F(µv)F(µvu) for all u 6 v in U , so by the universal property of colimits,the map given by the assignment ∑u∈U λ

u(xu) 7→∑u∈U F(µu)(xu) is the unique mor-phism that makes the following diagram in C(S) commutative for all u6 v,

F(Mv)

λv ((

F(µv)

&&

colimu∈U

F(Mu) // F(colimu∈U

Mu) .

F(Mu)

λu 66

F(µvu)

OO

F(µu)55

3.2.13 Definition. A functor F: C(R)→ C(S) preserves colimits if the morphism in3.2.12 is an isomorphism for every direct system µvu : Mu→Mvu6v in C(R).

REMARK. A functor that preserves colimits is also called cocontinuous.

While 3.2.13 is a condition on objects, it carries over to morphisms.

3.2.14. Let F: C(R)→ C(S) be a functor and let αu : Mu→ Nuu∈U be a morphismof U-direct systems in C(R). It is easy to see that there is a commutative diagram,

colimu∈U

F(Mu)

colimu∈U F(αu)// colim

u∈UF(Nu)

F(colimu∈U

Mu)F(colimu∈U αu)

// F(colimu∈U

Nu) ,

where the vertical morphisms are the canonical ones; see 3.2.12. Thus, if F preservescolimits, then the morphisms colimu∈U F(αu) and F(colimu∈U α

u) are isomorphic.

3.2.15 Example. It follows from 3.1.7 and the construction of colim that shift pre-serves colimits.

The tensor product, as a functor from C(R) to C(k), preserves colimits.

3.2.16 Proposition. Let µvu : Mu→Mvu6v be a U-direct system in C(Ro) and letN be an R-complex. The canonical map


3.2 Colimits 97

(3.2.16.1) colimu∈U

(Mu⊗R N)−→ (colimu∈U

Mu)⊗R N ,

given by ∑u∈U λu(tu) 7→ ∑u∈U (µ

u⊗R N)(tu), cf. 3.2.12, is an isomorphism in C(k).

PROOF. The map is a morphism of k-complexes by 3.2.12. There is a commutativediagram in C(k),∐

(u,v)∈∇(U)(M(u,v)⊗R N)

∆µ⊗N//

∼=

∐u∈U

(Mu⊗R N) //

∼=

colimu∈U

(Mu⊗R N) //

κ

0

( ∐(u,v)∈∇(U)

M(u,v))⊗R N∆µ⊗N

//( ∐

u∈UMu)⊗R N // (colim

u∈UMu)⊗R N // 0 ,

where κ is the canonical morphism (3.2.16.1), and the middle and left-most verticalmaps are the isomorphisms from 3.1.11. The rows are exact by the construction ofcolimits, 3.2.2, and right exactness of tensor products, 2.4.9, and it follows from theFive Lemma 2.1.40 that κ is an isomorphism.

3.2.17 Proposition. Let M be an Ro-complex and νvu : Nu→ Nvu∈U be a U-directsystem in C(R). The canonical map

(3.2.17.1) colimu∈U

(M⊗R Nu)−→M⊗R colimu∈U

Nu ,

given by ∑u∈U λu(tu) 7→ ∑u∈U (M⊗R ν

u)(tu), cf. 3.2.12, is an isomorphism in C(k).


PUSHOUTS

Simple non-trivial colimits arise from three term direct systems M← X → N.

3.2.18 Construction. Let U = u,v,w be a set, preordered as follows v > u 6 w.

Given a diagram M α←− Xβ−→ N in C(R), set

Mv = M , Mu = X , Mw = N ,

µvv = 1M , µvu = α , µuu = 1X , µwu = β , and µww = 1N .

This defines a U-direct system in C(R). It is straightforward to verify that the colimitof this system is the cokernel of the morphism (−α β) : X →M⊕N.

3.2.19 Definition. For a diagram M α←− Xβ−→N in C(R), the colimit of the U-direct

system constructed in 3.2.18 is called the pushout of (α,β) and denoted MtX N.

REMARK. As for the colimit, the notation for the pushout suppresses the morphisms. Other namesfor the pushout are fibered coproduct, fibered sum, Cocartesian square, and amalgamated product.

3.2.20. Given morphisms α :X →M and β :X → N, the pushouts of (α,β) and (β,α)are isomorphic via the map induced by the canonical isomorphism M⊕N ∼= N⊕M.



3.2.21 Construction. Given a diagram in M α←− Xβ−→ N in C(R), let

α′ : N −→ MtX N and β′ : M −→ MtX N

be the canonical morphisms (3.2.2.1); they are given by n 7→ [(0,n)]Im(−α β) andm 7→ [(m,0)]Im(−α β). There is a commutative diagram with exact rows and columns

(3.2.21.1)

Xβ

//

α

N //

α′

Cokerβ //

α

0

Mβ′//

MtX N //

Cokerβ′ // 0

Cokerαβ//

Cokerα′

0 0 ,

where β is the morphism induced by β′, it maps [m]Imα to [β′(m)]Imα′ , and α is themorphism induced by α′.

3.2.22. Given a diagram Mβ′′−→ Y α′′←− N in C(R) with β′′α = α′′β, it follows from

3.2.4 that the assignment

[(m,n)]Im(−α β) = α′(n)+β′(m) 7−→ α′′(n)+β′′(m)

defines the unique morphism that makes the next diagram commutative,

Xβ//

α

N

α′

α′′

Mβ′//

β′′

//

MtX N

##

Y .

3.2.23 Proposition. Let M α←− Xβ−→ N be a diagram in C(R). The following asser-

tions hold for the morphisms in (3.2.21.1).(a) If α is injective, then α′ is injective.(b) β is an isomorphism, whence α is surjective if and only if α′ is surjective.(c) If β is injective, then β′ is injective.(d) α is an isomorphism, whence β is surjective if and only if β′ is surjective.

PROOF. By symmetry, see 3.2.20, it is sufficient to prove parts (a) and (b).(a): If n is in Kerα′, then one has n = β(x) for some x in Kerα.(b): If β([m]Imα) is zero, then one has β′(m) = α′(n) for some n ∈ N. Thus

[(−m,n)]Im(−α β) = α′(n)− β′(m) is zero in MtX N and, consequently, m belongs


3.2 Colimits 99

to Imα; hence β is injective. For every element α′(n)+ β′(m) in M tX N one has[α′(n)+β′(m)]Imα′ = [β′(m)]Imα′ in Cokerα′, so β is surjective.

The following example shows that colimits are not left exact.

3.2.24 Example. The embeddings below form an injective morphism of pushoutsdiagrams; that is, an injective morphism of U-direct systems as in 3.2.8, where U isthe preordered set described in 3.2.18.

Z

// 0

2Z??

??

// 0??

??

Z

Z??

??

The colimit Zt2Z 0→ ZtZ 0 of this morphism is Z/2Z→ 0, which is not injective.

FILTERED COLIMITS

A colimit of a U-direct system is called filtered if the preordered set (U,6) is fil-tered. Every complex is a filtered colimit of bounded above complexes.

3.2.25 Example. Let M be an R-complex. The inclusions among subcomplexesMďu give rise to a direct system µvu : MďuMďvu6v in C(R). The embeddingsαu : MďuM satisfy αu =αvµvu for all u6 v, so by the universal property of colim-its, there is a morphism α : colimu∈ZMďu→M, given by ∑u∈Z µ

u(mu) 7→ ∑u∈Zmu.It is surjective by construction. Let m = ∑u∈Z µ

u(mu) be an element in Kerα andset v = maxu ∈ Z | mu 6= 0, then one has ∑u∈Z ε

vµvu(mu) = εv(∑u∈Zmu) = 0. In∐u∈ZMu one then has ∑u∈Z ε

u(mu) =∑u∈Z(εu(mu)−εvµvu(mu)). This element is in

the image of ∆µ, see 3.2.2, so m is zero in colimu∈ZMďu. Thus, α is an isomorphism.

3.2.26 Lemma. Let µvu : Mu→Mvu6v be a U-direct system in C(R). If U isfiltered, then the following assertions hold.

(a) For every m in colimu∈U Mu there is a v∈U and an mv ∈Mv with µv(mv) =m.(b) If mv in Mv has µv(mv) = 0, then there is a w∈U with v6w and µwv(mv) = 0.(c) If elements mu ∈Mu and mv ∈Mv satisfy µu(mu) = µv(mv), then there is a w

in U with u6 w and v6 w such that µwu(mu) = µwv(mv) holds.

PROOF. (a): Fix an element m in colimu∈U Mu; it has the form m = ∑u∈U µu(mu).

As U is filtered, one can choose v with v> u for all u∈U with mu 6= 0; then one has

m = ∑u6v

µvµvu(mu) = µv( ∑u6v

µvu(mu)) .

(b): If µv(mv) = 0 holds, then one has

(?) εv(mv) = ∑(t,u)∈∇(U)

εt(m(t,u))−εuµut(m(t,u))



for some element (m(t,u))(t,u)∈∇(U) in∐

(t,u)∈∇(U) M(t,u); see 3.2.2. Choose a wwith w > t for all t ∈ U with mt 6= 0. Now apply the morphism

∐u∈U Mu → Mw

given by ∑u∈U εu(mu) 7→ ∑u6w µ

wu(mu) to both sides in (?) to get µwv(mv) =

∑(t,u)∈∇(U)µwt(m(t,u))−µwuµut(m(t,u)) = 0.

(c): Choose w′ ∈U with u,v6w′, then one has µw′(µw′u(mu)−µw′v(mv)) = 0. Bypart (b) there is a now a w in U with w > w′ and 0 = µww′(µw′u(mu)−µw′v(mv)) =µwu(mu)−µwv(mv).

A classic application of colimits is to write an arbitrary module as a filteredcolimit of finitely generated modules.

3.2.27 Example. Let M be an R-complex and let U be the set of all bounded anddegreewise finitely generated subcomplexes of M. The set U is preordered underinclusion and filtered. For elements M′ ⊆ M′′ in U let µM′′M′ be the embedding,then µM′′M′ : M′M′′M′⊆M′′ is a U-direct system. The morphisms in the familyαM′ : M′MM′∈U of embeddings satisfy αM′ = αM′′µM′′M′ for all M′ ⊆M′′, soby the universal property of colimits there is a morphism

α : colimM′∈U

M′ −→ M ,

given by ∑M′∈U µM′(m′) 7→ ∑M′∈U α

M′(m′) = ∑M′∈U m′. For a homogeneous ele-ment m′ in M of degree d = |m′| consider the following subcomplex of M,

M′ = 0−→ Rm′∂M

d−−→ R∂Md (m′)−→ 0 .

Now m′ is in the image of αM′ , so α is surjective. For m∈ colimM′∈U M′ in the kernelof α there is, by 3.2.26, an element M′ in U and an m′ in M′ with µM′(m′) = m. InM one now has m′ = αM′(m′) = αµM′(m′) = α(m) = 0. Thus, α is an isomorphism.

Every subsequence of a convergent sequence of numbers is convergent with thesame limit; colimits behave similarly.

3.2.28 Definition. Let (U,6) be a preordered filtered set. A subset V of U is calledcofinal if for every u ∈U there is a v ∈V with u6 v.

3.2.29 Proposition. Let (U,6) be a preordered filtered set and let V be a cofinalsubset of U . For every U-direct system µvu : Mu→Mvu6v in C(R) there is anisomorphism colimv∈V Mv ∼= colimu∈U Mu.

PROOF. As usual, µu is the canonical morphism Mu→ colimu∈U Mu for every u∈U .For v ∈ V denote by µv the canonical morphism Mv → colimv∈V Mv. For v 6 w inV one has µv = µwµwv, so by the universal property of colimits there is a morphismµ : colimv∈V Mv→ colimu∈U Mu. For every m in colimu∈U Mu there is by 3.2.26 a uin U and an element mu in Mu with µu(mu) = m. As V is cofinal in U , there is av∈V with v> u, so one has m = µvµvu(mu). It follows that µ is surjective; see 3.2.5.

Let m be an element in Kerµ. As V is filtered, there exists a v∈V and an elementmv in Mv with µv(mv) = m. Now one has 0 = µ(m) = µµv(mv) = µv(mv). It follows


3.2 Colimits 101

from 3.2.26 that there is a w in U with µwv(mv) = 0. As V is cofinal in U we mayassume that w is in V , and then one has m = µv(mv) = µwµwv(mv) = 0.

3.2.30 Proposition. Let µvu : Mu→Mvu6v be a U-direct system in C(R) and letαu : Mu→ Nu∈U be a family of morphisms with αu = αvµvu for all u 6 v. If Uis filtered and the subset u ∈ U | αu is injective is cofinal, then the morphismα : colimu∈U Mu→ N from 3.2.4 is injective.

PROOF. Let m be an element in Kerα. By 3.2.26 there is a u ∈U and an mu ∈Mu

with µu(mu) = m; by assumption there is a v> u such that αv is injective. Now onehas m = µvµvu(mu) and 0 = α(m) = αµvµvu(mu) = αvµvu(mu), so m = 0 holds.

3.2.31 Example. Let Muu∈U be a family of R-complexes and letΥ be the set of allfinite subsets of U ; it is preordered under inclusion and filtered. For V ⊆W inΥ thereis a canonical morphism µWV :

⊕u∈V Mu→

⊕u∈W Mu, namely the embedding, and

it is elementary to verify that these morphisms form an Υ-direct system. For V ∈ Υlet αV :

⊕u∈V Mu→

∐u∈U Mu be the embedding. One has αV = αWµWV for all V ⊆

W in Υ, so there is a morphism α : colimV∈Υ(⊕

u∈V Mu)→∐

u∈U Mu. It is surjectiveby 3.2.5 and injective by 3.2.30. Thus, a coproduct is a filtered colimit.

The next result sums up as: filtered colimits are exact.

3.2.32 Proposition. Let αu : Mu→ Nuu∈U and βu : Nu→ Xuu∈U be morphismsof U-direct systems in C(R). If the sequence

0−→Mu αu−−→ Nu βu

−−→ Xu −→ 0

is exact for every u ∈U and U is filtered, then the next sequence is exact,

0−→ colimu∈U

Mu colimu∈U αu


Nu colimu∈U βu


Xu −→ 0 .

PROOF. By 3.2.11 it is sufficient to prove that α = colimu∈U αu is injective. We

write µuv : Mu→Mv and νuv : Nu→ Nv for the morphisms in the direct systems. Letm ∈ Kerα and choose by 3.2.26 a v in U and an element mv in Mv with µv(mv) =m, then one has 0 = αµv(mv) = νvαv(mv). By 3.2.26 there is a w ∈U with w > vand 0 = νwvαv(mv) = αwµwv(mv). Since αw is injective, one has µwv(mv) = 0 and,therefore, m = µv(mv) = µwµwv(mv) = 0.

3.2.33. Let µvu : Mu→Mvu6v be a U-direct system in C(R). Consider the canon-ical morphisms

µB : colimu∈U

B(Mu) −→ B(colimu∈U

Mu) and µZ : colimu∈U

Z(Mu) −→ Z(colimu∈U

Mu)

obtained by applying 3.2.12 to the functors B(–) and Z(–); see 2.2.12. By definition,µB and µZ map the coset of an element ∑u∈U ε

u(mu) to ∑u∈U µu(mu). Note that µB

is surjective; cf. (3.2.2.2). By (2.2.11.2) and 3.2.11 there is a commutative diagramwith exact rows



colimu∈U

B(Mu)%//

µB

colimu∈U

Mu //

=

colimu∈U

C(Mu) //

µC

0

0 // B(colimu∈U

Mu) // colimu∈U

Mu // C(colimu∈U

Mu) // 0 .

It follows from the Snake Lemma 2.1.42 that µC is an isomorphism. Observe thatif εv : C(Mv)

∐u∈U C(Mu) denotes the injection, then µC maps the image of

∑u∈U εu([mu]B(Mu)) in colimu∈U C(Mu) to [∑u∈U µ

u(mu)]B(colimMu).Assume now that U is filtered. By 3.2.32 the morphism % is injective, so µB is an

isomorphism; again by the Snake Lemma. Finally, a similar commutative diagrambased on (2.2.11.1) shows that µZ is an isomorphism.

Filtered colimits commute with homology.

3.2.34 Theorem. Let µvu : Mu→Mvu6v be a U-direct system in C(R). Thecanonical map

(3.2.34.1) colimu∈U

H(Mu)−→ H(colimu∈U

Mu) ,

given by ∑u∈U λu(hu) 7→∑u∈U H(µu)(hu), cf. 3.2.12, is a morphism of R-complexes.

If U is filtered, then (3.2.34.1) is an isomorphism.

PROOF. The map is a morphism of R-complexes by 3.2.12. In the following com-mutative diagram, the middle and left-hand vertical maps are the isomorphisms es-tablished in 3.2.33, and µH is the canonical morphism (3.2.34.1),

0 // colimu∈U

B(Mu) //

∼= µB

colimu∈U

Z(Mu) //

∼= µZ

colimu∈U

H(Mu) //

µH

0

0 // B(colimu∈U

Mu) // Z(colimu∈U

Mu) // H(colimu∈U

Mu) // 0 .

It follows from (2.2.11.3) and 3.2.32 that the rows in the diagram are exact, and thenthe Five Lemma 2.1.40 implies that µH is an isomorphism.

3.2.35 Corollary. Let µvu : Mu→Mvu6v be a U-direct system in C(R). If U isfiltered and each complex Mu is acyclic, then colimu∈U Mu is acyclic.

Under conditions on M the functor HomR(M,–) preserves filtered colimits.

3.2.36 Proposition. Let M be an R-complex and let νvu : Nu→ Nvu6v be a U-direct system in C(R). The canonical map

(3.2.36.1) colimu∈U

HomR(M,Nu)−→ HomR(M,colimu∈U

Nu) ,

given by ∑u∈U λu(ϑu) 7→∑u∈U HomR(M,νu)(ϑu) = ∑u∈U ν

uϑu, cf. 3.2.12, is a mor-phism of k-complexes. If U is filtered and M is bounded and degreewise finitelypresented, then (3.2.36.1) is an isomorphism.


3.2 Colimits 103

PROOF. The map is a morphism of k-complexes by 3.2.12. Assume that M isbounded and degreewise finitely presented. By 2.5.26 there is an exact sequenceL′→ L→M→ 0 where L and L′ are bounded complexes of finitely generated freeRo-modules. Consider the following commutative diagram, whose lower row is ob-tained by application of the functor HomR(–,colimNu) to this sequence,

0 // colimu∈U

HomR(M,Nu) //

κM

colimu∈U

HomR(L,Nu) //

κL

colimu∈U

HomR(L′,Nu)

κL′

0 // HomR(M,colimu∈U

Nu) // HomR(L,colimu∈U

Nu) // HomR(L′,colimu∈U

Nu) .

The rows are exact by left exactness of Hom, 2.3.10, and by exactness of filteredcolimits 3.2.32. The canonical morphism (3.2.36.1) yields the vertical morphisms.To prove that κM is an isomorphism, it suffices by the Five Lemma 2.1.40 to arguethat κL and κL′ are isomorphisms. To this end, consider the commutative diagram

∐(u,v)∈∇(U)

HomR(L,N(u,v))∆Hom(L,ν)

//

∼=

∐u∈U

HomR(L,Nu)

∼=

// colimu∈U

HomR(L,Nu) //

κL

0

HomR(L,∐

(u,v)∈∇(U)N(u,v))

Hom(L,∆ν)// HomR(L,

∐u∈U

Nu) // HomR(L,colimu∈U

Nu) // 0 ,

where the left-hand and middle vertical maps are the isomorphisms from 3.1.31.The upper row is exact by the construction of colimits 3.2.2. Exactness of the lowerrow is a consequence of the unique extension property, 1.3.5, of the free R-modulesLv. It follows from the Five Lemma 2.1.40 that κL is an isomorphism. Similarly, κL′

is an isomorphism, as desired.

The requirement in 3.2.36 that M be bounded and degreewise finitely presentedcannot be relaxed; see C.9.

TELESCOPES

A frequently occurring form of colimits stem from sequences of morphisms.

3.2.37 Construction. Let κu : Mu→Mu+1u∈Z be a sequence of morphisms inC(R). It determines a direct system µvu : Mu→Mvu6v as follows: set

µuu = 1Mufor all u in Z and µvu = κv−1 · · ·κu for all u < v in Z .

Given additional sequences λu : Nu→ Nu+1u∈Z and αu : Mu→ Nuu∈Z of mor-phisms, such that αu+1κu = λuαu holds for all u ∈ Z, it is elementary to verify thatαuu∈Z is a morphism of the direct systems determined by κuu∈Z and λuu∈Z .



3.2.38 Definition. A sequence κu : Mu→Mu+1u∈Z of morphisms in C(R) withMu = 0 for u 0 is called a telescope in C(R). The colimit colimu∈ZMu of theassociated direct system, see 3.2.37, is called the colimit of the telescope in C(R).

Given telescopes κu : Mu→Mu+1u∈Z and λu : Nu→ Nu+1u∈Z , a sequenceof morphisms αu : Mu→ Nuu∈Z that satisfy αu+1κu = λuαu for all u∈Z is called amorphism of telescopes. The morphism colimu∈Zα

u : colimu∈ZMu→ colimu∈ZNu,see 3.2.37 and 3.2.8, is called the colimit of αu : Mu→ Nuu∈Z .

3.2.39. Let κu : Mu→Mu+1u∈Z be a telescope and let µvu : Mu→Mvu6v bethe associated direct system in C(R). Given an R-complex N and a sequence ofmorphisms αu : Mu→ Nu∈Z that satisfy αu = αu+1κu for all u ∈ Z, one hasαu = αvµvu for all u6 v. By the universal property of colimits, there is a morphismα : colimu∈ZMu→ N in C(R) with properties as described in 3.2.4.

3.2.40 Example. Let M0 ⊆M1 ⊆M2 ⊆ ·· · be an ascending chain of R-complexes.The embeddings MuMu+1 define a telescope whose colimit is isomorphic to thecomplex

⋃u∈ZMu. This is immediate from 3.2.5 and 3.2.30.

3.2.41 Example. Let x be an element in k and let M be a k-module. Consider thefollowing commutative diagram of k-modules,

C

α

M

µ0,,

α0 =111

κ0 =x// M

µ1

..

α1 =1/x

//

κ1 =x// M

µ2

33

α2 =1/x2

++

κ2 =x// · · ·

Mx

The row is the telescope where each homomorphism κu is multiplication by x on M,the module C is the colimit of the telescope with canonical homomorphisms µu, Mxis the localization of M at the multiplicative set of powers of x, and αu multiplieswith 1/xu. As one has αu = αu+1κu there is a unique homomorphism α : C→Mxwith αµu = αu for all u; see 3.2.39. We argue that α is an isomorphism. Note thatαµu(m) = αu(m) = m/xu, so α is surjective. By 3.2.26 every element z ∈ C has theform z = µu(m) for some u and some m ∈M, so if z is in the kernel of α one has 0 =α(z) = αµu(m) = αu(m) = m/xu in Mx. This means that xvm = 0 in M for some v, andit follows that z = µu(m) = µu+vκu+v−1 · · ·κu+1κu(m) = µu+v(xvm) = µu+v(0) = 0,so α is injective.

3.2.42 Proposition. Let κu : Mu→Mu+1u∈Z be a telescope in C(R). The follow-ing assertions hold.

(a) If κu = 0 holds for infinitely many u > 0, then one has colimu∈ZMu = 0.(b) If there exists an integer w such that κu is bijective for all u > w, then the

canonical map Mw→ colimu∈ZMu is an isomorphism.

PROOF. Let µvu : Mu→Mvu∈Z be the direct system associated to the telescope.


3.2 Colimits 105

(a): By 3.2.26 every element m in colimu∈ZMu has the form m= µu(mu) for someu∈ Z and mu ∈Mu. It follows from the assumption that the map µvu is zero for somev> u, and consequently one has m = µu(mu) = µv(µvu(mu)) = 0.

(b): Define a sequence αu : Mu→Mwu∈Z of morphisms in C(R) as follows: setαw = 1Mw

; set αu = κw−1 · · ·κu for u < w; and set αu = (κu−1 · · ·κw)−1 for u > w. Byconstruction one has αu = αu+1κu for all u ∈ Z, so by 3.2.39 there is a morphismα : colimu∈ZMu→Mw, given by µu(mu) 7→ αu(mu). Evidently one has αµw = αw =1Mw

. It follows from 3.2.30 that α is injective, and hence it is the inverse of µw.

EXERCISES

E 3.2.1 Let µvu : Mu→Mvu6v be a U-direct system in C(R). Let µu : Mu→Cu∈U be afamily of morphisms that satisfy the following conditions. (1) One has µu = µvµvu forall u6 v. (2) For every family αu : Mu→ Nu∈U of morphisms with αu = αvµvu for allu6 v there exists a unique morphism α : C→ N with αµu = αu for all u ∈U . Show thatthere is an isomorphism ϕ : colimu∈U Mu→C with ϕµu = µu for every u∈U . Concludethat the universal property determines the colimit uniquely up to isomorphism.

E 3.2.2 (Cf. 3.2.6) Show that the colimit in C(R) of a direct system of morphisms of gradedR-modules is a graded R-module. Conclude, in particular, that Mgr(R) has colimits.

E 3.2.3 (Cf. 3.2.6) Show that the colimit in C(R) of a direct system of homomorphisms of R-modules is an R-module. Conclude, in particular, that the category M(R) has colimits.

E 3.2.4 Fix a preordered set U . Show that U-direct systems in C(R) and their morphisms form anAbelian category and that the colimit is a right exact functor from this category to C(R).

E 3.2.5 Generalize the result in E 3.1.6 by showing that every functor F: C(R)→C(S) that hasa right adjoint preserves colimits.

E 3.2.6 (Cf. 3.2.18) Verify the isomorphism colimu∈U Mu ∼= Coker(−α β) in 3.2.18.E 3.2.7 Show that the categories M(R) and Mgr(R) have pushouts.E 3.2.8 Adobt the notation from 3.2.21.

(a) Show that there is a commutative diagram with exact rows and columns:

0

0

0

0

0 // Z //

Kerα //

Kerα′ //

0 //

0

0 // Kerβ //

Xβ//

α

N

α′

// Cokerβ //

α

0

0 // Kerβ′ //

M

β′// MtX N

// Cokerβ′

// 0

0 // 0

// Cokerα

β// Cokerα′

// 0

// 0

0 0 0 0 .

(b) Show that Z in the diagram above can be non-zero. Hint: Let a and b be left idealsin R and let α : R R/a and β : R R/b be the canonical homomorphisms.

(c) In the following two diagrams, the solid parts are given. Show that they can becompleted to commutative diagrams with exact rows and columns, as depicted.



0

0

0 // X //

N //

D // 0

0 // M //

MtX N //

D // 0

C

C

0 0

and

0

0

0 // L // X //

N //

0

0 // L // M //

MtX N //

0

C

C

0 0 .

E 3.2.9 Show that a coproduct in C(R) is the filtered colimit of its finite sub-coproducts.E 3.2.10 Let µvu : Mu→Mvu6v be a U-direct system in C(R). Show that the inequalities

sup(colimu∈U Mu)6 supu∈UsupMu and inf(colimu∈U Mu)> infu∈UinfMuhold if U is filtered.

E 3.2.11 Let αu : Mu→ Nuu∈U and βu : Nu→ Xuu∈U be morphisms of U-direct systems inC(R). Show that if the sequence 0→Mu→ Nu→ Xu is exact for every u ∈U , then thesequence 0−→Ker∆µ→Ker∆ν→Ker∆χ is exact. Show also that if each morphism βu

is surjective and U is filtered, then the morphism Ker∆ν→ Ker∆χ.E 3.2.12 Let M be a finitely presented module, and let µvu : Mu→Mvu6v be a U-direct system

of finitely presented modules. Show that if U is filtered and there is a surjective homo-morphism colimu∈U Mu →M, then M is a direct summand of one of the modules Mu.Hint: 3.2.36

E 3.2.13 Show that an R-module F is flat if and only if the morphism ι⊗R F induced by theembedding b R is injective for every finitely generated right ideal b in R.

E 3.2.14 Let R be right coherent; show that for an injective R-module I the Ro-module Homk(I,E)is flat.

E 3.2.15 Show that every R-module is a filtered colimit of finitely presented modules. Hint: Letmuu∈U generate M, let L be free with basis euu∈U , and consider the canonical exactsequence 0→K→ L π−→M→ 0. For subsets V ⊆U set LV =R〈eu | u∈V 〉. Put a partialorder on the set of all pairs (V,H) where V ⊆U is finite and H ⊆ K ∩LV is a finitelygenerated submodule, such that it becomes filtered. Set M(V,H) = FV /H and show thatM is the colimit of a direct system of these modules.

E 3.2.16 Let R be left Noetherian. Show that every filtered colimit of injective R-modules is aninjective R-module.

E 3.2.17 Show that a filtered colimit of flat modules is a flat module.E 3.2.18 Show that the following conditions on R are equivalent. (i) R is von Neumann regular.

(ii) R/b is a flat Ro-module for every (finitely generated) right ideal b in R. (iii) EveryR-module is flat.

E 3.2.19 Show that a left Noetherian and von Neumann regular ring is semi-simple.E 3.2.20 As in 3.2.37 let κu : Mu→Mu+1u∈Z be a sequence (not necessarily a telescope) of

morphisms in C(R). Show that the colimit of the associated direct system does not de-pend on κu for u 0.

E 3.2.21 Show that every complex is the colimit of a telescope of bounded above complexes.E 3.2.22 Show that for a U-direct system µvu : Mu→Mvu6v, the morphism ∆µ is not injective.

E 3.2.23 Let κu : Mu→Mu+1u∈Z be a telescope. Show that colimu∈ZMu is the cokernelof the injective endomorphism of

∐u∈ZMu given by ∑u∈Z ε

u(mu) 7→ ∑u∈Z εu(mu)−

εu+1κu(mu).E 3.2.24 Let κu : Mu→Mu+1u∈Z be a telescope in C(R) with colimit M. Set Nu = Mu+1 and

λu = κu+1; show that λu : Nu→ Nu+1u∈Z is a telescope with colimit M and that themorphisms αu = κu from Mu to Nu form a morphism of telescopes with colimit 1M .


3.3 Limits 107

E 3.2.25 Let M be an R-module. Show that M is finitely generated if and only if (3.2.36.1) isinjective for every U-direct system νvu : Nu→ Nvu6v.

3.3 Limits

SYNOPSIS. Inverse system; limit; universal property; functor that preserves limits; pullback; tower;Mittag-Leffler Condition.

Limits are categorically dual to colimits; they are subcomplexes of products. Theirtheory starts out parallel to the theory for colimits, however, the two deviate at oneimportant point: While colimits often are exact, limits rarely are.

3.3.1 Definition. Let (U,6) be a preordered set. A U-inverse system in C(R) is afamily νuv : Nv→ Nuu6v of morphisms in C(R) with the following properties.

(1) νuu = 1Nufor all u ∈U .

(2) νuvνvw = νuw for all u6 v6 w in U .Any mention of a U-inverse system νuv : Nv→ Nuu6v includes the tacit assump-tion that (U,6) is a preordered set. A Z-inverse system is called an inverse system.

3.3.2 Construction. Let νuv : Nv→ Nuu6v be a U-inverse system in C(R). Wedescribe the subcomplex of the product

∏u∈U Nu generated by the elements (nu)u∈U

with nu = νuv(nv) for all u6 v as the kernel of a morphism between products in C(R).Indeed, let ∇(U) be as in 3.2.2 and set N(u,v) = Nu for all (u,v) ∈ ∇(U). The

assignment(nu)u∈U 7−→ (nu− νuv(nv))(u,v)∈∇(U)

defines a morphism of R-complexes

∆ν :∏u∈U

Nu −→∏

(u,v)∈∇(U)

N(u,v) .

The kernel of ∆ν is denoted limu∈U Nu.Note that for every u ∈U , restriction of the projection (3.1.13.1) yields a mor-

phism of R-complexes,

(3.3.2.1) νu : limu∈U

Nu −→ Nu ,

and one has νu = νuvνv for all u6 v.

REMARK. As for colimits, it is standard to use notation that suppresses the morphisms νuv, thoughthe complex limu∈U Nu depends on them.

The next definition is justified by 3.3.4, which shows that the complex limu∈U Mu

and the projections νu have the universal property that defines a limit. In any cate-gory, this property determines the limit uniquely up to isomorphism.

3.3.3 Definition. For a U-inverse system νuv : Nv→ Nuu6v in C(R) the complexlimu∈U Nu together with the canonical morphisms νuu∈U , constructed in 3.3.2, iscalled the limit of νuv : Nv→ Nuu6v in C(R).



REMARK. Other names for the limit defined above are inverse limit and projective limit; othersymbols used for this gadget are lim←− and proj lim.

3.3.4 Lemma. Let νuv : Nv→ Nuu6v be a U-inverse system in C(R). For everyfamily of morphisms αu : M→ Nuu∈U in C(R) with αu = νuvαv for all u6 v, thereis a unique morphism α that makes the next diagram commutative for all u6 v,

Nv

νuv

M

αu

//

αv

//

α// limu∈U

Nuνv

99

νu

%%

Nu .

The morphism α is given by m 7→ (αu(m))u∈U .

PROOF. Let αu : M→ Nuu∈U be a family of morphisms with αu = νuvαv for allu6 v. The equalities αu = νuvαv ensure that the morphism α : M→

∏u∈U Nu from

3.1.14, given by m 7→ (αu(m))u∈U , maps to the subcomplex limu∈U Nu.It is evident from the definition that α satisfies αu = νuα for all u ∈U . Moreover,

for any morphism α′ : M→ limu∈U Nu that satisfies αu = νuα′ for all u ∈U , one hasα′(m) = (νu(α′(m))u∈U = (αu(m))u∈U = α(m), so α is unique.

3.3.5. For νuv : Nv→ Nuu6v and αu : M→ Nuu∈U as in 3.3.4, notice that themorphism α : M→ limu∈U Nu is injective if one has

⋂u∈U Kerαu = 0.

3.3.6. It follows readily from 3.3.2 and 3.3.4 that the full subcategories M(R) andMgr(R) of C(R) are closed under limits.

3.3.7 Example. Let Nuu∈U be a family of R-complexes. Endowed with the dis-crete order, U is a preordered set, and νuu = 1Nuu∈U is a U-inverse system withlimu∈U Nu =

∏u∈U Nu and νu =$u for all u ∈U . Thus, every product is a limit.

Every complex is a limit of bounded below complexes.

3.3.8 Example. Let N be an R-complex. The canonical surjections among the quo-tient complexes Něu give rise to an inverse system νuv : Ně−v Ně−uu6v inC(R). The canonical maps βu : N Ně−u satisfy βu = νuvβv for all u 6 v, so bythe universal property of limits, there is a morphism β : N→ limu∈ZNě−u, givenby β(n) = (βu(n))u∈Z. It is injective by construction. To see that β is surjective, let(nu)u∈Z in limu∈ZNě−u be homogeneous of degree−w, and set n= nw considered asan element in the module N−w. Now one has βu(n) = 0= nu for u<w, and for u>wone has βu(n) = nw = nu as the homomorphisms νwu

−w : (Ně−u)−w→ (Ně−w)−w arethe identity map on N−w for w6 u. Thus, β is an isomorphism.

3.3.9 Definition. Let νuv : Nv→ Nuu6v and µuv : Mv→Muu6v be U-inversesystems in C(R). A family of morphisms βu : Nu→Muu∈U in C(R) that satisfiesβuνuv = µuvβv for all u6 v is called a morphism of U-inverse systems. Such a mor-phism is called injective (surjective) if each map βu is injective (surjective).


3.3 Limits 109

Given a morphism βu : Nu→Muu∈U of U-inverse systems, it follows from theuniversal property of limits that the map given by (nu)u∈U 7→ (βu(nu))u∈U is theunique morphism that makes the next diagram commutative for all u6 v in U ,

Nv βv//

νuv

Mv

µuv

limu∈U

Nu //

νvee

νu

yy

limu∈U

Mu

µv 88

µu

&&

Nu βu// Mu .

This morphism is called the limit of βu : Nu→Muu∈U and denoted limu∈U βu.

3.3.10. Let νuv : Nv→ Nuu6v be a U-inverse system in C(R) and let s be aninteger. It follows from 2.2.3 and 3.1.19 that Σs νuv : Σs Nv→ Σs Nuu6v is a U-inverse system with Σs limu∈U Nu = limu∈U Σ

s Nu. Moreover, if βu : Nu→Muu∈Uis a morphism of U-inverse systems, then one has Σs limu∈U β

u = limu∈U Σs βu.

3.3.11 Example. Let νuv : Nv→ Nuu6v be a U-inverse system. Because the mapsνuv are morphisms in C(R), the family ∂Nu

: Nu→ ΣNuu∈U is a morphism of U-inverse systems. From the definitions one has limu∈U ∂

Nu= ∂limu∈U Nu

.

3.3.12 Construction. Consider U-inverse systems in C(R),

χuv : Xv→ Xuu6v , νuv : Nv→ Nuu6v , and µuv : Mv→Muu6v .

Let αu : Xu→ Nuu∈U and βu : Nu→Muu∈U be morphisms of U-inverse sys-tems, such that the sequence Xu→ Nu→Mu is exact for every u ∈U . Let ∇(U) beas in 3.2.2, and for all (u,v)∈∇(U) set α(u,v) = αu and β(u,v) = βu . In view of 3.1.17and 3.3.9 there is a commutative diagram with exact columns and exact lower andmiddle rows

(3.3.12.1)

limu∈U Xu limu∈U αu//

limu∈U Nu limu∈U βu//

limu∈U Mu

∏u∈U

Xu

∆χ

∏u∈U αu

//∏

u∈UNu

∆ν

∏u∈U βu

//∏

u∈UMu

∆µ

∏(u,v)∈∇(U)

X (u,v)∏

(u,v)∈∇(U)α(u,v)

//∏

(u,v)∈∇(U)N(u,v)

∏(u,v)∈∇(U) β

(u,v)

//∏

(u,v)∈∇(U)M(u,v) ,

where the vertical morphisms ∆ are defined in 3.3.2.

The next statement sums up as: limits are left exact. Exactness of limits is a del-icate issue, not to say a rare occurrence. A sufficient condition for exactness certainlimits is given in 3.3.42, and an example of a non-exact limit is given in 3.3.43.



3.3.13 Lemma. Let αu : Xu→ Nuu∈U and βu : Nu→Muu∈U be morphisms ofU-inverse systems in C(R). If the sequence

0−→ Xu αu−−→ Nu βu

−−→Mu

is exact for every u ∈U , then the next sequence is exact,

0−→ limu∈U

Xu limu∈U αu

−−−−−−→ limu∈U

Nu limu∈U βu

−−−−−→ limu∈U

Mu .

PROOF. Consider the commutative diagram (3.3.12.1). Assuming that each se-quence 0→ Xu → Nu → Mu is exact, it follows from 3.1.17 that the morphisms∏

u∈U αu and

∏(u,v)∈∇(U)α

(u,v) are injective. Now a straightforward diagram chaseyields the desired exact sequence.

FUNCTORS THAT PRESERVE LIMITS

3.3.14. Let F: C(R)→ C(S) be a functor and let νuv : Nv→ Nuu6v be a U-inversesystem in C(R). It is elementary to verify that F(νuv) : F(Nv)→ F(Nu)u6v isa U-inverse system in C(S). For every u ∈ U let λu be the canonical morphismlimu∈U F(Nu) → F(Nu). For every u ∈ U , the canonical morphism (3.3.2.1) in-duces a morphism F(νu) : F(limu∈U Nu)→ F(Nu). These morphisms satisfy F(νu)=F(νuv)F(νv) for all u6 v in U , so by the universal property of limits, the map givenby the assignment x 7→ (F(νu)(x))u∈U is the unique morphism that makes the fol-lowing diagram in C(S) commutative for all u6 v,

F(Nv)

F(νuv)

F(limu∈U

Nu)

F(νu)

..

F(νv)

00

// limu∈U

F(Nu)λv

77

λu

''

F(Nu) .

3.3.15 Definition. A functor F: C(R)→ C(S) preserves limits if the morphism in3.3.14 is an isomorphism for every inverse system νuv : Nv→ Nuu6v in C(R).

REMARK. A functor that preserves limits is also called continuous.


3.3.16. Let F: C(R)→ C(S) be a functor and let βu : Nu→Muu∈U be a morphismof U-inverse systems in C(R). It is elementary to verify that there is a commutativediagram in C(S),

F(limu∈U

Nu)

F(limu∈U βu)// F(lim

u∈UMu)

limu∈U

F(Nu)limu∈U F(βu)

// limu∈U

F(Mu) ,


3.3 Limits 111

where the vertical morphisms are the canonical ones; see 3.3.14. Thus, if F preserveslimits, then the morphisms F(limu∈U β

u) and limu∈U F(βu) are isomorphic.

The functor HomR(M,–) from C(R) to C(k) preserves limits.

3.3.17 Proposition. Let νuv : Nv→ Nuu6v be a U-inverse system in C(R) and letM be an R-complex. The canonical map

(3.3.17.1) HomR(M, limu∈U

Nu)−→ limu∈U

HomR(M,Nu) ,

given by ϑ 7→ (HomR(M,νu)(ϑ))u∈U = (νuϑ)u∈U , is an isomorphism in C(k).

PROOF. The map is a morphism of k-complexes by 3.3.14. There is a commutativediagram in C(k),

0 // HomR(M, limu∈U

Nu) //

κ

HomR(M,∏

u∈UNu)

∼=

Hom(M,∆ν)// HomR(M,

∏(u,v)∈∇(U)

N(u,v))

∼=

0 // limu∈U

HomR(M,Nu) //∏

u∈UHomR(M,Nu)

∆Hom(M,ν)//

∏(u,v)∈∇(U)

HomR(M,N(u,v)) ,

where κ is the canonical morphism (3.3.17.1), and the middle and right-hand ver-tical maps are the isomorphisms from 3.1.23. The rows are exact by left exact-ness of Hom, 2.3.10, and the construction of limits, 3.3.2; it follows from the FiveLemma 2.1.40 that κ is an isomorphism.

3.3.18. Let F: C(R)op→ C(S) be a functor and let µvu : Mu→Mvu6v be a U-direct system in C(R). It is elementary to verify that F(µvu) : F(Mv)→ F(Mu)u6vis a U-inverse system in C(S). For every u ∈U let λu be the canonical morphismlimu∈U F(Mu) → F(Mu). The canonical morphism (3.2.2.1) induces a morphismF(µu) : F(colimMu)→ F(Mu) for every u ∈ U . By the universal property of lim-its, the map given by the assignment x 7→ (F(µu)(x))u∈U is the unique morphismthat makes the following diagram in C(S) commutative for all u6 v,

F(Mu)

F(µvu)

F(colimu∈U

Mu)

F(µv)

..

F(µu)

00

// limu∈U

F(Mu)λv

77

λu

''

F(Mu) .

3.3.19 Definition. A functor F: C(R)op→ C(S) preserves limits if the morphism in3.3.18 is an isomorphism for every direct system µvu : Mu→Mvu6v in C(R).


3.3.20. Let F: C(R)op→ C(S) be a functor and let αu : Mu→ Nuu∈U be a mor-phisms of U-direct systems in C(R). It is elementary to verify that there is a com-mutative diagram in C(S),



F(colimu∈U

Nu)

F(colimu∈U αu)// F(colim

u∈UMu)

limu∈U

F(Nu)limu∈U F(αu)

// limu∈U

F(Mu) ,

where the vertical morphisms are the canonical ones; see 3.3.18. Thus, if F preserveslimits, then the morphisms F(colimu∈U α

u) and limu∈U F(αu) are isomorphic.

Inverse systems and limits in C(R)op correspond to direct systems and colimits inC(R), so C(R)op has limits by 3.2.4. Together with 3.3.17 the next result, therefore,sums up as: the Hom functor preserves limits.

3.3.21 Proposition. Let µvu : Mu→Mvu6v be a U-direct system in C(R) and letN be an R-complex. The canonical map

(3.3.21.1) HomR(colimu∈U

Mu,N)−→ limu∈U

HomR(Mu,N) ,

given by ϑ 7→ (HomR(µu,N)(ϑ))u∈U = (ϑµu)u∈U , is an isomorphism in C(k).

PROOF. The map is a morphism of k-complexes by 3.3.18. There is a commutativediagram in C(k),

0 // HomR(colimu∈U

Mu,N) //

κ

HomR(∐

u∈UMu,N)

∼=

Hom(∆µ,N)// HomR(

∐(u,v)∈∇(U)

M(u,v),N)

∼=

0 // limu∈U

HomR(Mu,N) //∏

u∈UHomR(Mu,N)

∆Hom(µ,N)//

∏(u,v)∈∇(U)

HomR(M(u,v),N) ,

where κ is the canonical morphism (3.3.21.1), and the middle and right-hand verticalmaps the isomorphisms from 3.1.25. The rows are exact by left exactness of Hom,2.3.10, and the construction of limits, 3.3.2; it follows from the Five Lemma 2.1.40that κ is an isomorphism.

PULLBACKS

Simple non-trivial limits arise from three term direct systems N→ Y ←M.

3.3.22 Construction. Let U = u,v,w be a set, preordered as follows v > u 6 w.

Given a diagram Nβ−→ Y α←−M in C(R), set

Nv = N , Nu = Y, Nw = M ,

νvv = 1N , νuv = β , νuu = 1Y , νuw = α , and νww = 1M .

This defines a U-inverse system in C(R). It is straightforward to verify that the limitof this system is the kernel of the morphism (β −α)T : N⊕M→ Y .


3.3 Limits 113

3.3.23 Definition. For a diagram Nβ−→Y α←−M in C(R), the limit of the U-inverse

system constructed in 3.3.22 is called the pullback of (β,α) and denoted NuY M.

REMARK. As for the limit, the notation for the pullback suppresses the morphisms. Other namesfor the pullback are fibered product and Cartesian square.

3.3.24. Given morphisms α :M→ Y and β :N→ Y , the pullbacks of (β,α) and (α,β)are isomorphic via the map induced by the canonical isomorphism N⊕M ∼= M⊕N.

3.3.25 Construction. Given a diagram in Nβ−→ Y α←−M in C(R), let

α′ : NuY M −→ N and β′ : NuY M −→ M

be the canonical morphisms (3.3.2.1); they are given by (n,m) 7→ n and (n,m) 7→m.There is a commutative diagram with exact rows and columns

(3.3.25.1)

0

0

Kerα′β//

Kerα

0 // Kerβ′ //

α

NuY Mβ′//

α′

M

α

0 // Kerβ // Nβ

// Y ,

where α and β are restrictions of α′ and β′.

3.3.26. Given a diagram N α′′←− Xβ′′−→M in C(R) with βα′′ = αβ′′, it follows from

3.3.4 that the assignmentx 7−→ (α′′(n),β′′(m))

defines the unique morphism that makes the next diagram commutative,

X

##

β′′

α′′

--

NuY Mβ′//

α′

M

α

Nβ// Y .

3.3.27 Proposition. Let Nβ−→Y α←−M be a diagram in C(R). The following asser-

tions hold for the morphisms in (3.3.25.1).(a) If α is surjective, then α′ is surjective.(b) β is an isomorphism, whence α is injective if and only if α′ is injective.(c) If β is surjective, then β′ is surjective.



(d) α is an isomorphism, whence β is injective if and only if β′ is injective.

PROOF. By symmetry, see 3.3.24, it is sufficient to prove parts (a) and (b).(a): Assume that α is surjective. For every n in N there is then an m ∈ M with

α(m) = β(n). The pair (n,m) is, therefore, in NuY M and one has n = α′(n,m).(b): The kernel of α′ consists of all pairs (0,m) in N⊕M with α(m) = 0.

TOWERS

Inverse systems that are, essentially, sequences of morphisms play a special role forat least two reasons: they occur frequently and they come with a simple sufficientcondition for exactness.

3.3.28 Example. Let p be a prime. The sequence · · · Z/p3Z Z/p2Z Z/pZdetermines an inverse system whose limit is the Z-module Zp of p-adic integers.

3.3.29 Construction. Let λu : Nu→ Nu−1u∈Z be a sequence of morphisms inC(R). It determines an inverse system νuv : Nv→ Nuu6v as follows: set

νuu = 1Nufor all u in Z and νuv = λu+1 · · ·λv for all u < v in Z .

Given additional sequences κu : Mu→Mu−1u∈Z and βu : Nu→Muu∈Z of mor-phisms, such that βu−1λu = κuβu holds for all u ∈ Z, it is elementary to verify thatβuu∈Z is a morphism of the inverse systems determined by λuu∈Z and κuu∈Z .

3.3.30 Definition. A sequence λu : Nu→ Nu−1u∈Z of morphisms in C(R) withNu = 0 for u 0 is called a tower in C(R). The limit limu∈ZNu of the associatedinverse system, see 3.3.29, is called the limit of the tower in C(R).

Given towers λu : Nu→ Nu−1u∈Z and κu : Mu→Mu−1u∈Z in C(R), a se-quence of morphisms βu : Nu→Muu∈Z that satisfy βu−1λu = κuβu for all u ∈ Zis called a morphism of towers. The morphism limu∈Z β

u : limu∈ZNu→ limu∈ZMu,see 3.3.29 and 3.3.9, is called the limit of βu : Nu→Muu∈Z .

3.3.31. Let λu : Nu→ Nu−1n∈Z be a tower and let νuv : Nv→ Nuu6v be theassociated inverse system in C(R). Given an R-complex M and a sequence ofmorphisms αu : M→ Nuu∈Z that satisfy αu−1 = λuαu for all u ∈ Z, one hasαu = νuvαv for all u 6 v. By the universal property of limits, there is a morphismα : M→ limu∈ZNu in C(R) with properties as described in 3.3.4.

3.3.32 Proposition. Let λu : Nu→ Nu−1u∈Z be a tower in C(R). With

∆λ0 :∏u∈Z

Nu −→∏u∈Z

Nu given by (nu)u∈Z 7−→ (nu−λu+1(nu+1))u∈Z

one has limu∈ZNu = Ker∆λ0.

PROOF. As in 3.3.29 let νuv : Nv→ Nuu6v be the inverse system associated withthe tower and consider the morphism ∆ν; see 3.3.2. Since one has νuv = λu+1 · · ·λv,an element (nu)u∈Z ∈

∏u∈ZNu satisfies nu = λu+1(nu+1) for all u in Z if and only if

one has nu = νuv(nv) for all u6 v in Z, and hence Ker∆λ0 = Ker∆ν.


3.3 Limits 115

3.3.33 Example. Let N0 ⊇ N1 ⊇ N2 ⊇ ·· · be a descending chain of R-complexes.The embeddings λu : Nu Nu−1 define a tower. It is evident that the morphismfrom

⋂u∈ZNu to Ker∆λ0 that maps an element n to the sequence with nu = n for all

u> 0 is an isomorphism. Thus limu∈ZNu is the intersection⋂

u∈ZNu.

3.3.34 Example. Let Mii∈N be a sequence of R-complexes. Set Nu =⊕u

i=1 Mi

for u > 0 and Nu = 0 for u6 0, and let λu : Nu→ Nu−1 be the canonical projections.The limit of the tower λu : Nu→ Nu−1u∈Z is M =

∏i∈NMi. Indeed, the morphism

α : M→ limu∈ZNu determined by the canonical projections M Nu maps a se-quence (mi)i>1 to the family of truncated sequences ((x1, . . . ,xu))u>1; see 3.3.4. Itis injective, and from 3.3.32 evidently also surjective.

3.3.35 Proposition. Let λu : Nu→ Nu−1u∈Z be a tower in C(R). The followingassertions hold.

(a) If λu = 0 holds for infinitely many u > 0, then one has limu∈ZNu = 0.(b) If there exists an integer w such that λu is bijective for all u > w, then the

canonical map limu∈ZNu→ Nw is an isomorphism.

PROOF. (a): Let n = (nu)u∈Z be an element in limu∈ZNu; for w > u one then hasnu = λu+1 · · ·λw(nw). For every u ∈ Z there is by assumption an integer w > u withλw = 0, whence one has nu = 0 and, consequently, n = 0.

(b): Define a sequence of morphisms αu : Nw→ Nuu∈Z in C(R) as follows: setαw = 1Nw

, set αu = λu+1 · · ·λw for u < w, and set αu = (λw+1 · · ·λu)−1 for u > w.By construction one has αu−1 = λuαu for all u ∈ Z, so by 3.3.31 there is a mor-phism α : Nw→ limu∈ZNu given by α(n) = (αu(n))u∈Z. Evidently there are equal-ities νwα = αw = 1Nw

. For n = (nu)u∈Z in limu∈ZNu one has ανw(n) = α(nw) =(αu(nw))u∈Z = (nu)u∈Z = n, where the penultimate equality holds by definition ofthe maps αu and the complex limu∈ZNu.

3.3.36 Definition. A tower λu : Nu→ Nu−1u∈Z of R-complexes is said to satisfythe Mittag-Leffler Condition if for every u ∈ Z the descending chain

Imλu+1 ⊇ Im(λu+1λu+2)⊇ Im(λu+1λu+2λu+3)⊇ ·· ·

of subcomplexes of Nu becomes stationary, that is, there exists v > u such that forevery w> v one has Im(λu+1 · · ·λv · · ·λw) = Im(λu+1 · · ·λv). This subcomplex of Nu

is called the stable image of the tower at stage u.If for every u∈Z there exists v> u with λu+1 · · ·λv = 0, then the condition clearly

holds; in this case the tower is said to satisfy the trivial Mittag-Leffler Condition.

REMARK. A tower that satisfies the trivial Mittag-Leffler condition has a cofinal trivial subtower(i.e. all morphisms are zero); another word for such a tower is trump.

3.3.37 Example. A tower λu : Nu→ Nu−1u∈Z in which every morphism λu issurjective evidently satisfies the Mittag-Leffler Condition.

3.3.38 Lemma. Let Xu→ Xu−1u∈Z, Nu→Nu−1u∈Z, and Mu→Mu−1u∈Z betowers in C(R) and let αu : Xu→ Nuu∈Z and βu : Nu→Muu∈Z be morphisms oftowers such that the sequence




−−→Mu −→ 0

is exact for every u ∈U . The following assertions hold.(a) If the tower Nu→ Nu−1u∈Z satisfies the Mittag-Leffler Condition, then so

does Mu→Mu−1u∈Z.(b) If the towers Xu → Xu−1u∈Z and Mu → Mu−1u∈Z satisfy the Mittag-

Leffler Condition, then so does Nu→ Nu−1u∈Z.

PROOF. Write ξu : Xu→ Xu−1, λu : Nu→ Nu−1, and κu : Mu→Mu−1 for the mor-phisms in the given towers.

(a): For every u ∈ Z and every w > u surjectivity of βw yields

Im(κu+1 · · ·κw) = Im(κu+1 · · ·κwβw) = Im(βuλu+1 · · ·λw) = βu(Im(λu+1 · · ·λw)) .

(b): Given u ∈ Z choose v > u such that Im(ξu+1 · · ·ξv · · ·ξw) = Im(ξu+1 · · ·ξv)for all w > v and choose q > v such that Im(κv+1 · · ·κq · · ·κw) = Im(κv+1 · · ·κq) forall w> q. For every w> q one now has

βv(Im(λv+1 · · ·λq · · ·λw)) = Im(κv+1 · · ·κq · · ·κwβw) = Im(κv+1 · · ·κq) ,

where the last equality holds as βw is surjective and by the choice of q. In particular,one has βv(Im(λv+1 · · ·λq)) = βv(Im(λv+1 · · ·λq · · ·λw)) for w > q, and hence alsoIm(λv+1 · · ·λq)⊆ Im(λv+1 · · ·λq · · ·λw)+ Imαv. Applying λu+1 · · ·λv to this yields

(?) Im(λu+1 · · ·λq) ⊆ Im(λu+1 · · ·λq · · ·λw)+ Im(λu+1 · · ·λvαv) .

The choice of v explains the second equality below,

Im(λu+1 · · ·λvαv) = αu(Im(ξu+1 · · ·ξv))

= αu(Im(ξu+1 · · ·ξv · · ·ξw))

= Im(λu+1 · · ·λv · · ·λwαw)

⊆ Im(λu+1 · · ·λv · · ·λw) .

Thus (?) shows that Im(λu+1 · · ·λq)⊆ Im(λu+1 · · ·λq · · ·λw) holds for all w> q.

3.3.39 Lemma. If a tower λu : Nu→ Nu−1u∈Z satisfies the trivial Mittag-LefflerCondition, then the map ∆λ0 from 3.3.32 is bijective, in particular, limu∈ZNu = 0.

PROOF. Let n = (nu)u∈Z be an element in∏

u∈ZNu. For every u ∈ Z the sum

αu(n) = nu +λu+1(nu+1)+λu+1λu+2(nu+2)+ · · ·

is finite as the tower satisfies the trivial Mittag-Leffler Condition, and thus αu(n)is a well-defined element in Nu. The map α :

∏u∈ZNu→

∏u∈ZNu that sends n to

α(n) = (αu(n))u∈Z is an inverse of ∆λ0. Indeed, the uth component of ∆λ0α(n) is

αu(n)−λu+1(αu+1(n)) = nu +λu+1(nu+1)+λu+1λu+2(nu+2)+ · · ·−λu+1(nu+1 +λu+2(nu+2)+λu+2λu+3(nu+3)+ · · ·) ,

which is nu; and the uth component of α∆λ0(n) is


3.3 Limits 117

(nu−λu+1(nu+1))+λu+1(nu+1−λu+2(nu+2))+λu+1λu+2(nu+2−λu+3(nu+3))+ · · · ,

which is also nu. In particular, limu∈ZNu = 0 by 3.3.32.

3.3.40 Lemma. Let λu : Nu→ Nu−1u∈Z be a tower in C(R). If every morphismλu is surjective, then the map ∆λ0 from 3.3.32 is surjective.

PROOF. Given n = (nu)u∈Z in∏

u∈ZNu we seek an element x = (xu)u∈Z with∆λ0(x) = n, that is, a solution to the equations xu − λu+1(xu+1) = nu for u ∈ Z.By the definition of a tower, one has nu = 0 for u 0, so set xu = 0 for u 0.Now let v ∈ Z and assume that xu for u 6 v have been constructed such thatxu − λu+1(xu+1) = nu holds for all u < v. Surjectivity of λv+1 yields an elementxv+1 with xv−λv+1(xv+1) = nv. This procedure yields the desired element x.

3.3.41 Proposition. Let λu : Nu→ Nu−1u∈Z be a tower that satisfies the Mittag-Leffler Condition. There is a tower λu : Nu→ Nu−1u∈Z with every λu surjective,a tower λu : Nu→ Nu−1u∈Z that satisfies the trivial Mittag-Leffler Condition, andmorphisms of towers ιu : Nu Nuu∈Z and πu : Nu Nuu∈Z such that

0−→ Nu ιu−−→ Nu πu−−→ Nu −→ 0

is an exact sequence for every u ∈ Z.

PROOF. Let Nu be the stable image of the given tower at stage u; see 3.3.36. Notethat λu(Nu)= Nu−1 whence λu restricts to a surjective morphism Nu→ Nu−1; denoteit by λu. Let ιu : Nu Nu be the embedding. Set Nu = Nu/Nu, let πu : Nu Nu bethe quotient map, and let λu : Nu→ Nu−1 be the map induced by λu. For every u∈ Zone has Im(λu+1 · · ·λv) = Nu for some v > u, and hence λu+1 · · · λv = 0.

3.3.42 Theorem. Let Xu→ Xu−1u∈Z, Nu→ Nu−1u∈Z, and Mu→Mu−1u∈Zbe towers in C(R), and let αu : Xu→ Nuu∈Z and βu : Nu→Muu∈Z be mor-phisms of towers such that the sequence


−−→Mu −→ 0

is exact for every u ∈U . If the tower Xu→ Xu−1u∈Z satisfies the Mittag-LefflerCondition, then the following sequence is exact,

0−→ limu∈Z

Xu limu∈Z αu

−−−−−→ limu∈Z

Nu limu∈Z βu

−−−−−→ limu∈Z

Mu −→ 0 .

PROOF. Write ξu : Xu→ Xu−1, λu : Nu→ Nu−1, and κu : Mu→Mu−1 for the mapsin the given towers and consider the commutative diagram with exact rows,

0 //∏

u∈U Xu

∆ξ0

∏u∈U αu

//∏

u∈U Nu

∆λ0

∏u∈U βu

//∏

u∈U Mu

∆κ0

// 0

0 //∏

u∈U Xu∏

u∈U αu//∏

u∈U Nu∏

u∈U βu//∏

u∈U Mu // 0 .

By 3.3.32 and the Snake Lemma 2.1.42 there is an exact sequence



0−→ limu∈ZXu limαu−−−→ limu∈ZNu limβu

−−−→ limu∈ZMu δ−−→ Coker∆ξ0 .

If the tower ξu : Xu→ Xu−1u∈Z satisfies the trivial Mittag-Leffler Condition, or ifevery ξu is surjective, then Coker∆ξ0 = 0 holds by 3.3.39 and 3.3.40, and the asser-tion follows. In the general case, apply 3.3.41 to the tower ξu : Xu→ Xu−1u∈Z toget towers ξu : Xu→ Xu−1u∈Z with every ξu surjective and ξu : Xu→ Xu−1u∈Zthat satisfies the trivial Mittag-Leffler Condition, and to get morphisms of towerssuch that for every u ∈ Z the sequence

0−→ Xu ιu−−→ Xu πu−−→ Xu −→ 0

is exact. For u ∈ Z there is a commutative diagram in C(R) with exact rows,

(?)

0 // Xu

ιu

αuιu// Nu

=

%u// Cu

βu

// 0

0 // Xu αu// Nu βu

// Mu // 0 ,

where Cu is Coker(αuιu) and %u is the canonical map, while the morphism βu isinduced by βu. The morphisms λu : Nu→ Nu−1 induce morphisms Cu→Cu−1 thatdefine a tower Cu→Cu−1u∈Z, such that %uu∈Z and βuu∈Z are morphisms oftowers. The Snake Lemma 2.1.42 yields Ker βu ∼= Xu, and thus for every u ∈ Z thereis an exact sequence,

(‡) 0−→ Xu −→Cu βu

−−→Mu −→ 0 .

From the exact sequence in the upper row in (?), from the exact sequence (‡), andfrom the special cases established in the beginning of the proof, it follows that themorphisms limu∈Z %

u and limu∈Z βu are surjective. Since limu∈Z β

u is the compositeof these morphisms, it too is surjective; cf. 3.3.9.

REMARK. A U-inverse system χuv : Xv→ Xuu6v is said to satisfy the Mittag-Leffler Conditionif for every u there exists v> u such that for every w> v one has Imχuv = Imχuw. It can be provedthat the limit functor is exact on short exact sequences 0→Xu→Nu→Mu→ 0 of U-directsystems where Xu satisfies the Mittag-Leffler Condition. A special case of this result is 3.3.42.

3.3.43 Example. Let m,n > 1 be integers and relatively prime. Consider the fol-lowing commutative diagram of Z-modules


3.3 Limits 119

...

m

...

m

...

m

0 // Z

m

n// Z

m

// Z/nZ

m

// 0

0 // Z

m

n// Z

m

// Z/nZ

m

// 0

0 // Zn// Z // Z/nZ // 0 .

The rows are exact and the vertical maps define towers with limits 0, 0, and Z/nZ,respectively; cf. 3.3.35(b). Thus the sequence of limits 0→ 0→ 0→ Z/nZ→ 0 isnot exact.

3.3.44. Let νuv : Nv→ Nuu6v be a U-inverse system in C(R). By 3.3.14, appliedto the functors B(–) and Z(–), see 2.2.12, there are canonical morphisms

νB : B(limu∈U

Nu) −→ limu∈U

B(Nu) and νZ : Z(limu∈U

Nu) −→ limu∈U

Z(Nu) .

Notice that both maps are injective. By (2.2.11.1) and 3.3.13 there is a commutativediagram with exact rows

(3.3.44.1)

0 // Z(limu∈U

Nu) //

νZ

limu∈U

Nu //

=

ΣB(limu∈U

Nu) //

ΣνB

0

0 // limu∈U

Z(Mu) // limu∈U

Nu // Σ limu∈U

B(Nu) .

The Snake Lemma 2.1.42 implies that νZ is an isomorphism.

The Mittag-Leffler Condition facilitates computation of homology of limits.

3.3.45 Theorem. Let λu : Nu→ Nu−1u∈Z be a tower in C(R) and consider thecanonical morphism,

(3.3.45.1) H(limu∈Z

Nu)−→ limu∈Z

H(Nu) ,

given by h 7→ (H(νu)(h))u∈Z. If the tower λu : Nu→ Nu−1u∈Z satisfies the Mittag-Leffler Condition, then the following assertions hold.

(a) The morphism (3.3.45.1) is surjective.(b) Let v ∈ Z; if the tower Hv+1(λ

u) : Hv+1(Nu)→ Hv+1(Nu−1)u∈Z satisfiesthe Mittag-Leffler Condition, then the degree v component of the canonicalmap (3.3.45.1) is an isomorphism; so one has Hv(limu∈ZNu)∼= limu∈ZHv(Nu).

PROOF. (a): As λu : Nu→ Nu−1u∈Z satisfies the Mittag-Leffler Condition, so doesB(λu) : B(Nu)→ B(Nu−1)u∈Z by 3.3.38, so in the commutative diagram below,



where the vertical morphisms are the canonical ones, see 3.3.16, rows are exact by(2.2.11.3) and 3.3.42.

0 // B(limu∈ZNu) //

κB

Z(limu∈ZNu) //

∼= κZ

H(limu∈ZNu) //

κH

0

0 // limu∈ZB(Nu) // limu∈ZZ(Nu) // limu∈ZH(Nu) // 0

As κZ is surjective by 3.3.44, so is κH.(b): By the Snake Lemma 2.1.42 applied to the diagram above, proving that the

degree v component κHv is injective, and hence by part (a) an isomorphism, is equiva-

lent to showing that κBv is surjective. Consider the towers determined by the families

of morphisms Bv+1(λu)u∈Z, Zv+1(λ

u)u∈Z, and Hv+1(λu)u∈Z. The canonical

exact sequences,

0−→ Bv+1(Nu)−→ Zv+1(Nu)−→ Hv+1(Nu)−→ 0 ,

yield morphisms of towers. As in the proof of part (a), the tower Bv+1(λu)u∈Z

satisfies the Mittag-Leffler Condition, and so does Hv+1(λu)u∈Z by assumption.

Hence 3.3.38 implies that Zv+1(λu)i∈Z satisfies the Mittag-Leffler Condition. The

short exact sequences,

0−→ Zv+1(Nu)−→ Nuv+1 −→ Bv(Nu)−→ 0 ,

induce by 3.3.42 the lower exact row in the following commutative diagram,

0 // Zv+1(limu∈ZNu)

κZv+1

∼=

// (limu∈ZNu)v+1

=

// Bv(limu∈ZNu)

κBv

// 0

0 // limu∈ZZv+1(Nu) // limu∈Z(Nuv+1)

// limu∈ZBv(Nu) // 0 .

The upper row is exact by (2.2.11.1). As the middle map in this diagram is an equal-ity, see 3.3.6, it follows that κB

v is surjective, as desired.

For ease of reference, we record two frequently used corollaries to 3.3.45.

3.3.46 Corollary. Let λu : Nu→ Nu−1u∈Z be a tower in C(R). If this tower andH(λu) : H(Nu)→ H(Nu−1)u∈Z both satisfy the Mittag-Leffler Condition, then thecanonical map H(limu∈ZNu)→ limu∈ZH(Nu) from 3.3.45 is an isomorphism.

3.3.47 Corollary. Let λu : Nu→ Nu−1u∈Z be a tower in C(R). If this tower satis-fies the Mittag-Leffler Condition and if Nu is acyclic for all u ∈ Z, then the complexlimu∈ZNu is acyclic.

3.3.48 Example. The diagram in 3.3.43 can be interpreted as a tower of identicalacyclic complexes. The morphisms in the tower are multiplication by m, so it doesnot satisfy the Mittag-Leffler Condition and, indeed, the limit is not acyclic.


3.3 Limits 121

EXERCISES

E 3.3.1 Let νuv : Nv→ Nuu6v be a U-inverse system in C(R). Let νu : L→ Nuu∈U be afamily of morphisms that satisfy the next conditions. (1) One has νv = νuvνv for allu 6 v. (2) For every family αu : M→ Nuu∈U of morphisms with αu = νuvαv for allu6 v there exists a unique morphism α : M→ L with νuα= αu for all u ∈U . Show thatthere is an isomorphism ϕ : L→ limu∈U Nu with νuϕ = νu for every u ∈U . Concludethat the universal property determines the limit uniquely up to isomorphism.

E 3.3.2 (Cf. 3.3.6) Show that the limit in C(R) of an inverse system of morphisms of gradedR-modules is a graded R-module. Conclude, in particular, that Mgr(R) has limits.

E 3.3.3 (Cf. 3.3.6) Show that the limit in C(R) of a inverse system of homomorphisms of R-modules is an R-module. Conclude, in particular, that the category M(R) has limits.

E 3.3.4 Fix a preordered set U . Show that U-inverse systems in C(R) and their morphisms forman Abelian category and that the limit is a left exact functor from this category to C(R).

E 3.3.5 Generalize the result in E 3.1.13 by showing that every functor F: C(R)→C(S) thathas a left adjoint preserves limits.

E 3.3.6 (a) Show that U-inverse systems in C(R)op correspond to U-direct systems in C(R) andthat U-direct systems in C(R)op correspond to U-inverse systems in C(R). (b) Showlimits in C(R)op correspond to colimits in C(R) and that colimits in C(R)op correspondto limits in C(R).

E 3.3.7 Let U be a preordered filtered set and let U ′ be a cofinal subset of U ; that is, for ev-ery u ∈ U there exists an u′ ∈ U ′ with u′ > u. Show that for every U-inverse systemνuv : Nv→ Nuu6v in C(R) there is an isomorphism limu∈U Nu ∼= limu∈U ′ N

u.E 3.3.8 (Cf. 3.3.22) Verify the isomorphism limu∈U Nu ∼= Ker(β−α) in 3.3.22.E 3.3.9 Show that the categories M(R) and Mgr(R) have pullbacks.E 3.3.10 Adobt the notation from 3.3.25.

(a) Show that there is a commutative diagram with exact rows and columns:

0

0

0

0

0 // 0 //

Kerα′β//

Kerα //

0 //

0

0 // Kerβ′ //

α

NuY Mβ′//

α′

Mα

// Cokerβ′ //

0

0 // Kerβ //

N

β// Y

// Cokerβ

// 0

0 // 0

// Cokerα′

// Cokerα

// W

// 0

0 0 0 0 .

(b) Show that W in the diagram above can be non-zero. Hint: Let a and b be left idealsin R and let α : a R and β : b R be the canonical homomorphisms.

(c) In the following two diagrams, the solid parts are given. Show that they can becompleted to commutative diagrams with exact rows and columns, as depicted.



0

0

K

K

0 // L // NuY M //

M

// 0

0 // L // N

// Y

// 0

0 0

and

0

0

K

K

0 // NuY M //

M

// D // 0

0 // N

// Y

// D // 0

0 0 .

E 3.3.11 Consider a commutative diagram of R-complexes,

X

α

β// M

γ

Nδ// Y .

(a) Show that if Y is isomorphic to NtX M and α is injective, then one has X ∼= NuY M.(b) show that if X is isomorphic to NuY M and δ is surjective, then one has Y ∼= NtX M.

E 3.3.12 Consider a diagram of R-complexes, not a priori assumed to be commutative,

X

α

β// M

γ

Nδ// Y .

(a) Show that Y is isomorphic MtX N if and only if the next sequence is exact,

X

(βα

)//

M⊕N

(γ −δ)// Y // 0 .

(b) Show that X is isomorphic NuY M if and only if the next sequence is exact,

0 // X

(βα

)//

M⊕N

(γ −δ)// Y .

E 3.3.13 As in 3.3.29 let λu : Nu→ Nu−1u∈Z be a sequence (not necessarily a tower) of mor-phisms in C(R). Show that the limit of the associated inverse system does not dependon λu for u 0.

E 3.3.14 Show that every complex is the limit of a tower of bounded below complexes.E 3.3.15 Show that for a U-inverse system νuv : Nv→ Nuu6v the map ∆ν is not surjective.

E 3.3.16 Show that the p-adic integers Zp from 3.3.28 form a ring isomorphic to Z[[x]]/(x− p).E 3.3.17 Let k be a countable integral domain. Consider the set U = (u) | u ∈ k\0 of princi-

pal ideals ordered under reverse inclusion. Show that limU k/(u) is not countable. Hint:Enumerate the non-units in U and construct a totally ordered cofinal subsystem of U .

E 3.3.18 Let λu : Nu→ Nu−1u∈Z be a tower in C(R), let αu : M→ Nuu∈Z be a sequenceof morphisms with αu−1 = λuαu for all u ∈ Z, and denote by α : M→ limu∈ZNu thecanonical morphism. Assume that αu is surjective and Kerαu = Kerαu−1 holds foru 0. Show that α is surjective.

E 3.3.19 Let λu : Nu→ Nu−1u∈Z be a tower in C(R) with limit N. Set Mu = Nu−1 andκu = λu−1; show that κu : Mu→Mu−1u∈Z is a tower with colimit N and that themorphisms βu = λu from Nu to Mu form a morphism of towers with limit 1N .

E 3.3.20 Let λu : Nu→ Nu−1u∈Z be a tower in C(R) with λu surjective for all u ∈ Z. Assumethat Nu is acyclic for infinitely many u > 0. Show that limu∈ZNu is acyclic.


3.3 Limits 123

E 3.3.21 Let λu : Nu→ Nu−1u∈Z be a tower in C(R) with λu surjective for all u ∈ Z. Assumethat for every n ∈ Z there is a un > 0 such that Hn(Nu) = 0 for all u > un. Show thatlimu∈ZNu is acyclic.

E 3.3.22 Show that the conclusion in 3.3.47 may fail if the homomorphisms λu are surjectivefor infinitely many but not all u ∈ Z. Hint: Set N = 0 −→ Z n−→ Z −→ Z/nZ −→ 0 andconsider the tower given by · · · m−→ N =−→ N m−→ N =−→ N; cf. 3.3.43.

E 3.3.23 Let k be a field and let L be the category of commutative local k-algebras. (a) Showthat k is a zero object in L. (b) For R and S in L, show that the pullback Ruk S in thecategory of k-modules yields a product in L. (c) Show that the full subcategory of Lwhose objects are integral domains does not have a product.


Chapter 4Equivalences and Isomorphisms

This chapter opens with the mapping cone construction. It attaches to every mor-phism a complex that reflects essential properties of the morphism. The utility of thisconstruction stretches far beyond the category of complexes, indeed, it is the key todefining a triangulated structure on the derived category. Another notion that pointstowards the derived category is quasi-isomorphisms; these are morphisms that in-duce isomorphisms in homology. They are introduced in the second section, andhomotopy equivalences—an especially robust subclass of quasi-isomorphisms—aretreated in Sect. 4.3.

In Sects. 4.4 and 4.5 we extend the standard isomorphisms and evaluation ho-momorphisms for modules treated in Chap. 1 to the realm of complexes. To someextent this is an exercise in bootstrapping; but it is a critical one as these maps areamong our tools of choice when it comes to applications of homological algebra inring theory.

4.1 Mapping Cone

SYNOPSIS. Mapping cone (sequence); homotopy; Σ-functor.

The mapping cone of a continuous map f : X → Y between topological spaces is aspace glued together from X and Y via the map f . Here we explore the algebraic ver-sion of this construction. In Chap. 6 the mapping cone will be key to the constructionof triangulated structures on the homotopy and derived categories.

4.1.1 Definition. Let α : M→ N be a morphism of R-complexes. The mapping coneof α is the complex with underlying graded module

(Coneα)\ =N\

⊕ΣM\

and differential ∂Coneα =

(∂N αςΣM

−10 ∂ΣM

).

125

126 4 Equivalences and Isomorphisms

4.1.2. A straightforward computation shows that ∂Coneα is square zero,

∂Coneα∂Coneα =

(∂N∂N ∂NαςΣM

−1 +αςΣM−1 ∂

ΣM

0 ∂ΣM∂ΣM

)= 0 ;

it uses that α is a morphism and that ςΣM−1 is a degree −1 chain map.

4.1.3 Lemma. Let M be an R-complex. For every x ∈ k one has xH(ConexM) = 0.In particular, Cone1M is acyclic.

PROOF. The goal is to show that xz is a boundary for every homogeneous cycle z inthe complex ConexM . For such a cycle z = (m ςM

1 (m′))T one has ∂M(m)+ xm′ = 0in M, and hence −ςM

1 ∂M(m) = xςM

1 (m′) in ΣM. In the computation,(∂M xςΣM

−10 ∂ΣM

)(0

ςM1 (m)

)=

(xm

∂ΣMςM1 (m)

)=

(xm

−ςM1 ∂

M(m)

)=

(xm

xςM1 (m′)

)= xz ,

the second equality holds as ςM1 is a degree 1 chain map.

In 4.3.31 we show that Cone1M more than acyclic, namely so-called contractible.

4.1.4. For an R-module M the disc complex Dv(M) from 2.5.24 is Σv−1 Cone1M .

4.1.5. For every morphism of R-complexes α : M→ N there is a degreewise splitexact sequence of R-complexes, known as the mapping cone sequence,

(4.1.5.1) 0−→ N

(1N

0

)−−−→ Coneα

(0 1ΣM )−−−−−→ ΣM −→ 0 .

4.1.6 Theorem. Let α,α′ : M→ N be morphisms of R-complexes, and consider thefollowing diagram whose rows are the exact sequences (4.1.5.1),

0 // N // Coneα //

γ

ΣM // 0

0 // N // Coneα′ // ΣM // 0 .

The morphisms α and α′ are homotopic if and only if there exists a morphism γ thatmakes the diagram commutative; moreover, any such morphism is an isomorphism.

PROOF. It follows from the Five Lemma 2.1.40 that a morphism γ that makes thediagram commutative is an isomorphism. The assignment σ 7→ γσ, where

γσ =

(1N σςΣM

−10 1ΣM

),

is a 1-to-1 correspondence between degree 1 homomorphisms σ : M→ N and de-gree 0 homomorphisms that make the diagram commutative. Now one has

γσ∂Coneα−∂Coneα′γσ =

(1N σςΣM

−10 1ΣM

)(∂N αςΣM

−10 ∂ΣM

)−(∂N α′ςΣM

−10 ∂ΣM

)(1N σςΣM

−10 1ΣM

)=

(0 (α−σ∂M−∂Nσ−α′)ςΣM

−10 0

),


4.1 Mapping Cone 127

so γσ is a morphism if and only if α−α′ = ∂Nσ+σ∂M holds; that is, if and only ifσ is a homotopy from α to α′.

As an immediate consequence of the theorem we get a characterization of null-homotopic maps; similar characterizations of other classes of morphisms are givenin 4.2.15 and 4.3.30.

4.1.7 Corollary. A morphism α : M→ N of R-complexes is null-homotopic if andonly if the mapping cone sequence (4.1.5.1) is split exact.

PROOF. The mapping cone of the zero morphism is the direct sum N⊕ΣM, whencethe claim is immediate from 4.1.6 and 2.1.44.

Σ-FUNCTORS

This section closes with four technical results whose utility will only become fullyapparent in Chap. 6. The gist is that the mapping cone construction commutes withHom and tensor product in such a way that mapping cone sequences are preserved.

4.1.8 Definition. A functor F: C(R)→ C(S) is called a Σ-functor if it is additiveand there exists a natural isomorphism φ : FΣ→ ΣF of functors, such that for everymorphism α : M→ N in C(R) there exists an isomorphism α that makes the diagram

F(N)F(

1N

0

)// F(Coneα)

∼= α

F(0 1ΣM )// F(ΣM)

φM∼=

F(N)

(1F(N)

0

)// ConeF(α)

(0 1ΣF(M) )// ΣF(M)

in C(S) commutative.

4.1.9 Definition. Let E,F: C(R)→ C(S) be Σ-functors with associated natural iso-morphisms φ : EΣ→ ΣE and ψ : FΣ→ ΣF. A natural transformation τ : E→ F iscalled a Σ-transformation if the next diagram is commutative for every M in C(R),

E(ΣM)

τΣM

∼=φM// ΣE(M)

ΣτM

F(ΣM) ∼=ψM// ΣF(M) .

That is, τΣM and ΣτM are isomorphic, and the isomorphism is natural in M.

4.1.10 Definition. A functor G: C(R)op→ C(S) is called a Σ-functor if it is additiveand there is a natural isomorphism ψ : Σ−1 G→ GΣop of functors, such that for everymorphism α : M→ N in C(R) there exists an isomorphism α that makes the diagram



Σ−1 G(M)

ψM∼=

Σ−1(

1G(M)

0

)// Σ−1 ConeG(α)

α∼=

(0 1G(N) )// G(N)

G(ΣM)G(0 1ΣM )

// G(Coneα)G(

1N

0

)// G(N)

in C(S) commutative.

4.1.11 Definition. Let G,J : C(R)op→ C(S) be Σ-functors with associated naturalisomorphisms ψ : Σ−1 G→ GΣop and φ : Σ−1 J→ JΣop. A natural transformationτ : G→ J is called a Σ-transformation if the next diagram is commutative for ev-ery M in C(R),

Σ−1 G(M)

Σ−1 τM

∼=ψM// G(ΣM)

τΣM

Σ−1 J(M) ∼=φM// J(ΣM) .

That is, τΣM and ΣτM are isomorphic, and the isomorphism is natural in M.

4.1.12. The following assertions are immediate from the definitions of Σ-functors.

(a) Let C(Q)F1−→ C(R)

F2−→ C(S) be Σ-functors with associated natural isomor-phisms φ1 and φ2. The composite F2 F1 : C(Q)→ C(S) is a Σ-functor with as-sociated natural isomorphism φ where φM is the the composite

F2 F1(ΣM)F2(φ

M1 )

−−−−→ F2ΣF1(M)φ

F1(M)2−−−−→ ΣF2 F1(M) .

(b) Let C(Q)op G−→ C(R) F−→ C(S) be Σ-functors with associated natural isomor-phisms φ and ψ. The composite FG: C(Q)op→ C(S) is a Σ-functor with asso-ciated natural isomorphism ψ where ψM is the composite

Σ−1 FG(M)

Σ−1 φΣ−1 G(M)

−−−−−−−−→ FΣ−1 G(M)F(ψM)−−−−→ FG(ΣM) .

(c) Let C(Q)F−→ C(R) and C(R)op G−→ C(S) be Σ-functors with associated natural

isomorphisms φ and ψ. The composite GFop : C(Q)op→ C(S) is a Σ-functorwith associated natural isomorphism ψ where ψM is the composite

Σ−1 GFop(M)

ψF(M)

−−−→ GΣop Fop(M)G(φM)−−−−→ GFop

Σop(M) .

(d) Let C(Q)op G1−→ C(R) and C(R)op G2−→ C(S) be Σ-functors with associated nat-ural isomorphisms ψ1 and ψ2. The composite G2 Gop

1 : C(Q)→ C(S) is a Σ-functor with associated natural isomorphism φ where φM is the composite

G2 Gop1 (ΣM)

G2(ψM1 )

−−−−−→ G2Σ−1 Gop

1 (M)Σψ

Σ−1 G1(M)2−−−−−−−→ ΣG2 Gop

1 (M) .



4.1.13 Proposition. Let F: M(R)→M(S) be an additive functor. The extendedfunctor F: C(R)→ C(S) from 2.1.45 is a Σ-functor.

PROOF. The extended functor is additive, see 2.1.45, and there is an equalityF(ΣM) = ΣF(M) for every R-complex M; see 2.2.2. For a morphism α : M→ N,the definition of the extended functor yields the following identifications,

F(Coneα) = ConeF(α) , F(

1N

0

)=

(1F(N)

0

), and F

(0 1ΣM

)=(0 1ΣF(M)

).

Therefore, the identity maps ϕM : F(ΣM)→ ΣF(M) and α : F(Coneα)→ ConeF(α)make the diagram in 4.1.8 commutative, and thus F is a Σ-functor.

4.1.14 Proposition. Let G: M(R)op→M(S) be an additive functor. The extendedfunctor G: C(R)op→ C(S) from 2.1.45 is a Σ-functor.


We prove below that the Hom and tensor product functors are Σ-functors. Thehomology functor, H, is k-linear and commutes with Σ by 2.2.15; it is, however, nota Σ-functor, as the complexes H(Coneα) and ConeH(α) need not be isomorphic.Indeed, for every R-complex M one has H(Cone1M) = 0 by 4.1.3, but if M is notacyclic, then ConeH(1M) = Cone1H(M) is non-zero.

4.1.15 Proposition. Let M be an R-complex. The functor HomR(M,–) is a Σ-functorwith associated natural isomorphism φ : HomR(M,Σ–)→ ΣHomR(M,–) given by

φN = ςHomR(M,N)1 HomR(M,ςΣN

−1 )

for every R-complex N. In particular, there is an isomorphism of k-complexes,

Cone(HomR(M,β)) ∼= HomR(M,Coneβ) ,

for every morphism β of R-complexes.

PROOF. First note that by 2.3.10 and 2.3.15 the functor HomR(M,–) is additive.Evidently, φN is an isomorphism and natural in N. To prove that HomR(M,–) is a Σ-functor, it must be shown that for every morphism β : N→ N′ of R-complexes thereexists an isomorphism β that makes the following diagram in C(k) commutative.

(?)

HomR(M,N′)Hom(M,

(1N′

0

))// HomR(M,Coneβ)

∼= β

Hom(M,(0 1ΣN ))// HomR(M,ΣN)

φN∼=

HomR(M,N′)

(1Hom(M,N′)

0

)// ConeHomR(M,β)

(0 1ΣHom(M,N) )// ΣHomR(M,N) .

For reasons of simplicity, we shall construct an isomorphism

γ : ConeHomR(M,β) −→ HomR(M,Coneβ)

with γ−1 = β. To this end, observe that on the level of graded modules one has



HomR(M,Coneβ)\ = HomR(M\,(Coneβ)\) = HomR(M\,N′\⊕ΣN\) ,

and similarly,

(ConeHomR(M,β))\ = HomR(M\,N′\)⊕ΣHomR(M\,N\) .

These equalities, combined with the fact that the functor HomR(M\,–) is additiveand φN is an isomorphism, show that one gets an isomorphism of graded modules,γ : (ConeHomR(M,β))\→ HomR(M,Coneβ)\, by setting

(‡) γ

(ϑξ

)=

(ϑ

(φN)−1(ξ)

)for homogeneous elements ϑ ∈ HomR(M,N′) and ξ ∈ ΣHomR(M,N) of the samedegree, say, d. Notice that the left-hand column (ϑ ξ)T in (‡) is a pair of homo-morphisms, whereas the right-hand column is a homomorphism M→ N′⊕ΣN ofdegree d. To show that γ is a morphism, and hence an isomorphism of complexes,note first that the definition of φN and (2.2.4.1) yield

HomR(M,β)ςΣHom(M,N)−1 = HomR(M,βςΣN

−1 ) (φN)−1 .

Using this identity and the fact that (φN)−1 is a morphism of complexes, one gets

∂Hom(M,Coneβ)γ

(ϑξ

)=

(∂N′ βςΣN

−10 ∂ΣN

)(ϑ

(φN)−1(ξ)

)− (−1)d

(ϑ

(φN)−1(ξ)

)∂M

=

(∂N′ϑ+βςΣN

−1 (φN)−1(ξ)− (−1)dϑ∂M

∂ΣN(φN)−1(ξ)− (−1)d(φN)−1(ξ)∂M

)=

((∂N′ϑ− (−1)dϑ∂M)+βςΣN

−1 (φN)−1(ξ)

∂Hom(M,ΣN)((φN)−1(ξ))

)=

(∂Hom(M,N′)(ϑ)+Hom(M,βςΣN

−1 )((φN)−1(ξ))

(φN)−1(∂ΣHom(M,N)(ξ))

)= γ

(∂Hom(M,N′)(ϑ)+Hom(M,β)ς

ΣHomR(M,N)−1 (ξ)

∂ΣHom(M,N)(ξ)

)

= γ∂ConeHom(M,β)

(ϑξ

).

Thus γ is a morphism of complexes.It remains to verify that the diagram (?) is commutative. The computation

HomR(M,(0 1ΣN))γ

(ϑξ

)=(0 1ΣN

)( ϑ(φN)−1(ξ)

)= (φN)−1(ξ)

= (φN)−1 (0 1ΣHom(M,N))(ϑξ

)



shows that the right-hand square in (?) is commutative. A similar simple computa-tion shows that the left-hand square is commutative.

4.1.16 Proposition. Let N be an R-complex. The functor HomR(–,N) is a Σ-functor;with associated natural isomorphism ψ : Σ−1 HomR(–,N)→ HomR(Σ

op–,N) givenby

ψM = HomR(ςΣM−1 ,N)ς

Σ−1 HomR(M,N)1

for every R-complex M. In particular there is an isomorphism of k-complexes,

Σ−1 ConeHomR(α,N) ∼= HomR(Coneα,N) ,

for every morphism α of R-complexes.


4.1.17 Proposition. Let M be an Ro-complex. The functor M⊗R – is a Σ-functor; inparticular, there is an isomorphism of k-complexes,

Cone(M⊗R β) ∼= M⊗R Coneβ ,

for every morphism β of R-complexes.

PROOF. First note that by 2.4.9 and 2.4.12 the functor M⊗R – is additive, and thereis a natural isomorphism

φN : M⊗R ΣN −→ Σ(M⊗R N) ;

namely φN = ςM⊗RN1 (M⊗R ς

ΣN−1 ). To prove that M⊗R – is a Σ-functor, it must be

shown that for every morphism β : N→ N′ of R-complexes there exists an isomor-phism β that makes the following diagram in C(k) commutative.

(?)

M⊗R N′M⊗

(1N′

0

)// M⊗R Coneβ

∼= β

M⊗(0 1ΣN )// M⊗R ΣN

φN∼=

M⊗R N′

(1M⊗N′

0

)// Cone(M⊗R β)

(0 1Σ (M⊗N) )// Σ(M⊗R N) .

To define β, note that on the level of graded modules one has

(M⊗R Coneβ)\ = M\⊗R (Coneβ)\ = M\⊗R (N′\⊕ΣN\) ,

and(Cone(M⊗R β))

\ = (M\⊗R N′\)⊕Σ(M\⊗R N\) .

These equalities, combined with the fact that the functor M\⊗R – is additive andφN is an isomorphism, show that one defines an isomorphism of graded modules,β : (M⊗R Coneβ)\→ (Cone(M⊗R β))

\, by setting

β

(m⊗

(n′

n

))=

(m⊗n′

φN(m⊗n)

)



for an elementary tensor in M⊗R Coneβ with n ∈ ΣN and n′ ∈ N′. To show that β isa morphism, and hence an isomorphism of complexes, note first that the definitionof φN and (2.2.4.1) yield

(M⊗ β)ςΣ (M⊗N)−1 φN = M⊗ (βςΣN

−1 ) .

Using this identity and the fact that φN is a morphism of complexes, one gets

∂Cone(M⊗β)β

(m⊗

(n′

n

))=

(∂M⊗N′ (M⊗ β)ςΣ (M⊗N)

−10 ∂Σ (M⊗N)

)(m⊗n′

φN(m⊗n)

)=

(∂M⊗N′(m⊗n′)+(M⊗ (βςΣN

−1 ))(m⊗n)∂Σ (M⊗N)φN(m⊗n)

)=

(∂M(m)⊗ n′+(−1)|m|m⊗ ∂N′(n′)+(−1)|m|m⊗ βςΣN

−1 (n)φN∂M⊗ΣN(m⊗n)

)=

(∂M(m)⊗ n′+(−1)|m|m⊗ (∂N′(n′)+βςΣN

−1 (n))φN(∂M(m)⊗ n+(−1)|m|m⊗ ∂ΣN(n))

)= β

(∂M(m)⊗

(n′

n

)+(−1)|m|m⊗

(∂N′(n′)+βςΣN

−1 (n)∂ΣN(n)

))= β∂M⊗Coneβ

(m⊗

(n′

n

)).

Thus β is a morphism of complexes.It remains to verify that the diagram (?) is commutative. The computation(

0 1Σ (M⊗N))β

(m⊗

(n′

n

))=(0 1Σ (M⊗N)

)( m⊗n′

φN(m⊗n)

)= φN(m⊗n)

= φN(M⊗ (0 1ΣN))(

m⊗(

n′

n

))shows that the right-hand square in (?) is commutative. A similar simple computa-tion shows that the left-hand square is commutative.

4.1.18 Proposition. Let N be an R-complex. The functor –⊗R N is a Σ-functor; inparticular, there is an isomorphism of k-complexes,

Cone(α⊗R N) ∼= (Coneα)⊗R N ,

for every morphism α of Ro-complexes.



4.2 Quasi-isomorphisms 133

EXERCISES

E 4.1.1 Let α : M→ N be a homomorphism of modules; examine the homology of Coneα.E 4.1.2 Let α be a morphism of complexes; show that one has Cone(Σα)∼= ΣConeα.E 4.1.3 Examine for x1,x2 ∈ k the mapping cones K = Conexk1 and K′ = ConexK

2 of the homoth-eties xk1 and xK

2 . Compare with the Koszul complexes Kk(x1) and Kk(x1,x2); see 2.1.24.E 4.1.4 Show that truncations are not Σ-functors.E 4.1.5 Let α : M→ N be an isomorphism of complexes. Show that 1Coneα is null-homotopic.

4.2 Quasi-isomorphisms

SYNOPSIS. Quasi-isomorphism; functor that preserves quasi-isomorphisms; semi-simple module.

More often than not, it is the homology of a complex that one is interested in; moreso than the complex itself. For example, it is the homology of the singular chaincomplex S(X) that yields information about holes in the space X , and it is the ho-mology of the Koszul complex K(x) that tells whether the element x is a zerodivisor.

Quasi-isomorphisms are morphisms that preserve homology.

4.2.1 Definition. A morphism α : M→ N in C(R) is called a quasi-isomorphism ifthe induced morphism H(α) : H(M)→ H(N) is an isomorphism.

A quasi-isomorphism is marked by a ‘'’ next to the arrow.

REMARK. Other words for quasi-isomorphism are homology equivalence and homology isomor-phism.

Given a quasi-isomorphism of R-complexes α : M→ N there need not exist amorphism β : N→M with H(β) = H(α)−1. Moreover, for R-complexes M and Nwith H(M) ∼= H(N) there need not exist a quasi-isomorphism M→ N or N → M.Examples follow below.

4.2.2 Example. There is a quasi-isomorphism of Z-complexes

0 // Z2//

Z //

0

0 // 0 // Z/2Z // 0 ,

but there is not even a non-zero morphism in the opposite direction, as the zero mapis the only homomorphism from Z/2Z to Z.

4.2.3 Example. Set R = k[x,y]. The complexes

M = 0−→ R/(x)y−−→ R/(x)−→ 0

N = 0−→ R/(y) x−−→ R/(y)−→ 0

concentrated in degrees 1 and 0 have isomorphic homology H(M)∼= k∼= H(N), butthere are no non-zero morphisms between them and hence no quasi-isomorphism.



Quasi-isomorphisms between complexes allow for considerable leeway in thestructure of the underlying graded modules. Via quasi-isomorphisms one can thushope to replace a complex by one with better properties.

A quasi-isomorphism between complexes that originate from independent con-structions may have conceptual significance. E.g. for a smooth real manifold M it isa theorem that the embedding S∞(M) S(M) from 2.1.33 is a quasi-isomorphism;see [55]. If M is paracompact, then the morphism Ω(M)→ HomZ(S∞(M),R) from2.1.33 is also a quasi-isomorphism; piecing these two results together one arrivesat de Rham’s theorem: The de Rham cohomology, which is defined in terms of thesmooth structure on M, is isomorphic to the singular cohomology of M, whose def-inition involves only the structure of M as a topological space.

Soft truncations induce quasi-isomorphisms.

4.2.4. Let M be an R-complex. By 2.5.20 the embedding τMĚn : MĚnM is a quasi-

isomorphism for every n 6 infM, and the canonical morphism τMĎn : MMĎn is a

quasi-isomorphism for every n> supM.

Quasi-isomorphisms have the following convenient two-out-of-three property.

4.2.5 Proposition. Consider a commutative diagram in C(R) with exact rows,

0 // M′ //

ϕ′

M //

ϕ

M′′ //

ϕ′′

0

0 // N′ // N // N′′ // 0 .

If two of the morphisms ϕ′, ϕ, and ϕ′′ are quasi-isomorphisms, then so is the third.

PROOF. Immediate from 2.2.21 and the Five Lemma 2.1.40.

4.2.6 Proposition. Let 0 −→ M′ α′−→ M α−→ M′′ −→ 0 be an exact sequence of R-complexes. The following assertions hold.

(a) The complex M′ is acyclic if and only if α is a quasi-isomorphism.(b) The complex M′′ is acyclic if and only if α′ is a quasi-isomorphism.

PROOF. Immediate from the exact sequence (2.2.19.1) and the definition, 4.2.1, ofa quasi-isomorphism.

Recall from 2.2.12 that a morphism α : M→ N of complexes restricts to a mor-phism on cycles, Z(M)→ Z(N), and to a morphism on boundaries B(M)→ B(N).Surjectivity of a quasi-isomorphism can be detected on boundaries and on cycles.

4.2.7 Lemma. Let α be a quasi-isomorphism of R-complexes. The following con-ditions are equivalent.

(i) α is surjective.(ii) α is surjective on boundaries.

(iii) α is surjective on cycles.(iv) α is surjective on cycles and boundaries.



PROOF. It is immediate from 2.1.27 that a surjective morphism is surjective onboundaries, whence (i) implies (ii). An application of the Snake Lemma 2.1.42 tothe diagram (2.2.14.1) shows that conditions (ii) and (iii) are equivalent and, there-fore, that they both imply (iv). To prove that (iv) implies (i), apply the Snake Lemmato the following commutative diagram in C(R),

0 // Z(M) //

α

M∂M//

α

ΣB(M) //

Σα

0

0 // Z(N) // N ∂N// ΣB(N) // 0 .

REMARK. Injectivity of a quasi-isomorphism can, by a result dual to 4.2.7, be detected on bound-aries and on cokernels; see E 4.2.6.

PRESERVATION OF QUASI-ISOMORPHISMS

4.2.8 Example. The shift functor, Σ, preserves quasi-isomorphisms.

Recall from 2.5.19 that soft truncations are functors.

4.2.9 Proposition. For every integer n, the functors (–)Ďn and (–)Ěn preserve quasi-isomorphisms.

PROOF. Let α : M→ N be a quasi-isomorphism. In each degree v > n one hasHv(αĚn) = Hv(α), which is an isomorphism by assumption. For v < n the homo-morphisms Hv(αĚn) are isomorphisms as the truncated complexes are zero in thosedegrees. A similar argument shows that (–)Ďn preserves quasi-isomorphisms.

Several categorical constructions preserve quasi-isomorphisms.

4.2.10 Proposition. Let αu : Mu→ Nuu∈U be a family of morphisms in C(R). Ifαu is a quasi-isomorphism for every u ∈ U , then the coproduct

∐u∈U α

u and theproduct

∏u∈U α

u are quasi-isomorphisms.

PROOF. Homology preserves coproducts and products by 3.1.10 and 3.1.22, and theassertions now follow as a (co)product of isomorphisms is an isomorphism.

4.2.11 Proposition. Let αu : Mu→ Nuu∈U be a morphism of U-direct systems inC(R). If U is filtered and αu is a quasi-isomorphism for every u∈U , then the colimitcolimu∈U α

u is a quasi-isomorphism.

PROOF. A colimit of isomorphisms is an isomorphism. Homology, as a functor,preserves filtered colimits by 3.2.34, and the assertion now follows from 3.2.14.

4.2.12 Proposition. Let κu : Mu→Mu−1u∈Z and λu : Nu→ Nu−1u∈Z be towersin C(R), and let αu : Mu→ Nuu∈Z be a morphism of towers. If for all u ∈ Z themorphisms κu, λu, H(κu), and H(λu) are surjective and αu is a quasi-isomorphism,then the limit limu∈Zα

u is a quasi-isomorphism.



PROOF. A limit of isomorphisms in C(R) is an isomorphism. The canonical mor-phisms H(limu∈ZNu)→ limu∈ZH(Nu) and H(limu∈ZMu)→ limu∈ZH(Mu) are iso-morphisms by 3.3.46, and the assertion now follows from 3.3.16.

A functor F that commutes with homology clearly preserves quasi-isomorphisms.For ease of reference, we spell out an important special case.

4.2.13 Example. It is immediate from 2.2.18 that a functor on complexes extendedfrom an exact functor on modules preserves quasi-isomorphisms.

The Hom and tensor product functors do not commute with homology, and theyalso fail to preserve quasi-isomorphisms. Indeed, let α be the quasi-isomorphismof Z-complexes from 4.2.2. Not one of the induced morphism HomZ(α,Z/2Z),HomZ(Z/2Z,α), or α⊗Z Z/2Z is a quasi-isomorphism.

In the next section we shall see that there are non-trivial quasi-isomorphismsα with the property that every morphism Hom(α,X), Hom(X ,α), and α⊗ X is aquasi-isomorphism. Complexes X with the property that, say, Hom(α,X) is a quasi-isomorphism for every quasi-isomorphism α also exist; they are studied in Chap. 5.

THE MAPPING CONE OF A QUASI-ISOMORPHISM

4.2.14. Let α : M→ N be a morphism of R-complexes and consider the mappingcone sequence from 4.1.5,

0−→ N

(1N

0


(0 1ΣM )−−−−−→ ΣM −→ 0 .

By (2.2.19.1) there is an induced exact sequence,

(4.2.14.1) H(N)−→ H(Coneα)−→ H(ΣM)ð−−→ ΣH(N)−→ ΣH(Coneα) ,

where ð is the connecting morphism in homology. Note that for [z] in H(ΣM) onehas z = (0 1ΣM)(0 z)T and

ςConeα1 ∂Coneα

(0z

)=

(ςN

1 00 ςΣM

1

)(∂N αςΣM

−10 ∂ΣM

)(0z

)=

(ςN

1 αςΣM−1 (z)0

)= Σ

(1N

0

)((Σα)(z)) ,

and consequently, ð([z]) = [(Σα)(z)] by 2.2.19. Thus, one has ð= H(Σα) = ΣH(α).

The exact sequence (4.2.14.1) establishes a crucial connection between preser-vation of homology and vanishing of homology.

4.2.15 Theorem. A morphism α of R-complexes is a quasi-isomorphism if and onlyif the complex Coneα is acyclic.



PROOF. For a morphism α : M→ N the sequence (4.2.14.1) shows that Coneα isacyclic, i.e. H(Coneα) = 0, if and only if ð= ΣH(α) is an isomorphism.

SEMI-SIMPLE MODULES

A graded R-module is called semi-simple if every graded submodule is a gradeddirect summand.

4.2.16 Proposition. Let M be an R-complex. If the graded R-module M\ is semi-simple, then there is an isomorphism of R-complexes M ∼= H(M)⊕Cone1B(M).

PROOF. Set Z = Z(M), B = B(M), and H = H(M); recall from 2.2.11 the shortexact sequences of R-complexes,

0−→ B ι−−→ Z π−−→ H −→ 0 and 0−→ Z ε−−→MςB

1 ∂M

−−−→ ΣB−→ 0 .

By assumption M\ is semi-simple, and hence so is the graded submodule Z\. Itfollows that both sequences are degreewise split. In particular, there are morphismsσ : H\→ Z\ and τ : ΣB\→M\ with πσ= 1H\

and ςB1 ∂

Mτ= 1ΣB\. Thus, the map

H\⊕Cone1B\ = H\⊕B\⊕ΣB\ −→M\

given by (εσ ει τ) is an isomorphism of graded R-modules. In fact, it is an isomor-phism of complexes as one has

∂M (εσ ει τ)=(0 0 ∂Mτ

)=(εσ ει τ

)0 0 00 0 ςΣB

−10 0 0

=(εσ ει τ

)∂H⊕Cone1B

.

4.2.17 Corollary. Assume that R is semi-simple. For every R-complex M there arequasi-isomorphisms H(M)→M and M→ H(M).

PROOF. Follows from 4.2.16 as the complex Cone1B(M) is acyclic by 4.2.15.

EXERCISES

E 4.2.1 Show that a homomorphism of modules is an isomorphism if and only if it is a quasi-isomorphism when considered as a morphism of complexes.

E 4.2.2 Let α, β, and γ be morphisms of R-complexes. Show that if αβ and βγ are quasi-isomor-phisms, then α, β, and γ are quasi-isomorphisms.

E 4.2.3 Show that a sequence of modules 0−→M′ α′−→M α−→M′′ −→ 0 with αα′ = 0 is exact if

one of the morphisms of complexes defined by the diagrams

0 //

M′ //

α′

0

0 // Mα// M′′ // 0

and0 // M′

α′//

M

α

// 0

0 // M′′ // 0

is a quasi-isomorphism and only if they both are quasi-isomorphisms.



E 4.2.4 Show that the Z/4Z-complexes 0→ Z/4Z 2−→ Z/4Z→ 0 and 0→ Z/2Z 0−→ Z/2Z→ 0have isomorphic homology but that there is no quasi-isomorphism in either direction.

E 4.2.5 Show that there are surjective quasi-isomorphisms from the Koszul complex K =KR(x,y)to each of the complexes M and N in 4.2.3. Decide if there are quasi-isomorphisms in theopposite direction M→ K and N→ K.

E 4.2.6 Let α be a quasi-isomorphism of R-complexes. Show that the following conditions areequivalent: (i) α is injective; (ii) α is injective on boundaries; (iii) α is injective on coker-nels; (iv) α is injective on boundaries and on cokernels.

E 4.2.7 Show that the Koszul complexes KZ(2,3) and KZ(4,5) are acyclic. Decide if there is anon-zero quasi-isomorphism KZ(2,3)→ KZ(4,5) or KZ(4,5)→ KZ(2,3); cf. E 2.1.5.

E 4.2.8 Let x1, . . . ,xm and y1, . . . ,yn be elements in k. Show that if the Koszul complexesKk(x1, . . . ,xm) and Kk(y1, . . . ,yn) have isomorphic homology, then they are acyclic orone has (x1, . . . ,xm) = (y1, . . . ,yn) and m = n.

E 4.2.9 Show that a graded R-module M is semi-simple if and only if each Mv is semi-simple.

4.3 Homotopy Equivalences

SYNOPSIS. Homotopy equivalence; functor that preserves homotopy; contractible complex; map-ping cylinder (sequence).

4.3.1 Definition. A morphism of R-complexes α : M→ N is called a homotopy equi-valence if there is a morphism β : N→M such that 1M− βα and 1N −αβ are null-homotopic, that is, βα∼ 1M and αβ∼ 1N ; such a morphism β is called a homotopyinverse of α. A homotopy equivalence is marked by a ‘u’ next to the arrow.

If there exists a homotopy equivalence α : M→ N, then the complexes M and Nare called homotopy equivalent.

4.3.2 Example. Let M be an R-module and v be an integer. Consider the disc com-plex from 2.5.24; the morphism Dv(M)→ 0 is a homotopy equivalence by 2.2.24.

A homotopy inverse is unique up to homotopy.

4.3.3 Lemma. Let α : M→ N and β,β′ : N→M be morphisms in C(R). If 1M ∼ βαand 1N ∼αβ′ hold, then β and β′ are homotopy inverses of α and they are homotopic.

PROOF. By 2.2.25 one has β = β1N ∼ βαβ′ ∼ 1Mβ′ = β′. Another application of2.2.25 yields 1N ∼ αβ′ ∼ αβ, so β is a homotopy inverse of α; a similar computationshows that β′ is a homotopy inverse of α.

4.3.4. Let α : M→ N be a morphism of R-complexes. If α is an isomorphism, thenα−1 is a homotopy inverse of α. If α is a homotopy equivalence with homotopyinverse β, then 2.2.26 implies that H(α) is an isomorphism with inverse H(β).Thus, every homotopy equivalence is a quasi-isomorphism. Notice that the quasi-isomorphism in 4.2.2 is not a homotopy equivalence.

Homotopy equivalences constitute an important class of quasi-isomorphisms;one reason is that they are often simpler to detect than general quasi-isomorphisms.


4.3 Homotopy Equivalences 139

Indeed, to confirm that a morphism is a quasi-isomorphism, one has to computehomology: either to check directly that the induced morphism in homology is anisomorphism or to check that the mapping cone is acyclic. To detect a homotopyequivalence, one only has to find a homotopy inverse.

A quasi-isomorphism between complexes with different levels of complicationcan simplify the task of computing homology. To compute the singular homology of,say, some convex subset X of Rn directly from the definition could be complicated.The next example, however, shows that the singular chain complexes S(pt) andS(X) are homotopy equivalent, in particular they have isomorphic homology, andthe singular homology of the one-point space pt was computed with ease in 2.2.9.

4.3.5 Example. Let X be a contractible topological space. That is, there is a pointx0 in X such that the identity map 1X : X → X and the constant map c : X → X withvalue x0 are homotopic, meaning that there is a continuous map H : X× [0,1]→ Xwith H(x,0) = x and H(x,1) = x0 for all x ∈ X . The continuous map π : X →x0induces a morphism of singular chain complexes, see 2.1.25 and 2.2.9,

S(X) = · · ·

π∗

// Z〈C(∆2,X)〉

π–

∂X2// Z〈C(∆1,X)〉

π–

∂X1// Z〈C(∆0,X)〉

π–

// 0

S(x0) = · · · // Z〈σ2〉∂x02

// Z〈σ1〉∂x01

// Z〈σ0〉 // 0 .

The morphism π∗ is a homotopy equivalence with homotopy inverse ι∗ induced bythe embedding ι : x0 X . In particular, the spaces X and x0 have isomorphicsingular homology by 4.3.4. Moreover, it follows from 2.2.9 and 4.3.2 that the com-plex S(x0), and hence also S(X), is homotopy equivalent to the complex with Zin degree 0 and zero elsewhere.

To describe the functors that preserve homotopy equivalences, we use a technicalconstruction known as the mapping cylinder. Before starting down that road wenotice that homotopy equivalences are preserved by (co)products.

4.3.6 Proposition. Let αu : Mu→ Nuu∈U be a family or morphisms in C(R). Ifαu is a homotopy equivalence for every u ∈U , then the coproduct

∐u∈U α

u and theproduct

∏u∈U α

u are homotopy equivalences.

PROOF. Let βu be a homotopy inverse to αu. It follows from 3.1.6 that the morphism∐u∈U (1Mu −βuαu) is null-homotopic, and by 3.1.4 one has∐

u∈U(1Mu −βuαu) =

∐u∈U

1Mu −∐

u∈Uβuαu = 1

∐u∈U Mu − (

∐u∈U

βu)(∐

u∈Uαu) .

By symmetry, the morphism 1∐

u∈U Nu − (∐

u∈U αu)(∐

u∈U βu) is null-homotopic, so∐

u∈U αu is a homotopy equivalence with homotopy inverse

∐u∈U β

u.A parallel argument based on 3.1.18 and 3.1.16 shows that the morphism

∏u∈U α

u

is a homotopy equivalence.



THE MAPPING CYLINDER

4.3.7 Definition. Let α : M→ N be a morphism of R-complexes. The mappingcylinder of α is the complex with underlying graded module

(Cylα)\ =

N\

⊕ΣM\

⊕M\

and differential ∂Cylα =

∂N αςΣM−1 0

0 ∂ΣM 00 −ςΣM

−1 ∂M

.

4.3.8. A straightforward computation shows that ∂Cylα is square zero,

∂Cylα∂Cylα =

∂N∂N ∂NαςΣM−1 +αςΣM

−1 ∂ΣM 0

0 ∂ΣM∂ΣM 00 −ςΣM

−1 ∂ΣM−∂MςΣM

−1 ∂M∂M

= 0 ;

it uses that α is a morphism and that ςΣM−1 is a degree −1 chain map.

4.3.9. For every morphism of R-complexes α : M→ N there is a degreewise splitexact sequence of R-complexes, known as the mapping cylinder sequence,

(4.3.9.1) 0−→M

( 00

1M

)−−−−→ Cylα

(1N 0 00 1ΣM 0

)−−−−−−−−→ Coneα−→ 0 .

Indeed, the embedding of M into Cylα is evidently a morphism, and an elementarycomputation shows that the surjection onto Coneα is a morphism as well.

4.3.10 Lemma. For every short exact sequence 0 −→ M α−→ Nβ−→ X −→ 0 of R-

complexes, there is a commutative diagram with exact rows,

0 // Mι=

( 00

1M

)// Cylα

π=

(1N 0 00 1ΣM 0

)//

α=(1N 0 α)

Coneα //

β=(β 0)

0

0 // M α// N

β// X // 0 .

The morphism α is a homotopy equivalence with homotopy inverse ε = (1N 0 0)T

and β is a quasi-isomorphism.

PROOF. The upper row is the exact sequence (4.3.9.1) and the lower row is exact byassumption. The next computations show that α and β are morphisms. One has,

∂N (1N 0 α)=(∂N 0 ∂Nα

)=(∂N 0 α∂M) = (

1N 0 α)∂N αςΣM

−1 00 ∂ΣM 00 −ςΣM

−1 ∂M

and

∂X (β 0)=(∂Xβ 0

)=(β∂N 0

)=(β 0)(∂N αςΣM

−10 ∂ΣM

).



As the composite βα is zero, it is evident that the diagram is commutative. It isevident that ε is a morphism; to see that it is a homotopy inverse of α, consider thedegree 1 homomorphism,

% : Cylα −→ Cylα given by

0 0 00 0 ςM

10 0 0

.

A simple calculation yields

∂Cylα%+%∂Cylα =

0 0 α0 −1ΣM 00 0 −1M

= εα−1Cylα ;

together with the identity αε= 1N it shows that ε is a homotopy inverse of α.Note that β is surjective with (Ker β)\ = (Imα)\⊕ΣM\. Let ϑ be the inverse to α

considered as an isomorphism M→ Imα. The degree 1 homomorphism,

σ : Ker β −→ Ker β given by(

0 0ςM

1 ϑ 0

),

is a contraction of Ker β. Indeed, there are equalities

∂Ker βσ+σ∂Ker β =

(αϑ 0

∂ΣMςM1 ϑ+ςM

1 ϑ∂N ςM

1 ϑαςΣM−1

)= 1Ker β ;

the last one follows as ςM1 ϑ is a chain map of degree 1 and one has αϑ = 1Imα and

ϑα= 1ΣM . Thus, Ker β is contractible, in particular acyclic, whence it follows from4.2.6 that β is a quasi-isomorphism.

REMARK. If the exact sequence 0−→M α−→ Nβ−→ X −→ 0 is degreewise split, then the morphism

β in 4.3.11 i a homotopy equivalence; see E 4.3.20.

4.3.11 Proposition. For every commutative diagram of R-complexes,

0 // M α//

ϕ

Nβ//

ψ

X //

χ

0

0 // M′ α′// N′

β′// X ′ // 0 ,

with exact rows, there is a commutative diagram with exact rows,

0 // M ι//

ϕ

Cylα π//(

ψ 0 00 Σϕ 00 0 ϕ

)= α

αu

Coneα //(ψ 00 Σϕ

)= β

β'

0

0 // M′ ι′// Cylα′ π′

//

α′u

Coneα′ //

β′'

0

0 // M α//

ϕ

Nβ

//

ψ

X //

χ

0 ,

0 // M′ α′// N′

β′// X ′ // 0



where the upper rows and vertical morphisms are as in 4.3.10.

PROOF. It is straightforward to verify that the maps α and β are morphisms ofR-complexes, and the “top” is evidently a commutative diagram. The “front” and“back” are commutative diagrams by 4.3.10, and the “bottom” is a commutative di-agram by assumption. It follows immediately from the definitions of α, α′, β, and β′

that the “walls” are commutative as well.

FUNCTORS THAT PRESERVE HOMOTOPY

4.3.12 Definition. A functor F: C(R)→ C(S) is said to preserve homotopy if forevery pair of morphisms α∼ β in C(R) one has F(α)∼ F(β) in C(S).

Preserving the homotopy relation is equivalent to preserving homotopy equiva-lences, and for an additive functor it is the same as preserving null-homotopy.

4.3.13 Proposition. For every functor F: C(R)→ C(S), conditions (i) and (ii) beloware equivalent; furthermore if F is additive, then they are equivalent to (iii).

(i) F preserves homotopy.(ii) For every homotopy equivalence α in C(R) the morphism F(α) in C(S) is a

homotopy equivalence.(iii) For every null-homotopic morphism α in C(R) the morphism F(α) in C(S) is

null-homotopic.

PROOF. (i)=⇒ (ii): Let α : M→ N be a homotopy equivalence with homotopy in-verse β. The assumptions yield F(β)F(α)∼ 1F(M) and F(α)F(β)∼ 1F(N). Thus, F(α)is a homotopy equivalence with homotopy inverse F(β).

(ii)=⇒ (i): First, consider the following morphisms in C(R),

M

ε=

(1M

00

)//

ι=

( 00

1M

) // Cyl(1M)π=(1M 0 1M )

// M .

Note that one has πε= 1M = πι and, consequently, F(π)F(ε) = 1F(M) = F(π)F(ι) inC(S). The morphism π is by 4.3.10 a homotopy equivalence, and hence so is F(π)by the assumption. Now, 4.3.3 yields F(ε)∼ F(ι).

Let α,β : M→ N be homotopic morphisms of R-complexes and % be a homotopyfrom α to β; that is, one has α− β = ∂N%+ %∂M . The degree 0 homomorphismγ = (α %ςΣM

−1 β) : Cyl(1M)→ N is a morphism by the following computation,

∂N(α %ςΣM−1 β

)=(α∂M αςΣM

−1 +%ςΣM−1 ∂

ΣM−βςΣM−1 β∂M

)=(α %ςΣM

−1 β)∂M ςΣM

−1 00 ∂ΣM 00 −ςΣM

−1 ∂M

,



which uses 2.2.4. Finally, from the equalities α= γε and γι= β and from 2.2.25 onegets F(α) = F(γ)F(ε)∼ F(γ)F(ι) = F(β).

Assuming that F is additive, it follows immediately from the definition 2.2.23 of’∼’ that conditions (i) and (iii) are equivalent.

4.3.14 Definition. A morphism M′→M in C(R)op is called null-homotopic (a quasi-isomorphism, a homotopy equivalence) if the corresponding morphism M→M′ inC(R) is null-homotopic (a quasi-isomorphism, a homotopy equivalence) as definedin 2.2.23 (in 4.2.1, in 4.3.1). Accordingly, morphisms α,β : M′→M in C(R)op arecalled homotopic, and we write α∼ β, if the corresponding morphisms M→M′ inC(R) are homotopic per 2.2.23.

A functor G: C(R)op→ C(S) is said to preserve homotopy if for every pair ofmorphisms α∼ β in C(R)op one has G(α)∼ G(β) in C(S).

4.3.15 Proposition. For every functor G: C(R)op→ C(S), conditions (i) and (ii)below are equivalent; furthermore if G is additive, then they are equivalent to (iii).

(i) G preserves homotopy.(ii) For every homotopy equivalence α in C(R)op the morphism G(α) in C(S) is a

homotopy equivalence.(iii) For every null-homotopic morphism α in C(R)op the morphism G(α) in C(S)

is null-homotopic.


4.3.16 Example. The shift functor, Σ, preserves homotopy.

4.3.17 Example. The homology functor, H, preserves homotopy by 2.2.26.

4.3.18 Example. A functor on complexes that is extended from an additive functoron modules, as described in 2.1.45, preserves homotopy.

See 4.3.28 for examples of functors do not preserve homotopy.

4.3.19 Proposition. Let M and N be R-complexes. The functors HomR(M,–) andHomR(–,N) preserve homotopy.

PROOF. Immediate from 2.3.8.

4.3.20 Proposition. Let M be an Ro-complex and let N be an R-complex. The func-tors M⊗R – and –⊗R N preserve homotopy.


4.3.21 Proposition. For all n ∈ Z, the functors (–)Ďn and (–)Ěn preserve homotopy.

PROOF. Let α : M→ N be null-homotopic and let σ : M→ N be a degree 1 homo-morphism with α = ∂Nσ+σ∂M . The degree 1 homomorphism σ : MĚn→ NĚn de-fined by σv = σv for v > n and σn = σn|Zn(M), and of course σv = 0 for v < n,satisfies αĚn = ∂NĚnσ+ σ∂MĚn ; thus αĚn is null-homotopic.

A similar argument shows that αĎn is null-homotopic.



CONTRACTIBLE COMPLEXES

An acyclic complex A is characterized by the unique morphism A→ 0 being a quasi-isomorphism. Next we consider complexes for which it is a homotopy equivalence.

4.3.22 Definition. An R-complex M is called contractible if the identity morphism1M is null-homotopic; a homotopy between 1M and 0 is called a contraction of M.

REMARK. Other words for contractible are split and homotopically trivial.

4.3.23 Example. For every R-module M and every integer v, the disc complexDv(M) is contractible; see 4.3.2.

The disc complexes are atomic contractible complexes in the sense that everysuch complex a coproduct of countably many disc complexes. By 4.3.6 a coproductof disc complexes is contractible, and the converse comes in 4.3.32; cf. 4.1.4.

4.3.24. The complex L constructed in 2.5.24 is a coproduct∐

v∈ZDv(Fv). Thus, forevery R-complex M there is by 2.5.25 a surjective morphism L→ M where L is acontractible complex of free R-modules.

4.3.25 Example. Let x1,x2 be elements in k with (x1,x2) = k and choose l1, l2 ∈ kwith l1x1 + l2x2 = 1. Set K = Kk(x1,x2); cf. 2.1.24. The degree 1 homomorphismσ : K→ K whose non-zero components σ0 and σ1 are given by

1 7−→ l1e1 + l2e2 ande1 7−→ −l2e1∧ e2

e2 7−→ l1e1∧ e2

satisfies ∂Kσ+σ∂K = 1K , whence K is contractible.

4.3.26. If M is a contractible complex, then one has 1H(M) = H(1M) = 0, see 2.2.26,so M is acyclic.

4.3.27. Every additive functor F: C(R)→ C(S), or G: C(R)op→ C(S), that pre-serves homotopy preserves contractability of complexes; see 4.3.13 and 4.3.15.

4.3.28 Example. The functors B,C,Z: C(R)→ C(R), see 2.2.12, are k-linear butdo not preserve homotopy. Indeed, let M 6= 0 be an R-module and consider thecontractible complex D = D1(M) from 4.3.23. The complexes B(D) = Z(D) = Mand C(D) = ΣM are not acyclic and hence not contractible by 4.3.26. The assertionnow follows from 4.3.27.

The Hom functor not only preserves contractability, it detects it in a strong sense.

4.3.29 Proposition. For an R-complex M, the following conditions are equivalent.(i) M is contractible.

(ii) HomR(K,M) is contractible for every R-complex K.(iii) HomR(K,M) is acyclic for every R-complex K.(iv) HomR(M,N) is contractible for every R-complex N.



(v) HomR(M,N) is acyclic for every R-complex N.(vi) HomR(M,M) is acyclic.

(vii) H0(HomR(M,M)) = 0.

PROOF. Condition (i) implies (ii), and (iv) by 4.3.19. The implications (ii)=⇒ (iii)and (iv)=⇒ (v) are evident, see 4.3.26, and (iii)=⇒ (vi), (v)=⇒ (vi), and (vi)=⇒ (vii)are trivial. It follows from (vii) that 1M is a boundary in HomR(M,M), whence it isnull-homotopic; see 2.3.3. This proves (vii)=⇒ (i).

REMARK. Given a contractible R-complex M it follows from 4.3.20 that N⊗R M is contractiblefor every Ro-complex N, but the converse fails. Indeed, let 0→M2→M1→M0→ 0 be a sequencein M(R) that is pure exact but not split. Considered as a complex M of R-modules it is thus acyclicbut not contractible, see 2.2.27. For every Ro-complex N the k-complex N⊗R M is acyclic by E.2and A.14, so if k is semi-simple, e.g. a field, then N⊗R M is contractible; see E 4.3.3.

THE MAPPING CONE OF A HOMOTOPY EQUIVALENCE

Homotopy equivalences are a robust type of quasi-isomorphisms and their mappingcones are likewise acyclic for a prominent reason.

4.3.30 Theorem. A morphism α of R-complexes is a homotopy equivalence if andonly if the complex Coneα is contractible.

PROOF. Let α : M→ N be a morphism and set C = Coneα.If α is a homotopy equivalence, then by 4.3.19 so is HomR(C,α). In particular,

HomR(C,α) is a quasi-isomorphism by 4.3.4, and it follows from 4.2.15 that thecomplex ConeHomR(C,α) is acyclic. Since HomR(C,C)∼= ConeHomR(C,α) holdsby 4.1.15, the complex HomR(C,C) is acyclic; thus C is contractible by 4.3.29.

For the converse, assume that Coneα is contractible and let υ : Coneα→ Coneαbe a degree 1 homomorphism with 1Coneα = ∂Coneαυ+υ∂Coneα. It has the form

υ =

(ν στ Σµ

)for degree 1 homomorphisms ν : N→ N,σ : ΣM→ N, τ : N→ ΣM, and µ : M→M.There are equalities,(

1N 00 1ΣM

)=

(∂N αςΣM

−10 ∂ΣM

)(ν στ Σµ

)+

(ν στ Σµ

)(∂N αςΣM

−10 ∂ΣM

)=

(∂Nν+αςΣM

−1 τ+ ν∂N ∂Nσ+αςΣM−1 Σµ+ ναςΣM

−1 +σ∂ΣM

∂ΣMτ+τ∂N ∂ΣM Σµ+ταςΣM−1 +(Σµ)∂ΣM

).

Comparison of entries yields

0 = ∂ΣMτ+τ∂N ,

1N = ∂Nν+αςΣM−1 τ+ ν∂N , and

1ΣM = ∂ΣMΣµ+ταςΣM

−1 +(Σµ)∂ΣM .



The first equality shows that τ is a chain map, whence ςΣM−1 τ : N→M is a morphism.

The second equality yields 1N ∼ αςΣM−1 τ. An application of Σ−1 to the third equality

yields 1M = −∂Mµ+ ςΣM−1 τα− µ∂M; that is, 1M ∼ ςΣM

−1 τα, whence the morphismςΣM−1 τ is a homotopy inverse of α.

4.3.31 Corollary. If α is an isomorphism of R-complexes, then the complex Coneαis contractible.

While a morphism with contractible mapping cone need not be an isomorphism,every contractible complex is isomorphic to the mapping cone of an isomorphism.

4.3.32 Proposition. For an R-complex M, the following conditions are equivalent.(i) M is contractible.

(ii) There is a graded R-module N with M ∼= Cone1N .(iii) There exist graded R-modules M′ and M′′ with M\ = M′⊕M′′ and ∂M|M′′ = 0,

and such that ∂M|M′ yields an isomorphism M′ ∼= Σ(∂M(M′))∼= ΣM′′.

PROOF. Condition (ii) implies (i) by 4.3.31.(i)=⇒ (iii): By assumption there is a homomorphism σ : M→M of degree 1

such that ∂Mσ+σ∂M = 1M holds. The endomorphism ε= σ∂M of M\ satisfies

ε2 = (σ∂M)(1M−∂Mσ) = ε and 1M−ε = ∂Mσ ,

whence there is an equality M\ = M′⊕M′′ with M′ = Imε and M′′ = Im∂Mσ. Evi-dently, one has ∂M|M′′ = 0 and, therefore,

M′′ ⊆ B(M) = ∂M(M′) = ∂Mσ∂M(M)⊆ ∂Mσ(M) = M′′ .

It follows that ∂M|M′ is a surjective homomorphism M′ → M′′ of degree −1.To see that ∂M|M′ is injective, let m′ = ε(m) be an element in M′ with 0 =∂M(m′) = ∂Mσ∂M(m). Now one has 0 = σ∂Mσ∂M(m) = ε2(m) = ε(m) = m′. Thus,∂M|M′ : M′→ ΣM′′ is an isomorphism of graded modules.

(iii)=⇒ (ii): It is straightforward to verify that the map given by the assignment(m′,m′′) 7→ (m′′,∂M(m′)) is an isomorphism of R-complexes M→ Cone1M′′ .

EXERCISES

E 4.3.1 Let α, β, and γ be morphisms of R-complexes. Show that if αβ and βγ are homotopyequivalences, then α, β, and γ are homotopy equivalences.

E 4.3.2 Let α be a morphism in K(R). Show that if α considered in K(k) is a quasi-isomorphism,then α is a quasi-isomorphism. If α considered in K(k) is an isomorphism, is α then anisomorphism?

E 4.3.3 Let R be semi-simple. Show that every acyclic R-complex is contractible and concludethat every quasi-isomorphism of R-complexes is a homotopy equivalence.

E 4.3.4 Let f : X → Y be a continuous map of topological spaces. The mapping cone, Cone f , isdefined as the quotient space of (X × [0,1])

⊎Y with respect to the equivalence relation

(x,0)∼ (x′,0) and (x,1)∼ f (x) for all x,x′ ∈ X . Denote by S(–) the singular chain com-plex functor, cf. 2.1.25 and E 2.1.7. Show that the complexes S(Cone f ) and ConeS( f )are homotopy equivalent.


4.4 Standard Isomorphisms for Complexes 147

E 4.3.5 Show that every morphism that is homotopic to a homotopy equivalence is a homotopyequivalence.

E 4.3.6 Let α be a morphism of complexes; show that one has Cyl(Σα)∼= ΣCylα.E 4.3.7 Show that hard truncations do not preserve homotopy.E 4.3.8 Consider a commutative diagram of R-complexes,

0 // M′ //

ϕ′

M //

ϕ

M′′ //

ϕ′′

0

0 // N′ // N // N′′ // 0 ,

with exact rows. Show that if two of the morphisms ϕ′, ϕ, and ϕ′′ are homotopy equiva-lences, the third may not be a homotopy equivalence.

E 4.3.9 (Cf. 4.3.18) Show that every functor on complexes that is extended from an additivefunctor on modules preserves homotopy.

E 4.3.10 Show that an R-complex may be contractible as a k-complex but not as an R-complex.

E 4.3.11 Show that the Z/6Z-complex · · · → Z/6Z 2−→ Z/6Z 3−→ Z/6Z 2−→ ·· · is contractible.E 4.3.12 Let x1, . . . ,xn be elements in k such that (x1, . . . ,xn) = k. Show that the Koszul complex

Kk(x1, . . . ,xn) is contractible.E 4.3.13 Let M be a contractible complex. Show that if M is a complex of projective/injective/flat

modules, then so is the subcomplex B(M) = Z(M) and the quotient complex C(M).E 4.3.14 Let M be an R-complex such that the exact sequence 0→ Z(M)→M→ ΣB(M)→ 0 is

degreewise split. Show that M is contractible if and only if it is acyclic. Conclude that anacyclic complex M is contractible if B(M) = Z(M) is a complex of projective modulesor a complex of injective modules.

E 4.3.15 Let R be left hereditary. (a) Show that every acyclic complex of projective R-modules iscontractible. (b) Show that every acyclic complex of injective R-modules is contractible.

E 4.3.16 Show that a bounded below complex of projective modules is contractible if and only ifit is acyclic.

E 4.3.17 Show that a bounded above complex of injective modules is contractible if and only if itis acyclic.

E 4.3.18 Show that for every complex M there is an injective morphism ι : M→C with C a con-tractible complex and a surjective morphism π : C′→M with C′ a contractible complex.

E 4.3.19 Let α : M→ N be a morphism of complexes. (a) Show that α factors as M ι−→M′ β−→N,where ι is an injective homotopy equivalence and β is surjective. (b) Show that α factorsas M γ−→ N′ π−→ N, where γ is injective and π is a surjective homotopy equivalence.

E 4.3.20 Let 0 −→ M α−→ N β−→ X −→ 0 be a degreewise split exact sequence of R-complexes.Show that the morphism β in 4.3.11 is a homotopy equivalence with homotopy inverse(σ −ςM

1 %∂Nσ)T where % : N→M and σ : X → N are the splitting homomorphisms.

4.4 Standard Isomorphisms for Complexes


Categories and functors were introduced in a 1945 paper by Eilenberg and MacLane[23]. The title “General Theory of Natural Equivalences” is suggestive and MacLanewent on to write: “category” has been defined in order to be able to define “functor”



and “functor” has been defined in order to be able to define “natural transforma-tion” [54, p. 18]. While category theory has evolved to become much more than alanguage of “abstract nonsense”, natural transformations have not lost standing.

The standard isomorphisms from Sect. 1.2 are natural transformations of functorson M(k); in this section they are extended to C(k).

UNITOR AND COUNITOR

4.4.1. For every R-complex M, there are isomorphisms in C(R),

µMR : R⊗R M −→ M given by µM

R (r⊗m) = rm and(4.4.1.1)

εMR : M −→ HomR(R,M) given by εM

R (m)(r) = rm ,(4.4.1.2)

where m ∈ M is a homogeneous element. These isomorphisms are called the uni-tor and counitor, and they are both natural in M. Moreover, if M is a complex ofR–So-bimodules, then these maps are isomorphisms in C(R–So). Finally, as naturaltransformations of functors µR and εR are Σ-transformations.

COMMUTATIVITY

The next construction and the proposition that follows establish a commutativityisomorphism for tensor products of complexes. It is based on, and it extends, theisomorphism from 1.2.2.

4.4.2 Construction. Let M be an Ro-complex and N be an R-complex. The com-mutativity isomorphism for modules 1.2.2 induces a natural isomorphism of gradedk-modules, M⊗R N −→ N⊗Ro M, with the isomorphism in degree v given by

(M⊗R N)v =∐i∈Z

Mi⊗R Nv−i

∐i(−1)(v−i)iυMiNv−i−−−−−−−−−−−−→

∐i∈Z

Nv−i⊗Ro Mi = (N⊗Ro M)v .

This isomorphism is also denoted υMN . For homogeneous elements m ∈M and n ∈N it is given by

(4.4.2.1) υMN(m⊗n) = (−1)|n||m|n⊗m .

Note that (4.4.2.1) agrees with the definition in 1.2.2 for modules M and N.

4.4.3 Proposition. Let M be an Ro-complex and N be an R-complex. The commu-tativity map defined in 4.4.2,

υMN : M⊗R N −→ N⊗Ro M ,

is an isomorphism in C(k), and it is natural in M and N. Moreover, if M is inC(Q–Ro) and N is in C(R–So), then υMN is an isomorphism in C(Q–So). Finally,as a natural transformation of functors, υ is a Σ-transformation in each variable.



PROOF. By construction, υMN is an isomorphism of graded k-modules and naturalin M and N. If M is in C(Q–Ro) and N is in C(R–So), then υMN is a natural isomor-phism of graded Q–So-bimodules. This follows from 1.2.2 and the construction. Forhomogeneous elements m ∈M and n ∈ N one has

υMN(∂M⊗RN(m⊗n))

= υMN(∂M(m)⊗n+(−1)|m|m⊗∂N(n))

= (−1)|n|(|m|−1)n⊗∂M(m)+(−1)|m|+(|n|−1)|m|∂N(n)⊗m

= (−1)|n||m|(∂N(n)⊗m+(−1)|n|n⊗∂M(m))

= (−1)|n||m|(∂N⊗RoM(n⊗m))

= ∂N⊗RoM(υMN(m⊗n)) .

Thus, υMN is a morphism, and hence an isomorphism, of complexes.

4.4.4 Example. For elements x1,x2 ∈ k it is immediate from 2.4.3 and commutativ-ity 4.4.3 that the Koszul complexes KR(x1,x2) and KR(x2,x1) are isomorphic.

ASSOCIATIVITY

4.4.5 Construction. Let M be an Ro-complex, X be a complex of R–So-bimodules,and N be an S-complex. The associativity isomorphism for modules 1.2.3 inducesa natural isomorphism (M⊗R X)⊗S N→M⊗R (X⊗S N) of graded k-modules. Thecomponent in degree v is induced by

∐i∈Z∐

j∈ZωM jXi− jNv−i . Indeed, it maps

((M⊗R X)⊗S N)v =∐i∈Z

(∐j∈Z

M j⊗R Xi− j)⊗S Nv−i ∼=

∐i∈Z

∐j∈Z

(M j⊗R Xi− j)⊗S Nv−i ,

where the isomorphism follows from 3.1.11, isomorphically to

(M⊗R (X⊗S N))v =∐j∈Z

M j⊗R(∐

i∈ZXi− j⊗S Nv−i

) ∼= ∐i∈Z

∐j∈Z

M j⊗R (Xi− j⊗S Nv−i) ,

where the isomorphism follows from 3.1.12. The resulting isomorphism of gradedmodules (M⊗R X)⊗S N→M⊗R (X⊗S N) is also denotedωMXN . On homogeneouselements m ∈M, x ∈ X , and n ∈ N it is given by

(4.4.5.1) ωMXN((m⊗ x)⊗n) = m⊗ (x⊗n) .

Note that (4.4.5.1) agrees with the definition in 1.2.3 for modules M, X , and N.

4.4.6 Proposition. Let M be an Ro-complex, X be a complex of R–So-bimodules,and N be an S-complex. The associativity map defined in 4.4.5,


is an isomorphism in C(k), and it is natural in M, X , and N. Moreover, if M is inC(Q–Ro) and N is in C(S–T o), then ωMXN is an isomorphism in C(Q–T o). Finally,as a natural transformation of functors, ω is a Σ-transformation in each variable.



PROOF. By construction, ωMXN is an isomorphism of graded k-modules and naturalin M, X , and N. If M is in C(Q–Ro) and N is in C(S–T o), then ωMXN is a naturalisomorphism of graded Q–T o-bimodules. This follows from 1.2.3 and the construc-tion. A straightforward computation, similar to the one in the proof of 4.4.3, showsthat ωMXN is a morphism, and hence an isomorphism, of complexes.

For elements x1, . . . ,xn in k the notation Kk(x1)⊗k · · ·⊗kKk(xn) is unambiguousby associativity 4.4.6. The next proposition generalizes 2.4.3.

4.4.7 Proposition. For every sequence x1, . . . ,xn in k, there is an isomorphism ofk-complexes

Kk(x1, . . . ,xn) ∼= Kk(x1)⊗k · · ·⊗kKk(xn) .

PROOF. By induction on n. The case n = 1 is trivial, so it suffices for n > 1 to provethat there is an isomorphism

(?) ψ : Kk(x1, . . . ,xn−1)⊗kKk(xn) −→ Kk(x1, . . . ,xn) .

For clarity, consider Kk(x1, . . . ,xn−1) and Kk(xn) to be generated by the free mod-ules k〈e1, . . . ,en−1〉 and k〈en〉 while Kk(x1, . . . ,xn) is generated by k〈 f1, . . . , fn〉. Let0 6 v 6 n and recall from 2.1.24 that Kkv(x1, . . . ,xn) is a free k-module of rank

(nv

)with basis elements fh1 ∧·· ·∧ fhv . The module in degree v in the left-hand complexin (?) is a direct sum

(Kkv(x1, . . . ,xn−1)⊗k k)⊕ (Kkv−1(x1, . . . ,xn−1)⊗k k〈en〉)

of free k-modules of rank(n−1

v

)and

(n−1v−1

), so of total rank

(nv

). Let ψv be given by

(eh1 ∧·· ·∧ ehv)⊗1 7−→ fh1 ∧·· ·∧ fhv and(eh1 ∧·· ·∧ ehv−1)⊗ en 7−→ fh1 ∧·· ·∧ fhv−1 ∧ fn

and extended by k-linerarity. It is straightforward to verify that ψ = (ψv)06v6n is amorphism of k-complexes, and each component ψv has an obvious inverse, so ψ isan isomorphism.

SWAP

4.4.8 Construction. Let M be an R-complex, X be a complex of R–So-bimodules,and N be an So-complex. By 3.1.23 one has

HomR(M,HomSo(N,X))v =∏i∈Z

HomR(Mi,

∏j∈Z

HomSo(N j,X j+i+v))

∼=∏i∈Z

∏j∈Z

HomR(Mi,HomSo(N j,Xi+ j+v)) ,

and similarly,

HomSo(N,HomR(M,X))v ∼=∏i∈Z

∏j∈Z

HomSo(N j,HomR(Mi,Xi+ j+v)) .



It follows from swap for modules 1.2.4 that the map

HomR(M,HomSo(N,X))−→ HomSo(N,HomR(M,X))

with degree v component induced by∏

i∈Z∏

j∈Z (−1)i jζMiXi+ j+vN j is a natural iso-morphism of graded k-modules; it is denoted by ζMXN . On homogeneous elementsψ ∈ HomR(M,HomSo(N,X)), m ∈M, and n ∈ N it is given by

(4.4.8.1) ζMXN(ψ)(n)(m) = (−1)|m||n|ψ(m)(n) .


4.4.9 Proposition. Let M be an R-complex, X be a complex of R–So-bimodules,and N be an So-complex. The swap map defined in 4.4.8,


is an isomorphism in C(k), and it is natural in M, X , and N. Moreover, if M is inC(R–Qo) and N is in C(T –So), then ζMXN is an isomorphism in C(Q–T o). Finally,as a natural transformation of functors, ζ is a Σ-transformation in each variable.

PROOF. By construction, ζMXN is an isomorphism of graded k-modules and naturalin M, X , and N. If M is in C(R–Qo) and N is in C(T –So), then ζMXN is an isomor-phism of graded Q–T o-bimodules; this follows from the construction and 1.2.4. Forhomogeneous elements ψ ∈ HomR(M,HomSo(N,X)), m ∈M, and n ∈ N one has

ζMXN(∂HomR(M,HomSo(N,X))(ψ))(n)(m)

= ζMXN(∂HomSo(N,X)ψ− (−1)|ψ|ψ∂M)(n)(m)

= (−1)|m||n|(∂HomSo(N,X)(ψ(m))− (−1)|ψ|ψ(∂M(m))

)(n)

= (−1)|m||n|(∂Xψ(m)− (−1)|ψ(m)|ψ(m)∂N− (−1)|ψ|ψ(∂M(m))

)(n)

= (−1)|m||n|(∂X (ψ(m)(n))− (−1)|ψ|+|m|ψ(m)(∂N(n))− (−1)|ψ|ψ(∂M(m))(n)

)and(

∂HomSo(N,HomR(M,X))(ζMXN(ψ)))(n)(m)

=(∂HomR(M,X)ζMXN(ψ)− (−1)|ζ

MXN(ψ)|ζMXN(ψ)∂N)(n)(m)

=(∂HomR(M,X)(ζMXN(ψ)(n))− (−1)|ψ|ζMXN(ψ)(∂N(n))

)(m)

= ∂X (ζMXN(ψ)(n)(m))− (−1)|ζMXN(ψ)(n)|ζMXN(ψ)(n)(∂M(m))

− (−1)|ψ|+|m|(|n|−1)ψ(m)(∂N(n))

= (−1)|m||n|∂X (ψ(m)(n))− (−1)|ψ|+|n|+(|m|−1)|n|ψ(∂M(m))(n)

− (−1)|ψ|+|m||n|−|m|ψ(m)(∂N(n)) .

Thus, ζMXN is a morphism, hence an isomorphism, of complexes.The functors HomR(–,HomSo(–,–)) and HomSo(–,HomR(–,–)) are Σ-functors

in each variable; see 4.1.12. Fix complexes N and X . To verify that ζ–XN is a Σ-transformation, recall from 2.3.13 that the natural isomorphism



ψM1 : Σ−1 HomR(M,HomSo(N,X))→ HomR(ΣM,HomSo(N,X)) ,

upon suppression of the degree shifting chain maps, is given by ϑ 7→ (−1)|ϑ|ϑ fora homogeneous element ϑ in HomR(M,HomSo(N,X)). Similarly, upon suppressionof the degree shifting chain maps, the composite

ψM2 : Σ−1 HomSo(N,HomR(M,X))−→ HomSo(N,Σ−1 HomR(M,X))

−→ HomSo(N,HomR(ΣM,X))

maps a homogeneous element ϕ in HomSo(N,HomR(M,X)) to the homomorphismgiven by n 7→ (−1)|ϕ|+|n|ϕ(n) for n ∈ N. Thus, for homogeneous elements m ∈M,n ∈ N, and ϑ in HomR(M,HomSo(N,X)) one has

(ζ(ΣM)XNψM1 )(ϑ)(n)(m) = ζ(ΣM)XN((−1)|ϑ|ϑ)(n)(m)

= (−1)|ϑ|(−1)(|m|+1)|n|ϑ(m)(n)

= (−1)|ζMXN(ϑ)|+|n|(−1)|m||n|ϑ(m)(n)

= (ψM2 Σ−1 ζMXN)(ϑ)(n)(m) .

Thus, ζ–XN is a Σ-transformation, and similar arguments show that swap is a Σ-transformation in the other variables as well.

ADJUNCTION

4.4.10 Construction. Let M be an R-complex, X be a complex of R–So-bimodules,and N be an S-complex. By 3.1.25 one has

HomR(X⊗S N,M)v =∏h∈Z

HomR(∐

j∈ZX j⊗S Nh− j,Mh+v

)∼=∏h∈Z

∏j∈Z

HomR(X j⊗S Nh− j,Mh+v)

=∏i∈Z

∏j∈Z

HomR(X j⊗S Ni,Mi+ j+v) ,

and by 3.1.23 one has

HomS(N,HomR(X ,M))v =∏i∈Z

HomS(Ni,∏j∈Z

HomR(X j,M j+i+v))

=∏i∈Z

∏j∈Z

HomS(Ni,HomR(X j,Mi+ j+v)) .

It follows from adjunction for modules 1.2.5 that the map

HomR(X⊗S N,M)−→ HomS(N,HomR(X ,M))

with degree v component induced by∏

i∈Z∏

j∈Z (−1)i jρMi+ j+vX jNi is a natural iso-morphism of graded k-modules; this isomorphism is denoted by ρMXN . On homo-geneous elements ψ ∈ HomR(X⊗S N,M), n ∈ N, and x ∈ X it is given by



(4.4.10.1) ρMXN(ψ)(n)(x) = (−1)|x||n|ψ(x⊗n) .


4.4.11 Proposition. Let M be an R-complex, X be a complex of R–So-bimodules,and N be an S-complex. The adjunction map defined in 4.4.10,


is an isomorphism in C(k), and it is natural in M, X , and N. Moreover, if M is inC(R–T o) and N is in C(S–Qo), then ρMXN is an isomorphism in C(Q–T o). Finally,as a natural transformation of functors, ρ is a Σ-transformation in each variable.

PROOF. By construction, ρMXN is an isomorphism of graded k-modules and naturalin M, X , and N. If M is in C(R–T o) and N is in C(S–Qo), then ρMXN is an isomor-phism of graded Q–T o-bimodules; this follows from 1.2.5 and the construction. Forhomogeneous elements ψ ∈ HomR(X⊗S N,M), n ∈ N, and x ∈ X one has

ρMXN(∂HomR(X⊗SN,M)(ψ))(n)(x)

= ρMXN(∂Mψ− (−1)|ψ|ψ∂X⊗SN)(n)(x)= (−1)|x||n|

(∂Mψ(x⊗n)− (−1)|ψ|ψ∂X⊗SN(x⊗n)

)= (−1)|x||n|∂Mψ(x⊗n)− (−1)|ψ|+|x||n|ψ

(∂X (x)⊗n+(−1)|x|x⊗∂N(n)

)and(

∂HomS(N,HomR(X ,M))(ρMXN(ψ)))(n)(x)

=(∂HomR(X ,M)ρMXN(ψ)− (−1)|ρ

MXN(ψ)|ρMXN(ψ)∂N)(n)(x)= ∂M(ρMXN(ψ)(n)(x))− (−1)|ρ

MXN(ψ)(n)|ρMXN(ψ)(n)(∂X (x))

− (−1)|ψ|ρMXN(ψ)(∂N(n))(x)

= (−1)|x||n|∂Mψ(x⊗n)− (−1)|ψ|+|n|+(|x|−1)|n|ψ(∂X (x)⊗n)

− (−1)|ψ|+|x|(|n|−1)ψ(x⊗∂N(n))

= (−1)|x||n|∂Mψ(x⊗n)− (−1)|ψ|+|x||n|ψ(∂X (x)⊗n+(−1)|x|x⊗∂N(n)

).

Thus, ρMXN is a morphism, and hence an isomorphism, of complexes.

EXERCISES

E 4.4.1 Apply 2.4.12 and commutativity 4.4.3 to give a proof of 2.4.13.E 4.4.2 Apply 3.2.16 and commutativity 4.4.3 to give a proof of 3.2.17.E 4.4.3 Let X be a complex of R–So-bimodules. Show that the functor X⊗S –: C(S)→C(R) is

left adjoint for HomR(X ,–).E 4.4.4 Let M be an R-complex and consider the map R→ HomR(M,M) that maps r to multi-

plication by r on M. (a) Show that it is a morphism of k-complexes. (b) Show that it is amorphism of k-algebras.



E 4.4.5 Let X be a complex of R–So-bimodules and consider the map R→ HomSo(X ,X) thatmaps r to multiplication by r on M. Show that it is a morphism in C(R–Ro).

E 4.4.6 Let X be an R–So-bimodule. Show that the functor HomR(–,X) : C(R)op→C(So) is aright adjoint for HomSo(–,X).

E 4.4.7 Let F,G: C(R)→C(R) be functors. Assume that F has a right adjoint F∗ and that thereis a natural isomorphism ζ : F∗G→ GF∗. Show that there is a canonical natural transfor-mation θ : FG→ GF.

Let M and N be k-complexes. Use adjunction 4.4.11 and swap 4.4.9 to show that theresult above applies with F = M⊗k –, G = Homk(N,–), and R = k.

E 4.4.8 Let F: C(R)→C(R) and G: C(R)op→C(R) be functors. Assume that F has a right ad-joint F∗ and that there is a natural isomorphism ρ : GFop→ F∗G. Show that there is acanonical natural transformation η : FG→ GFop

∗ .Let M and N be k-complexes. Use adjunction 4.4.11 to show that the result above

applies with F = M⊗k –, G = Homk(–,N), and R = k.

4.5 Evaluation Morphisms for Complexes


In this section, the evaluation homomorphisms from Sect. 1.4 are extended to mor-phisms of complexes.

BIDUALITY

The next construction and the results that follow it extend 1.4.2 to complexes.

4.5.1 Construction.Let M be an R-complex and X be a complex of R–So-bimodules.For every v ∈ Z one has

HomSo(HomR(M,X),X)v =∏i∈Z

HomSo(∏

j∈ZHomR(M j,Xi+ j),Xi+v

).

To define a map from Mv to HomSo(HomR(M,X),X)v it suffices, in view of 3.1.14,to define, for every integer i, a map

Mv −→ HomSo(∏

j∈ZHomR(M j,Xi+ j),Xi+v

).

This is achieved by postcomposing biduality 1.4.2, adjusted by a sign,

Mv

(−1)ivδMvXi+v−−−−−−−→ HomSo(HomR(Mv,Xi+v),Xi+v) ,

with the map induced by the projection∏

j∈ZHomR(M j,Xi+ j) HomR(Mv,Xi+v).The map of complexes M → HomSo(HomR(M,X),X), defined hereby, is denotedδM

X . It follows from 1.4.2 that it is a natural morphism of graded R-modules. Onhomogeneous elements m ∈M and ψ ∈ HomR(M,X) it is given by

(4.5.1.1) δMX (m)(ψ) = (−1)|ψ||m|ψ(m) .


4.5 Evaluation Morphisms for Complexes 155

Note that (4.5.1.1) agrees with the definition in 1.4.2 for modules M and X .

4.5.2 Proposition. Let X be a complex of R–So-bimodules. For an R-complex Mthe biduality map defined in 4.5.1,

δMX : M −→ HomSo(HomR(M,X),X) ,

is a morphism in C(R), and it is natural in M. Moreover, if M is in C(R–T o), thenδM

X is a morphism in C(R–T o). Finally, as a natural transformation of functors, δ isa Σ-transformation.

PROOF. By construction, δMX is a morphism of graded R-modules and natural in

M. If M is in C(R–T o), then δMX is a morphism of graded R–T o-bimodules; this

follows from 1.4.2 and the construction above. For homogeneous elements m ∈Mand ψ ∈ HomR(M,X) one has(

∂HomSo(HomR(M,X),X)δMX (m)

)(ψ)

=(∂XδM

X (m)− (−1)|δMX (m)|δM

X (m)∂HomR(M,X))(ψ)

= (−1)|ψ||m|∂Xψ(m)− (−1)|m|δMX (m)

(∂Xψ− (−1)|ψ|ψ∂M)

= (−1)|ψ||m|∂Xψ(m)− (−1)|m|+(|ψ|−1)|m|(∂Xψ(m)− (−1)|ψ|ψ∂M(m))

= (−1)|ψ|(|m|−1)ψ∂M(m)

= δMX (∂M(m))(ψ) .

Thus, δMX is a morphism of complexes.

4.5.3 Proposition. Let M be an R-complex and X be a complex of R–So-bimodules.If one module Xp is faithfully injective as an R-module, then biduality 4.5.2,


is injective.

PROOF. It is sufficient to show that δMX is injective on homogeneous elements. Let

m 6= 0 be homogeneous of degree q. By assumption, Xp is faithfully injective asan R-module, whence there is a non-zero homomorphism from the submodule Rmof Mq to Xp. By the lifting property 1.3.22 there is then a homomorphism ψ inHomR(Mq,Xp) with ψ(m) 6= 0. Let ψ : M→ X be the degree p− q homomorphismwith ψq = ψ and ψv = 0 for v 6= q. One now has δM

X (m)(ψ) = ψ(m) = ψ(m) 6= 0, soδM

X (m) is non-zero.

4.5.4 Proposition. For every complex P of finitely generated projective R-modules,HomR(P,R) is a complex of finitely generated projective Ro-modules, and biduality

δPR : P −→ HomRo(HomR(P,R),R)

is an isomorphism.

PROOF. For all v ∈ Z one has HomR(P,R)v = HomR(P−v,R) and (δPR)v = (−1)v2

δPvR ;

see 4.5.1. The assertions now follow from 1.4.3.



TENSOR EVALUATION

The next construction and the results that follow it extend 1.4.4 and 1.4.6 to com-plexes.

4.5.5 Construction. Let M be an R-complex, X be a complex of R–So-bimodules,and N be an S-complex. There are equalities

(HomR(M,X)⊗S N)v =∐i∈Z

(∏h∈Z

HomR(Mh,Xh+i))⊗S Nv−i and(4.5.5.1)

HomR(M,X⊗S N)v =∏j∈Z

HomR(M j,

∐k∈Z

Xk⊗S N j+v−k).(4.5.5.2)

To define a map from (HomR(M,X)⊗S N)v to HomR(M,X⊗S N)v it suffices, inview of 3.1.2 and 3.1.14, to define, for all integers ı and , a map(∏

h∈ZHomR(Mh,Xh+ı)

)⊗S Nv−ı −→ HomR

(M ,

∐k∈Z

X +k⊗S Nv−k).

This is done by pre- and postcomposing tensor evaluation 1.4.4, adjusted by a sign,

HomR(M ,X +ı)⊗S Nv−ı(−1) (v−ı)θM X +ıNv−ı−−−−−−−−−−−−−→ HomR(M ,X +ı⊗S Nv−ı) ,

with the map induced by the projection∏

h∈ZHomR(Mh,Xh+ı) HomR(M ,X +ı)and the map induced by the injection X +ı⊗S Nv−ı

∐k∈ZX +k⊗S Nv−k; cf. 3.1.1

and 3.1.13. The map of complexes defined hereby, HomR(M,X)⊗S N→HomR(M,X⊗S N),is denoted θMXN . It follows from 1.4.4 that it is a natural morphism of graded k-mo-dules. On homogeneous elements ψ∈HomR(M,X), m∈M, and n∈N it is given by

(4.5.5.3) θMXN(ψ⊗n)(m) = (−1)|m||n|ψ(m)⊗n .


4.5.6 Proposition. Let M be an R-complex, X be a complex of R–So-bimodules,and N be an S-complex. The tensor evaluation map defined in 4.5.5,


is a morphism in C(k), and it is natural in M, X , and N. Moreover, if M is in C(R–Qo)and N is in C(S–T o), then θMXN is a morphism in C(Q–T o). Finally, as a naturaltransformation of functors, θ is a Σ-transformation in each variable.

PROOF. By construction, θMXN is a morphism of graded k-modules and natural inM, X , and N. If M is in C(R–Qo) and N is in C(S–T o), then θMXN is a morphism ofgraded Q–T o-bimodules; this follows from 1.4.4 and the construction. For homoge-neous elements ψ ∈ HomR(M,X), m ∈M, and n ∈ N one has(

∂HomR(M,X⊗SN)(θMXN(ψ⊗n)

))(m)



=(∂X⊗SNθMXN(ψ⊗n)− (−1)|θ

MXN(ψ⊗n)|θMXN(ψ⊗n)∂M)(m)

= (−1)|m||n|∂X⊗SN(ψ(m)⊗n)− (−1)|ψ|+|n|+(|m|−1)|n|ψ∂M(m)⊗n

= (−1)|m||n|(∂Xψ(m)⊗n+(−1)|ψ(m)|ψ(m)⊗∂N(n)

)− (−1)|ψ|+|m||n|ψ(∂M(m))⊗n

= (−1)|m||n|∂Xψ(m)⊗n+(−1)|m||n|+|ψ|+|m|ψ(m)⊗∂N(n)

− (−1)|ψ|+|m||n|ψ∂M(m)⊗n

and

θMXN(∂HomR(M,X)⊗SN(ψ⊗n))(m)

= θMXN(∂HomR(M,X)(ψ)⊗n+(−1)|ψ|ψ⊗∂N(n))(m)

= θMXN((∂Xψ− (−1)|ψ|ψ∂M)⊗n+(−1)|ψ|ψ⊗∂N(n))(m)

= (−1)|m||n|∂Xψ(m)⊗n− (−1)|ψ|+|m||n|ψ∂M(m)⊗n

+(−1)|ψ|+|m|(|n|−1)ψ(m)⊗∂N(n) .

These two computations show that θMXN is a morphism of complexes.

4.5.7 Theorem. Let M be an R-complex, X be a complex of R–So-bimodules, andN be an S-complex. Tensor evaluation 4.5.6,


is an isomorphism if the complexes satisfy one of the conditions (1)–(3) and one ofthe conditions (a)–(c).

(1) M is bounded below, and X and N are bounded above.(2) M is bounded above, and X and N are bounded below.(3) Two of the complexes M, X , and N are bounded.(a) M or N is a complex of finitely generated projective modules.(b) M is a complex of projective modules and N is a complex of finitely presented

modules.(c) M is a complex of finitely presented modules and N is a complex of flat mod-

ules.Furthermore, θMXN is an isomorphism if M or N is a bounded complex of finitely

presented modules and one of the following conditions is satisfied.(d) M is a complex of projective modules.(e) N is a complex of flat modules.

PROOF. Under any one of the conditions (a)–(c), each homomorphism θMhXiN j is anisomorphism of modules by 1.4.6. To prove the first assertion, it is now sufficientto show that under each of the boundedness conditions (1)–(3), every component ofθMXN is given by a direct sum of homomorphisms θMhXiN j .

The products and coproducts in (4.5.5.1) and (4.5.5.2) are finite under any one ofthe conditions (1)–(3). Indeed, under (1), assume without loss of generality that one



has Mv = 0 for all v < 0 and Xv = 0 = Nv for all v > 0; cf. 2.3.13, 2.3.15, 2.4.12, and2.4.13. By 2.5.9 and 2.5.13 one has (HomR(M,X)⊗S N)v = 0=HomR(M,X⊗S N)vfor all v > 0. For v6 0 equation (4.5.5.1) yields

(HomR(M,X)⊗S N)v =∐i>v

(−i∏j=0

HomR(M j,X j+i))⊗S Nv−i

∼=0⊕

i=v

−i⊕j=0

HomR(M j,X j+i)⊗S Nv−i ,

and from (4.5.5.2) one gets

HomR(M,X⊗S N)v =∏j>0

HomR(M j,− j∐i=v

X j+i⊗S Nv−i)

∼=−v⊕j=0

− j⊕i=v

HomR(M j,X j+i⊗S Nv−i)

=0⊕

i=v

−i⊕j=0

HomR(M j,X j+i⊗S Nv−i) .

In particular, the vth component of the morphism θMXN is given by

θMXNv =

0⊕i=v

−i⊕j=0

(−1) j(v−i)θM jX j+iNv−i .

Parallel arguments apply under conditions (2) and (3). Thus, θMXN is an isomor-phism when one of (1)–(3) and one of (a)–(c) holds.

If M or N is a bounded complex of finitely presented modules, then under eitherone of the conditions (d) and (e), each homomorphism θMhXiN j is an isomorphismof modules by 1.4.6. To prove the second assertion, it is now sufficient to prove thatevery component of θMXN is a product or a coproduct of homomorphisms θMhXiN j .

First, let M be a bounded complex of finitely presented modules and assumewithout loss of generality that one has Mv = 0 for all v < 0 and for all v > u, forsome u> 0. From (4.5.5.1) one gets

(HomR(M,X)⊗S N)v =∐i∈Z

(u∏

j=0HomR(M j,X j+i))⊗S Nv−i

∼=∐i∈Z

u⊕j=0


and (4.5.5.2) yields

HomR(M,X⊗S N)v =u∏

j=0HomR(M j,

∐i∈Z

X j+i⊗S Nv−i)

∼=∐i∈Z

u⊕j=0

HomR(M j,X j+i⊗S Nv−i) ,

where the isomorphism follows from 3.1.31, as M is bounded and the modules M jare finitely generated. It follows that the vth component of the morphism θMXN isgiven by



θMXNv =

∐i∈Z

u⊕j=0


Finally, let N be a bounded complex of finitely presented modules and assumewithout loss of generality that one has Nv = 0 for all v < 0 and for all v > u, forsome u> 0. Now (4.5.5.1) yields

(HomR(M,X)⊗S N)v =v∐

i=v−u(∏j∈Z

HomR(M j,X j+i))⊗S Nv−i

∼=∏j∈Z

v⊕i=v−u


where the isomorphism follows from 3.1.29, as the modules Nv−i are finitely pre-sented. Further, (4.5.5.2) yields

HomR(M,X⊗S N)v =∏j∈Z

HomR(M j,v∐

i=v−uX j+i⊗S Nv−i)

∼=∏j∈Z

v⊕i=v−u

HomR(M j,X j+i⊗S Nv−i) .

It follows that the vth component of the morphism θMXN is given by

θMXNv =

∏j∈Z

v⊕i=v−u



The next construction and the results that follow it extend 1.4.7 and 1.4.9 to com-plexes.

4.5.8 Construction. Let M be an R-complex, X be a complex of R–So-bimodules,and N be an So-complex. There are equalities

(N⊗S HomR(X ,M))v =∐i∈Z

Ni⊗S∏h∈Z

HomR(Xh,Mh+v−i) and(4.5.8.1)

HomR(HomSo(N,X),M)v =∏j∈Z

HomR(∏

k∈ZHomSo(Nk,Xk+ j),M j+v

).(4.5.8.2)

To define a map from (N⊗S HomR(X ,M))v to HomR(HomSo(N,X),M)v it suffices,in view of 3.1.2 and 3.1.14, to define, for all integers ı and , a map

Nı⊗S∏h∈Z

HomR(Xh,Mh+v−ı)−→ HomR(∏

k∈ZHomSo(Nk,Xk+ ),M +v

).

This is done by precomposing homomorphism evaluation 1.4.7, adjusted by a sign,

Nı⊗S HomR(Xı+ ,M +v)(−1)(v−ı+ )ıηM +vXı+ Nı

−−−−−−−−−−−−−−→ HomR(HomSo(Nı,Xı+ ),M +v) ,

with the map induced by∏

h∈ZHomR(Xh,Mh+v−ı) HomR(Xı+ ,M +v) and post-composing by the map induced by

∏k∈ZHomSo(Nk,Xk+ ) HomSo(Nı,Xı+ ). The



map of complexes, N⊗S HomR(X ,M)→ HomR(HomSo(N,X),M), defined hereby,is denoted ηMXN . It follows from 1.4.7 that it is a natural morphism of graded k-modules. On homogeneous elements n∈N, ψ∈HomR(X ,M), and ϑ∈HomSo(N,X)it is given by

(4.5.8.3) ηMXN(n⊗ψ)(ϑ) = (−1)(|ψ|+|ϑ|)|n|ψϑ(n) .

Note that (4.5.8.3) agrees with the definition in 1.4.7 for modules M, X and N.

4.5.9 Proposition. Let M be an R-complex, X be a complex of R–So-bimodules,and N be an So-complex. The homomorphism evaluation map defined in 4.5.8,


is a morphism in C(k), and it is natural in M, X , and N. Moreover, if M is in C(R–T o)and N is in C(Q–So), then ηMXN is a morphism in C(Q–T o). Finally, as a naturaltransformation of functors, η is a Σ-transformation in each variable.

PROOF. By construction, ηMXN is a morphism of graded k-modules and natural inM, X , and N. If M is in C(R–T o) and N is in C(Q–So), then ηMXN is a morphism ofgraded Q–T o-bimodules; this follows from 1.4.4 and the construction. For homoge-neous elements n ∈ N, ψ ∈ HomR(X ,M), and ϑ ∈ HomSo(N,X) one has(∂HomR(HomSo(N,X),M)(ηMXN(n⊗ψ))

)(ϑ)

=(∂MηMXN(n⊗ψ)− (−1)|η

MXN(n⊗ψ)|ηMXN(n⊗ψ)∂HomSo(N,X))(ϑ)

= (−1)(|ψ|+|ϑ|)|n|∂Mψϑ(n)

− (−1)|n|+|ψ|ηMXN(n⊗ψ)(∂Xϑ− (−1)|ϑ|ϑ∂N)

= (−1)(|ψ|+|ϑ|)|n|∂Mψϑ(n)

− (−1)|n|+|ψ|+(|ψ|+|ϑ|−1)|n|ψ(∂Xϑ(n)− (−1)|ϑ|ϑ∂N(n))

= (−1)(|ψ|+|ϑ|)|n|(∂Mψϑ(n)− (−1)|ψ|ψ∂Xϑ(n)+(−1)|ψ|+|ϑ|ψϑ∂N(n)

)and

ηMXN(∂N⊗SHomR(X ,M)(n⊗ψ))(ϑ)

= ηMXN(∂N(n)⊗ψ+(−1)|n|n⊗∂HomR(X ,M)(ψ))(ϑ)

= (−1)(|ψ|+|ϑ|)(|n|−1)ψϑ∂N(n)+(−1)|n|ηMXN(n⊗ (∂Mψ− (−1)|ψ|ψ∂X ))(ϑ)

= (−1)(|ψ|+|ϑ|)|n|((−1)|ψ|+|ϑ|ψϑ∂N(n)+∂Mψϑ(n)− (−1)|ψ|ψ∂Xϑ(n)

).

These two computations show that ηMXN is a morphism of complexes.

4.5.10 Theorem. Let M be an R-complex, X be a complex of R–So-bimodules, andN be an So-complex. Homomorphism evaluation 4.5.9,


is an isomorphism if the complexes satisfy one of the conditions (1)–(3) and one ofthe conditions (a)–(b).



(1) M and N are bounded below, and X is bounded above.(2) M and N are bounded above, and X is bounded below.(3) Two of the complexes M, X , and N are bounded.(a) N is a complex of finitely generated projective modules.(b) N is a complex of finitely presented modules and M is a complex of injective

modules.Furthermore, ηMXN is an isomorphism if N is a bounded complex of finitely

presented modules and one of the following conditions are satisfied.(c) N is a complex of projective modules.(d) M is a complex of injective modules.

PROOF. Under either condition (a) or (b), each homomorphism ηM jXhNi is an iso-morphism of modules by 1.4.9. To prove the first assertion, it is now sufficient toshow that under any one of the boundedness conditions (1)–(3), every componentof ηMXN is given by a direct sum of homomorphisms ηM jXhNi .

The products and coproducts in (4.5.8.1) and (4.5.8.2) are finite under any oneof the conditions (1)–(3). Indeed, under (1), assume without loss of generalitythat one has Mv = 0 = Nv for all v < 0 and Xv = 0 for all v > 0; cf. 2.3.13,2.3.15, 2.4.12, and 2.4.13. It follows that one has (N⊗S HomR(X ,M))v = 0 =HomR(HomSo(N,X),M)v for all v < 0. For v> 0 equation (4.5.8.1) yields

(N⊗S HomR(X ,M))v =∐

i>0Ni⊗S (

0∏j=i−v

HomR(X j,M j+v−i))

∼=v⊕

i=0

0⊕j=i−v

Ni⊗S HomR(X j,M j+v−i) ,

and from (4.5.8.2) one gets

HomR(HomSo(N,X),M)v =∏

h>−vHomR(

−h∏i=0

HomSo(Ni,Xi+h),Mh+v)

∼=0⊕

h=−v

−h⊕i=0

HomR(HomSo(Ni,Xi+h),Mh+v)

∼=v⊕

i=0

−i⊕h=−v


∼=v⊕

i=0

0⊕j=i−v

HomR(HomSo(Ni,X j),M j+v−i) .

In particular, the vth component of the morphism ηMXN is

ηMXNv =

v⊕i=0

0⊕j=i−v

(−1)( j+v)iηM j+v−iX jNi .

Parallel arguments apply under conditions (2) and (3). Thus, ηMXN is an isomor-phism when one of (1)–(3) and one of (a)–(b) holds.



If M is a bounded complex of finitely presented modules, then under either oneof the conditions (c) and (d), each homomorphism ηMpXqNr is an isomorphism ofmodules by 1.4.9. To prove the second assertion, it is now sufficient to prove thatevery component of ηMXN is given by a product of homomorphisms ηMpXqNr . As-sume without loss of generality that one has Nv = 0 for all v < 0 and for all v > u,for some u> 0. From (4.5.8.1) one gets

(N⊗S HomR(X ,M))v =u∐

i=0(Ni⊗S

∏j∈Z

HomR(X j,M j+v−i))

∼=u⊕

i=0

∏j∈Z

Ni⊗S HomR(X j,M j+v−i) ,

where the isomorphism follows from 3.1.28, as N is a bounded complex of finitelypresented modules. Further, (4.5.8.2) yields

HomR(HomSo(N,X),M)v =∏

h∈ZHomR(

u∏i=0

HomSo(Ni,Xi+h),Mh+v)

∼=u⊕

i=0

∏h∈Z


∼=u⊕

i=0

∏j∈Z

HomR(HomSo(Ni,X j),M j+v−i) .

It follows that the vth component of the morphism ηMXN is

ηMXNv =

u⊕i=0

∏j∈Z

(−1)( j+v)iηM j+v−iX jNi .

EXERCISES

E 4.5.1 Let L be a complex of finitely generated free R-modules. Show that δHomR(L,R)R is an

isomorphism of Ro-complexes with inverse HomR(δLR,R).

E 4.5.2 Let P be a projective R-module, let X be a complex of R–So-bimodules, and let N be acomplex of finitely presented S-modules; show that θPXN is injective.

E 4.5.3 Let N be a complex of finitely presented So-modules, let X be a complex of R–So-bimodules, and let E be an injective R-module; show that ηEXN is injective.

E 4.5.4 Let M be an R-complex. Show that biduality δMR is an isomorphism if and only if homo-

morphism evaluation ηMRR is an isomorphism.E 4.5.5 Let M be an R-complex. Show that if the Ro-complex Homk(M,E) is contractible, then

biduality δME is null-homotopic.

E 4.5.6 Show that the natural transformation M⊗kHomk(N,–)→ Homk(N,M⊗k –) in E 4.4.7is tensor evaluation, up to an application of commutativity 4.4.3.

E 4.5.7 Show that the natural transformation M⊗kHomk(–,N) → Homk(Homk(–,M),N) inE 4.4.8 is homomorphism evaluation.

E 4.5.8 Show that biduality 4.5.2 yields the unit of the adjunction from E 4.4.6.E 4.5.9 Let X be a complex of R–So-bimodules. Show that the unit of and counit of the adjunction

E 4.4.3 are the unique morphisms that make the following diagrams commutative



S⊗S N

∼=µN

χ⊗N// HomR(X ,X)⊗S N

θXXN

N // HomR(X ,X⊗S N)

X⊗S HomR(X ,M)

ηXXM

// M

HomR(HomSo(X ,X),M)Hom(χ,M)

// HomR(R,M) .

εM∼=OO

Here χ is the morphism from E 4.4.5.


Chapter 5Resolutions

A resolution of a complex is a quasi-isomorphism between the given complex andone that is distinguished by amenable homological properties. Resolutions come inseveral flavors; in Chap. 6 they will be used to compute derived functors, and in laterchapters they will be used to attach homological invariants to complexes.

5.1 Semi-freeness

SYNOPSIS. Semi-basis; semi-free complex; semi-free resolution; free resolution of module.

There are several ways in which the notion of freeness for modules can be extendedto complexes; the simplest of these, graded-freeness of the underlying module, wasintroduced in Sect. 2.5. The goal of this section is to show that the homological dataof any complex can be encoded into one whose underlying module is graded-free.

5.1.1 Definition. An R-complex L is called semi-free if the graded R-module L\ isgraded-free with a graded basis E that can be written as a disjoint union

E =⊎

n>0

En with E0 ⊆ Z(L) and ∂L(En)⊆ R〈⋃n−1

i=0 E i〉 for every n> 1 .

Such a basis is called a semi-basis for L.

5.1.2. Let L be a graded R-module. A graded basis for L is trivially a semi-basisfor the R-complex L. Thus, a graded R-module is graded-free if and only if it issemi-free as an R-complex.

5.1.3 Example. Let L be a bounded below complex of free R-modules and set w =infL\. For every n > 0, let En be a basis for the free module Lw+n, then

⋃n>0 En is

a semi-basis for L. Thus L is semi-free.

5.1.4 Example. The Dold complex from 2.1.22 is a complex of free Z/4Z-modules.It has no semi-basis, as no graded basis for this complex contains a cycle.

165

166 5 Resolutions

5.1.5. Notice that an R-complex L is semi-free if and only if ΣL is semi-free.

EXISTENCE OF SEMI-FREE RESOLUTIONS

5.1.6 Definition. A semi-free resolution of an R-complex M is a quasi-isomorphismL−→M of R-complexes where L is semi-free.

The next theorem achieves the goal of this section.

5.1.7 Theorem. Every R-complex M has a semi-free resolution π : L '−−→M withLv = 0 for all v < infM\. Moreover, π can be chosen surjective.

The proof relies on the next construction and follows after the proof of 5.1.9.

5.1.8 Construction. Given an R-complex M 6= 0, we shall construct a commutativediagram in C(R),

(5.1.8.1)L0 // ι0

//

π0((

· · · // // Ln−1 // ιn−1//

πn−1

Ln // ιn//

πn

||

· · · // L .

πooM

For n = 0, choose a set Z0 of homogeneous cycles in M whose homology classesgenerate H(M). Let E0 = ez | |ez| = |z|, z ∈ Z0 be a graded set and define anR-complex L0 as follows:

(5.1.8.2) (L0)\ = R〈E0〉 and ∂L0= 0 .

To see that the map π0 : L0→M given by the assignment ez 7→ z is a morphism ofcomplexes, notice that the differential on L0 is 0 and that π0 maps to Z(M), thekernel of ∂M .

Let n> 1 and let a morphism πn−1 : Ln−1→M be given. Choose a set Zn of ho-mogeneous cycles in Ln−1 whose homology classes generate the kernel of H(πn−1).Let En = ez | |ez|= |z|+1, z ∈ Zn be a graded set and set

(Ln)\ = (Ln−1)\⊕R〈En〉 and

∂Ln(x+ ∑

z∈Znrzez)= ∂Ln−1

(x)+ ∑z∈Zn

rzz .(5.1.8.3)

This defines an R-complex. For each z ∈ Zn choose an element mz ∈ M such thatπn−1(z) = ∂M(mz). It is elementary to verify that the map πn : Ln→M defined by

(5.1.8.4) πn(x+ ∑z∈Zn

rzez)= πn−1(x)+ ∑

z∈Znrzmz

is a morphism of R-complexes. Moreover, it agrees with πn−1 on the subcomplexLn−1 of Ln. That is, there is an equality of morphisms πn−1 = πnιn−1, where ιn−1 isthe embedding of Ln−1 into Ln; cf. (5.1.8.3).

For n < 0 set Ln = 0, ιn = 0, and πn = 0, then the family ιn : Ln→ Ln+1n∈Zis a telescope in C(R), and one has πn = πn+1ιn for all n ∈ Z. Set L = colimn∈ZLn,


5.1 Semi-freeness 167

by 3.2.39 there is a morphism of R-complexes π : L→M, such that the diagram(5.1.8.1) is commutative.

5.1.9 Proposition. Let M 6= 0 be an R-complex. The sets, morphisms, and com-plexes constructed in 5.1.8 have the following properties.

(a) Each set En consists of homogeneous elements of degree at least n+ infM\.(b) Each complex Ln is semi-free with semi-basis

⊎ni=0 E i.

(c) The complex L is semi-free with semi-basis E =⊎

n>0 En.(d) The morphism π : L→M is a quasi-isomorphism.(e) If πn is surjective for some n> 0, then π is surjective.

PROOF. Parts (a) and (b) are immediate from the definition of the sets En and(5.1.8.3); part (e) follows from commutativity of the diagram (5.1.8.1).

(c): The morphisms ιn are embeddings, so L = colimn∈ZLn is by 3.2.40 simplythe union

⋃n>0 Ln; in particular,

⊎n>0 En is a graded basis for L\. By (5.1.8.3) there

are containments ∂L(En) = ∂Ln(En) ⊆ R〈

⋃n−1i=0 E i〉 for n > 1, and (5.1.8.2) yields

∂L(E0) = ∂L0(E0) = 0, so E0 consists of cycles.

(d): For each n> 0 there is a commutative diagram

H(L0) //

H(π0) ""

H(Ln) //

H(πn)

H(L)

H(π)||

H(M) ,

induced from (5.1.8.1) By the choice of Z0, the morphism H(π0) is surjective andhence so is H(π). To see that H(π) is injective, let y be a cycle in L and assume thatH(π)([y]) = 0. Choose an integer n such that y ∈ Ln−1; now one has

0 = H(π)([y]) = [π(y)] = [πn−1(y)] = H(πn−1)([y]) ,

so [y] is in KerH(πn−1). By the choice of Zn there exists a element x ∈ Ln−1 suchthat one has

y = ∑z∈Zn

rzz+∂Ln−1(x) = ∂Ln(

x+ ∑z∈Zn

rzez),

where the second equality follows from (5.1.8.3). It follows that [y] is 0 in H(Ln)and hence also in H(L). Thus, H(π) is injective, and π is a quasi-isomorphism.

PROOF OF 5.1.7. The identity morphism 0−→ 0 is a semi-free resolution of the zerocomplex. For an R-complex M 6= 0, apply the construction 5.1.8. It follows fromparts (c) and (d) in 5.1.9 that π : L '−→M is a semi-free resolution. Parts (a) and (c)ensure that Lv = 0 holds for all v < infM\. Finally, notice that choosing Z0 as a setof generators for Z(M) makes the morphism π0 is surjective on cycles, and then π issurjective on cycles by commutativity of (5.1.8.1). As π is a quasi-isomorphism, itfollows from 4.2.7 that it is surjective.

5.1.10 Proposition. Let R→ S be a ring homomorphism. If L is a semi-free R-complex, then the S-complex S⊗R L is semi-free.


168 5 Resolutions

PROOF. Let E =⊎

n>0 En = euu∈U be a semi-basis for L. One then has L\ =∐u∈U Reu, and by 3.1.12 there is an isomorphism (S⊗R L)\ ∼=

∐u∈U S⊗R Reu. It is

elementary to see that the graded S-module∐

u∈U S⊗R Reu is graded-free with basisE ′ = 1⊗ euu∈U , and it follows from 2.4.1 that E ′ is a semi-basis for S⊗R L.

5.1.11 Proposition. If L is a semi-free S-complex and L′ is a semi-free k-complex,then the S-complex L⊗k L′ is semi-free.

PROOF. Let E =⊎

n>0 En be a semi-basis for L and F =⊎

n>0 Fn be a semi-basisfor L′. For n > 0 set Gn = e⊗ f | e ∈ E i, f ∈ F j, i+ j = n. It is elementary toverify that the graded S-module (L⊗k L′)\ is graded-free with basis G =

⊎n>0 Gn,

and it follows from 2.4.1 that G is a semi-basis for L⊗k L′.

BOUNDEDNESS AND FINITENESS

A suitably bounded and/or finite complex has a semi-free resolution with similarproperties. The construction of such a resolution could also be performed degree-wise, thus resembling the classic construction of a free resolution of a module.

5.1.12 Theorem. Every R-complex M has a semi-free resolution L '−→M with Lv =0 for all v < infM.

PROOF. If M is acyclic, then the morphism 0 '−→ M is the desired resolution. IfH(M) is not bounded below, then any semi-free resolution of M has the desiredproperty. Assume now that H(M) is bounded below and set w = infM. By 4.2.4there is a quasi-isomorphism MĚw

'−→M, and by 5.1.7 the truncated complex MĚwhas a semi-free resolution L '−→MĚw with Lv = 0 for v < w. The desired semi-freeresolution is the composite L '−→MĚw

'−→M.

5.1.13 Theorem. Let R be left Noetherian. Every bounded below degreewise finitelygenerated R-complex M has a semi-free resolution π : L '−→M with L degreewisefinitely generated and Lv = 0 for all v< infM\. Moreover, π can be chosen surjective.

PROOF. The identity morphism 0−→ 0 is a semi-free resolution of the zero complex.Assume now that M is non-zero and set w= infM\; it is an integer by the assumptionon M. Apply the construction 5.1.8 to M and notice the following.

As R is left Noetherian, and M is a complex of finitely generated R-modules,the set Z0 can be chosen such that it contains only finitely many elements of eachdegree. Doing so ensures that E0 = ez | |ez| = |z|, z ∈ Z0 contains only finitelymany elements of each degree v and no elements of degree v < w; see 5.1.9(a).

As R is left Noetherian, it follows by induction that KerH(πn−1) is degreewisefinitely generated for every n > 1. Choosing the set Zn such that it has only finitelymany elements of each degree ensures that the set En = ez | |ez|= |z|+1, z ∈ Zncontains only finitely many elements of each degree v and no elements of degreev < w+n; see 5.1.9(a).

From 5.1.9 it follows that π : L '−→M is a semi-free resolution of M, and that E =⊎n>0 En is a semi-basis for L. For each v ∈ Z the subset Ev ⊆ E of basis elements

of degree v is a basis for Lv, and it is finite as one has


5.1 Semi-freeness 169

Ev =( ⊎

n>0En)

v =v−w⊎n=0

(En)v .

Thus, each free module Lv is finitely generated, and Lv = 0 holds for all v < w.Finally, as R is left Noetherian, one can choose as Z0 a set of generators for

Z(M) with the additional property that it contains only finitely many elements ofeach degree. With this choice, the quasi-isomorphism π is surjective on cycles by5.1.9(e) and, therefore, surjective by 4.2.7.

5.1.14 Theorem. Let R be left Noetherian and let M be an R-complex. If H(M) isbounded below and degreewise finitely generated, then M has a semi-free resolutionL '−→M with L degreewise finitely generated and Lv = 0 for all v < infM.

PROOF. If M is acyclic, then the morphism 0 '−→ M is the desired resolution. As-sume now that M is not acyclic. Set w = infM and apply 5.1.13 to the truncatedcomplex MĚw to obtain a semi-free resolution L '−→ MĚw with each module Lvfinitely generated and Lv = 0 for all v < w. By 4.2.4 there is a quasi-isomorphismMĚw

'−→M, and the desired resolution is the composite L '−→MĚw'−→M.

The requirement in the theorem that H(M) be bounded below cannot be relaxed;an example is provided in ??.

THE CASE OF MODULES

Semi-free resolutions of complexes subsume the classic notion of free resolutionsof modules [16, V.§1].

5.1.15. It follows from 5.1.2 that an R-module is free if and only if it is semi-freeas an R-complex.

5.1.16 Theorem. For every R-module M there is an exact sequence of R-modules

· · · −→ Lv −→ Lv−1 −→ ·· · −→ L0 −→M −→ 0

where each module Lv is free.

PROOF. Choose by 5.1.7 a surjective semi-free resolution π : L '−→M with Lv = 0for all v < 0. The displayed sequence of R-modules is the complex Σ−1 Coneπ; inparticular, the map L0M is the homomorphism−π0. The cone is acyclic becauseπ is a quasi-isomorphism; see 4.2.15.

5.1.17 Definition. Let M be an R-module. Together, the surjective homomorphismL0→M and the R-complex · · · → Lv→ Lv−1→ ··· → L0→ 0 in 5.1.16 is called afree resolution of M.

5.1.18. By 5.1.3 a free resolution of an R-module M is a semi-free resolution of Mas an R-complex.


170 5 Resolutions

5.1.19 Theorem. Assume that R is left Noetherian. Every finitely generated R-module M has a free resolution

· · · −→ Lv −→ Lv−1 −→ ·· · −→ L0 −→M −→ 0

where each module Lv is finitely generated.

PROOF. Choose by 5.1.13 a surjective semi-free resolution π : L '−→M, where eachfree module Lv is finitely generated and Lv = 0 holds for all v < 0. The displayedsequence is the acyclic complex Σ−1 Coneπ; cf. the proof of 5.1.16.

EXERCISES

E 5.1.1 A semi-free filtration of an R-complex L is a sequence · · · ⊆ Lu−1 ⊆ Lu ⊆ Lu+1 ⊆ ·· ·of subcomplexes, such that the graded module underlying each quotient Lu/Lu−1 isgraded-free and one has L =

⋃u∈Z Lu, L−1 = 0, and ∂L(Lu)⊆ Lu−1 for all u ∈ Z. Show

that an R-complex is semi-free if and only if it admits a semi-free filtration.E 5.1.2 Show that a complex L of free modules is semi-free if one has ∂L

v = 0 for v 0.E 5.1.3 Let L be a semi-free R-complex; show that every subcomplex Lďn is semi-free.E 5.1.4 Show that a coproduct of semi-free complexes is semi-free.E 5.1.5 Show that a direct summand of a semi-free complex may not be semi-free.E 5.1.6 Let L be a semi-free R-complex and let N be a bounded and degreewise finitely generated

subcomplex of L. Show that N is contained in bounded and degreewise finitely generatedsemi-free subcomplex of L.

E 5.1.7 Let L be a semi-free R-complex and α : L→ N be a morphism in C(R). Show thatfor every surjective quasi-isomorphism β : M→ N there exists a morphism γ : L→Mwith α = βγ. Hint: For each n let Ln be the semi-free subcomplex of L with semi-basis

⊎ni=0 E i and construct morphisms γn : Ln→M compatible with the embeddings

Ln Ln+1.E 5.1.8 Apply 5.1.8 to construct a semi-free resolution of the Z/4Z-module Z/2Z.E 5.1.9 Construct semi-free resolutions of the complexes in 4.2.3.E 5.1.10 Show that every morphism α : M→ N of R-complexes admits factorizations in C(R),

M

ι

α// N

Xπ

'CC CC

andM

ε'

α// N

Yϕ

DD DD

where ι and ε are injective with semi-free cokernels; π and ϕ are surjective; and π and εare quasi-isomorphisms. Hint: (1) Modify the first step in 5.1.8. (2) 4.3.24.

E 5.1.11 Give alternative proofs of 5.1.16 and 5.1.19 based on 1.3.11.

5.2 Semi-projectivity

SYNOPSIS. Graded-projective module; complex of projective modules; semi-projective complex;semi-projective resolution; lifting property; projective resolution of module.

For a module P, the functor Hom(P,–) preserves short exact sequences of complexesif and only if it preserves acyclicity of complexes, and these properties characterize


5.2 Semi-projectivity 171

projective modules. Any complex P of projective modules has the first of theseproperties, see 2.3.17, but not the second; adding the requirement that the functorHom(P,–) preserves acyclicity of complexes leads to the notion of semi-projectivity.We start by studying complexes of projective modules.

5.2.1. Lifting properties are a central theme in this section, and several key resultscan be interpreted in terms of the diagram

(5.2.1.1)

P

~~

M // N

where the solid arrows represent given maps of certain sorts, and a lifting propertyof P ensures the existence of a dotted map of a specific sort such that the diagram iscommutative, or commutative up to homotopy.

COMPLEXES OF PROJECTIVE MODULES

Part (iii) below can be interpreted in terms of the diagram (5.2.1.1).

5.2.2 Proposition. For an R-complex P, the following conditions are equivalent.(i) Each R-module Pv is projective.

(ii) The functor HomR(P,–) is exact.(iii) For every homomorphism α : P→ N and every surjective homomorphism

β : M→ N, there exists a homomorphism γ : P→M such that α= βγ holds.(iv) Every exact sequence 0→M′→M→ P→ 0 in C(R) is degreewise split.(v) The graded module P\ is a graded direct summand of a graded-free R-module.

PROOF. (i)=⇒ (ii): Immediate from 2.3.17.(ii)=⇒ (iii): The homomorphism β yields a morphism ςN

−|β|β : M\→ Σ−|β|N\

and by exactness of the functor HomR(P,–) there exists a homomorphism γ ∈HomR(P,M\) with ςN

−|β|α= ςN−|β|βγ and hence α= βγ.

(iii)=⇒ (iv): Let β denote the morphism M P. By (iii) there exists a homo-morphism γ : P→M such that 1P = HomR(P,β)(γ) = βγ. As β is a morphism ofcomplexes, also the degree of γ must be 0, that is, γ is a morphism of the underlyinggraded modules. Hence, the sequence 0→M′→M→ P→ 0 is degreewise split.

(iv)=⇒ (v): Choose by 5.1.7 a surjective semi-free resolution π : L→ P and apply(iv) to the associated exact sequence 0→ Kerπ→ L→ P→ 0 in C(R). It followsthat P\ is a graded direct summand of the graded-free R-module L\.

(v)=⇒ (i): Each module Pv is a direct summand of a free R-module and, there-fore, projective by 1.3.15.

REMARK. The complexes described in 5.2.2 are not the projective objects in the category C(R);see E 5.2.1 and E 5.2.5.


172 5 Resolutions

5.2.3 Corollary. Let 0→ P′→ P→ P′′→ 0 be an exact sequence of R-complexes.If P′′ is a complex of projective modules, then P is a complex of projective modulesif and only if P′ is a complex of projective of modules.

PROOF. If P′′ is a complex of projective modules, then 0→ P′ → P→ P′′ → 0 isdegreewise split, and the assertion follows from 1.3.20.

5.2.4 Definition. A graded R-module P is called graded-projective if the R-complexP satisfies the conditions in 5.2.2.

EXISTENCE OF SEMI-PROJECTIVE RESOLUTIONS

5.2.5 Definition. An R-complex P is called semi-projective if HomR(P,β) is a sur-jective quasi-isomorphism for every surjective quasi-isomorphism β in C(R).

REMARK. Another word for semi-projective is DG-projective.

5.2.6. Notice that by 2.3.13 an R-complex P is semi-projective if and only if ΣP issemi-projective.

5.2.7 Example. A contractible complex of projective modules is semi-projective by5.2.2 and 4.3.29.

5.2.8 Example. Let P be a bounded below complex of projective R-modules, andlet β be a surjective quasi-isomorphism. The morphism HomR(P,β) is surjective by5.2.2. The complex Coneβ is acyclic by 4.2.15, and hence so is HomR(Pv,Coneβ)for every v ∈ Z. As P is bounded below, it follows from 4.1.15 and A.10 thatConeHomR(P,β) ∼= HomR(P,Coneβ) is acyclic, whence β is a quasi-isomorphism.Thus, P is is semi-projective.

The next lemma shows that a semi-free complex is semi-projective; see 5.2.11.

5.2.9 Lemma. For a semi-free R-complex L, the functor HomR(L,–) is exact andpreserves acyclicity of complexes.

PROOF. Let L be a semi-free R-complex; by 2.5.22 it is a complex of free R-modules, so the functor HomR(L,–) is exact by 5.2.2. Choose a semi-basis E =⊎

i>0 E i for L. For n < 0 set Ln = 0. For n> 0 let Ln be the semi-free subcomplex ofL with semi-basis

⊎ni=0 E i. Now one has L =

⋃n>0 Ln ∼= colimn∈ZLn, and for every

n> 0 there is an exact sequence

(?) 0−→ Ln−1 −→ Ln −→ Ln/Ln−1 −→ 0 .

The induced differential on the subquotient Ln/Ln−1 is 0, so it is isomorphic to thegraded-free R-module R〈En〉. In particular, (?) is degreewise split by 5.2.2.

Let A be an acyclic R-complex. For every n > 0 the complex HomR(R〈En〉,A)is acyclic; cf. 3.1.25. It follows by induction that HomR(Ln,A) is acyclic for all



n> 0. The morphisms in the tower HomR(Ln,A)→ HomR(Ln−1,A)n∈Z are sur-jective because the sequence (?) is degreewise split, and the Hom functor preservesdegreewise split sequences; see 2.3.12. Now it follows from 3.3.21 and 3.3.47 that

HomR(L,A) = HomR(colimn∈Z

Ln,A) ∼= limn∈Z

HomR(Ln,A)

is an acyclic complex.

The next result offers useful characterizations of semi-projective complexes. Thelifting property in part (iii) can be interpreted in terms of the diagram (5.2.1.1).

5.2.10 Proposition. For an R-complex P, the following conditions are equivalent.(i) P is semi-projective.

(ii) The functor HomR(P,–) is exact and preserves quasi-isomorphisms.(iii) For every chain map α : P→ N and for every surjective quasi-isomorphism

β : M→ N there exists a chain map γ : P→M such that α= βγ holds.(iv) Every exact sequence 0→M′→M→ P→ 0 in C(R) with M′ acyclic is split.(v) For every morphism α : N→ P and for every surjective quasi-isomorphism

β : M→ P there exists a morphism γ : N→M such that α= βγ holds.(vi) P is a direct summand of a semi-free R-complex.

(vii) P is a complex of projective R-modules, and the functor HomR(P,–) preservesacyclicity of complexes.

PROOF. The implication (ii)=⇒ (i) is evident.(i)=⇒ (iii): The morphism HomR(P,β) is a surjective quasi-isomorphism. In par-

ticular, it is surjective on cycles by 4.2.7. Thus, in view of 2.3.3 there exists a chainmap γ : P→M such that α= HomR(P,β)(γ) = βγ.

(iii)=⇒ (iv): By 4.2.6 the morphism β : M P is a quasi-isomorphism, so thereexists a chain map γ : P→M with 1P = βγ. As β is of degree 0, so is γ. That is, γ isa morphism in C(R), whence the sequence is split.

(iv)=⇒ (v): By (iv) there is a morphism σ : P→M with βσ = 1P. The desiredmorphism is thus γ = σα.

(v)=⇒ (vi): By 5.1.7 there exists a surjective quasi-isomorphism β : L→ P whereL is a semi-free complex. By (v) there is a morphism γ : P→ L with 1P = βγ, so Pis a direct summand of L by 2.1.44.

(vi)=⇒ (vii): Immediate from 5.2.2 and 5.2.9, as the Hom functor is additive.(vii)=⇒ (ii): The functor HomR(P,–) is exact by 5.2.2. For a quasi-isomorphism

β, the complex Coneβ is acyclic by 4.2.15. Hence the complex HomR(P,Coneβ)∼=ConeHomR(P,β) is acyclic; here the isomorphism follows from 4.1.15. Thus, themap HomR(P,β) is a quasi-isomorphism.

5.2.11 Corollary. A semi-free R-complex is semi-projective.

REMARK. A semi-projective complex of free modules may not be semi-free; see E 5.2.4.

5.2.12 Corollary. A graded R-module is graded-projective if and only if it is semi-projective as an R-complex.


174 5 Resolutions

PROOF. Let P be a graded R-module. If P is semi-projective as an R-complex, theneach module Pv is projective, whence P is graded-projective.

If P is graded-projective, then it is a graded direct summand of a graded-free R-module; see 5.2.2. A graded-free R-module is semi-free as an R-complex by 5.1.2,and then P is semi-free as an R-complex by 5.2.10.

5.2.13 Definition. A semi-projective resolution of an R-complex M is a quasi-isomorphism P→M of R-complexes where P is semi-projective.

Existence of semi-free resolutions implies the existence of semi-projective ones.

5.2.14 Theorem. Every R-complex M has a semi-projective resolution π : P '−→Mwith Pv = 0 for all v < infM\. Moreover, π can be chosen surjective.


5.2.15 Theorem. Every R-complex M has a semi-projective resolution P '−→ Mwith Pv = 0 for all v < infM.


PROPERTIES OF SEMI-PROJECTIVE COMPLEXES

The class of semi-projective complexes over a ring is closed under extensions, ker-nels of surjective morphisms, direct summands, and coproducts. Semi-projectivityis also preserved under base change.

5.2.16 Proposition. Let 0 → P′ → P → P′′ → 0 be an exact sequence of R-complexes. If P′′ is semi-projective, then P′ is semi-projective if and only if P issemi-projective.

PROOF. First note that since P′′ is a complex of projective modules, it follows from5.2.3 that P is a complex of projective modules if and only if P′ is so. Next, let A bean acyclic R-complex. The sequence 0→ P′→ P→ P′′→ 0 is degreewise split by5.2.2, and hence so is the induced sequence

0−→ HomR(P′′,A)−→ HomR(P,A)−→ HomR(P′,A)−→ 0 ;

see 2.3.12. As P′′ is semi-projective, the complex HomR(P′′,A) is acyclic by theequivalence of (i) and (vi) in 5.2.10. It follows from 2.2.20 that HomR(P,A) isacyclic if and only if HomR(P′,A) is acyclic. Now 5.2.10 yields the desired con-clusion.

5.2.17 Proposition. Let Puu∈U be a family of R-complexes. The coproduct∐u∈U Pu is semi-projective if and only if each complex Pu is semi-projective.

PROOF. Let β : M→ N be a surjective quasi-isomorphism. There is a commutativediagram in C(k),



HomR(∐

u∈UPu,M)

Hom(∐

u∈U Pu,β)//

∼=

HomR(∐

u∈UPu,N)

∼=∏

u∈UHomR(Pu,M)

∏u∈U Hom(Pu,β)

//∏

u∈UHomR(Pu,N) ,

where the vertical maps are the canonical isomorphisms from 3.1.23. It follows thatHomR(

∐u∈U Pu,β) is a surjective quasi-isomorphism if and only if each morphism

HomR(Pu,β) is a surjective quasi-isomorphism.

Also the next result can be interpreted in terms of the diagram (5.2.1.1).

5.2.18 Proposition. Let P be a semi-projective R-complex, let α : P→ N be a chainmap, and let β : M→ N be a quasi-isomorphism. There exists a chain map γ : P→Msuch that α∼ βγ. Moreover, γ is homotopic to any other chain map γ′ with α∼ βγ′.PROOF. Recall from 2.3.3 the characterization of (null-homotopic) chain maps as(boundaries) cycles in Hom complexes. By 5.2.10 the induced morphism HomR(P,β)is a quasi-isomorphism, so there exists a γ ∈ Z(HomR(P,M)) such that

[α] = H(HomR(P,β))([γ]) = [βγ] ;

that is, α−βγ is in B(HomR(P,N)). Given another morphism γ′ such that α∼ βγ′,one has [α] = [βγ′] and, therefore 0 = [β(γ−γ′)] = H(HomR(P,β))([γ−γ′]). It fol-lows that the homology class [γ−γ′] is 0 as H(HomR(P,β)) is an isomorphism, soγ−γ′ is in B(HomR(P,M)). That is, γ and γ′ are homotopic.

REMARK. Existence and uniqueness of lifts up to homotopy, as described in 5.2.18, is an importantproperty of semi-projective complexes, but it does not characterize them. Complexes with thisproperty are called K-projective or homotopically projective; see E 5.2.15.

5.2.19 Corollary. Let P be a semi-projective R-complex and let β : M→ P be aquasi-isomorphism. There exists a quasi-isomorphism γ : P→M with 1P ∼ βγ.

PROOF. By 5.2.18 there is a chain map γ : P→M with 1P ∼ βγ; comparison ofdegrees shows that γ is a morphism. Moreover, by 2.2.26 one has 1H(P)=H(β)H(γ),whence H(γ) is an isomorphism.

Recall from 4.3.4 that every homotopy equivalence is a quasi-isomorphism. Thenext result is a partial converse.

5.2.20 Corollary. A quasi-isomorphism of semi-projective R-complexes is a homo-topy equivalence.

PROOF. Let β : P′→ P be a quasi-isomorphism of semi-projective R-complexes. By5.2.19 there are morphisms γ : P→ P′ and β′ : P′→ P such that 1P ∼ βγ and 1P′ ∼γβ′ hold. It now follows from 4.3.3 that β is a homotopy equivalence.

5.2.21 Proposition. Let P be a T -complex and W a complex of Q–T o-bimodules. IfP is semi-projective and W is semi-projective over Q, then the Q-complex W ⊗T Pis semi-projective.


176 5 Resolutions

PROOF. Adjunction 4.4.11 yields a natural isomorphism,

HomQ(W ⊗T P,–) ∼= HomT (P,HomQ(W,–)) ,

of functors from C(Q) to C(k). It follows from the assumptions on P and W that thefunctor HomT (P,HomQ(W,–)) is exact and preserves quasi-isomorphisms.

5.2.22 Corollary. Let R→ S be a ring homomorphism.(a) For a semi-projective R-complex P the S-complex S⊗R P is semi-projective.(b) If S is projective as an R-module, then a semi-projective S-complex is semi-

projective over R.

PROOF. For (a) apply 5.2.21 with Q = S, T = R, and W = SSR. For (b) use Q = R,T = S, and W = RSS and note that W ⊗S – is the forgetful functor C(S)→ C(R).

THE CASE OF MODULES

Semi-projective resolutions of complexes subsume the classic projective resolutionsof modules. Notice that specialization of 5.2.2 to modules recovers 1.3.15.

5.2.23 Proposition. Let 0→ P′→ P→ P′′→ 0 be an exact sequence of R-modules.Assume that P′′ is projective, then P′ is projective if and only if P is projective.

PROOF. Immediate by specialization of 5.2.3 to modules.

5.2.24. It follows from 5.2.12 that an R-module is projective if and only if it issemi-projective as an R-complex. Thus one recovers 1.3.20 from 5.2.17.


· · · −→ Pv −→ Pv−1 −→ ·· · −→ P0 −→M −→ 0

where each module Pv is projective.


5.2.26 Definition. Let M be an R-module. Together, the surjective homomorphismP0→M and the R-complex · · · → Pv→ Pv−1→ ··· → P0→ 0 in 5.2.25 is called aprojective resolution of M.

5.2.27. By 5.2.8 a projective resolution of an R-module M is a semi-projective re-solution of M as an R-complex.

EXERCISES

E 5.2.1 Show that a graded R-module is graded-projective if and only if it is a projective objectin the category Mgr(R).

E 5.2.2 Show that a graded R-module is graded-projective if and only if it is projective as anR-module.



E 5.2.3 Let R be left hereditary. Show that for every complex P of projective R-modules there isa quasi-isomorphism P '−→ H(P). Hint: E 1.3.17

E 5.2.4 The Z/6Z-complex · · · −→ Z/6Z 2−→ Z/6Z 3−→ Z/6Z 2−→ ·· · is contractible; see E 4.3.11.Show that it is a semi-projective complex of free modules but not semi-free.

E 5.2.5 For an R-complex P, show that the following conditions are equivalent. (i) P is a projec-tive object in the category C(R). (ii) P is a contractible complex of projective R-modules.(iii) P is semi-projective and acyclic. Conclude from 4.3.24 that the category C(R) hasenough projectives.

E 5.2.6 Show that the Dold complex from 5.1.4 is acyclic but not contractible; conclude that itis not semi-projective.

E 5.2.7 Show that a complex P of projective R-modules is semi-projective if HomR(P,A) isacyclic for every acyclic R-complex A that is bounded above.

E 5.2.8 Show that the mapping cone of a morphism between semi-projective complexes is semi-projective.

E 5.2.9 Let M be an R-complex. Show that for every semi-projective replacement P of M themodules Bv(P) are projective for v < infM.

E 5.2.10 Show that every complex over a semi-simple ring is semi-projective.E 5.2.11 Show that a complex of projective modules over a left hereditary ring is semi-projective.E 5.2.12 Let L be a bounded complex of finitely generated projective R-modules and P be a com-

plex of R–So-bimodules that are all projective over So. Show that HomR(L,P) is a com-plex of projective So-modules; conclude that if P is bounded below, then HomR(L,P) issemi-projective.

E 5.2.13 Let 0→ P′ → P→ P′′ → 0 be a degreewise split exact sequence of complexes. Showthat if two of the complexes are semi-projective, then so is the third.

E 5.2.14 Show that the following conditions are equivalent for an R-complex P. (i) P is semi-projective. (ii) The complex Pďn is semi-projective for every n ∈ Z. (iii) P is a complexof projective R-modules and Pďn is semi-projective for some n ∈ Z.

E 5.2.15 Show that the following conditions are equivalent for an R-complex X . (i) For everychain map α : X → N and every quasi-isomorphism β : M→ N there exists a chain mapγ : X →M, unique up to homotopy, such that α∼ βγ. (ii) For every quasi-isomorphismβ the induced morphism HomR(X ,β) is a quasi-isomorphism. (iii) For every acycliccomplex A, the complex HomR(X ,A) is acyclic.

A complex with these properties is called K-projective or homotopically projective.E 5.2.16 Let K be an acyclic K-projective R-complex. Show that HomR(K,M) is acyclic for every

R-complex M.E 5.2.17 Show that a complex is semi-projective if and only if it is a complex of projective mod-

ules and K-projective in the sense of E 5.2.15. Give an example of a K-projective com-plex that is not semi-projective.

E 5.2.18 Show that a graded R-module is graded-projective if and only if it is K-projective as anR-complex, in the sense of E 5.2.15.

E 5.2.19 Show that the class of K-projective R-complexes, in the sense of E 5.2.15, is closed underhomotopy equivalence. Is the same true for the class of semi-projective R-complexes?

E 5.2.20 Let α : M→ N be a morphism in C(R). Show that for any two semi-projective resolu-tions π : P '−→M and λ : L '−→ N there is a morphism α : P→ L such that the diagram

Pπ

'//

α

M

α

Lλ

'// N

is commutative up to homotopy. Show also that if λ is surjective, then α can be chosensuch that the diagram is commutative.


178 5 Resolutions

E 5.2.21 Let x be a central element in R and let P '−→ M be a semi-projective resolution of R-complexes. Show that if the homothety xM is null-homotopic, then xP is null-homotopic.

E 5.2.22 Let 0−→M′ α′−→M α−→M′′ −→ 0 be an exact sequence of R-complexes. Show that there

is a commutative diagram in C(R) in which the columns are semi-projective resolutions,

0 // P′α′//

π′'

Pα//

π'

P′′ //

π′′'

0

0 // M′α′// M

α// M′′ // 0 .

This is known as the Horseshoe Lemma (for semi-projective resolutions).E 5.2.23 Let P be a complex of projective modules. Show that for every n > supP the truncated

complex Pěn yields a projective resolution of the module Cn(P).

5.3 Semi-injectivity

SYNOPSIS. Character complex; graded-injective module, complex of injective modules; semi-injective complex; semi-injective resolution; lifting property; injective resolution of module.

Semi-injectivity is dual to semi-projectivity and will be defined in terms of the func-tor HomR(–, I) from C(R)op to C(k). The goal of this chapter is show that everycomplex has a semi-injective resolution.

CHARACTER COMPLEXES

Semi-injective resolutions come from semi-free resolutions via character com-plexes. Recall from 1.3.32 that E=HomZ(k,Q/Z) is a faithfully injective k-module.

5.3.1 Definition. Let M be an R-complex. The Ro-complex Homk(M,E) is calledthe character complex of M. The graded R-module Homk(M\,E) = Homk(M,E)\ iscalled the character module of the graded R-module M\.

5.3.2 Lemma. If L is a semi-free Ro-complex, then Homk(L,E) is a complex ofinjective R-modules, and the functor HomR(–,Homk(L,E)) preserves acyclicity ofcomplexes.

PROOF. By 1.3.44 each module Homk(L,E)v = Homk(L−v,E) is an injective R-module. Let A be an acyclic R-complex; adjunction 4.4.11 and commutativity 4.4.3yield isomorphisms

HomR(A,Homk(L,E)) ∼= Homk(L⊗R A,E) ∼= HomRo(L,Homk(A,E)) ,

and HomRo(L,Homk(A,E)) is acyclic by 5.2.9 and exactness of Homk(–,E).

5.3.3 Construction. Let M be an R-complex and choose by 5.1.7 a semi-free resolu-tion π : L '−→ Homk(M,E) with Lv = 0 for all v < infHomk(M,E)\. Precompose theinduced morphism Homk(π,E) : Homk(Homk(M,E),E)→ Homk(L,E) with bidu-ality δM

E : M→ Homk(Homk(M,E),E) to get a morphism of R-complexes


5.3 Semi-injectivity 179

εM = Homk(π,E)δME : M −→ E ,

where E is the character complex Homk(L,E).

5.3.4 Proposition. Let M be an R-complex. The morphisms and complexes con-structed in 5.3.3 have the following properties.

(a) E is a complex of injective R-modules with Ev = 0 for v > supM\, and thefunctor HomR(–,E) preserves acyclicity of complexes.

(b) The morphism H(εM) is injective.(c) The morphism π can be chosen such that εM is injective.

PROOF. (a): By 2.5.6(b) one has infHomk(M,E)\ = −supM\ and, therefore, Ev =Homk(L−v,E) = 0 for all v > supM\. The other assertions follow from 5.3.2.

(b): One has H(εM) = H(Homk(π,E))H(δME ), and the map H(Homk(π,E)) is

an isomorphism by 4.2.13. It follows from 2.2.18 and 4.5.1 that the map H(δME ) is

biduality δH(M)E , which is injective by 4.5.3. Thus H(εM) is injective.

(c): By 5.1.7 one can choose π surjective, and then it follows by exactness ofHomk(–,E) that the morphism Homk(π,E) is injective. By 4.5.3 the morphism δM

E

is injective, and hence so is the composite εM .

COMPLEXES OF INJECTIVE MODULES

5.3.5. Lifting properties are also central to this section; key results can be inter-preted in terms of the diagram

(5.3.5.1)

K //

M

~~

I

where the solid arrows represent given maps of certain sorts, and a lifting propertyof I ensures the existence of a dotted map of a specific sort such that the diagram iscommutative, or commutative up to homotopy.

Part (iii) below can be interpreted in terms of the diagram (5.3.5.1).

5.3.6 Proposition. For an R-complex I, the following conditions are equivalent.(i) Each R-module Iv is injective.

(ii) The functor HomR(–, I) is exact.(iii) For every homomorphism α : K→ I and for every injective homomorphism

β : K→M, there exists a homomorphism γ : M→ I such that γβ= α holds.(iv) Every exact sequence 0→ I→M→M′′→ 0 in C(R) is degreewise split.(v) The graded module I\ is a graded direct summand of the character module

Homk(L,E) of a graded-free Ro-module L.

PROOF. (i)=⇒ (ii): Immediate from 2.3.19.


180 5 Resolutions

(ii)=⇒ (iii): The homomorphism β yields a morphism β(ςK|β|)−1 : Σ|β|K\→M\

and by exactness of the functor HomR(–, I) there exists a homomorphism γ ∈HomR(M\, I) with γβ(ςK

|β|)−1 = α(ςK

|β|)−1 and hence γβ= α.

(iii)=⇒ (iv): Let β denote the morphism I M. By (iii) there exists a homo-morphism γ : M→ I such that 1I = HomR(β, I)(γ) = γβ holds. As β is a morphism,also the degree of γ must be 0; that is, γ is a morphism of the underlying gradedmodules. Hence, the sequence 0→ I→M→M′′→ 0 is degreewise split.

(iv)=⇒ (v): Choose by 5.3.4 an injective morphism ε : I→ E, where E is thecharacter complex Homk(L,E) of a semi-free Ro-complex L. Apply (iv) to the ex-act sequence 0 → I → E → Cokerε → 0 in C(R). It follows that I\ is a gradeddirect summand of the graded module E\ = Homk(L\,E), and L\ is a graded-freeRo-module by 2.5.22.

(v)=⇒ (i): The character module of a free Ro-module is an injective R-moduleby 1.3.44. A direct summand of an injective module is injective by additivity of theHom functor. Thus, each module Iv is an injective R-module.

REMARK. The complexes described in 5.3.6 are not the injective objects in the category C(R); seeE 5.3.1 and E 5.3.3.

5.3.7 Corollary. Let 0→ I′→ I→ I′′→ 0 be an exact sequence of R-complexes.If I′ is a complex of injective modules, then I is a complex of injective of modulesif and only if I′′ is a complex of injective of modules.

PROOF. If I′ is a complex of injective modules, then 0→ I′ → I → I′′ → 0 is de-greewise split, and the assertion follows from 1.3.23.

5.3.8 Definition. A graded R-module I is called graded-injective if the R-complexI satisfies the conditions in 5.3.6.

EXISTENCE OF SEMI-INJECTIVE RESOLUTIONS

5.3.9 Definition. An R-complex I is called semi-injective if HomR(α, I) is an sur-jective quasi-isomorphism for every injective quasi-isomorphism α in C(R).

REMARK. Another word for semi-injective is DG-injective.

5.3.10. Notice that by 2.3.15 an R-complex I is semi-injective if and only if Σ I issemi-injective.

5.3.11 Example. A contractible complex of injective modules is semi-injective by5.3.6 and 4.3.29.

5.3.12 Example. Let I be a bounded above complex of injective R-modules, andlet β be an injective quasi-isomorphism. The morphism HomR(β, I) is surjective by5.3.6, and it follows from 4.2.15, 4.1.16, and A.12 that it is a quasi-isomorphism;cf. 5.2.8. Thus I is semi-injective.



5.3.13 Definition. A semi-injective resolution of an R-complex M is a quasi-isomorphism M→ I of R-complexes where I is semi-injective.

REMARK. Sometimes, not here, a semi-injective resolution is called a semi-injective coresolution.

The goal of this section is obtained by 5.3.19; it relies on the next construction.

5.3.14 Construction. Given an R-complex M, we shall construct a commutativediagram in C(R),

(5.3.14.1)

Mι0

((

ιn−1

ιn

ι

I // · · · // // In πn// // In−1 πn−1

// // · · · // // I0 .

For n = 0 choose by 5.3.4 an injective morphism ι0 : M→ I0, where I0 is the char-acter complex of a semi-free Ro-complex.

Let n > 1 and let a morphism ιn−1 : M→ In−1 be given. Choose by 5.3.4 aninjective morphism εn : CokerH(ιn−1)→ En, where En is the character complex ofa semi-free Ro-complex. The induced morphism Z(In−1)→En is zero on boundariesand on ιn−1(Z(M)); see (5.3.14.2) below. It extends by 5.3.6 to a homomorphismδn : In−1→ En with B(In−1)+ ιn−1(Z(M)) contained in Kerδn∩Z(In−1).

(5.3.14.2)

Z(In−1) // //

H(In−1) // // CokerH(ιn−1)

εn

In−1 δn// En .

Conversely, let z be a cycle in In−1. If z is in Kerδn, then the element [z]+ImH(ιn−1)in CokerH(ιn−1) is in the kernel of εn, whence the homology class [z] is in the imageof H(ιn−1). Thus, there is an equality

(5.3.14.3) B(In−1)+ ιn−1(Z(M)) = Kerδn ∩ Z(In−1) .

Consider δn as a degree −1 homomorphism: In−1→ Σ−1 En; cf. 2.2.4. Set

(5.3.14.4) (In)\ = (In−1)\⊕ (Σ−1 En)\ and ∂In(i+ e) = ∂In−1

(i)+δn(i) .

This defines an R-complex, as δn is zero on boundaries in In−1. Notice that theprojection πn : In In−1 is a morphism of complexes.

For each boundary b ∈ B(M) choose a preimage mb. The assignment

(5.3.14.5) b 7−→ δnιn−1(mb)

is independent of the choice of preimage. Indeed, if mb is another preimage of b,then mb − mb is a cycle in M, and (5.3.14.3) yields the first equality in the nextcomputation

0 = δnιn−1(mb− mb) = δnιn−1(mb)−δnιn−1(mb) .

Thus, (5.3.14.5) defines a (degree 0) homomorphism from B(M) to Σ−1 En. It ex-tends by 5.3.6 to a homomorphism σn : M→ Σ−1 En, and there is an equality


182 5 Resolutions

(5.3.14.6) σn∂M = δnιn−1 .

Define a map ιn : M→ In as follows:

(5.3.14.7) ιn(m) = ιn−1(m)+σn(m) .

The next computation shows that it is a morphism of R-complexes; the penultimateequality uses (5.3.14.6).

∂Inιn = ∂In

(ιn−1 +σn) = ∂In−1ιn−1 +δnιn−1 = ιn−1∂M +σn∂M = ιn∂M .

For n < 0 set In = 0, ιn = 0, and πn+1 = 0, then the family πn : In→ In−1n∈Z isa tower in C(R), and one has ιn−1 = πnιn for all n ∈ Z. Set I = limn∈Z In, by 3.3.31there is a morphism of R-complexes ι : M→ I, given by m 7→ (ιn(m))n∈Z.

5.3.15 Proposition. Let M be an R-complex. The complexes and morphisms con-structed in 5.3.14 have the following properties.

(a) Each In is a complex of injective R-modules with Inv = 0 for v > supM\.

(b) I is a complex of injective R-modules with Iv = 0 for all v > supM\, and thefunctor HomR(–, I) preserves acyclicity of complexes.

(c) The morphism ι : M→ I is an injective quasi-isomorphism.

PROOF. Part (a) follows from 5.3.4 and (5.3.14.4).(b): One has Iv = 0 for all v > supM\ by part (a) and the definition 3.3.2 of limits.

Let 0→ K →M→ N → 0 be an exact sequence in C(R). For every n > 0 there isan exact sequence

0−→ HomR(N, In)−→ HomR(M, In)−→ HomR(K, In)−→ 0 ;

this follows from (a) and 5.3.6. Because of the degreewise split exact sequences

(?) 0−→ (Σ−1 En)\ −→ In πn−−→ In−1 −→ 0 ,

the morphisms in the tower HomR(N,πn) : HomR(N, In)→ HomR(N, In−1)n∈Zare surjective; see 2.3.12. It now follows from 3.3.42 that the lower row in the com-mutative diagram below is exact.

HomR(M, I) //

∼=

HomR(K, I)

∼=

limn∈Z

HomR(M, In) // limn∈Z

HomR(K, In) // 0

The vertical maps are the isomorphisms from 3.3.17; the diagram shows that thefunctor HomR(–, I) is exact, whence I is a complex of injective modules by 5.3.6.

Let A be an acyclic R-complex. As above the morphisms in the induced towerHomR(A,πn) : HomR(A, In)→ HomR(A, In−1)n∈Z are surjective. By 5.3.4, thefunctors HomR(–,(En)\) preserve acyclicity, so by induction it follows from (?) and2.2.20 that the functors HomR(–, In) preserve acyclicity; in particular HomR(A, In)is acyclic for every n. By 3.3.17 and 3.3.47 the complex



HomR(A, I) = HomR(A, limn∈Z

In) ∼= limn∈Z

HomR(A, In)

is acyclic.(c): As ι0 is injective, commutativity of (5.3.14.1) shows that ι is injective as well.

By 5.3.4 the morphism H(ι0) is injective, and the commutative diagram

(‡)

H(M)

H(ι)

||H(ιn)

H(ι0)

""

H(I) // H(In) // H(I0) ,

which is induced from (5.3.14.1), shows that H(ι) is injective. To see that it is sur-jective, let z = (zn)n∈Z be a cycle in I; the goal is to show that there exist elementsm ∈ Z(M) and i = (in)n∈Z in I with z = ∂I(i)+ ι(m). From (5.3.14.4) one gets

() 0 = ∂I(z) = (∂I0(z0), . . . ,∂In−1

(zn−1)+δn(zn−1),∂In(zn)+δn+1(zn), . . .) .

It follows for each n> 1 that the element zn−1 is a cycle in In−1 with δn(zn−1) = 0,whence zn−1 belongs B(In−1)+ ιn−1(Z(M)) by (5.3.14.3).

Choose elements j2 in I2 and m∈ Z(M) such that z2 = ∂I2( j2)+ ι2(m) holds. The

sequence (in)n∈Z is constructed by induction. Set i1 = π2( j2) and i0 = π1(i1), thenthere are equalities z1 = π2(z2) = ∂I1

(i1)+ ι1(m) and z0 = π1(z1) = ∂I0(i0)+ ι0(m).

Set in = 0 for n < 0. Fix n> 2 and assume that elements iu ∈ Iu for u < n and jn ∈ In

have been constructed, such that one has

zn = ∂In( jn)+ ιn(m) and zu = ∂Iu

(iu)+ ιu(m) for u < n ;

πn( jn) = in−1 and πu(iu) = iu−1 for u < n .

Choose j′ in In+1 and m′ ∈ Z(M) with zn+1 = ∂In+1( j′)+ ιn+1(m′). The equalities

∂In( jn)+ ιn(m) = zn = πn+1(zn+1) = ∂In

(πn+1( j′))+ ιn(m′)

show that ιn(m′−m) is a boundary in In. It follows from commutativity of the dia-gram (‡) that H(ιn) is injective, so m′−m is in B(M) and, therefore, ιn+1(m′−m)

is in B(In+1). Thus, there exists j′′ ∈ In+1 with ∂In+1( j′′) = ∂In+1

( j′)+ ιn+1(m′−m)

and, therefore, zn+1 = ∂In+1( j′′)+ ιn+1(m). The equalities

∂In( jn)+ ιn(m) = zn = πn+1(zn+1) = ∂In

(πn+1( j′′))+ ιn(m)

show that jn− πn+1( j′′) is a cycle in In and, therefore, πn( jn− πn+1( j′′)) is a cy-cle in In−1. Now (5.3.14.4) yields δn(πn( jn− πn+1( j′′))) = 0, and it follows from(5.3.14.3) that there are elements i′ ∈ In−1 and c ∈ Z(M) with

(§) πn( jn−πn+1( j′′)) = ∂In−1(i′)+ ιn−1(c) .

Choose i′′ ∈ In+1 with πnπn+1(i′′) = i′ and set jn+1 = j′′+ ∂In+1(i′′)+ ιn+1(c). As

∂In+1(i′′)+ ιn+1(c) is a cycle in In+1, the equality


184 5 Resolutions

(#) zn+1 = ∂In+1( jn+1)+ ιn+1(m)

holds. Set in = πn+1( jn+1); then (#) yields zn = πn+1(zn+1) = ∂In(in)+ ιn(m), and

the third equality below follows from (§),

πn(in) = πnπn+1( jn+1) = πnπn+1( j′′)+∂In−1πnπn+1(i′′)+ ιn−1(c) = πn( jn) = in−1 .

Thus, for u < n+1 one has

(??) zu = ∂Iu(iu)+ ιu(m) and iu−1 = πu(iu) .

From (#) and (??) it now follows that the desired element i = (in)n∈Z in I withz = ι(m)+∂I(i) exists.

The next result offers useful characterizations of semi-injective complexes. Thelifting property in part (iii) can be interpreted in terms of the diagram (5.3.5.1).

5.3.16 Proposition. For an R-complex I, the following conditions are equivalent.(i) I is semi-injective.

(ii) The functor HomR(–, I) is exact and preserves quasi-isomorphisms.(iii) For every chain map α : K→ I and for every injective quasi-isomorphism

β : K→M there exists a chain map γ : M→ I such that γβ= α holds.(iv) Every exact sequence 0→ I→M→M′′→ 0 in C(R) with M′′ acyclic is split.(v) For every morphism α : I→ K and for every injective quasi-isomorphism

β : I→M there exists a morphism γ : M→ K such that γβ= α holds.(vi) I is a complex of injective R-modules, and the functor HomR(–, I) preserves

acyclicity of complexes.

PROOF. The implication (ii)=⇒ (i) is evident.(i)=⇒ (iii): The morphism HomR(β, I) is a surjective quasi-isomorphism. In par-

ticular, it is surjective on cycles; see 4.2.7. Thus, in view of 2.3.3 there exists a chainmap γ : M→ I such that α= HomR(β, I)(γ) = γβ holds.

(iii)=⇒ (iv): By 4.2.6 the morphism β : IM is a quasi-isomorphism, so thereexists a chain map γ : M→ I with γβ= 1I . As β is of degree 0, so is γ. That is, γ isa morphism in C(R), whence the sequence is split.

(iv)=⇒ (v): By (iv) there is a morphism % : M→ I with %β = 1I . The desiredmorphism is thus γ = α%.

(v)=⇒ (vi): Chose by 5.3.15 an injective quasi-isomorphism β : I→ I′, whereI′ is a complex of injective modules such that the functor HomR(–, I′) preservesacyclicity of complexes. By (v) there is a morphism γ : I′→ I with γβ= 1I . Thus Iis a direct summand of I′, see 2.1.44, and by additivity of the Hom functor, I is acomplex of injective modules and HomR(–, I) preserves acyclicity of complexes.

(vi)=⇒ (ii): The functor HomR(–, I) is exact by 5.3.6. For a quasi-isomorphismα, the complex Coneα is acyclic by 4.2.15, and hence so is HomR(Coneα, I). By4.1.16 the latter complex is isomorphic to ΣConeHomR(α, I), and it follows thatHomR(α, I) is a quasi-isomorphism.



5.3.17 Corollary. Let P be an Ro-complex. If P is semi-projective, then the R-complex Homk(P,E) is semi-injective.

PROOF. It follows from 1.3.43 that Homk(P,E) is a complex of injective R-modules.By adjunction 4.4.11 and commutativity 4.4.3 there are natural isomorphisms

HomR(–,Homk(P,E)) ∼= Homk(P⊗R –,E) ∼= HomRo(P,Homk(–,E))

of functors from C(R)op to C(k). By assumption, HomRo(P,Homk(–,E)) preservesacyclicity of complexes. Thus, Homk(P,E) is semi-injective.

5.3.18 Corollary. A graded R-module is graded-injective if and only if it is semi-injective as an R-complex.

PROOF. Let I be a graded R-module. If I is semi-injective as an R-complex, theneach module Iv is injective and hence I is graded-injective.

If I is a graded-injective R-module, then by 5.3.6 it is a direct summand of thecharacter module of a graded-free Ro-module. By 5.1.2 and 5.2.11 a graded-freemodule is semi-projective. Thus 5.3.17 shows that I is a direct summand of a semi-injective R-complex and hence semi-injective by additivity of the Hom functor.

5.3.19 Theorem. Every R-complex M has a semi-injective resolution ι : M '−→ Iwith Iv = 0 for all v > supM\. Moreover, ι can be chosen injective.

PROOF. Apply 5.3.15 to get an injective quasi-isomorphism ι : M→ I. The complexI has Iv = 0 for v > supM\, and it is semi-injective by 5.3.15 and 5.3.16.

PROPERTIES OF SEMI-INJECTIVE COMPLEXES

The class of semi-injective complexes over a ring is closed under extensions, cok-ernels of injective morphisms, direct summands, and products. Semi-injectivity isalso preserved under cobase change.

5.3.20 Proposition. Let 0→ I′→ I→ I′′→ 0 be an exact sequence of R-complexes.If I′ is semi-injective, then I is semi-injective if and only if I′′ is semi-injective.

PROOF. First note that since I′ is a complex of injective modules, it follows from5.3.7 that I is a complex of injective modules if and only if I′′ is so. Next, let A be anacyclic R-complex. The sequence 0→ I′→ I→ I′′→ 0 is degreewise split by 5.3.6,so by applying HomR(A,–) one obtains an exact sequence; see 2.3.12. As I′ is semi-injective, the complex HomR(A, I′) is acyclic by the equivalence of (i) and (v) in5.3.16. It follows from 2.2.20 that HomR(A, I) is acyclic if and only if HomR(A, I′′)is acyclic. Now 5.3.16 yields the desired conclusion.

5.3.21 Proposition. Let Iuu∈U be a family of R-complexes. The product∏

u∈U Iu

is semi-injective if and only if each complex Iu is semi-injective.

PROOF. Let β : K→M be an injective quasi-isomorphism. There is a commutativediagram in C(k),


186 5 Resolutions

HomR(M,∏

u∈UIu)

Hom(β,∏

u∈U Iu)//

∼=

HomR(K,∏

u∈UIu)

∼=∏

u∈UHomR(M, Iu)

∏u∈U Hom(β,Iu)

//∏

u∈UHomR(M, Iu) ,

where the vertical maps are the canonical isomorphisms from 3.1.25. It follows thatHomR(β,

∏u∈U Iu) is a surjective quasi-isomorphism if and only if each morphism

HomR(β, Iu) is a surjective quasi-isomorphism.

Also the next result can be interpreted in terms of the diagram (5.3.5.1).

5.3.22 Proposition. Let I be a semi-injective R-complex, let α : K→ I be a chainmap, and let β : K→M be a quasi-isomorphism. There exists a chain map γ : M→ Isuch that γβ∼ α. Moreover, γ is homotopic to any other chain map γ′ with γ′β∼ α.

PROOF. Recall from 2.3.3 the characterization of (null-homotopic) chain maps as(boundaries) cycles in Hom complexes. By 5.3.16 the induced morphism HomR(β, I)is a quasi-isomorphism, so there exists a γ ∈ Z(HomR(M, I)) such that

[α] = H(HomR(β, I))([γ]) = [γβ] ;

that is, α−γβ is in B(HomR(K, I)). Given another morphism γ′ such that γ′β∼ α,one has [α] = [γ′β] and, therefore 0 = [(γ−γ′)β] = H(HomR(β, I))([γ−γ′]). It fol-lows that the homology class [γ−γ′] is 0 as H(HomR(β, I)) is an isomorphism, soγ−γ′ is in B(HomR(M, I)). That is, γ and γ′ are homotopic.

REMARK. Existence and uniqueness of lifts up to homotopy, as described in 5.3.22, is an impor-tant property of semi-injective complexes, but it does not characterize them. Complexes with thisproperty are called K-injective or homotopically injective; see E 5.3.15.

5.3.23 Corollary. Let I be a semi-injective R-complex and let β : I→M be a quasi-isomorphism. There exists a quasi-isomorphism γ : M→ I such that γβ∼ 1I .

PROOF. By 5.3.22 there exists a chain map γ : M→ I with γβ ∼ 1I ; comparison ofdegrees shows that γ is a morphism. Moreover, by 2.2.26 one has H(γ)H(β) = 1H(I),whence H(γ) is an isomorphism.

Recall from 4.3.4 that every homotopy equivalence is a quasi-isomorphism. Thenext result is a partial converse and akin to 5.2.20.

5.3.24 Corollary. A quasi-isomorphism of semi-injective R-complexes is a homo-topy equivalence.

PROOF. Let β : I→ I′ is a quasi-isomorphism of semi-injective R-complexes. By5.3.23 there are morphisms γ : I′→ I and β′ : I→ I′ such that γβ∼ 1I and β′γ ∼ 1I′

hold. It now follows from 4.3.3 that β is a homotopy equivalence.

5.3.25 Proposition. Let P be a Q-complex and W a complex of Q–T o-bimodules.If P is semi-projective and W is semi-injective over T o, then HomQ(P,W ) is a semi-injective T o-complex.



PROOF. By swap 4.4.9 there is a natural isomorphism

HomT o(–,HomQ(P,W )) ∼= HomQ(P,HomT o(–,W ))

of functors from C(T o)op to C(k). It follows from the assumptions on P and W thatthe functor HomQ(P,HomT o(–,W )) is exact and preserves quasi-isomorphisms.

BOUNDEDNESS

A complex with homology bounded above can be resolved by a bounded abovesemi-injective complex. Such a resolution could, in fact, be constructed degreewise,mimicking the classic construction of an injective resolution of a module.

5.3.26 Theorem. Every R-complex M has a semi-injective resolution M '−→ I withIv = 0 for all v > supM.

PROOF. If M is acyclic, then the morphism M '−→ 0 is the desired resolution. IfH(M) is not bounded above, then any semi-injective resolution of M has the desiredproperty. Assume now that H(M) is bounded above and set u = supM. By 4.2.4there is a quasi-isomorphism M '−→ MĎu. By 5.3.19 the truncated complex MĎuhas a semi-injective resolution MĎu

'−→ I with Iv = 0 for v > u. The desired semi-injective resolution is the composite M '−→MĎu

'−→ I.

THE CASE OF MODULES

Semi-injective resolutions of complexes subsume the classic notion of injective res-olutions of modules. The next result is dual to 1.3.11. A homomorphism M I asbelow is called an injective preenvelope of M.

5.3.27 Proposition. For every R-module M there is an injective homomorphism ofR-modules M→ I where I is injective.


Part (ii) in the next result recovers the lifting property of injective modules 1.3.22;it can be interpreted in terms of the diagram (5.3.5.1).

5.3.28 Proposition. For an R-module I, the following conditions are equivalent.(i) I is injective.

(ii) For every homomorphism α : K→ I and for every injective homomorphismβ : K→M, there exists a homomorphism γ : M→ I such that γβ= α holds.

(iii) Every exact sequence 0→ I→M→M′′→ 0 of R-modules is split.(iv) I is a direct summand of the character module of a free Ro-module.


5.3.29 Proposition. Let 0→ I′→ I→ I′′→ 0 be an exact sequence of R-modules.Assume that I′ is injective, then I is injective if and only if I′′ is injective.


188 5 Resolutions


5.3.30. It follows from 5.3.18 that an R-module is injective if and only if it is semi-injective as an R-complex. Thus one recovers 1.3.23 from 5.3.21.


0−→M −→ I0 −→ ·· · −→ Iv −→ Iv−1 −→ ·· ·

where each module Iv is injective.

PROOF. Choose by 5.3.19 a semi-injective resolution ι : M '−→ I with Iv = 0 for allv > 0 and ι injective. The displayed sequence of R-modules is the complex Cone ι;in particular, the map M0 I0 is the homomorphism ι0. The cone is acyclic becauseι is a quasi-isomorphism; see 4.2.15.

5.3.32 Definition. Let M be an R-module. Together, the injective homomorphismM→ I0 and the R-complex 0→ I0 → ··· → Iv → Iv−1 → ··· in 5.3.31 is called aninjective resolution of M.

5.3.33. By 5.3.12 an injective resolution of an R-module M is a semi-injective re-solution of M as an R-complex.

EXERCISES

E 5.3.1 Show that a graded R-module is graded-injective if and only if it is an injective object inthe category Mgr(R).

E 5.3.2 Show that a graded R-module is graded-injective if it is injective as an R-module. Is theconverse true?

E 5.3.3 For an R-complex I, show that the following conditions are equivalent. (i) I is an injectiveobject in the category C(R). (ii) I is a contractible complex complex of injective R-modules. (iii) I is semi-injective and acyclic. Show that the category C(R) has enoughinjectives.

E 5.3.4 Show that the category C(R) has enough injectives. That is, for every R-complex Mthere is an injective morphism M → I where I is a contractible complex of injectiveR-modules; cf. E 5.3.3.

E 5.3.5 Show that the Dold complex from 5.1.4 is an acyclic complex of injective modules.Show that it is not contractible and conclude that it is not semi-injective.

E 5.3.6 Show that a complex I of injective R-modules is semi-injective if HomR(A, I) is acyclicfor every acyclic R-complex A that is bounded below.

E 5.3.7 Let R→ S be a righ homomoorphism. Show that for a semi-injective R-complex I, theS-complex HomR(S, I) is semi-injective.

E 5.3.8 Let I be a complex of Q–Ro-bimodules that are all injective over Q and F be a complexof flat R-modules. Show that if Q is left Noetherian, then I⊗R F is a complex of injectiveQ-modules; conclude that if I and F are bounded above, then I⊗R F is semi-injective.

E 5.3.9 Show that the mapping cone of a morphism between semi-injective complexes is semi-injective.

E 5.3.10 Let M be an R-complex. Show that for every semi-injective replacement I of M themodules Bv(P) are injective for v > supM.

E 5.3.11 Show that every complex over a semi-simple ring is semi-injective.



E 5.3.12 Let R be left hereditary. Show that for every complex I of injective R-modules there is aquasi-isomorphism H(I) '−→ I. Hint: See E 1.4.8.

E 5.3.13 Show that a complex of injective modules over a left hereditary ring is semi-injective.E 5.3.14 Let 0→ I′ → I→ I′′ → 0 be a degreewise split exact sequence of R-complexes. Show

that if two of the complexes I, I′, and I′′ are semi-injective, then so is the third.E 5.3.15 Show that the following conditions are equivalent for an R-complex Y . (i) For every

chain map α : K→ Y and every quasi-isomorphism β : K→M there exists a chain mapγ : M→ Y , unique up to homotopy, such that γβ ∼ α. (ii) For every quasi-isomorphismβ the induced morphism HomR(β,Y ) is a quasi-isomorphism. (iii) For every acycliccomplex A, the complex HomR(A,Y ) is acyclic.

A complex with these properties is called K-injective or homotopically injective.E 5.3.16 Let K be an acyclic K-injective R-complex. Show that HomR(M,K) is acyclic for every

R-complex M.E 5.3.17 Show that a complex is semi-injective if and only if it is a complex of injective modules

and K-injective in the sense of E 5.3.15. Give an example of a K-injective complex thatis not semi-injective.

E 5.3.18 Show that a graded R-module is graded-injective if and only if it is K-injective as anR-complex, in the sense of E 5.3.15.

E 5.3.19 Show that the class of K-injective R-complexes, in the sense of E 5.3.15, is closed underhomotopy equivalence. Is the same true for the class of semi-injective R-complexes?

E 5.3.20 Let α : M→ N be a morphism in C(R). Show that for any two semi-injective resolutionsι : M '−→ I and ε : N '−→ E there is a morphism α : I→ E such that the diagram

Mι

'//

α

I

α

Nε

'// E

is commutative up to homotopy. Show also that if ι is injective, then α can be chosensuch that the diagram is commutative.

E 5.3.21 Let x be a central element in R and let M '−→ I be a semi-injective resolution of R-complexes. Show that if the homothety xM is null-homotopic, then xI is null-homotopic.

E 5.3.22 Show that every morphism α : M→ N of R-complexes admits factorizations in C(R),

M

ε

α// N

Xϕ

'CC CC

andM

ι'

α// N

Yπ

DD DD

where ε and ι are injective; ϕ and π are surjective with semi-injective kernels; and ϕ andι are quasi-isomorphisms. Hint: (1) E 5.3.4. (2) Modify the first step in 5.3.14

E 5.3.23 Give an alternative proof of 5.3.31 based on 5.3.27.

E 5.3.24 Let 0−→M′ α′−→M α−→M′′ −→ 0 be an exact sequence of R-complexes. Show that there

is a commutative diagram in C(R) in which the columns are semi-injective resolutions,

0 // M′α′//

ι′'

Mα//

ι'

M′′ //

ι′′'

0

0 // I′α′// I

α// I′′ // 0 .

This is known as the Horseshoe Lemma for semi-injective resolutions.E 5.3.25 Let I be a complex of injective modules. Show that for every n 6 inf I the truncated

complex Iďn yields an injective resolution of the module Zn(I).


190 5 Resolutions

5.4 Semi-flatness

SYNOPSIS. Graded-flat module; complex of flat modules; semi-flat complex; perfect ring.

For every complex of flat modules, the functor –⊗ F preserves short exact se-quences of complexes, see 2.4.16, but not necessarily acyclicity of complexes.Adding this as a requirement, one arrives at the notion of semi-flatness.

COMPLEXES OF FLAT MODULES

5.4.1 Proposition. For an R-complex F , the following conditions are equivalent.(i) Each R-module Fv is flat.

(ii) The functor –⊗R F is exact.(iii) For every exact sequence 0→M′→M→ F → 0 in C(R) the exact sequence

0→ Homk(F,E)→ Homk(M,E)→ Homk(M′,E)→ 0 is degreewise split.(iv) The character complex Homk(F,E) is a complex of injective Ro-modules.

PROOF. Conditions (i) and (iv) are equivalent by 1.3.43.(ii)⇐⇒ (iv): By adjunction 4.4.11 and commutativity 4.4.3 there is a natural

isomorphism of functors from C(Ro)op to C(k),

HomRo(–,Homk(F,E)) ∼= Homk(–⊗R F ,E) .

By 5.3.6 the functor on the left-hand side is exact if and only if Homk(F,E) is acomplex of injective Ro-modules. As E is faithfully injective, the functor on theright-hand side is exact if and only if –⊗R F is exact.

(iv)=⇒ (iii): Immediate from 5.3.6.(iii)=⇒ (i): Choose by 5.1.7 a surjective semi-free resolution π : L '−→ F and

consider the associated short exact sequence 0→ Kerπ→ L→ F → 0. By 5.3.2the complex Homk(L,E) consists of injective Ro-modules, so it follows from 5.3.6and split exactness of the sequence

0−→ Homk(F,E)\ −→ Homk(L,E)\ −→ Homk(Kerπ,E)\ −→ 0

that each module Homk(F,E)−v = Homk(Fv,E) is an injective Ro-module, whenceeach Fv is a flat R-module by 1.3.43.

5.4.2 Corollary. Let 0→ F ′→ F→ F ′′→ 0 be an exact sequence of R-complexes.If F ′′ is a complex of flat modules, then F is a complex of flat modules if and onlyif F ′ is a complex of flat modules.

PROOF. Apply 5.4.1 and 5.3.7 to the exact sequence of Ro-complexes

0−→ Homk(F ′′,E)−→ Homk(F,E)−→ Homk(F ′,E)−→ 0 .

5.4.3 Corollary. Let 0→M′→M→ F→ 0 be an exact sequence in C(R). If F is acomplex of flat modules, then the sequence 0→N⊗R M′→N⊗R M→N⊗R F→ 0is exact for every Ro-complex N.


5.4 Semi-flatness 191

PROOF. Assume that F is a complex of flat R-modules, and let N be an Ro-complex.As the tensor product is right exact, it is sufficient to show that the induced mor-phism N⊗R M′→ N⊗R M is injective. There is a commutative diagram in C(k),

Homk(M⊗Ro N,E) //

ρNME∼=

Homk(M′⊗Ro N,E) //

ρNM′E∼=

0

HomRo(N,Homk(M,E)) // HomRo(N,Homk(M′,E)) // 0 .

The vertical maps are adjunction isomorphisms. The lower row is exact by 5.4.1and the fact that the functor HomR(N,–) preserves degreewise split exactness ofsequences; cf. 2.3.12. By commutativity of the diagram, the upper row is also exact.As E is faithfully injective, this implies that the sequence 0→M′⊗Ro N→M⊗Ro Nis exact, and commutativity 4.4.3 finishes the proof.

5.4.4 Definition. A graded R-module F is called graded-flat if the R-complex Fsatisfies the conditions in 5.4.1.

EXISTENCE OF SEMI-FLAT COMPLEXES

5.4.5 Definition. An R-complex F is called semi-flat if β⊗R F is an injective quasi-isomorphism for every injective quasi-isomorphism β in C(Ro).

REMARK. Another word for semi-flat is DG-flat.

5.4.6. Note that by 2.4.12 an R-complex F is semi-flat if and only if ΣF is semi-flat.

5.4.7 Example. A contractible complex of flat modules is semi-flat by 5.4.1, 4.3.20,and 4.3.27.

5.4.8 Example. Let F be a bounded below complex of flat R-modules and let β bean injective quasi-isomorphism in C(Ro). The morphism β⊗R F is injective by 5.4.1.The complex Coneβ is acyclic by 4.2.15, and hence so is (Coneβ)⊗R Fv for everyv ∈ Z. As F is bounded below, it follows from 4.1.18 and A.14 that the complexCone(β⊗R F) ∼= (Coneβ)⊗R F is acyclic, whence β⊗R F is a quasi-isomorphism.Thus, F is is semi-flat.

Semi-projectivity and semi-injectivity are categorically dual notions. By adjoint-ness of Hom and tensor product, semi-flatness is also, in a different sense, dualto semi-injectivity. The next result gives useful characterizations of semi-flat com-plexes.

5.4.9 Proposition. For an R-complex F , the following conditions are equivalent.(i) F is semi-flat.

(ii) The functor –⊗R F is exact and preserves quasi-isomorphisms.(iii) The character complex Homk(F,E) is a semi-injective Ro-complex.


192 5 Resolutions

(iv) F is a complex of flat R-modules and the functor –⊗R F preserves acyclicityof complexes.

PROOF. The implication (ii)=⇒ (i) is trivial. By commutativity 4.4.3 and adjunc-tion 4.4.11 there is a natural isomorphism of functors from C(Ro)op to C(k),

Homk(–⊗R F ,E) ∼= HomRo(–,Homk(F,E)) .

Thus it follows from 2.5.6b and 5.3.16 and that (ii) and (iii) are equivalent.(iii)=⇒ (iv): It follows from 5.3.16 that Homk(F,E) is a complex of injective

Ro-modules, so F is a complex of flat R-modules by 5.4.1. Let A be an acyclic Ro-complex; by adjunction 4.4.11 and commutativity 4.4.3 there is an isomorphismHomRo(A,Homk(F,E)) ∼= Homk(A⊗R F ,E). The left-hand complex is acyclic by5.3.16, so it follows by faithfulness of the functor Homk(–,E) that A⊗R F is acyclic.

(iv)=⇒ (i): Let β be an injective quasi-isomorphism. The morphism β⊗R F isthen injective by 5.4.1. The complex Coneβ is acyclic by 4.2.15, and hence so isthe complex (Coneβ)⊗R F ∼= Cone(β⊗R F), where the isomorphism comes from4.1.18. It follows that β⊗R F is a quasi-isomorphism.

REMARK. The characterizations of semi-projective 5.2.10 and semi-injective 5.3.16 complexesinclude lifting statements, but 5.4.9 does not. As far as it is possible, C.3 makes up for this.

5.4.10 Corollary. Every semi-projective R-complex is semi-flat; in particular, everysemi-free R-complex is semi-flat.

PROOF. Immediate in view of 5.3.17 and 5.2.9.

REMARK. It is a fact, but not obvious, that a semi-flat complex of projective modules is semi-projective. It follows from work of Neeman [64] as explained by Murfet [61]; see also E C.7.

5.4.11 Corollary. A graded R-module is graded-flat if and only if it is semi-flat asan R-complex.

PROOF. If F is semi-flat as a complex, then each module Fv is flat by 5.4.9, whenceF is graded-flat. If F is graded-flat, then the character module Homk(F,E) is graded-injective by 5.4.1 and hence semi-injective as an Ro-complex; see 5.3.18. As anR-complex, F is then semi-flat.

PROPERTIES OF SEMI-FLAT COMPLEXES

The class of semi-flat complexes over a ring is closed under extensions, kernels ofsurjective morphisms, direct summands, and filtered colimits. Semi-flatness is alsopreserved under base change.

5.4.12 Proposition. Let 0 → F ′ → F → F ′′ → 0 be an exact sequence of R-complexes. If F ′′ is semi-flat, then F is semi-flat if and only if F ′ is semi-flat.

PROOF. Apply 5.4.9 and 5.3.20 to the exact sequence of Ro-complexes

0−→ Homk(F ′′,E)−→ Homk(F,E)−→ Homk(F ′,E)−→ 0 .



5.4.13 Proposition. Let µvu : Fu→ Fvu6v be a U-direct system of semi-flat R-complexes. If U is filtered, then colimu∈U Fu is semi-flat.

PROOF. Let β : K→M be an injective quasi-isomorphism in C(Ro). There is a com-mutative diagram in C(k),

colimu∈U

(K⊗R Fu)colimu∈U (β⊗Fu)

//

∼=

colimu∈U

(M⊗R Fu)

∼=

K⊗R (colimu∈U

Fu)β⊗(colimu∈U Fu)

// M⊗R (colimu∈U

Fu) ,

where the vertical maps are the canonical isomorphisms (3.2.17.1). In view of 3.2.32and 3.2.34 it follows that β⊗R (colimFu) is an injective quasi-isomorphism if eachmap β⊗R Fu is an injective quasi-isomorphism.

Direct sums and direct summands of semi-flat complexes are semi-flat, and soare coproducts of semi-flat complexes.

5.4.14 Corollary. Let Fuu∈U be a family of R-complexes. The coproduct∐

u∈U Fu

is semi-flat if and only if each complex Fu is semi-flat.

PROOF. It is immediate from 3.2.31 and 5.4.13 that a coproduct of semi-flat com-plexes is semi-flat, and it is evident from the definition that a direct summand of asemi-flat complex is semi-flat.

Contrary to the situation for semi-projective and semi-injective complexes, aquasi-isomorphism of semi-flat R-complexes need not be a homotopy equivalence.

5.4.15 Example. It follows from 1.3.10, 1.3.11, and 5.1.18 that the Z-module Q hasa semi-free resolution π : L '−→ Q with Lv = 0 for v 6= 0,1. Both Z-complexes Q andL are semi-flat. Suppose γ : Q→ L were a homotopy inverse of π, then one wouldhave 1Q ∼ πγ, and hence 1Q = πγ as ∂Q = 0. This would imply that Q is a directsummand of L0 and hence a free Z-module, but it is not.

Quasi-isomorphisms between semi-flat complexes still exhibit some robustness;two important instances are captured by the next proposition and C.24.

5.4.16 Proposition. Let α : F → F ′ be a quasi-isomorphism of semi-flat R-complexes.For every Ro-complex M, the morphism M⊗R α is a quasi-isomorphism.

PROOF. By 5.3.16 and 5.4.9 the morphism Homk(α,E) is a quasi-isomorphism ofsemi-injective Ro-complexes and hence a homotopy equivalence by 5.3.24. There-fore, the upper horizontal map in the following commutative diagram is also a ho-motopy equivalence; see 4.3.19.


194 5 Resolutions

HomRo(M,Homk(F ′,E))Hom(M,Hom(α,E))

u// HomRo(M,Homk(F,E))

Homk(F ′⊗Ro M,E)

ρMF ′E ∼=

OO

Hom(υMF ′ ,E) ∼=

Homk(F⊗Ro M,E)

ρMFE∼=

OO

Hom(υMF ,E)∼=

Homk(M⊗R F ′,E)Hom(M⊗α,E)

// Homk(M⊗R F ,E)

The diagram shows that Homk(M⊗R α,E) is a quasi-isomorphism, and by faithfulinjectivity of E it follows that M⊗R α is a quasi-isomorphism; cf. 4.2.13.

5.4.17 Proposition. Let F be a T -complex and W a complex of Q–T o-bimodules.If F is semi-flat and W is semi-flat over Q, then the Q-complex W ⊗T F is semi-flat.

PROOF. Associativity 4.4.6 yields a natural isomorphism,

–⊗Q (W ⊗T F) ∼= (–⊗Q W )⊗T F ,

of functors from C(Qo) to C(k). By the assumptions on F and W , the functor(–⊗Q W )⊗T F is exact and preserves quasi-isomorphisms.

5.4.18 Corollary. Let R→ S be a ring homomorphism.(a) For a semi-flat R-complex F the S-complex S⊗R F is semi-flat.(b) If S is flat as an R-module, then a semi-flat S-complex is semi-flat over R.

PROOF. For (a) apply 5.4.17 with Q = S, T = R, and W = SSR. For (b) use Q = R,T = S, and W = RSS and note that W ⊗S – is the forgetful functor C(S)→ C(R).

5.4.19 Proposition. Let I be a Q-complex and W a complex of Q–T o-bimodules.If I is semi-injective and W is semi-flat over T o, then the T -complex HomQ(W, I) issemi-injective.

PROOF. Adjunction 4.4.11 yields a natural isomorphism

HomT (–,HomQ(W, I)) ∼= HomQ(W ⊗T –, I)

of functors from C(T )op to C(k). It follows from the assumptions on I and W thatthe functor HomQ(W ⊗T –, I) is exact and preserves quasi-isomorphisms.

5.4.20 Corollary. Let R→ S be a ring homomorphism.(a) For a semi-injective R-complex I the S-complex HomR(S, I) is semi-injective.(b) If S is flat as an Ro-module, then a semi-injective S-complex is semi-injective

over R.

PROOF. For (a) apply 5.4.19 with Q = R, T = S, and W = RSS. For (b) use Q = S,T =R, W = SSR and note that HomS(W,–) is the forgetful functor C(S)→C(R).



THE CASE OF MODULES

5.4.21 Proposition. For an R-module F , the following conditions are equivalent.(i) F is flat.

(ii) For every exact sequence 0→ M′ → M → F → 0 of R-modules, the exactsequence 0→ Homk(F,E)→ Homk(M,E)→ Homk(M′,E)→ 0 is split.

(iii) The character module Homk(F,E) is an injective Ro-module.


5.4.22 Corollary. Let 0→ F ′→ F → F ′′→ 0 be an exact sequence of R-modules.Assume that F ′′ is flat, then F ′ is flat if and only if F is flat.


5.4.23. It follows from 5.4.11 that an R-module is flat if and only if it is semi-flat asan R-complex.

5.4.24 Proposition. Let µvu : Fu→ Fvu6v be a U-direct system of flat R-modules.If U is filtered, then colimu∈U Fu is flat.


REMARK. In contrast to 5.4.24, Bergman [13] shows that every module is a filtered limit of injec-tive modules.

5.4.25 Proposition. Let Fuu∈U be a family of R-modules. The coproduct∐

u∈U Fu

is flat if and only if each module Fu is flat.


PERFECT RINGS

We end this chapter with a homological characterization of perfect rings.

5.4.26 Lemma. Assume that every flat R-module is projective. For every sequence(ai)i∈N in R there exists n> 1 such that for every j > n there is an equality of rightideals, (a1 · · ·a j−1)R = (a1 · · ·a j−1a j)R.

PROOF. For 1 6 i < j let α ji be the homothety given by right multiplication on Rwith ai · · ·a j−1 and set αii = 1R. These maps form a direct system of R-modules;let A denote its colimit. By 5.4.24 the module A is flat; hence it is projective bythe assumption on R. Set L = R(N) and let ιi : R L be the embedding into the ith

component; note that ιi(1)i∈N is a basis for L. Set fi = ιi(1)−aiιi+1(1) for every

i ∈ N. The elements fii∈N in L are linearly independent as one has

r1 f1 + r2 f2 + · · ·+ rm fm

= r1ι1(1)+(r2− r1a1)ι

2(1)+ · · ·+(rm− rm−1am−1)ιm(1)− rmamι

m+1(1)


196 5 Resolutions

for r1,r2, . . . ,rm in R. Denote by F the free submodule of L generated by fii∈N.Since one has ιi(r)− ι jα ji(r) = r fi + rai fi+1 + · · ·+ rai · · ·a j−2 f j−1, it follows from3.2.2 that there is an exact sequence 0 → F → L → A → 0. As A is projective,the embedding F L has a right inverse π : L→ F ; cf. 1.3.15. Write π(ιi(1)) =∑ j>1 bi j f j; then one has

fi = π( fi) = π(ιi(1)−aiιi+1(1)) = ∑

j>1(bi j−aib(i+1) j) f j

and, therefore, bii−aib(i+1)i = 1 and bi j−aib(i+1) j = 0 for all j 6= i. Thus, for everyj > 1 there are equalities,

b1 j = a1b2 j = a1a2b3 j = · · · = a1a2 · · ·a j−1b j j = a1a2 · · ·a j−1(1+a jb( j+1) j) ,

and hence a1 · · ·a j−1 = b1 j− a1 · · ·a j−1a jb( j+1) j. Since there exists an n such thatone has b1 j = 0 for all j > n, the desired assertion follows.

5.4.27 Lemma. If the Jacobson radical of R is zero and every descending chain ofprincipal right ideals in R becomes stationary, then R is semi-simple.

PROOF. First note that every right ideal a 6= 0 in R contains a minimal right idealb; indeed, take b minimal among the non-zero principal right ideals contained in a.Furthermore, every minimal right ideal b in R has a complement. Indeed, as theJacobson radical of R is zero, one has b*M for some maximal right ideal M; andsince b is minimal, b∩M= 0 follows. Consequently, R = b⊕M holds.

Now, let b1 be a minimal right ideal in R and write R = b1⊕ a1 for some rightideal a1. If a1 = 0 then the Ro-module R = b1 is simple. Otherwise, let b2 be aminimal right ideal contained in a1, and write a1 = b2⊕ a2 for some right ideala2; then one has R = b1⊕ b2⊕ a2. If a2 = 0 then the Ro-module R = b1⊕ b2 issemi-simple. If a2 6= 0 one can continue the process, which after n iterations yieldsminimal right ideals b1,b2, . . . ,bn and right ideals a1 ⊃ a2 ⊃ ·· · ⊃ an such that R =b1⊕·· ·⊕bn⊕an. Each right ideal an is principal, as it is a direct summand of R, sothe process terminates with an = 0 for some n. Thus the Ro-module R = b1⊕·· ·⊕bnis semi-simple.

5.4.28 Theorem. The ring R is left perfect if an only if every flat R-module isprojective.

PROOF. Let J be the Jacobson radical of R.“Only if”: Let F be a flat R-module. By B.51 it has projective cover, so there is

a projective R-module P with a superfluous submodule K such that P/K is flat. By1.3.39 one has JK = JP∩K. As K is a superfluous submodule of P it is containedin JP by B.35. Thus one has JK = K and, therefore, K = 0 by B.47.

“If”: To show that J is left T-nilpotent, let (ai)i∈N be a sequence in J. By 5.4.26there exist n> 1 and r ∈ R such that one has a1 · · ·an = a1 · · ·anan+1r, and thereforea1 · · ·an(1− an+1r) = 0. Since an+1 is in J, the element 1− an+1r is a unit, and itfollows that a1 · · ·an = 0. It remains to show that the ring k = R/J is semi-simple;to this end apply 5.4.27. Clearly, the Jacobson radical of k is zero. A descendingchain of principal right ideals in k has the form (a1)k ⊇ (a1a2)k ⊇ (a1a2a3)k ⊇ ·· ·



with ai ∈ R. It follows from 5.4.26 that the descending chain (a1)R ⊇ (a1a2)R ⊇(a1a2a3)R⊇ ·· · in R becomes stationary, and hence so does the chain in k.

REMARK. The statement of the previous theorem is part of Bass’ Theorem P [10] from 1960. Withthe existence of flat covers for all modules, which was only proved in 2001 by Bican, El Bashir,and Enochs [14], a short proof of the “if” part in 5.4.28 became available. Indeed, every R-moduleM has a flat cover π : F M, so if every flat R-module is projective, then π is a projective cover.

EXERCISES

E 5.4.1 Show that a graded R-module is graded-flat if and only if it is flat as an R-module.E 5.4.2 Show that the Dold complex from 2.1.22 is not semi-flat.E 5.4.3 Show that a complex of flat modules over a right hereditary ring is semi-flat.E 5.4.4 Let S be a faithfully flat k-algebra. Show that a k-complex F is semi-flat if and only if

the S-complex S⊗k F is semi-flat.E 5.4.5 Let I be a complex of R–So-bimodules that are all injective over So and J be a complex

of injective R-modules. Show that if S is right Noetherian, then HomR(I,J) is a complexof flat S-modules; conclude that if I is bounded above and J is bounded below, thenHomR(I,J) is semi-flat.

E 5.4.6 Let P be a complex of projective R-modules and F be a complex of R–So-bimodulesthat are all flat over So. Show that if S is left Noetherian, then HomR(P,F) is a complexof flat So-modules; conclude that if P is bounded above and F is bounded below, thenHomR(P,F) is semi-flat.

E 5.4.7 Show that a complex F of flat R-modules is semi-flat if A⊗R F is acyclic for everyacyclic Ro-complex A that is bounded below.

E 5.4.8 Show that the mapping cone of a morphism between semi-flat complexes is semi-flat.E 5.4.9 Let F ′′ be a complex of flat R-modules and let 0→ F ′ → F → F ′′ → 0 be an exact

sequence of R-complexes. Show that if two of the complexes F , F ′, and F ′′ are semi-flat, then so is the third.

E 5.4.10 Show that the following conditions are equivalent for an R-complex F . (i) F is semi-flat.(ii) The complex Fďn is semi-flat for every n ∈ Z. (iii) F is a complex of flat R-modulesand Fďn is semi-flat for some n ∈ Z.

E 5.4.11 Use E 5.3.3 to give a shorter proof of 5.4.16.E 5.4.12 Find an exact, but not split, sequence 0→M′→M→M′′→ 0 of R-modules with M′′

not flat such that 0→ Homk(M′′,E)→ Homk(M,E)→ Homk(M′,E)→ 0 is split.E 5.4.13 Show that the following conditions are equivalent for an R-complex Z. (i) For every

quasi-isomorphism β in C(Ro) the induced morphism β⊗R Z is a quasi-isomorphism.(ii) For every acyclic Ro-complex A, the complex A⊗R Z is acyclic.

A complex with these properties is called K-flat or homotopically flat.E 5.4.14 Let K be an acyclic K-flat R-complex. Show that M⊗R K is acyclic for every Ro-

complex M.E 5.4.15 Show that a complex is semi-flat if and only if it is a complex of flat modules and K-flat

in the sense of E 5.4.13. Give an example of a K-flat complex that is not semi-flat.E 5.4.16 Show that a graded R-module is graded-flat if and only if it is K-flat as an R-complex,

in the sense of E 5.4.13.E 5.4.17 Show that the class of K-flat R-complexes, in the sense of E 5.4.13, is closed under

homotopy equivalence. Is the same true for the class of semi-flat R-complexes?

E 5.4.18 Let k be a field and set R= kN. Show that a= k(N) is and ideal in R and that the complexF = 0→ a→ R→ R/a→ 0 is semi-flat; cf. 1.3.40. Show that F is not contractible.


198 5 Resolutions

E 5.4.19 Consider the following subsets of M2×2(R),

R =

(x y0 x

)∣∣∣∣ x ∈ Q and y ∈ R

and J =

(0 y0 0

)∣∣∣∣ y ∈ R

.

(a) Show that R is a commutative ring with Jacobson radical J. (b) Show that R is perfectbut not Artinian. Hint: Pick an infinite descending sequence of Q-submodules of R.


Chapter 6The Derived Category

From any Abelian category U one can obtain a triangulated category, called the de-rived category of U. In this chapter, we construct the derived category of the modulecategory M(R). This category, D(R), which is also called the derived category overR, provides an efficacious environment for homological studies of R-modules. Theobjects in D(R) are familiar, they are simply all R-complexes, but the morphismsmay appear odd on first encounter: They are not maps but equivalence classes ofcertain diagrams. This part of the construction is technically involved but conceptu-ally simple: the goal is to turn all quasi-isomorphisms into isomorphisms, which isachieved by a procedure that emulates the construction of a ring of fractions.

A three-step process leads from M(R) to D(R): The first step was taken inChap. 2 with the introduction of the category of R-complexes, and what followsimmediately below is the second step.

6.1 Construction of the Homotopy Category KKK

SYNOPSIS. Homotopy category; quasi-isomorphism; product; coproduct; homotopy invariantfunctor; universal property; functor that preserves homotopy.

Let U be a category and let ≈ be a congruence relation on U. The quotient categoryU/≈ has the same objects as U, and for two such objects M, N the hom-set in U/≈is the set U(M,N)/≈ of equivalence classes. The canonical functor Q: U→ U/≈has the following universal property: If F : U→ V is any functor that is invariantunder ≈, that is, F(α) = F(β) holds whenever α≈ β, then there is a unique functorF that makes the next diagram commutative,

U

Q

F// V

U/≈ .

F

==

199

200 6 The Derived Category

This section is focused on a particular quotient category: the homotopy categoryK(R). It is the quotient of C(R) modulo homotopy. While C(R) is Abelian, thecategory K(R) is, in general, not. However, it is a triangulated category; cf. Appn. D.

OBJECTS AND MORPHISMS

The next definition is justified by the fact that homotopy is a congruence relation inC(R); details are recalled in 6.1.2.

6.1.1 Definition. The homotopy category K(R) has the same objects as C(R), i.e.R-complexes; the morphisms in K(R) are homotopy classes of morphisms in C(R).

6.1.2. For R-complexes M and N there is, by 2.3.10, an equality of k-modulesK(R)(M,N) = H0(HomR(M,N)). In accordance with 2.2.14 we write [α] for thehomotopy class of a morphism α in C(R). If L is also an R-complex, then the com-position K(R)(M,N)×K(R)(L,M)→K(R)(L,N) maps ([α], [β]) to [αβ]; it followsfrom 2.2.25 that [αβ] does not depend on the choice of representatives for [α] and[β].

6.1.3. There is a canonical full functor Q: C(R)→K(R); it is the identity on objectsand it maps a morphism α in C(R) to its homotopy class Q(α) = [α].

We rephrase part of 2.5.7 in the language of the homotopy category.

6.1.4. For R-complexes M and N with Mv = 0 for all v < 0 and Nv = 0 for all v > 0the homomorphism C(R)(M,N)→K(R)(M,N) induced by Q is an isomorphism.

6.1.5 Proposition. The restriction to M(R) of the functor Q: C(R)→K(R) yieldsan isomorphism between the module category M(R) and the full subcategory ofK(R) whose objects are all R-complexes concentrated in degree 0.


6.1.6. The homotopy class [α] of a morphism in C(R) is an isomorphism in K(R)if and only if α is a homotopy equivalence; see 4.3.1. Thus the symbol ‘u’ is alsoused for isomorphisms in K(R).

An isomorphism α in C(R) yields an isomorphism [α] in K(R); such isomor-phisms in K(R) may still be marked by the symbol ‘∼=’.

6.1.7. The zero complex is evidently a zero object in the homotopy category, so amorphism [α] in K(R) is zero if and only if α is null-homotopic; see 2.2.23.

The next result identifies the zero objects in K(R). Further characterizations ofthese complexes are given in 4.3.32.

6.1.8 Proposition. An R-complex is a zero object in K(R) if and only if it is con-tractible.


6.1 Construction of K 201

PROOF. Let M be an R-complex. If M is isomorphic to 0 in K(R), then K(R)(M,M)consists of a single element. In particular, [1M] = [0] holds, so 1M is null-homotopic,i.e. M is contractible. Conversely, if M is contractible then the morphism M→ 0 inC(R) is a homotopy equivalence, whence it represents an isomorphism in K(R).

A conspicuous consequence of 4.3.30 and 6.1.8 is that the homotopy class [α] ofa morphism in C(R) is an isomorphism in K(R) if and only if the complex Coneαis isomorphic to 0 in K(R).

PRODUCTS AND COPRODUCTS

6.1.9 Lemma. Let U and V be k-prelinear categories that have the same objects,and let F: U→ V be a k-linear functor that is the identity on objects. Let M and Nbe objects; if the tuple (M⊕N,$M,εM,$N ,εN) is a biproduct in U, then the tuple(M⊕N,F($M),F(εM),F($N),F(εN)) is a biproduct in V. In particular, if every pairof objects has a biproduct in U then every pair of objects has a biproduct in V.

PROOF. All assertions follow immediately from the definitions.

Recall that a category is said to have products/coproducts if all set-indexed prod-ucts/coproducts exist in the category.

6.1.10 Theorem. The homotopy category K(R) and the functor Q: C(R)→K(R)are k-linear. For every family Muu∈U of R-complexes the next assertions hold.

(a) If M with injections εu : MuMu∈U is the coproduct of Muu∈U in C(R),then M with the morphisms [εu]u∈U is the coproduct of Muu∈U in K(R).

(b) If M with projections $u : MMuu∈U is the product of Muu∈U in C(R),then M with the morphisms [$u]u∈U is the product of Muu∈U in K(R).

In particular, the homotopy category K(R) has products and coproducts, and thecanonical functor Q preserves products and coproducts.

PROOF. It is straightforward to verify that the category K(R) is k-prelinear and thatthe canonical functor Q is k-linear. The zero complex is a zero object in K(R), see6.1.7, and K(R) has biproducts by 6.1.9. Thus K(R) is a k-linear category.

(a): Let [αu] : Mu→ Nu∈U be morphisms in K(R). The task is to show thatthere exists a unique morphism [α] : M→ N in K(R) with [αεu] = [αu] for all u ∈U .Existence is straightforward; indeed, by the universal property of coproducts inC(R), there is a morphism α : M→ N with αεu = αu for all u ∈U . Applying Q tothese identities one gets [αεu] = [αu]. For uniqueness, assume that [αεu] = [0] holdsfor every u ∈U ; it must be shown that [α] is [0], and that is immediate from 3.1.6.

(b): Similar to the proof of part (a); only this time appeal to 3.1.18.By construction, the canonical functor Q preserves products and coproducts.

6.1.11. Let [αu] : Mu→ Nuu∈U and [βu] : Nu→Muu∈U be families of mor-phisms in K(R). By the universal properties of (co)products there are unique mor-phisms [α] and [β] that make the following diagrams commutative for every u ∈U ,



Mu

[αu]

//∐

u∈UMu

[α]

Nu //∐

u∈UNu

and

∏u∈U

Nu

[β]//∏

u∈UMu

Nu [βu]// Mu .

The horizontal morphisms in the left-hand diagram are injections and the verticalmorphisms in the right-hand diagram are projections. The morphism [α] is calledthe coproduct in K(R) of [αu]u∈U and denoted

∐u∈U [α

u]. Similarly, [β] is calledthe product in K(R) of [βu]u∈U and denoted

∏u∈U [β

u]. From the construction ofproducts and coproducts in K(R) given in 6.1.10, it follows that one has∐

u∈U

[αu] =[∐

u∈U

αu] and∏u∈U

[βu] =[∏

u∈U

βu] ,where

∐u∈U α

u and∏

u∈U βu are the coproduct of αuu∈U in C(R) and the product

of βuu∈U in C(R); see 3.1.4 and 3.1.16.

6.1.12 Definition. A morphism [α] in K(R) is called a quasi-isomorphism if some,equivalently every, morphism in C(R) that represents the homotopy class [α] is aquasi-isomorphism; cf. 2.2.26. Notice that by 4.3.4 and 6.1.6 every isomorphism inK(R) is a quasi-isomorphism. Quasi-isomorphisms in K(R) are also marked by thesymbol ‘'’; cf. 4.2.1.

A morphism in K(R)op is called a quasi-isomorphism if the corresponding mor-phism in K(R) is a quasi-isomorphism as defined above; this is in line with 4.3.14.

6.1.13 Proposition. Let [αu] : Mu→ Nuu∈U be a family of morphisms in K(R).If [αu] is a quasi-isomorphism for every u∈U , then the coproduct

∐u∈U [α

u] and theproduct

∏u∈U [α

u] are quasi-isomorphisms.


UNIVERSAL PROPERTY

6.1.14 Definition. A functor F: C(R)→ U is called homotopy invariant if one hasF(α) = F(β) for all homotopic morphisms α∼ β in C(R).

A functor G: C(R)op→ V is called homotopy invariant if G(α) = G(β) holds forall homotopic morphisms α∼ β in C(R)op; cf. 4.3.14.

Notice that a functor C(R)op→V is homotopy invariant if and only if the oppositefunctor C(R)→ Vop is homotopy invariant as defined in the first part of 6.1.14.

6.1.15 Example. Homology H: C(R)→ C(R) is homotopy invariant by 2.2.26.

The canonical functor Q: C(R)→K(R) from 6.1.3 has the universal propertydescribed in the next theorem.



6.1.16 Theorem. Let U be a category and let F: C(R)→ U be a functor. If F ishomotopy invariant, then there exists a unique functor F that makes the followingdiagram commutative,

C(R)

Q

F// U

K(R) .F

==

For every R-complex M there is an equality F(M) = F(M), and for every morphism[α] in K(R) one has F([α]) = F(α). Furthermore, the following assertions hold.

(a) If U is k-prelinear and F is k-linear, then F is k-linear.(b) If U has products/coproducts and F preserves products/coproducts, then F

preserves products/coproducts.

PROOF. Uniqueness of F follows as Q is the identity on objects and full. For exis-tence, set F(M) = F(M) for every R-complex M and F([α]) = F(α) for every mor-phism α of R-complexes; the latter assignment is well-defined as F is homotopyinvariant. Evidently, F is a functor with FQ = F. It remains to prove (a) and (b).

(a): If F is k-linear, then so is F as the equalities

F(x[α]+ [β]) = F([xα+β]) = F(xα+β) = xF(α)+F(β) = x F([α])+ F([β])

hold for every pair α,β of parallel morphisms in C(R) and every element x in k.(b): Let Muu∈U be a family of R-complexes, and let M be its coproduct in C(R)

with injections εu : MuMu∈U . It follows from 6.1.10 that M with injections[εu]u∈U is the coproduct in K(R). Thus there are commutative diagrams

F(Mu) //

F(εu)

∐u∈U

F(Mu)

ϕzz

F(M)

and

F(Mu) //

F([εu])

∐u∈U

F(Mu)

ϕzz

F(M) ,

where ϕ and ϕ are the canonical morphisms. By assumption, ϕ is an isomorphism,and as the two diagrams are identical, ϕ is an isomorphism as well. That is, F pre-serves coproducts. The assertion about products is proved similarly.

REMARK. Theorem 6.1.16 shows that K(R) has the universal property of a quotient categorydiscussed in the beginning of this section. Together with E 6.1.1 the theorem also shows thatK(R) is the localization of C(R) with respect to the collection of homotopy equivalences. See alsothe introduction to Sect. 6.4 where we treat the further localization of K(R) with respect to thecollection of quasi-isomorphisms; this leads to the derived category.

By the universal property above, certain functors on C(R) induce functors onK(R), and natural transformations follow along.

6.1.17 Proposition. Let E,F: C(R)→ U be homotopy invariant functors and con-sider the induced functors E, F : K(R)→ U; see 6.1.16. Every natural transformation



τ : E→ F induces a natural transformation τ : E→ F given by τM = τM for every R-complex M.

PROOF. For every R-complex M one has E(M) = E(M) and F(M) = F(M), whenceτM = τM is a morphism E(M)→ F(M). Let [α] : M→ N be a morphism in K(R).As τ : E→ F is a natural transformation of functors C(R)→ U there are equalities,

τN E([α]) = τN E(α) = F(α)τM = F([α])τM ,

which show that τ : E→ F is a natural transformation of functors K(R)→ U.

To parse and prove the next result, recall that if F : U→ V is a functor betweencategories with products (coproducts), then Fop : Uop→ Vop is a functor betweencategories with coproducts (products), and F preserves products (coproducts) if andonly if Fop preserves coproducts (products).

6.1.18 Theorem. Let V be a category and let G: C(R)op→V be a functor. If G ishomotopy invariant, then there exists a unique functor G that makes the followingdiagram commutative,

C(R)op

Qop

G// V

K(R)op .

G

<<

For every R-complex M there is an equality G(M) = G(M), and for every morphism[α] in K(R)op one has G([α]) = G(α). Furthermore, the following assertions hold.

(a) If V is k-prelinear and G is k-linear, then G is k-linear.(b) If V has products/coproducts and G preserves products/coproducts, then G


PROOF. Apply 6.1.16 to the functor Gop : C(R)→ Vop.

6.1.19 Proposition. Let G,J : C(R)op→ V be homotopy invariant functors and con-sider the induced functors G, J : K(R)op→ V; see 6.1.18. Every natural transfor-mation τ : G→ J induces a natural transformation τ : G→ J given by τM = τM forevery R-complex M.

PROOF. Apply 6.1.17 to the natural transformation τop : Jop→ Gop of functors fromC(R) to Vop.

A SPECIAL CASE OF THE UNIVERSAL PROPERTY

It is evident that a homotopy invariant functor C(R)→ C(S) preserves homotopy;see 4.3.12. On the other hand, if a functor C(R)→ C(S) preserves homotopy, thenthe composite C(R)→ C(S)→K(S) is homotopy invariant. We apply the homotopycategory’s universal property in this important special case.



6.1.20 Theorem. Let F: C(R)→ C(S) be a functor. If F preserves homotopy, thenthere is a unique functor F that makes the next diagram commutative,

C(R) F//

QR

C(S)

QS

K(R) F// K(S) .

For every R-complex M there is an equality F(M) = F(M), and for every morphism[α] in K(R) one has F([α]) = [F(α)]. Furthermore, the following assertions hold.

(a) If F is k-linear, then F is k-linear.(b) If F preserves products/coproducts, then F preserves products/coproducts.

PROOF. As F preserves homotopy, the composite functor QS F is homotopy invari-ant. Thus the existence and the uniqueness of F follow from 6.1.16. In symbols onehas F = (QS F)˙. The value of F on an R-complex M is F(M) = QS F(M) = F(M)as QS is the identity on objects. The value of F on a morphism [α] in K(R) isF([α]) = QS F(α) = [F(α)].

By 6.1.10 the functor QS is k-linear and preserves products/coproducts. Thus, ifF has any of these properties, then so does QS F, and the assertions in parts (a) and(b) now follow from the corresponding parts in 6.1.16.

6.1.21 Proposition. Let E,F: C(R)→ C(S) be functors that preserve homotopy andconsider the induced functors E, F : K(R)→K(S); see 6.1.20. Every natural trans-formation τ : E→ F induces a natural transformation τ : E→ F given by τM = [τM]for every R-complex M.

PROOF. Evidently, application of the canonical functor Q: C(S)→K(S) to the nat-ural transformation τ yields a natural transformation Qτ= [τ] : QE→ QF of func-tors C(R)→ K(S). By definition, E and F are the functors K(R)→ K(S) inducedby QE and QF; see 6.1.16. Thus 6.1.17 gives the desired conclusion.

6.1.22 Theorem. Let G: C(R)op→ C(S) be a functor. If G preserves homotopy,then there is a unique functor G that makes the next diagram commutative,

C(R)op G//

QopR

C(S)

QS

K(R)op G// K(S) .

For every R-complex M there is an equality G(M) = G(M), and for every morphism[α] in K(R)op one has G([α]) = [G(α)]. Furthermore, the following assertions hold.

(a) If G is k-linear, then G is k-linear.(b) If G preserves products/coproducts, then G preserves products/coproducts.

PROOF. Analogous to the proof of 6.1.20, only apply 6.1.18 in place of 6.1.16.



6.1.23 Proposition. Let G,J : C(R)op→ C(S) be functors that preserve homotopyand consider the induced functors G, J : K(R)op→K(S); see 6.1.22. Every naturaltransformation τ : G→ J induces a natural transformation τ : G→ J given by τM =[τM] for every R-complex M.


We shall often abuse notation and write F and G for the induced functors F andG from 6.1.20 and 6.1.22.

6.1.24 Lemma. Let C(R) E−→ C(S) F−→ UT−→ V be functors where E preserves ho-

motopy and F is homotopy invariant. The functor TFE is then homotopy invariant,and the induced functor K(R)→ V is TFE; in symbols, (TFE)˙ = TFE.

In particular, one has (TF)˙ = TF and (FE)˙ = F E.

PROOF. Evidently, TFE is homotopy invariant, and the induced functor (TFE)˙ isthe unique functor with (TFE)˙QR = TFE. As one has TF EQR = TFQS E = TFE,the assertion follows.

6.1.25 Lemma. Let C(Q)E−→ C(R) F−→ C(S) be functors that preserve homotopy.

The functor K(Q)→K(S) induced by FE is F E; in symbols, (FE)¨ = F E.

PROOF. The induced functor (FE)¨ is the unique functor with (FE)¨ QQ = QS FE.As one has F EQQ = FQR E = QS FE, the assertion follows.

EXERCISES

E 6.1.1 Show that a functor F: C(R)→U is homotopy invariant if and only if F(α) is an iso-morphism in U for every homotopy equivalence α in C(R). Hint: Proof of 4.3.13.

E 6.1.2 Let P be a semi-projective R-complex and let I be a semi-injective R-complex. Showthat if α is a quasi-isomorphism, then K(R)(P,α) and K(R)(α, I) are isomorphisms.

E 6.1.3 Show that the homotopy category K(Z) is not Abelian.E 6.1.4 Let R be semi-simple. Show that the categories K(R) and Mgr(R) are equivalent and

conclude that K(R) is Abelian.E 6.1.5 Show that Mgr(R) is isomorphic to a full subcategory of K(R).E 6.1.6 Let α : M→ N be a morphism in K(R). Show that for any two semi-projective resolu-

tions π : P '−→M and λ : L '−→ N there is a unique morphism α : P→ L in K(R) withαπ= λα.

E 6.1.7 Let α : M→ N be a morphism in K(R). Show that for any two semi-injective resolutionsι : M '−→ I and ε : N '−→ E there is a unique morphism α : I→ E in K(R) with εα= αι.

E 6.1.8 Show that homology, H, and shift, Σ, induce commuting functors on K(R).E 6.1.9 Consider the full subcategories of K(R) defined by specifying their objects as follows,

K(PrjR) = P ∈K(R) | P is a complex of projective modules and

Kprj(R) = P ∈K(R) | P is semi-projective .Show that both of these categories have coproducts and that Kprj(R) has products.

E 6.1.10 Consider the full subcategories of K(R) defined by specifying their objects as follows,K(InjR) = I ∈K(R) | I is a complex of injective modules and

Kinj(R) = I ∈K(R) | I is semi-injective .Show that both of these categories have products and that Kinj(R) has coproducts.


6.2 Triangulation of K 207

6.2 Triangulation of KKK

SYNOPSIS. Strict triangle; distinguished triangle; quasi-triangulated functor; universal property;homology.

The mapping cone construction is the key to the triangulated structure on the homo-topy category. The definition of a triangulated category is recalled in Appn. D

6.2.1. By 4.3.16 and 6.1.20 there is a unique k-linear endofunctor Σ on K(R) thatmakes the following diagram commutative,

C(R) Σ//

Q

C(R)

Q

K(R) //Σ// K(R) .

By 6.1.25 it is an isomorphism with inverse induced by Σ−1 : C(R)→ C(R). Bythe usual abuse of notation, the functor Σ is written Σ or, occasionally, ΣK. For amorphism [α] in K(R) one has ΣK[α] = [Σα].

Consider the k-linear category K(R), see 6.1.10, equipped with the k-linear au-tomorphism Σ= ΣK. One may now speak of candidate triangles in K(R); cf. D.1.

6.2.2 Lemma. Let α : M→ N be a morphism in C(R). The image of the diagram

M α−−→ N

(1N

0


(0 1ΣM )−−−−−→ ΣM

under the canonical functor Q: C(R)→K(R) is a candidate triangle in K(R).

PROOF. We must prove that the three composites in C(R),

λ=

(1N

0

)α=

(α0

), µ=

(0 1ΣM

)(1N

0

)= 0 , and ν= (Σα)

(0 1ΣM

)=(0 Σα

)are null-homotopic. Since µ is even zero in C(R), we are left to consider λ and ν.Define degree 1 homomorphisms % : M→ Coneα and τ : Coneα→ ΣN by

% =

(0ςM

1

)and τ =

(ςN

1 0).

It follows from commutativity of the diagram (2.2.4.1) that there are equalities∂Coneα%+%∂M = λ and ∂ΣNτ+τ∂Coneα = ν. Indeed, one has(

∂N αςΣM−1

0 ∂ΣM

)(0ςM

1

)+

(0ςM

1

)∂M =

(α0

)and

∂ΣN (ςN1 0)+(ςN

1 0)(∂N αςΣM

−10 ∂ΣM

)=(0 Σα

).



6.2.3 Definition. A candidate triangle in K(R) of the form considered in 6.2.2 iscalled a strict triangle. A candidate triangle in K(R) that is isomorphic, in the senseof D.1, to a strict triangle is called a distinguished triangle.

6.2.4 Theorem. The homotopy category K(R), equipped with the automorphism Σ

and the collection of distinguished triangles defined in 6.2.3, is triangulated.

PROOF. We verify the axioms from D.3.(TR0): Evidently, the collection of distinguished triangles is closed under iso-

morphisms. Furthermore, it follows from 4.3.31 and 6.1.8 that application of thecanonical functor Q: C(R)→K(R) to the following diagram in C(R),

M 1M−−→M

(1M

0

)−−−−→ Cone(1M)

(0 1ΣM )−−−−−→ ΣM ,

yields, up to isomorphism in K(R), the candidate triangle M 1M−→ M −→ 0 −→ ΣM

which, therefore, is distinguished.(TR1): By the definition of morphisms in K(R), every morphism in this category

fits into a strict, and hence distinguished, triangle; see 6.2.2.(TR2’): By D.5 it it sufficient to verify that (TR2) holds. Thus, let

∆ = M′ α′−−→ N′β′−−→ X ′

γ′−−→ ΣM′

be a distinguished triangle in K(R). We must argue that the candidate triangles

∆+ =N′

β′−→Xγ′−−→ΣM′ −Σα

′−−−→ΣN′ and ∆

−=Σ−1 X ′

−Σ−1 γ′−−−−−→M′ α′−−→N′β′−−→X ′

are distinguished. Up to isomorphism, ∆ is given by application of the canonicalfunctor Q to a diagram in C(R) of the form,

M α−−→ N

(1N

0


(0 1ΣM )−−−−−→ ΣM .

Thus, the candidate triangles ∆+ and ∆− are, up to isomorphism, given by applica-tion of Q to the following diagrams in C(R),

andN

(1N

0

)// Coneα

(0 1ΣM )// ΣM

−Σα// ΣN ,

Σ−1 Coneα(0 −1M )

// M α// N

(1N

0

)// Coneα .

These two diagrams in C(R) are the top rows in (?) and (‡) below. By definition,the bottom rows in (?) and (‡) give strict triangles in K(R) when the functor Qis applied; see 6.2.3. Thus, to show that ∆+ and ∆− are distinguished triangles inK(R), it suffices to argue that (?) and (‡) are commutative up to homotopy, and thatthe vertical morphisms in both diagrams are homotopy equivalences.



(?)

Nε=(

1N

0

)// Coneα

(0 1ΣM )// ΣM

ϕ=

(0

1ΣM

−Σα

)

−Σα// ΣN

Nε=(

1N

0

)// Coneα

(1N 00 1ΣM

0 0

)// Coneε

(0 0 1ΣN )// ΣN

(‡)

Σ−1 Coneαπ=(0 −1M )

// M α// N

ψ=

( 01N

0

)

(1N

0

)// Coneα

Σ−1 Coneαπ=(0 −1M )

// M

(1M

00

)// Coneπ

(0 1N 00 0 1ΣM

)// Coneα

First consider the diagram (?) and let ϑ : Coneε→ ΣM be the map (0 1ΣM 0).Note that ϕ and ϑ are morphisms, as one has

∂Coneεϕ =

∂N αςΣM−1 ςΣN

−10 ∂ΣM 00 0 ∂ΣN

01ΣM

−Σα

=

01ΣM

−Σα

∂ΣM = ϕ∂ΣM

and

∂ΣMϑ =(0 ∂ΣM 0

)=(0 1ΣM 0

)∂N αςΣM−1 ςΣN

−10 ∂ΣM 00 0 ∂ΣN

= ϑ∂Coneε .

Next we argue that ϕ is a homotopy equivalence with homotopy inverse ϑ. Evi-dently one has ϑϕ= 1ΣM , so it remains to show that the morphism

1Coneε−ϕϑ =

1N 0 00 1ΣM 00 0 1ΣN

− 0

1ΣM

−Σα

(0 1ΣM 0)=

1N 0 00 0 00 Σα 1ΣN

is null-homotopic. The degree 1 homomorphism

σ : Coneε −→ Coneε given by σ =

0 0 00 0 0ςN

1 0 0

is the desired homotopy. Indeed, one has∂N αςΣM

−1 ςΣN−1

0 ∂ΣM 00 0 ∂ΣN

0 0 00 0 0ςN

1 0 0

+

0 0 00 0 0ςN

1 0 0

∂N αςΣM−1 ςΣN

−10 ∂ΣM 00 0 ∂ΣN

=

1N 0 00 0 00 Σα 1ΣN

;

that is, ∂Coneεσ+σ∂Coneε = 1Coneε−ϕϑ holds. Thus ϑ is a homotopy inverse of ϕ.Now we turn to the issue of commutativity of (?). The left- and right-hand squares

in (?) are even commutative in C(R). For the commutativity, up to homotopy, of themiddle square, it must be proved that the morphism β : Coneα→ Coneε, given by



β =

1N 00 1ΣM

0 0

− 0

1ΣM

−Σα

(0 1ΣM)=

1N 00 00 Σα

,

is null-homotopic. Consider the degree 1 homomorphism,

τ : Coneα −→ Coneε given by τ =

0 00 0ςN

1 0

.

It is straightforward to verify the equality∂N αςΣM−1 ςΣN

−10 ∂ΣM 00 0 ∂ΣN

0 00 0ςN

1 0

+

0 00 0ςN

1 0

(∂N αςΣM−1

0 ∂ΣM

)=

1N 00 00 Σα

;

that is, ∂Coneετ+τ∂Coneα = β holds, and hence β is null-homotopic.Similar arguments show that the diagram (‡) is commutative up to homotopy and

that ψ is a homotopy equivalence with homotopy inverse ξ = (α 1N 0).(TR4’): Consider a diagram in C(R), commutative up to homotopy,

()M

ϕ

α// N

ψ

(1N

0

)// Coneα

(0 1ΣM )// ΣM

Σϕ

M′ α′// N′

(1N′

0

)// Coneα′

(0 1ΣM′ )// ΣM′ .

To verify (TR4’) one has, in view of the definition of distinguished triangles inK(R), to establish a morphism χ : Coneα→ Coneα′ with the following properties.In the first place, χ makes the diagram () commutative up to homotopy; observethat (Q(ϕ),Q(ψ),Q(χ)) is then a morphism of distinguished triangles in K(R). Sec-ondly, the mapping cone candidate triangle of (Q(ϕ),Q(ψ),Q(χ)), must be a distin-guished triangle in K(R); it is given by application of Q to the diagram in C(R),

(§)M′

⊕N

(α′ ψ

0 −1N

0 0

)//

N′

⊕Coneα

1N′ χ11 χ12

0 χ21 χ22

0 0 −1ΣM

//

Coneα′

⊕ΣM

(0 1ΣM′ Σϕ0 0 −Σα

)//

ΣM′

⊕ΣN

,

where the morphisms χi j are the entries in χ considered as a 2× 2 matrix. Firstwe construct a morphism χ that makes () commutative up to homotopy. By as-sumption, there is a degree 1 homomorphism σ : M→ N′ such that the equalityψα−α′ϕ= ∂N′σ+σ∂M holds. Consider the degree 0 homomorphism,

χ : Coneα −→ Coneα′ given by χ =

(ψ σςΣM

−10 Σϕ

).

It is straightforward verify that it is a morphism; that is, one has

∂Coneα′χ =

(∂N′ α′ςΣM′

−10 ∂ΣM′

)(ψ σςΣM

−10 Σϕ

)=

(ψ σςΣM

−10 Σϕ

)(∂N αςΣM

−10 ∂ΣM

)= χ∂Coneα .



Notice that χ makes the middle and right-hand squares in () commutative.Finally, to see that application of the functor Q to (§) yields a distinguished tri-

angle in K(R), note that (§) is the top row in following diagram, and that the bottomrow yields a strict triangle in K(R) when Q is applied. Thus, it suffices to argue thatthe diagram below is commutative up to homotopy, and that the vertical morphismsare homotopy equivalences.

M′

⊕N

θ=

(α′ ψ

0 −1N

0 0

)//

N′

⊕Coneα

1N′ ψ σςΣM−1

0 0 Σϕ

0 0 −1ΣM

//

Coneα′

⊕ΣM

(0 1ΣM′ Σϕ0 0 −Σα

)//

ξ=

1N′ 0 σςΣM

−10 0 00 0 −1ΣM

0 1ΣM′ Σϕ0 0 −Σα

ΣM′

⊕ΣN

M′

⊕N

θ=

(α′ ψ

0 −1N

0 0

)//

N′

⊕Coneα

1N′ 0 00 1N 00 0 1ΣM

0 0 00 0 0

// Coneθ (

0 0 0 1ΣM′ 00 0 0 0 1ΣN

) // ΣM′

⊕ΣN

The differentials on the complexes Coneα′⊕ΣM and Coneθ are given by

∂N′ α′ςΣM′−1 0

0 ∂ΣM′ 00 0 ∂ΣM

and

∂N′ 0 0 α′ςΣM′

−1 ψςΣN−1

0 ∂N αςΣM−1 0 −ςΣN

−10 0 ∂ΣM 0 00 0 0 ∂ΣM′ 00 0 0 0 ∂ΣN

.

Consider the degree 0 homomorphism,

η : Coneθ −→ Coneα′⊕ΣM given by η =

1N′ ψ σςΣM−1 0 0

0 0 Σϕ 1ΣM′ 00 0 −1ΣM 0 0

.

It is straightforward to verify that ξ and η are morphisms. Evidently there is an equal-ity ηξ = 1Coneα′⊕ΣM . Furthermore, the morphism ξη−1Coneθ is null-homotopic, asthe degree 1 homomorphism

τ : Coneθ −→ Coneθ given by τ =

0 0 0 0 00 0 0 0 00 0 0 0 00 0 0 0 00 ςN

1 0 0 0

satisfies the identity ∂Coneθτ+ τ∂Coneθ = ξη−1Coneθ. Hence ξ is a homotopy equi-valence with homotopy inverse η.

The left-hand and right-hand squares in the diagram are commutative. The dia-gram’s middle square is commutative up to homotopy; indeed, the difference mor-phism γ : N′⊕Coneα→ Coneθ, given by



γ =

1N′ 0 σςΣM

−10 0 00 0 −1ΣM

0 1ΣM′ Σϕ0 0 −Σα

1N′ ψ σςΣM

−10 0 Σϕ0 0 −1ΣM

−

1N′ 0 00 1N 00 0 1ΣM

0 0 00 0 0

=

0 ψ 00 −1N 00 0 00 0 00 0 Σα

,

is null-homotopic. This follows as

% : N′⊕Coneα −→ Coneθ given by % =

0 0 00 0 00 0 00 0 00 ςN

1 0

is a degree 1 homomorphism with ∂Coneθ%+%∂N′⊕Coneα = γ.

The triangulated structure on K(R) is by 6.2.3 based on triangles that come frommapping cone sequences; the next lemma shows that starting from mapping cylindersequences would yield the same structure. This will be used later in the chapter toshow that every short exact sequence of complexes can be completed to a trianglein the derived category.

6.2.5 Lemma. Let α : M→ N be a morphism in C(R). The image of the diagram

Mι=

( 00

1M

)// Cylα

α=(1N 0 α) u

π=

(1N 0 00 1ΣM 0

)// Coneα

$=(0 1ΣM )// ΣM

M α// N

ε=(

1N

0

)// Coneα

$=(0 1ΣM )// ΣM

under the canonical functor Q: C(R)→K(R) is an isomorphism of candidate trian-gles in K(R). In particular, the image of the top row is a distinguished triangle.

PROOF. The image of the bottom row is by definition a distinguished triangle inK(R). The diagram’s left- and right-hand squares are commutative. Consider

σ : Cylα −→ Coneα given by(

0 0 00 0 −ςM

1

).

It is a degree 1 homomorphism, and one has

∂Coneασ+σ∂Cylα =

(0 0 −α0 1ΣM 0

)= π−εα ,

so the middle square is commutative up to homotopy. Finally, α is a homotopyequivalence by 4.3.10, so Q(α) is an isomorphism.



QUASI-TRIANGULATED FUNCTORS

We introduce a condition that ensures that a functor from C(R) to a triangulatedcategory induces a triangulated functor on K(R).

6.2.6 Definition. Let (U,ΣU) be a triangulated category. A functor F: C(R)→ U

is called quasi-triangulated if it is additive and there exists a natural isomorphismφ : FΣ→ ΣU F such that the diagram

F(M)F(α)−−−→ F(N)

F(

1N

0

)−−−−→ F(Coneα)

φM F(0 1ΣM )−−−−−−−−→ ΣU F(M)

is a distinguished triangle in U for every morphism α : M→ N of R-complexes.

6.2.7 Example. The canonical functor Q: C(R)→K(R) is quasi-triangulated by6.1.10, 6.2.2, and 6.2.3.

6.2.8. Let U be a triangulated category and let F : C(R)→ U be a homotopy invari-ant functor. It follows from 6.1.24 and 6.2.1 that the functors K(R)→U induced byFΣ and ΣU F are FΣK and ΣU F.

6.2.9 Theorem. Let U be a triangulated category and F: C(R)→ U be a homotopyinvariant functor. If F is quasi-triangulated with natural isomorphism φ : FΣ→ ΣU F,then the induced functor F : K(R)→ U, see 6.1.16, is triangulated with natural iso-morphism φ : FΣ→ ΣU F.

PROOF. As F is additive, so if F by 6.1.16. It follows from 6.1.17 and 6.2.8 that φinduces a natural isomorphism φ : FΣ→ ΣU F. By 6.2.3 every distinguished trianglein K(R) is isomorphic to a strict triangle, i.e. to a candidate triangle of the form

MQ(α)−−−→ N

Q(

1N

0

)−−−−→ Coneα

Q(0 1ΣM )−−−−−−→ ΣM ,

where α : M→ N is a morphism in C(R). Thus, it suffices to show that the followingcandidate triangle in U is distinguished,

F(M)FQ(α)−−−−→ F(N)

FQ(

1N

0

)−−−−−→ F(Coneα)

φM FQ(0 1ΣM )−−−−−−−−−→ ΣU F(M) .

However, this diagram is identical to the one in 6.2.6, which is a distinguished tri-angle in U as F is assumed to be quasi-triangulated.

6.2.10 Definition. Let (U,ΣU) be a triangulated category, and let E,F: C(R)→ U

be quasi-triangulated functors with associated natural isomorphisms φ : EΣ→ ΣU Eand ψ : FΣ→ ΣU F. A natural transformation τ : E→ F is called quasi-triangulatedif the following diagram is commutative for every R-complex M,



E(ΣM)

τΣM

∼=φM// ΣU E(M)

ΣτM

F(ΣM) ∼=ψM// ΣU F(M) .

Thus, τΣM and ΣU τM are isomorphic, and the isomorphism is natural in M.

6.2.11 Proposition. Let U be a triangulated category, let E,F: C(R)→ U be quasi-triangulated and homotopy invariant functors, and let τ : E→ F be a natural transfor-mation. If τ is quasi-triangulated, then the induced natural transformation τ : E→ Fof triangulated functors, see 6.1.17 and 6.2.9, is triangulated.

PROOF. By assumption the functors E and F are quasi-triangulated; denote the asso-ciated natural isomorphisms by φ and ψ, respectively. By 6.2.9 the induced functorsE and F are triangulated with associated natural isomorphisms φ and ψ. For and R-complex M, the equality ψM τΣM = (Σ τM)φM holds as the left-hand side is ψMτΣM ,the right-hand side (ΣτM)φM , and those two composites agree by assumption.

The next definition ensures that also Qop : C(R)op→K(R)op is quasi-triangulated.

6.2.12 Definition. Let (V,ΣV) be a triangulated category. A functor G: C(R)op→ V

is called quasi-triangulated if the functor Gop from C(R) to the triangulated cate-gory (Vop,Σ−1

V ), see D.6, is quasi-triangulated in the sense of 6.2.6. Explicitly, thismeans that G is an additive functor with a natural isomorphism ψ : Σ−1

V G→ GΣop

of functors C(R)op→ V, such that the diagram

Σ−1V G(M)

G(0 1ΣM )ψM

−−−−−−−−→ G(Coneα)G(

1N

0

)−−−−→ G(N)

G(α)−−−→ G(M)

is a distinguished triangle in V for every morphism α : M→ N of R-complexes.

6.2.13 Theorem. Let V be a triangulated category and let G: C(R)op→ V be ahomotopy invariant functor. If G is quasi-triangulated with natural isomorphismψ : Σ−1

V G→ GΣop, then the functor G : K(R)op→ V, see 6.1.18, is triangulated withnatural isomorphism ψ : Σ−1

V G→ GΣop.

PROOF. Apply 6.2.9 to the functor Gop : C(R)→ Vop.

6.2.14 Definition. Let (V,ΣV) be a triangulated category, and let G,J : C(R)op→ V

be quasi-triangulated functors with natural isomorphisms ψ : Σ−1V G→ GΣop and

φ : Σ−1V J→ JΣop. A natural transformation τ : G→ J is called quasi-triangulated if

the following diagram is commutative for every R-complex M,

Σ−1V G(M)

Σ−1 τM

∼=ψM// G(ΣM)

τΣM

Σ−1V J(M) ∼=

φM// J(ΣM) .



Thus, τΣM and Σ−1V τM are isomorphic, and the isomorphism is natural in M.

For the next statement, recall from D.12 the notion of a triangulated transforma-tion of triangulated functors.

6.2.15 Proposition. Let V be a triangulated category, let G,J : C(R)op→ V be quasi-triangulated and homotopy invariant functors, and let τ : G→ J be a natural transfor-mation. If τ is quasi-triangulated, then the induced natural transformation τ : G→ Jof triangulated functors, see 6.1.19 and 6.2.13, is triangulated.

PROOF. Apply 6.2.11 to the natural transformation τop : Jop→ Gop of functors fromC(R) to Vop.

THE UNIVERSAL PROPERTY REVISITED

We begin with a result that extends 6.2.7.

6.2.16 Lemma. Let F: C(R)→ C(S) be a functor. If F is a Σ-functor with associatednatural isomorphism φ : FΣ→ ΣF, then the functor QS F: C(R)→K(S) is quasi-triangulated with associated natural isomorphism QS φ.

PROOF. As F is additive, so is QS F, and evidently QS φ is a natural isomorphismfrom (QS F)Σ to QSΣF = Σ(QS F), see 6.2.1. It remains to argue that for every mor-phism α : M→ N of R-complexes, the diagram

QS F(M)QS F(α)−−−−→ QS F(N)

QS F(

1N

0

)−−−−−−→ QS F(Coneα)

QS(φM) QS F(0 1ΣM )−−−−−−−−−−−−→ ΣQS F(M)

is a distinguished triangle in K(S). By assumption, see 4.1.8, there exists an isomor-phism α in C(S) that makes the following diagram commutative,

(?)F(M)

F(α)// F(N)

F(

1N

0

)// F(Coneα)

∼= α

φM F(0 1ΣM )// ΣF(M)

F(M)F(α)

// F(N)

(1F(N)

0

)// ConeF(α)

(0 1ΣF(M) )// ΣF(M) .

By application of the functor QS to (?), the top row becomes the diagram in questionand the bottom row becomes a strict triangle by definition; see 6.2.3. It follows thatthe relevant diagram is a (candidate and even a) distinguished triangle.

The next theorem justifies the notion of Σ-functors.

6.2.17 Theorem. Let F: C(R)→ C(S) be a functor that preserves homotopy. IfF is a Σ-functor with natural isomorphism φ : FΣ→ ΣF, then the induced functorF : K(R)→K(S), see 6.1.20, is triangulated with natural isomorphism φ : FΣ→ Σ F.

PROOF. It follows from 6.2.16 that QS F: C(R)→K(S) is a quasi-triangulated func-tor with associated natural isomorphism QS φ. Thus 6.2.9 implies that the functorF = (QS F)˙ is triangulated with associated natural isomorphism φ = (QS φ)˙ wherethe equalities follow from 6.1.20 and 6.1.21.



6.2.18 Proposition. Let E,F: C(R)→ C(S) be functors that preserve homotopy andlet τ : E→ F be a natural transformation. If E and F are Σ-functors and τ is a Σ-transformation, then the induced natural transformation τ : E→ F of triangulatedfunctors, see 6.1.21 and 6.2.17, is triangulated.

PROOF. Let E and F be Σ-functors with associated natural isomorphisms φ and ψ.By 6.2.16 the functors QS E,QS F: C(R)→K(S) are quasi-triangulated with natu-ral isomorphisms QS φ and QSψ. The natural transformation QS τ : QS E→ QS F isquasi-triangulated, indeed, if one applies the functor QS to the commutative diagramin 4.1.9, then by 6.2.1 the outcome is the required commutative diagram; see 6.2.10.Now 6.2.11 yields that the natural transformation τ = (QS τ)˙ from E = (QS E)˙ toF = (QS F)˙ is triangulated; the equalities follow from 6.1.20 and 6.1.21.

6.2.19 Theorem. Let G: C(R)op→ C(S) be a functor that preserves homotopy. If Gis a Σ-functor with natural isomorphism ψ : Σ−1 G→ GΣop, then the induced functorG, see 6.1.22, is triangulated with natural isomorphism ψ : Σ−1 G→ GΣop.

PROOF. Analogous to the proof of 6.2.17, only one applies 6.1.22, 6.1.23, and6.2.13 in place of 6.1.20, 6.1.21, and 6.2.9.

6.2.20 Proposition. Let G,J : C(R)op→ C(S) be functors that preserve homotopyand let τ : G→ J be a natural transformation. If G and J are Σ-functors and τ is a Σ-transformation, then the induced transformation τ : G→ J of triangulated functors,see 6.1.23 and 6.2.19, is triangulated.

PROOF. Analogous to the proof of 6.2.18, only one applies 6.1.22, 6.1.23, and6.2.15 in place of 6.1.20, 6.1.21, and 6.2.11.

HOMOLOGY

Homology induces a functor on the homotopy category. It is the primary exampleof a so-called homological functor; that is, it maps distinguished triangles in K(R)to exact sequences in C(R).

6.2.21 Proposition. The homology functor H: C(R)→ C(R) maps homotopy equi-valences to isomorphisms. The induced functor H: K(R)→ C(R) from 6.1.16 isk-linear and it preserves products and coproducts. Moreover, one has HΣ= ΣH.

PROOF. The first assertion follows from 4.3.4 and thus H: C(R)→ C(R) inducesby 6.1.16 a functor H : K(R)→ C(R), which outside of this proof is denoted H. By2.2.15, 3.1.10, and 3.1.22, the functor H is k-linear and it preserves products andcoproducts. It follows from 6.1.16 that the induced functor H has the same prop-erties. Now 6.2.1, the definition of H, and 2.2.15 yield equalities HΣQ = HQΣ =HΣ = ΣH = Σ HQ, where Q: C(R)→K(R) is the canonical functor from 6.1.3. Itnow follows from the uniqueness assertion in 6.1.16 that one has HΣ= Σ H.

In the balance of this chapter, we use Greek letters for morphisms in K(R). Un-less otherwise specified α,β,γ, . . . denote homotopy classes of morphisms in C(R).



The last assertion in the next theorem is akin to 4.2.5 and the Five Lemma D.15in K(R). The conclusion is weaker than the Five Lemma’s, so is the hypothesis.

6.2.22 Theorem. For every morphism of distinguished triangles in K(R),

M

ϕ

α// N

ψ

β// X

χ

γ// ΣM

Σϕ

M′ α′// N′

β′// X ′

γ′// ΣM′ ,

there is a commutative diagram in C(R) with exact rows,

(6.2.22.1)

H(M)

H(ϕ)

H(α)// H(N)

H(ψ)

H(β)// H(X)

H(χ)

H(γ)// ΣH(M)

ΣH(ϕ)

ΣH(α)// ΣH(N)

ΣH(ψ)

H(M′)H(α′)

// H(N′)H(β′)

// H(X ′)H(γ′)

// ΣH(M′)ΣH(α′)

// ΣH(N′) .

In particular, if two of the morphisms ϕ, ψ, and χ in K(R) are quasi-isomorphisms,then so is the third.

PROOF. By 6.2.21 the functor H: K(R)→ C(R) satisfies the identity HΣ = ΣH; itis, therefore, evident that (6.2.22.1) is commutative. We argue that the upper row isexact; a parallel argument shows that the lower row is exact as well. By the defini-tion of distinguished triangles 6.2.3, there is a morphism α : M→ N in C(R) and anisomorphism of candidate triangles in K(R),

M

u

α// N

u

β// X

u

γ// ΣM

u

MQ(α)// N

Q(ε)// Cone α

Q($)// ΣM ,

where ε and $ denote (1N 0)T and (0 1Σ M). Thus, the upper row in (6.2.22.1) isisomorphic to the sequence

H(M)HQ(α)−−−−→ H(N)

HQ(ε)−−−−→ H(Cone α)HQ($)−−−−→ ΣH(M)

ΣHQ(α)−−−−−→ ΣH(N) .

This sequence is nothing but

H(M)H(α)−−−→ H(N)

H(ε)−−−→ H(Cone α)H($)−−−→ ΣH(M)

ΣH(α)−−−−→ ΣH(N) ,

which is exact by 4.2.14.The final assertion about quasi-isomorphisms follows from commutativity of the

diagram (6.2.22.1) and the Five Lemma 2.1.40.

EXERCISES

E 6.2.1 Show that for every split exact sequence 0−→M α−→N β−→X −→ 0 in C(R), the diagram



M[α]−−→ N

[β]−−→ X 0−−→ ΣM

is a distinguished triangle in K(R).

E 6.2.2 Show that an exact sequence 0−→M α−→N β−→ X −→ 0 of R-modules is split if and only

if there exists a distinguished triangle M[α]−→ N

[β]−→ X −→ ΣM in K(R). Hint: D.17.

E 6.2.3 Show that every degreewise split exact sequence 0 −→ M α−→ N β−→ X −→ 0 in C(R)can be completed to a distinguished triangle in K(R),

M[α]−−→ N

[β]−−→ Xγ−−→ ΣM ,

with γ = [−ςM1 %∂

Nσ], where % : N→M and σ : X → N are the splitting homomor-phisms. Compare to E 4.3.8. Hint: See E 4.3.20.

E 6.2.4 Let (T,Σ) be a triangulated category and let S be a subcategory of T that is closedunder isomorphisms. Show that S is a triangulated subcategory if and only if (S,Σ) is atriangulated category and the embedding functor S→ T is full and triangulated.

E 6.2.5 Let S be a triangulated subcategory of (T,Σ) and let M→ N → X → ΣM be a distin-guished triangle in T. Show that if two of the objects M, N, and X are in S, then thethird object is also in S.

E 6.2.6 Show that the full subcategory of K(R) whose objects are all acyclic R-complexes istriangulated.

E 6.2.7 Show that the full subcategories of K(R) defined by specifying their objects as follows:K@(R) = M ∈K(R) | there is a bounded above complex M′ with M u M′ ,K@A(R) = M ∈K(R) | there is a bounded complex M′ with M u M′ , and

KA(R) = M ∈K(R) | there is a bounded below complex M′ with M u M′are triangulated subcategories of K(R).

E 6.2.8 Let S be a triangulated subcategory of a triangulated category (T,Σ). A morphismα : M→ N is called S-trivial if in some, equivalently in every, distinguished triangle,

M α−−→ N −→ X −→ ΣM ,

the object X belongs to S. Describe the S-trivial morphisms in the category K(R) if Sconsists of all acyclic R-complexes; cf. E 6.2.6.

E 6.2.9 Show that K(PrjR), see E 6.1.9, is a triangulated category but not a triangulated subcat-egory of K(R). Show that the inclusion functor K(PrjR)→K(R) is triangulated.

E 6.2.10 Show that Kprj(R) is a triangulated subcategory of K(PrjR); see E 6.1.9.E 6.2.11 Show that K(InjR), see E 6.1.10, is a triangulated category but not a triangulated sub-

category of K(R). Show that the inclusion functor K(InjR)→K(R) is triangulated.E 6.2.12 Show that Kinj(R) is a triangulated subcategory of K(InjR); see E 6.1.10.E 6.2.13 Give a proof of the Five Lemma in K(R) without using D.15.

6.3 Resolutions

SYNOPSIS. Unique lifting property; functoriality of resolution.

To all intents and purposes, the non-uniqueness inherent in the resolutions of com-plexes and liftings of morphisms disappears in the homotopy category.

Recall that Greek letters, unless otherwise specified, denote morphisms in K(R).


6.3 Resolutions 219

UNIQUE LIFTINGS

We rephrase Proposition 5.2.18 and its corollaries 5.2.19 and 5.2.20 in the languageof the homotopy category.

6.3.1. Let P be a semi-projective R-complex, let α : P→ N be a morphism, and letβ : M→ N be a quasi-isomorphism in K(R). There exists a unique morphism γ thatmakes the following diagram in K(R) commutative,

Pγ

~~

α

Mβ

'// N .

6.3.2. Let β : M→ P be a quasi-isomorphism in K(R). If the complex P is semi-projective, then β has a right inverse in K(R) which is also a quasi-isomorphism.

6.3.3. Let β : P→ P′ be a quasi-isomorphism in K(R). If the complexes P and P′

are semi-projective, then β is an isomorphism in K(R).

6.3.4 Lemma. Let P be a semi-projective R-complex. For morphisms in K(R),

Pα//

β// M

ϕ

'// N ,

where ϕ is a quasi-isomorphism, one has α= β if ϕα= ϕβ holds.


We also recast 5.3.22–5.3.24 in the language of the homotopy category.

6.3.5. Let I be a semi-injective R-complex, let α : M→ I be a morphism let, andβ : M→ N be a quasi-isomorphism in K(R). There exists a unique morphism γ thatmakes the following diagram in K(R) commutative,

M

α

β

'// N

γ~~

I .

6.3.6. Let β : I→M be a quasi-isomorphism in K(R). If the complex I is semi-injective, then β has a left inverse in K(R) which is also a quasi-isomorphism.

6.3.7. Let β : I→ I′ be a quasi-isomorphism in K(R). If the complexes I and I′ aresemi-injective, then β is an isomorphism in K(R).

The next lemma is dual to 6.3.4.

6.3.8 Lemma. Let I be a semi-injective R-complex. For morphisms in K(R),



Mϕ

'// N

α//

β// I ,

where ϕ is a quasi-isomorphism, one has α= β if αϕ= βϕ holds.


FUNCTORIALITY OF RESOLUTIONS

In the homotopy category, all semi-projective resolutions of a given complex areisomorphic, and a morphism of complexes lifts uniquely to their resolutions. This issufficient to make the process of taking semi-projective resolutions functorial.

We apply the terminology from 5.2.13 and 5.3.13 to quasi-isomorphisms in thehomotopy category. That is, a quasi-isomorphism P→M in K(R), where P is semi-projective is called a semi-projective resolution of M; similarly a quasi-isomorphismM→ I, where I is semi-injective, is called a semi-injective resolution of M.

6.3.9 Construction. For every R-complex M, choose a semi-projective resolutionπM : P(M)

'−→M in K(R). For every morphism α : M→ N in K(R) there exists by6.3.1 a unique morphism in K(R), which we denote P(α), such that the diagram

(6.3.9.1)

P(M) 'πM//

P(α)

M

α

P(N) 'πN// N

is commutative. Similarly, there is a unique morphism in K(R), which we denote byφM , such that the following diagram commutative,

(6.3.9.2)

P(ΣM)

φM

zz

πΣM'

ΣP(M)ΣπM

'// ΣM .

6.3.10 Theorem. The assignments M 7→ P(M) and α 7→ P(α) from 6.3.9 define anendofunctor on K(R). This functor P is k-linear, it preserves coproducts, and it mapsquasi-isomorphisms to isomorphisms. Furthermore, π is a natural transformationP→ IdK(R), and φ is a natural isomorphism PΣ→ ΣP such that P is triangulated.

PROOF. For morphisms α : M→ N and β : L→M in K(R), both morphisms P(αβ)and P(α)P(β) make the following diagram commutative,

P(L) 'πL//

L

αβ

P(N) 'πN// N ,


6.3 Resolutions 221

and hence they are identical: P(αβ)= P(α)P(β). Similarly one finds that the equalityP(1M) = 1P(M) holds for every R-complex M and that P(xα+ β) = xP(α)+ P(β)holds for every pair α,β of parallel morphisms in K(R) and every element x in k.Thus, P is k-linear functor.

Commutativity of (6.3.9.1) implies that if α is a quasi-isomorphism in K(R),then so is P(α). It now follows from 6.3.3 that P(α) is an isomorphism; thus P mapsquasi-isomorphisms to isomorphisms. Commutativity of (6.3.9.1) also shows that πis a natural transformation from P to IdK(R).

To see that P preserves coproducts, let Muu∈U be a family of R-complexes andnotice that there is a commutative diagram in K(R),∐

u∈UP(Mu)

∐πMu

'

ϕ// P(

∐u∈U

Mu)

π∐

Mu'

∐u∈U

Mu ,

where ϕ is the canonical morphism; cf. 3.1.9. The morphism π∐

Muis a quasi-

isomorphism by construction and∐

u∈U πMu

is a quasi-isomorphism by 6.1.13;therefore, ϕ is a quasi-isomorphism. By 5.2.17 the complex

∐u∈U P(Mu) is semi-

projective, so it follows from 6.3.3 that ϕ is an isomorphism.It remains to show that φ is a natural isomorphism such that P is triangulated.

Commutativity of (6.3.9.2) shows that φM is a quasi-isomorphism, and since itsdomain and codomain are semi-projective R-complexes, 6.3.3 implies that φM isan isomorphism. To show that φ is natural, it must be argued that φN P(Σα) =(ΣP(α))φM holds for every morphism α : M→ N in K(R). Since P(ΣM) is semi-projective and ΣπN is a quasi-isomorphism, it suffices by 6.3.4 to argue that(ΣπN)φN P(Σα) = (ΣπN)(ΣP(α))φM holds. And that is a straightforward compu-tation using the commutativity of (6.3.9.1) and (6.3.9.2),

(ΣπN)φN P(Σα) = πΣN P(Σα) = (Σα)πΣM = (Σα)(ΣπM)φM = (ΣπN)(ΣP(α))φM .

Finally, we argue that P with the natural isomorphism φ : PΣ→ ΣP is triangulated.By the definition 6.2.3 of distinguished triangles in K(R), it is enough to argue thatP maps every strict triangle in K(R) to a distinguished one. Consider a morphismα : M→ N in C(R) and the associated strict triangle in K(R),

M[α]−−→ N ε−−→ Coneα $−−→ ΣM ,

where ε and $ are the homotopy classes of the morphisms (1N 0)T and (0 1ΣM) inC(R). It must be shown that the upper candidate triangle in the following diagram isdistinguished,



(?)

P(M)

πM

'

P([α])// P(N)

πN

'

P(ε)// P(Coneα)

χ′

πConeα

'

φM P($)// ΣP(M)

ΣπM

'

M[α]

// N ε// Coneα $

// ΣM

P(M)πM

'??

[α]// P(N)

πN

'??

ε// Cone α

χ

'??

$// ΣP(M) .

ΣπM

'??

Commutativity of (6.3.9.1) and (6.3.9.2) shows that the “upper face” in (?) is com-mutative. Let α be a morphism in C(R) that represents the homotopy class P([α]),i.e. one has P([α]) = [α]. The lower row in (?) is the strict triangle in K(R) associ-ated to α. As K(R) is triangulated category there exists a morphism χ that makes the“lower face” in (?) commutative; see D.2. As πM and πN are quasi-isomorphisms,so is χ by 6.2.22. By 6.3.1 there exists a morphism χ′ in K(R) that makes the third“wall” in (?) commutative; that is, one has χχ′ = πConeα. Note that χ′ is a quasi-isomorphism as χ and πConeα are so. We now argue that the “back face” in (?) iscommutative, i.e. that the equalities hold,

(‡) ε = χ′P(ε) and $χ′ = φM P($) .

As the domains of these morphisms are semi-projective complexes, and since χ andΣπM are quasi-isomorphisms, it follows from 6.3.4 that the equalities in (‡) hold ifand only if the next equalities hold,

() χε = χχ′P(ε) and (ΣπM)$χ′ = (ΣπM)φM P($) .

That these equalities hold follows from the parts of the diagram (?) that are alreadyknown to be commutative. Indeed, one has χε= επN = πConeαP(ε) = χχ′P(ε) and(ΣπM)$χ′ =$χχ′ =$πConeα = (ΣπM)φM P($).

Application of 5.2.16 to the exact sequence 0→ P(N)→ Cone α→ ΣP(M)→0 from 4.1.5 shows that the complex Cone α is semi-projective. Therefore χ′ is aquasi-isomorphism whose domain and codomain are semi-projective complexes. Itfollows from 6.3.3 that χ′ is an isomorphism in K(R), and hence the “back face”in (?) is an isomorphism of candidate triangles in K(R). Since the lower candidatetriangle is strict, the upper candidate triangle is distinguished, as desired.

6.3.11. Consider the endofunctor P on K(R) from 6.3.10. For every R-complex Mthere is a commutative diagram,

PP(M) 'πP(M)

//

P(πM)

P(M)

πM'

P(M) 'πM

// M ,

and it follows from 6.3.4 that the equality P(πM) = πP(M) holds.


6.3 Resolutions 223

6.3.12. Let P and P′ be endofunctors on K(R) defined by making possibly differentchoices of semi-projective resolutions πM : P(M)

'−→M and πM : P′(M)'−→M in

K(R) for every R-complex M; cf. 6.3.10. By 6.3.1 there exists for every M a uniquemorphism ϕM in K(R) such that the following diagram is commutative,

P(M)

ϕM

πM'

P′(M)πM

'// M .

Since πM and πM are quasi-isomorphisms, so is ϕM , and hence 6.3.3 yields that ϕM

is an isomorphism in K(R). In fact, ϕ : P→ P′ is a natural isomorphism. Indeed, fora morphism α : M→ N in K(R), both ϕN P(α) and P′(α)ϕM make the diagram

P(M) 'πM//

M

α

P′(N) 'πN// N

commutative, whence one has ϕN P(α) = P′(α)ϕM by 6.3.4.

6.3.13 Construction. For every R-complex M, choose an injective semi-injectiveresolution ιM : M '−→ I(M) in K(R). For every morphism α : M→ N in K(R) thereis by 6.3.5 a unique morphism in K(R), which we denote I(α), such that the diagram

(6.3.13.1)

M 'ιM//

α

I(M)

I(α)

N 'ιN// I(N)

is commutative. Similarly, there is a unique morphism in K(R), which we denote byφM , such that the following diagram commutative,

(6.3.13.2)

ΣM ιΣM

'//

Σ ιM '

I(ΣM)

φMzz

Σ I(M) .

6.3.14 Theorem. The assignments M 7→ I(M) and α 7→ I(α) from 6.3.13 define anendofunctor on K(R). This functor I is k-linear, it preserves products, and it mapsquasi-isomorphisms to isomorphisms. Furthermore, ι is a natural transformationIdK(R)→ I, and φ is a natural isomorphism IΣ→ Σ I such that I is triangulated.

PROOF. Analogous to the proof of 6.3.10.

6.3.15. Consider the endofunctor I on K(R) from 6.3.14. Proceeding as in 6.3.11one shows that for every R-complex M there is an equality I(ιM) = ιI(M).



6.3.16. Let I and I′ be endofunctors on K(R) defined by making possibly differentchoices of semi-injective resolutions in K(R) for every R-complex. As in 6.3.12 oneshows that the functors I and I′ are naturally isomorphic.

The homology functor on K(R) plays a key role in the final results of this section.

6.3.17 Lemma. Let P be an R-complex with Pv = 0 for all v < 0 and let M be anR-module. If α,β : P→M are morphisms in K(R) with H(α) = H(β), then α= β.

PROOF. The canonical morphism % : P→ H0(P) satisfies the identities H0(α)%= αand H0(β)%= β. As H(α) = H(β) holds by assumption, one has α= β.

6.3.18 Proposition. For R-modules M and N the map,

K(R)(M,N)−→K(R)(P(M),P(N)) given by α 7−→ P(α) ,

is an isomorphism of k-modules with inverse given by β 7→ H(πN)H(β)H(πM)−1.

PROOF. The map given by α 7→ P(α) is k-linear as P is a k-linear functor; see6.3.10. For morphisms α : M→ N and β : P(M)→ P(N), set Φ(α) = P(α) andΨ(β) = H(πN)H(β)H(πM)−1. Apply the functor H to the identity πN P(α) = απM

from (6.3.9.1) to get H(πN)H(P(α)) = H(α)H(πM) = αH(πM) and, consequently,ΨΦ(α) = α. By construction, see 6.3.9, the map ΦΨ(β) = P(Ψ(β)) is the uniquemorphism that makes the following diagram commutative,

P(M) 'πM//

M

Ψ(β)

P(N) 'πN// N .

Thus, to show thatΦΨ(β) = β holds, it suffices to argue that one has πNβ=Ψ(β)πM .As M is a module we may by 5.2.14 and 6.3.12 assume that one has P(M)v = 0 for allv < 0. As one has H(Ψ(β)) = Ψ(β) = H(πN)H(β)H(πM)−1 the following equalityholds, H(πNβ) = H(Ψ(β)πM). Finally, 6.3.17 yields πNβ= Ψ(β)πM .

EXERCISES

E 6.3.1 Give a proof of 6.3.4.E 6.3.2 Give a proof of 6.3.8.E 6.3.3 Give a proof of 6.3.14.E 6.3.4 Let α be a morphism in K(R). Show that there exists a quasi-isomorphism ϕwith ϕα= 0

if and only if there exists a quasi-isomorphism ψ with αψ= 0.E 6.3.5 Let α be a morphism in K(R). (a) Show that α is a quasi-isomorphism only if P(α) is an

isomorphism. (b) Show that α is a quasi-isomorphism only if I(α) is an isomorphism.E 6.3.6 Let M and N be R-complexes. Show that the map K(R)(M,N)→K(R)(P(M),P(N))

given by α 7→ P(α) needs neither be injective nor surjective.E 6.3.7 Let M and N be R-modules. Show that the map K(R)(M,N)→K(R)(I(M), I(N)) given

by α 7→ I(α) is an isomorphism.


6.4 Construction of D 225

E 6.3.8 Let M and N be R-complexes. Show that the map K(R)(M,N)→K(R)(I(M), I(N))given by α 7→ I(α) needs neither be injective nor surjective.

E 6.3.9 Let M and N be R-complexes. Establish isomorphisms,K(R)(P(M),P(N))

∼=−−→K(R)(P(M), I(N))∼=←−−K(R)(I(M), I(N)) ,

and describe the inverse to both of these isomorphisms.E 6.3.10 Show that the endofunctor P on K(R) is left adjoint for I.

6.4 Construction of the Derived Category DDD

SYNOPSIS. Fraction; derived category; product; coproduct; universal property.

Let U be a category and let X be a collection of morphisms in U. One may seeka category X−1U—called the localization of U with respect to X—together witha functor L: U→ X−1U that has the following universal property: The functor Lmaps morphisms in X to isomorphisms in X−1U, and for every functor F: U→ V

that maps morphisms in X to isomorphisms in V there exists a unique functor F thatmakes the following diagram commutative,

UL//

F

X−1U

F||

V .

There is a formal way to construct X−1U; however, it may yield a “category” inwhich the hom-sets are not sets but proper classes. Thus, the localization of U withrespect to X may not exist. An early motivation for the development of the theoryof model categories was to circumvent such set theoretic problems.

We proceed to localize the homotopy category K(R) with respect to the collec-tion of quasi-isomorphisms. The outcome is a category D(R)—semi-projective res-olutions can be harnessed to show that the hom-sets in D(R) are actual sets—calledthe derived category of R; it inherits a triangulated structure from K(R).

FRACTIONS OF MORPHISMS

Recall that we use Greek letters for morphisms in the homotopy category.

6.4.1 Definition. Let M and N be R-complexes. A left prefraction from M to N isa pair (α,ϕ) of morphisms in K(R) such that α and ϕ have the same domain, thecodomain of ϕ is M, the codomain of α is N, and ϕ is a quasi-isomorphism:

Mϕ←−−'

U α−−→ N .

Two left prefractions (α1,ϕ1) and (α2,ϕ2) from M to N are equivalent, in symbols(α1,ϕ1) ≡ (α2,ϕ2), if there exist a third left prefraction (α,ϕ) from M to N andmorphisms µ1 and µ2 that make the following diagram in K(R) commutative,



(6.4.1.1)

U1

ϕ1

α1

M Uϕoo

α//

µ1

OO

µ2

N .

U2ϕ2

__

α2

??

Note that the morphisms µ1 and µ2 in (6.4.1.1) are quasi-isomorphisms.The relation on left prefractions defined above is an equivalence relation, and it

can be described conveniently by way of the resolution functors from Sect. 6.3.

6.4.2 Proposition. Let M and N be R-complexes. For left prefractions (α1,ϕ1) and(α2,ϕ2) from M to N the following conditions are equivalent.

(i) One has (α1,ϕ1)≡ (α2,ϕ2).(ii) P(α1)P(ϕ1)−1 and P(α2)P(ϕ2)−1 are identical morphisms P(M)→ P(N).

(iii) I(α1) I(ϕ1)−1 and I(α2) I(ϕ2)−1 are identical morphisms I(M)→ I(N).In particular, the relation≡ is an equivalence relation on the class of left prefractionsfrom M to N.

PROOF. Recall from 6.3.10 and 6.3.14 that P and I map quasi-isomorphisms to iso-morphisms; this is already implicit in the statements.

(i)=⇒ (iii): If one has (α1,ϕ1) ≡ (α2,ϕ2), then there exists a commutative dia-gram of the form (6.4.1.1). Application of I yields a commutative diagram,

I(U1)

I(ϕ1)

I(α1)

I(M) I(U)I(ϕ)oo

I(α)//

I(µ1)

OO

I(µ2)

I(N) ,

I(U2)

I(ϕ2)

__

I(α2)

??

from which it follows that I(α1) I(ϕ1)−1 = I(α2) I(ϕ2)−1 holds.(iii)=⇒ (ii): Let β : X → Y be a morphism in K(R). One has ιYβ = I(β)ιX by

6.3.13, and an application of P yields P(β) = P(ιY )−1 PI(β)P(ιX ). Thus one has

P(α1)P(ϕ1)−1 =(

P(ιN)−1 PI(α1)P(ιU1))(

P(ιU1)−1 PI(ϕ1)−1 P(ιM)

)= P(ιN)−1 P

(I(α1) I(ϕ1)−1)P(ιM) ,

and the equality P(α2)P(ϕ2)−1 = P(ιN)−1 P(

I(α2) I(ϕ2)−1)

P(ιM) is proved simi-larly. Thus I(α1) I(ϕ1)−1 = I(α2) I(ϕ2)−1 implies P(α1)P(ϕ1)−1 = P(α2)P(ϕ2)−1.

(ii)=⇒ (i): Assume that P(α1)P(ϕ1)−1 and P(α2)P(ϕ2)−1 are identical, and de-note this morphism by α. Set µi = πU i

P(ϕi)−1 for i = 1,2; the commutative diagram



U1

ϕ1

α1

M P(M)πM

'oo

πNα//

µ1

OO

µ2

N

U2

ϕ2

__

α2

??

shows that (α1,ϕ1) and (α2,ϕ2) are equivalent.

6.4.3 Definition. Let M and N be R-complexes. For a left prefraction (α,ϕ) from Mto N, denote by α/ϕ the equivalence class containing (α,ϕ). The class α/ϕ is called aleft fraction from M to N, and the collection of all such is denoted D(R)(M,N).

OBJECTS AND MORPHISMS

The notation introduced in 6.4.3 is suggestive and, indeed, we shall shortly provethat there is a category D(R) whose objects are all R-complexes and in which ahom-set D(R)(M,N) is the collection of all left fractions from M to N.

6.4.4. Consider a diagram in K(R) of the form

M Uϕ

'oo

α// N

V ,

'ψ

OO

where ϕ and ψ are quasi-isomorphisms. The commutative diagram

Uϕ

α

M Vϕψoo

αψ//

ψ

OO

N

Vϕψ

__

αψ

??

shows that the left prefractions (α,ϕ) and (αψ,ϕψ) are equivalent; whence there isan equality α/ϕ= (αψ)/(ϕψ).

6.4.5 Lemma. Let M and N be R-complexes and let α1/ϕ1 and α2/ϕ2 be left fractionsfrom M to N. There exist morphisms α1, α2 and a quasi-isomorphism ϕ in K(R) suchthat the equalities α1/ϕ1 = α1/ϕ and α2/ϕ2 = α2/ϕ hold.

PROOF. Consider the diagrams



M U1ϕ1

'oo

α1// N

P(M)

'πU1P(ϕ1)−1

OO

and

M U2ϕ2

'oo

α2// N

P(M) .

'πU2P(ϕ2)−1

OO

For i = 1,2 one has ϕiπU iP(ϕi)−1 = πM by 6.3.9. Hence 6.4.4 implies that with

αi = αiπU iP(ϕi)−1 and ϕ= πM the equality αi/ϕi = αi/ϕ holds.

The collection of all left prefractions from M to N is a proper class (i.e. not a set);the collection of equivalence classes of these left prefractions is, however, a set.

6.4.6 Proposition. Let M and N be R-complexes. The map

K(R)(P(M),P(N))−→D(R)(M,N) given by β 7−→ (πNβ)/πM

is a bijection with inverse given by α/ϕ 7→ P(α)P(ϕ)−1. In particular, the collectionD(R)(M,N) of left fractions from M to N is a set.

PROOF. Denote by Φ the map given by β 7→ (πNβ)/πM . Set Ψ(α/ϕ) = P(α)P(ϕ)−1;this is by 6.4.2 a well-defined map from D(R)(M,N) to K(R)(P(M),P(N)). Com-mutativity of (6.3.9.1) and 6.4.4 yield

ΦΨ(α/ϕ) = (πN P(α)P(ϕ)−1)/πM = (πN P(α))/(πM P(ϕ)) = (απU )/(ϕπU ) = α/ϕ ,

where U is the common domain of α and ϕ. Similarly, 6.3.11 and (6.3.9.1) yield

ΨΦ(β) = P(πNβ)P(πM)−1 = P(πN)P(β)P(πM)−1 = πP(N) P(β)(πP(M))−1 = β .

ThusΦ is bijective with inverseΨ . In particular, the class D(R)(M,N) is in bijectivecorrespondence with the set K(R)(P(M),P(N)), whence D(R)(M,N) is a set.

6.4.7 Lemma. Let M and N be R-complexes. The set D(R)(M,N) of left fractionsfrom M to N is a k-module with addition and k-multiplication defined as follows.• For α1/ϕ1 and α2/ϕ2 in D(R)(M,N) set

α1/ϕ1 +α2/ϕ2 = (α1 + α2)/ϕ

for any choice of left prefractions (αi,ϕ) with αi/ϕi = αi/ϕ for i = 1,2; cf. 6.4.5.• For x in k and α/ϕ in D(R)(M,N) set

x(α/ϕ) = (xα)/ϕ .

The equivalence class 0/1M containing the left prefraction M 1M←−M 0−→ N is the zero

element in the k-module D(R)(M,N). Finally, with this k-module structure, the mapK(R)(P(M),P(N))→D(R)(M,N) from 6.4.6 is an isomorphism of k-modules.

PROOF. Consider the bijection Φ : K(R)(P(M),P(N))→D(R)(M,N) from 6.4.6.As K(R)(P(M),P(N)) is a k-module, one turns D(R)(M,N) into a k-module, andΦ into an isomorphism, by defining addition and k-multiplication as follows,

α1/ϕ1 +α2/ϕ2 = Φ(Φ−1(α1/ϕ1)+Φ−1(α2/ϕ2)

)and



x(α/ϕ) = Φ(xΦ−1(α/ϕ)

).

Evidently, with this k-module structure on D(R)(M,N), the zero element is

Φ(0) = (πN0)/πM = (0πM)/πM = 0/1M .

We will show that these operations on D(R)(M,N) agree with the ones given in thelemma; in particular, the operations given in the lemma are well-defined.

For the addition operation, assume that one has α1/ϕ1 = α1/ϕ and α2/ϕ2 = α2/ϕ. Itmust be argued that the equality

(?) Φ−1(α1/ϕ1)+Φ−1(α2/ϕ2) = Φ−1((α1 + α2)/ϕ)

holds. Additivity of the functor P and 6.4.2 yield

P(α1)P(ϕ1)−1 +P(α2)P(ϕ2)−1 = P(α1)P(ϕ)−1 +P(α2)P(ϕ)−1

= P(α1 + α2)P(ϕ)−1 ,

which shows (?). A similar argument takes care of k-multiplication.

COMPOSITION OF FRACTIONS

Our aim is to make fractions the morphisms in a category, namely the derived cate-gory. To this end, it must be defined how fractions compose.

6.4.8. Let β : M→V be a morphism and let ψ : N→V be a quasi-isomorphism inK(R). There exists a morphism α and a quasi-isomorphism ϕ such that the followingdiagram in K(R) is commutative,

(6.4.8.1)

U

ϕ '

α// N

ψ'

Mβ// V .

Indeed, choose by 5.2.14 a semi-projective resolution ϕ : U '−→M and apply 6.3.1to get the morphism α. Similarly, it follows from the existence 5.3.19 and liftingproperty 6.3.5 of semi-injective resolutions that given α and ϕ there exist β and ψsuch that (6.4.8.1) is commutative.

REMARK. One does not need semi-projective and semi-injective resolutions to prove the asser-tions in 6.4.8; in fact, they may be proved using only that the homotopy category is triangulated;see E 6.2.8–E 6.5.7.

6.4.9 Proposition. Let L, M, and N be R-complexes. There is a composition rule,

D(R)(M,N)×D(R)(L,M)−→D(R)(L,N) ,

given by(α/ϕ, β/ψ) 7−→ (α/ϕ)(β/ψ) = (αγ)/(ψχ) ,

where γ/χ is any left fraction that makes the diagram



(6.4.9.1)

Wχ

'

γ

Vψ

'

β

Uϕ

'

α

L M N

in K(R) commutative, cf. 6.4.8.The composition rule for left fractions defined above is k-bilinear and associative.

Moreover, one has the following special cases

(6.4.9.2) (α/1M)(β/ψ) = (αβ)/ψ and (α/ϕ)(1M/ψ) = α/(ψϕ) .

PROOF. Consider the isomorphism ΦMN : K(R)(P(M),P(N))→D(R)(M,N) of k-modules from 6.4.7. As composition in K(R) is k-bilinear and associative, one evi-dently obtains a k-bilinear and associative composition rule

D(R)(M,N)×D(R)(L,M)−→D(R)(L,N) ,

by the assignment

(α/ϕ, β/ψ) 7−→ ΦLN(Φ−1

MN(α/ϕ)Φ−1LM(β/ψ)

).

We will show that this composition of left fractions agrees with the one given in thelemma; in particular, the composition rule given in the lemma is well-defined.

Let γ/χ be any left fraction that makes the diagram (6.4.9.1) commutative. It mustbe argued that the equality

Φ−1LN((αγ)/(ψχ)

)= Φ−1

MN(α/ϕ)Φ−1LM(β/ψ)

holds, and that follows from the commutativity of (6.4.9.1),

P(αγ)P(ψχ)−1 = P(α)P(γ)P(χ)−1 P(ψ)−1 = P(α)P(ϕ)−1 P(β)P(ψ)−1 .

The special cases in (6.4.9.2) are evident.

The following definition is justified by the subsequent proposition.

6.4.10 Definition. The derived category D(R) has the same objects as C(R) andK(R), that is, R-complexes. For R-complexes M and N, the hom-set D(R)(M,N) isthe set of all left fractions from M to N; cf. 6.4.3. Composition in D(R) is given bythe map in 6.4.9. Isomorphisms in D(R) are marked by the symbol ’'’.

6.4.11 Proposition. The derived category D(R) is a category. The identity mor-phism in D(R) for an R-complex M is 1M/1M .

PROOF. It follows from 6.4.6 that D(R)(M,N) is a set for every pair M,N of R-complexes. Composition in D(R) is associative by 6.4.9. That 1M/1M is the identityfor M with respect to the composition in D(R) is a consequence of (6.4.9.2).

6.4.12. There is a canonical functor V: K(R)→D(R); it is the identity on objectsand it maps a morphism α : M→ N in K(R) to the left fraction α/1M . Indeed, given



yet another morphism β : L→M in K(R), it follows from (6.4.9.2) that the equali-ties V(αβ) = (αβ)/1L = (α/1M)(β/1L) = V(α)V(β) hold. Furthermore, for every R-complex M, the morphism V(1M) = 1M/1M in D(R) is the identity for M.

For a morphism α in K(R) one conveniently writes V(α) = α/1; that is, the (un-specified) domain of the morphism is omitted from the symbol.

The localization process that leads to the derived category adds just enough mor-phisms to make all quasi-isomorphisms invertible. A morphism of modules is aquasi-isomorphism only if it is an isomorphism, so no new morphisms are needed.The next results make this precise.

6.4.13 Proposition. For R-modules M and N the homomorphism of k-modulesK(R)(M,N)→D(R)(M,N) induced by V is an isomorphism.

PROOF. The homomorphism in question is the composite of the isomorphisms

K(R)(M,N)−→K(R)(P(M),P(N))−→D(R)(M,N)

from 6.3.18 and 6.4.6. Indeed, for a morphism α in K(R) one has (πN P(α))/πM =(απM)/πM = α/1M , where the last equality follows from (6.4.9.2).

6.4.14 Theorem. The restriction to M(R) of the functor VQ: C(R)→D(R) yieldsan isomorphism between the module category M(R) and the full subcategory ofD(R) whose objects are all R-complexes concentrated in degree 0.

PROOF. By 6.1.4 and 6.4.13 the functor VQ: M(R)→D(R) is full and faithful, andhence it yields an isomorphism from M(R) to its image in D(R).

REMARK. The full subcategory M(R) of D(R) is not closed under isomorphisms. The smallestfull subcategory of D(R) that contains M(R) and is closed under isomorphisms is the one whoseobjects are complexes with homology concentrated in degree 0, and it is equivalent to M(R); seeE 6.4.5 and see also E 7.6.3.

ISOMORPHISMS

6.4.15 Lemma. Let α, β, and γ be morphisms in K(R). If αβ and βγ are quasi-isomorphisms then α, β, and γ are quasi-isomorphisms.

PROOF. Since H(α)H(β) = H(αβ) is an isomorphism, H(β) has a left-inverse, andas H(β)H(γ) = H(βγ) is an isomorphism, H(β) has a right-inverse. Consequently,H(β) is an isomorphism. It follows that also H(α) and H(γ) are isomorphisms.

6.4.16. If (α,ϕ) and (α′,ϕ′) are equivalent left prefractions, then α is a quasi-isomorphism if and only if α′ is a quasi-isomorphism; cf. (6.4.1.1).

6.4.17 Proposition. A morphism α/ϕ in D(R) is an isomorphism if and only if α isa quasi-isomorphism, in which case one has (α/ϕ)−1 = ϕ/α.



PROOF. Let α/ϕ be a morphism from M to N.If α is a quasi-isomorphism, then ϕ/α is a morphism from N to M in D(R). The

first equality in the computation (ϕ/α)(α/ϕ) = ϕ/ϕ = (1Mϕ)/(1Mϕ) = 1M/1M is ele-mentary to verify; the last one follows from 6.4.4. Similarly, one has (α/ϕ)(ϕ/α) =α/α= (1Nα)/(1Nα)= 1N/1N , whence α/ϕ is an isomorphism with inverse ϕ/α.

Conversely, assume that α/ϕ is an isomorphism. Denote by U the common do-main of α and ϕ. By the arguments above, 1U/ϕ is an isomorphism, and since one hasα/ϕ = (α/1U )(1U/ϕ) by (6.4.9.2), it follows that α/1U is an isomorphism; denote byβ/ψ its inverse. From 6.4.16 and the equalities 1N/1N = (α/1U )(β/ψ) = (αβ)/ψ it fol-lows that αβ is a quasi-isomorphism. Furthermore, one has 1U/1U = (β/ψ)(α/1U ) =(βγ)/χ for some morphism γ and quasi-isomorphism χ. Another application of6.4.16 gives that βγ is a quasi-isomorphism, and hence α is a quasi-isomorphismby 6.4.15.

6.4.18 Corollary. A morphism M α−→N in C(R) is a quasi-isomorphism if and onlyif V(α) = α/1M is an isomorphism in D(R), in which case the inverse is 1M/α.

6.4.19. In view of the preceding results, it is natural to mark isomorphisms in D(R)by the symbol ‘'’, which is used for quasi-isomorphisms in C(R) and K(R).

An isomorphism α in C(R) yields an isomorphism α/1 in D(R); such isomor-phisms in D(R) may still be marked by the symbol ‘∼=’.

For complexes with certain lifting properties there is a conceptual converse to thecorollary. That is, isomorphisms in D(R) yield quasi-isomorphisms of complexes.

6.4.20 Proposition. Let P and M be R-complexes. If P is semi-projective and Mand P are isomorphic in D(R), then there is a quasi-isomorphism P→M in C(R).

PROOF. If P and M are isomorphic, then by 6.4.17 there are quasi-isomorphismsP←U →M. By 6.3.2 there is a quasi-isomorphism P→U which composed withU →M yields the desired quasi-isomorphism.

6.4.21 Proposition. Let I and M be R-complexes. If I is semi-injective and M andI are isomorphic in D(R), then there is a quasi-isomorphism M→ I in C(R).

PROOF. If M and I are isomorphic, then there are quasi-isomorphisms M←U→ I;see 6.4.17. The lifting property 6.3.5 now yields a quasi-isomorphism M→ I.

REMARK. Existence of an isomorphism M → N in D(R) does not imply that there is even anon-zero morphism M→ N in C(R); see E 6.4.6.

It follows from 6.4.17 that complexes that are isomorphic in the derived categoryhave isomorphic homology. As the next example shows, the converse is not true.

6.4.22 Example. Over the ring Z/4Z, consider the complexes

P = 0 // Z/4Z 2// Z/4Z // 0 and

M = 0 // Z/2Z 0// Z/2Z // 0



concentrated in degrees 1 and 0. Evidently, one has H(P) ∼= M ∼= H(M). The com-plex P is semi-projective by 5.2.8, so if P and M were isomorphic in D(Z/4Z), thenby 6.4.20 there would exist a quasi-isomorphism P→M. However, it is evident thatevery morphism α : P→M in C(Z/4Z) has H1(α) = 0.

6.4.23 Lemma. Assume that R is a principal left ideal domain. For every R-complexL of free modules there is a quasi-isomorphism L−→ H(L).

PROOF. For every v ∈ Z the exact sequence 0→Zv(L)→ Lv→Bv−1(L)→ 0 is splitin M(R). Indeed, Bv−1(L) is a submodule of the free module Lv−1, so by 1.3.10 it isitself free and, in particular, projective. Thus, by 1.3.15 there are isomorphisms

(?) Lv −→ Zv(L)⊕Bv−1(L) .

For every v let Kv be the complex 0→ Bv(L)→ Zv(L)→ 0 concentrated in de-grees v+ 1 and v; notice that one has H(Kv) = Hv(L). The isomorphisms (?) yieldan isomorphism of complexes L→

∐v∈ZKv = K, and there is an obvious quasi-

isomorphism K −→ H(K)∼= H(L); cf. 4.2.10.

6.4.24 Proposition. Assume that R is a principal left ideal domain. For every R-complex M there is an isomorphism M ' H(M) in D(R).

PROOF. Pick by 5.1.7 a semi-free resolution π : L '−→M. It follows from 6.4.23 thatthere there is a quasi-isomorphism α : L→ H(L), and now (H(π)α)/π is an isomor-phism in D(R) from M to H(M) by 6.4.17.

REMARK. Notice from the proof of 6.4.23 that any complex of projective modules over a principalleft ideal domain is isomorphic to a coproduct of bounded complexes of projective modules andhence semi-projective by 5.2.8 and 5.2.17.

ZERO OBJECTS

The complexes that are isomorphic in the homotopy category to the complex 0 areprecisely the contractible complexes. The next result explains what is means for acomplex to be isomorphic to 0 in the derived category.

6.4.25 Proposition. An R-complex is a zero object in D(R) if and only if it isacyclic.

PROOF. In the homotopy category one has P(0) u 0. Thus for every R-complex P,each of the sets K(R)(P(0),P) and K(R)(P,P(0)) consists of a single element. Itfollows from 6.4.6 that for every R-complex M, each of the sets D(R)(0,M) andD(R)(M,0) consists of a single element. Thus the complex 0 is both an initial anda terminal object in D(R), i.e. a zero object.

If M is acyclic, then M→ 0 is a quasi-isomorphism in K(R), and it follows from6.4.18 that M and 0 are isomorphic in D(R). Conversely, if M is isomorphic to 0 inD(R), then by 6.4.17 there exist quasi-isomorphisms M '←−U '−→ 0 in K(R), andhence M is acyclic.



For a morphism α in C(R), it follows from 6.4.17, 4.2.15, and 6.4.25 that VQ(α)is an isomorphism in D(R) if and only if Coneα is isomorphic to 0 in D(R).

PRODUCTS AND COPRODUCTS

6.4.26 Theorem. The derived category D(R) and the functor V: K(R)→D(R) arek-linear. For every family Muu∈U of R-complexes the following assertions hold.

(a) If M with injectionss εu :Mu→Mu∈U is the coproduct of Muu∈U in K(R),then M with the morphisms εu/1Muu∈U is the coproduct of Muu∈U in D(R).

(b) If M with projections $u : M→Muu∈U is the product of Muu∈U in K(R),then M with the morphisms $u/1Mu∈U is the product of Muu∈U in D(R).

In particular, the derived category D(R) has products and coproducts, and the canon-ical functor V preserves products and coproducts.

PROOF. By 6.4.7 the hom-sets in D(R) are k-modules, and composition of mor-phisms in D(R) is k-bilinear by 6.4.9. Thus the category D(R) is k-prelinear. Thefunctor V is k-linear. Indeed, for parallel morphisms α,β in K(R) and x in k, itfollows from 6.4.7 that one has

V(xα+β) = (xα+β)/1 = x(α/1)+(β/1) = xV(α)+V(β) .

To show that D(R) is k-linear, it must be argued that it has biproducts and a zeroobject. That D(R) has a zero object follows from 6.4.25. As K(R) has biproducts,see 6.1.10, it follows from 6.1.9, applied to the canonical functor V: K(R)→D(R),that D(R) has biproducts as well.

(a): Let αu/ϕu : Mu→ Nu∈U be morphisms in D(R). It must be shown thatthere is a unique morphism α/ϕ : M→ N in D(R) with (α/ϕ)(ε

u/1Mu) = αu/ϕu forall u ∈ U . The functor P: K(R)→K(R) preserves coproducts, see 6.3.10; hencethe coproduct

∐v∈U P(Mv) and the uth injection P(Mu)→

∐v∈U P(Mv) are naturally

identified with the complex P(M) and the morphism P(εu) : P(Mu)→ P(M).For existence, consider the family P(αu)P(ϕu)−1 : P(Mu)→ P(N)u∈U of mor-

phisms in K(R). By the universal property of coproducts in K(R) there is a mor-phism ψ : P(M)→ P(N) with ψP(εu) = P(αu)P(ϕu)−1 for all u ∈U . Now one has

((πNψ)/πM)(εu/1Mu) = (πNψP(εu))/πMu = αu/ϕu ,

where the first equality is by commutativity of the diagram

P(Mu)πMu

'

P(εu)

Mu

1Mu

εu

P(M)πM

'

πNψ

Mu M N ,



and the second equality follows by applying the bijection from 6.4.6 to both sides ofthe identity ψP(εu) = P(αu)P(ϕu)−1. Thus, α/ϕ = (πNψ)/πM has the desired prop-erty.

For uniqueness, assume that (α/ϕ)(εu/1Mu) is zero in D(R) for all u ∈ U ; it

must be shown that α/ϕ is zero. By 6.4.9 the composite (α/ϕ)(εu/1Mu) has the form

(αγu)/χu where γu is a morphism and χu is a quasi-isomorphism with εuχu = ϕγu.As each (αγu)/χu is assumed to be zero, it follows that for every u ∈U one has

0 = P(αγu)P(χu)−1 = P(α)P(γu)P(χu)−1 = P(α)P(ϕ)−1 P(εu) .

By the universal property of coproducts in K(R), it follows that P(α)P(ϕ)−1 = 0holds. Evidently, one also has P(0)P(1M)−1 = 0, and thus 6.4.2 yields α/ϕ= 0/1M .

(b): Let αu/ϕu : N→Muu∈U be morphisms in D(R). It must be shown that thereis a unique morphism α/ϕ : N→M in D(R) with ($u/1M)(α/ϕ) = αu/ϕu for all u∈U .The functor I : K(R)→K(R) preserves products, see 6.3.14, and hence the product∏

v∈U I(Mv) and the uth projection∏

v∈U I(Mv)→ I(Mu) are naturally identified withthe complex I(M) and the morphism I($u) : I(M)→ I(Mu).

For existence, consider the family I(αu) I(ϕu)−1 : I(N)→ I(Mu)u∈U of mor-phisms in K(R). By the universal property of products in K(R) there is a morphismψ : I(N)→ I(M) with I($u)ψ = I(αu) I(ϕu)−1 for all u ∈ U . Consider the mor-phism α= πM P(ιM)−1 P(ψ)P(ιN) from P(N) to M. By (6.4.9.2) there is an equality($u/1M)(α/πN) = ($uα)/πN ; we will show that this left fraction is equal to αu/ϕu. By6.4.2 this is equivalent to showing that one has I($uα) I(πN)−1 = I(αu) I(ϕu)−1, i.e.

(?) I($u) I(πM) IP(ιM)−1 IP(ψ) IP(ιN) I(πN)−1 = I(αu) I(ϕu)−1 .

As ψ satisfies I($u)ψ= I(αu) I(ϕu)−1, the equality (?) will follow if the identity

I(πM) IP(ιM)−1 IP(ψ) IP(ιN) I(πN)−1 = ψ

holds, which by 6.3.8 happens if and only if one has

(‡) I(πM) IP(ιM)−1 IP(ψ) IP(ιN) I(πN)−1ιNπN = ψιNπN .

The commutative diagram, cf. 6.3.13,

N

ιN'

P(N)πN

oo

ιP(N)'

P(ιN)// PI(N)

ιPI(N)'

P(ψ)// PI(M)

ιPI(M)'

P(M)P(ιM)oo

ιP(M)'

πM// M

ιM'

I(N) IP(N)I(πN)oo

IP(ιN)// IPI(N)

IP(ψ)// IPI(M) IP(M)

IP(ιM)oo

I(πM)// I(M) ,

shows that the left-hand side of (‡) is equal to ιMπM P(ιM)−1 P(ψ)P(ιN), and thus itmust be argued that there is an equality ιMπM P(ιM)−1 P(ψ)P(ιN) = ψιNπN . How-ever, this follows from the commutative diagram, cf. 6.3.9,



P(N)

πN'

P(ιN)// PI(N)

πI(N)'

P(ψ)// PI(M)

πI(M)'

P(M)

πM'

P(ιM)oo

N ιN// I(N)

ψ// I(M) M .

ιMoo

For uniqueness, assume that the fraction ($u/1M)(α/ϕ) = ($uα)/ϕ is zero for allu ∈U ; it must be shown that α/ϕ is zero. As I($u) I(α) I(ϕ)−1 = I($uα) I(ϕ)−1 = 0holds for every u ∈U , see 6.4.2, it follows from the universal property of productsin K(R) that one has I(α) I(ϕ)−1 = 0. Evidently, one also has I(0) I(1N)−1 = 0, andthus 6.4.2 yields α/ϕ= 0/1N .

By construction, the canonical functor V preserves products and coproducts.

REMARK. Let M and N be R-complexes. A right prefraction from M to N is a diagram in K(R),

(∗) Mβ−−→V

ψ←−−'

N ,

where ψ is a quasi-isomorphism. Dually to 6.4.1 one can define an equivalence relation on thecollection of right prefractions from M to N; the equivalence class containing the right prefraction(∗) is denoted ψ\β and called a right fraction. Like D(R)(M,N), the collection D′(R)(M,N) ofall right fractions from M to N is a set. The collection of all such sets provide the hom-sets fora category D′(R) whose objects are all R-complexes. There is a functor D′(R)→D(R); it isthe identity on objects and it maps a right fraction ψ\β to the left fraction α/ϕ for any choice ofmorphism α and quasi-isomorphism ϕ such that the diagram (6.4.8.1) in K(R) is commutative.The functor D′(R)→D(R) yields an equivalence and, consequently, the derived category mayjust as well be constructed using right fractions. We shall soon prove that D(R) is a triangulatedcategory; similarly so is D′(R), and the equivalence between D′(R) and D(R) is actually anequivalence of triangulated categories.

We shall not pursue the right fraction point of view beyond this remark, even though it doeshave certain advantages. For example, as the proof of 6.4.26 reveals, the argument for existenceof coproducts in D(R) is more straightforward than the one proving existence of products. This isbecause left fractions mesh better with coproducts than with products. Dually, it is straightforwardto show existence of products in D′(R), but slightly more involved to establish the existence ofcoproducts. Had we proved the equivalence between D′(R) and D(R), existence of products inD(R) would follow immediately from the existence of products in D′(R).

UNIVERSAL PROPERTY

The canonical functor V: K(R)→D(R) from 6.4.12 has a universal property de-scribed in the next theorem.

6.4.27 Theorem. Let U be a category and let F: K(R)→ U be a functor. If Fmaps quasi-isomorphisms to isomorphisms, then there exists a unique functor Fthat makes the following diagram commutative,

K(R) V//

F

D(R)

F

U .



For every R-complex M there is an equality F(M) = F(M), and for every morphismα/ϕ in D(R) one has F(α/ϕ) = F(α)F(ϕ)−1. Furthermore, the next assertions hold.

(a) If U is k-prelinear and F is k-linear, then F is k-linear.(b) If U has products/coproducts and F preserves products/coproducts, then F


PROOF. Assume that there is a functor F with FV = F. As V is the identity onobjects, one has F(M)= F(M) for every R-complex M. A morphism α/ϕ in D(R) canby (6.4.9.2) and 6.4.17 be written α/ϕ= (α/1)(1/ϕ) = (α/1)(ϕ/1)−1 = V(α)V(ϕ)−1,and thus there are equalities,

F(α/ϕ) = F(V(α)V(ϕ)−1) = (FV(α))(FV(ϕ))−1 = F(α)F(ϕ)−1 .

Consequently, the functor F is uniquely determined by F.For existence, notice that if α1/ϕ1 = α2/ϕ2 holds in D(R), then there is an equality

F(α1)F(ϕ1)−1 = F(α2)F(ϕ2)−1 in U. Indeed, if one has α1/ϕ1 = α2/ϕ2, then by 6.4.1there exist a morphism α and quasi-isomorphisms ϕ, µ1, and µ2 in K(R) such thatα1µ1 = α= α2µ2 and ϕ1µ1 = ϕ= ϕ2µ2 hold. It follows that F(α1)F(µ1) = F(α) andF(ϕ1)F(µ1) = F(ϕ) hold and, consequently, there are equalities,

F(α1)F(ϕ1)−1 = F(α)F(µ1)−1 F(µ1)F(ϕ)−1 = F(α)F(ϕ)−1 .

Similarly one finds F(α2)F(ϕ2)−1 = F(α)F(ϕ)−1. Thus, one can set F(M) = F(M)for R-complexes M and F(α/ϕ) = F(α)F(ϕ)−1 for morphisms α/ϕ in D(R). With thisdefinition, one evidently has FV = F.

In order for F to be a functor, it must preserve identity morphisms and respectcomposition. By definition, F(1M/1M) = F(1M)F(1M)−1 = 1F(M) = 1F(M) holds forevery R-complex M. Let α/ϕ and β/ψ be composable morphisms in D(R). By 6.4.9the composition (α/ϕ)(β/ψ) is (αγ)/(ψχ) for any choice of morphism γ and quasi-isomorphism χ in K(R) with βχ= ϕγ. Thus there are equalities,

F((α/ϕ)(β/ψ)) = F((αγ)/(ψχ))

= F(αγ)F(ψχ)−1

= F(α)F(γ)F(χ)−1 F(ψ)−1

= F(α)F(ϕ)−1 F(β)F(ψ)−1

= F(α/ϕ) F(β/ψ) .

(a): Assume that F is k-linear, let α1/ϕ1 and α2/ϕ2 be parallel morphisms and letx be an element in k. Write α1/ϕ1 = α1/ϕ and α2/ϕ2 = α2/ϕ for morphisms α1, α2 anda quasi-isomorphism ϕ, see 6.4.5. Now 6.4.7 yields

F(x(α1/ϕ1)+α2/ϕ2) = F((xα1 + α2)/ϕ)

= F(xα1 + α2)F(ϕ)−1

= (xF(α1)+F(α2))F(ϕ)−1

= xF(α1)F(ϕ)−1 +F(α2)F(ϕ)−1



= x F(α1/ϕ)+ F(α2/ϕ)

= x F(α1/ϕ1)+ F(α2/ϕ2) ,

and hence F is k-linear.(b): The proof of 6.1.16(b) applies to prove part (b) in this theorem; only one has

to replace the functor Q by V and the reference to 6.1.10 by one to 6.4.26.

By the universal property above, certain types of functors on K(R) induce func-tors on D(R), natural transformations follow along.

6.4.28 Proposition. Let E,F: K(R)→ U be functors that map quasi-isomorphismsto isomorphisms and consider the induced functors E, F : D(R)→ U from 6.4.27.Every natural transformation τ : E→ F induces a natural transformation τ : E→ Fgiven by τM = τM for every R-complex M.

PROOF. For every R-complex M one has E(M) = E(M) and F(M) = F(M) by6.4.27, whence τM = τM is a morphism E(M)→ F(M). Let α/ϕ : M→ N be a mor-phism in D(R) and denote by U the common domain of α and ϕ. Since τ : E→ F isa natural transformation of functors K(R)→ U there are equalities,

τN E(α/ϕ) = τN E(α)E(ϕ)−1 = F(α)τU E(ϕ)−1 = F(α)F(ϕ)−1τM = F(α/ϕ)τM ,

which show that τ : E→ F is a natural transformation of functors D(R)→ U.

6.4.29 Theorem. Let V be a category and let G: K(R)op→ V be a functor. If Gmaps quasi-isomorphisms to isomorphisms, then there exists a unique functor Gthat makes the following diagram commutative,

K(R)op Vop//

G

D(R)op

Gyy

V .

For every R-complex M there is an equality G(M) = G(M), and for every morphismα/ϕ in D(R)opone has G(α/ϕ)=G(ϕ)−1 G(α). Furthermore, the next assertions hold.

(a) If V is k-prelinear and G is k-linear, then G is k-linear.(b) If V has products/coproducts and G preserves products/coproducts, then G


PROOF. Apply 6.4.27 to the functor Gop : K(R)→ Vop.

6.4.30 Proposition. Let G,J : K(R)op→ V be functors that map quasi-isomorphismsto isomorphisms and consider the induced functors G, J : D(R)op→ V from 6.4.29.Every natural transformation τ : G→ J induces a natural transformation τ : G→ Jgiven by τM = τM for every R-complex M.

PROOF. Apply 6.4.28 to the natural transformation τop : Jop→ Gop of functors fromK(R) to Vop.



A SPECIAL CASE OF THE UNIVERSAL PROPERTY

We apply the derived category’s universal property in an important special case.

6.4.31 Theorem. Let F: K(R)→K(S) be a functor. If F preserves quasi-isomorphisms,then there is a unique functor ´F that makes the next diagram commutative,

K(R)

F

VR// D(R)

´F

K(S)VS// D(S) .

For every R-complex M there is an equality ´F(M) = F(M), and for every morphismα/ϕ in D(R) one has ´F(α/ϕ) = F(α)/F(ϕ). Furthermore, the following assertions hold.

(a) If F is k-linear, then ´F is k-linear.(b) If F preserves products/coproducts, then ´F preserves products/coproducts.

PROOF. As F preserves quasi-isomorphisms, the composite functor VS F mapsquasi-isomorphisms in K(R) to isomorphisms in D(S); see 6.4.18. Hence, the exis-tence and uniqueness of ´F follow from 6.4.27. In symbols one has ´F = (VS F)´. Thevalue of ´F on an R-complex M is ´F(M) = VS F(M) = F(M) since VS is the identityon objects. By 6.4.27, 6.4.17, and (6.4.9.2) the value of ´F on a morphism α/ϕ is

´F(α/ϕ)=(VS F(α))(VS F(ϕ))−1=(F(α)/1)(F(ϕ)/1)−1=(F(α)/1)(1/F(ϕ))=F(α)/F(ϕ) .

By 6.4.26 the functor VS is k-linear and preserves products/coproducts. Thus, if Fhas any of these properties, then so does VS F, and the assertions in parts (a) and (b)now follow from the corresponding parts in 6.4.27.

6.4.32 Proposition. Let E,F: K(R)→K(S) be functors that preserve quasi-isomor-phisms and consider the induced functors É, ´F : K(R)→K(S); see 6.4.31. Everynatural transformation τ : E→ F induces a natural transformation ´τ : É→ ´F givenby ´τM = τM/1 for every R-complex M.

PROOF. Evidently, application of the canonical functor V: K(S)→D(S) to the nat-ural transformation τ yields a natural transformation Vτ= τ/1 : VE→ VF of func-tors K(R)→D(S). By definition, É and ´F are the functors D(R)→D(S) inducedby VE and VF; see 6.4.27. Thus 6.4.28 gives the desired conclusion.

6.4.33 Theorem. Let G: K(R)op→K(S) be a functor. If G preserves quasi-isomor-phisms, then there is a unique functor ´G that makes the next diagram commutative,

K(R)op

G

VopR// D(R)op

´G

K(S)VS

// D(S) .

For every R-complex M one has ´G(M) = G(M), and for every morphism β/ψ inD(R)op one has ´G(β/ψ) = (1/G(ψ))(G(β)/1). Furthermore, the next assertions hold.



(a) If G is k-linear, then ´G is k-linear.(b) If G preserves products/coproducts, then ´G preserves products/coproducts.


6.4.34 Proposition. Let G,J : K(R)op→K(S) be functors that preserve quasi-isomorphisms and consider the induced functors ´G, ´J : K(R)op→K(S); see 6.4.33.Every natural transformation τ : G→ J induces a natural transformation ´τ : ´G→ ´Jgiven by ´τM = τM/1 for every R-complex M.

PROOF. Analogous to the proof of 6.4.32, only apply 6.4.29 and 6.4.30 in place of6.4.27 and 6.4.28.

We shall often abuse notation and write F and G for the induced functors ´F and´G from 6.4.31 and 6.4.33.

6.4.35 Lemma. Let K(R) E−→ K(S) F−→ UT−→ V be functors where E preserves

quasi-isomorphisms and F maps quasi-isomorphisms to isomorphisms. The functorD(R)→ V induced by TFE is TF É; in symbols, (TFE)´ = TF É.

In particular, one has (TF)´ = TF and (FE)´ = F É.

PROOF. The induced functor (TFE)´ is the unique functor with (TFE)´VR = TFE.As one has TF ÉVR = TFVS E = TFE, the assertion follows.

6.4.36 Lemma. Let K(Q)E−→ K(R) F−→ K(S) be functors that preserve quasi-

isomorphisms. The functor D(Q)→D(S) induced by FE is ´F É; i.e. (FE) ´= ´F É.

PROOF. The induced functor (FE) ´ is the unique functor with (FE) ´VQ = VS FE.As one has ´F ÉVQ = ´FVR E = VS FE, the assertion follows.

EXERCISES

E 6.4.1 Let M and N be R-complexes. Show that the collection of left prefractions from M to Nis a proper class; that is, not a set.

E 6.4.2 Show that the isomorphism in 6.4.6 is a natural isomorphism of k-modules.E 6.4.3 For R-complexes M and N show that there are natural isomorphisms K(R)(P(M),N)∼=

D(R)(M,N) and K(R)(M, I(N))∼=D(R)(M,N).E 6.4.4 Write down explicitly the inverse of the functor M(R)→ VQ(M(R)).E 6.4.5 Let D0(R) be the full subcategory of D(R) whose objects are all complexes that are

isomorphic in D(R) to one in the full subcategory M(R). (a) Show that a complex Mis in D0(R) if and only if H(M) is concentrated in degree 0. (b) Show that D0(R) isequivalent to M(R).

E 6.4.6 Show that the complexes in 4.2.3 are isomorphic in D(R). Hint: E 5.1.9.E 6.4.7 Show that a left fraction α/ϕ is zero if and only if there exists a quasi-isomorphism µ

with αµ= 0; see also E 6.3.4.E 6.4.8 Consider k as a module over the polynomial algebra k[x] with the trivial x-action. Show

that there is a non-zero morphism κ : k→ Σk in D(k[x]) with H(κ) = 0.E 6.4.9 Show that Mgr(R) is isomorphic to a full subcategory of D(R).E 6.4.10 Let R be left hereditary. Show that there is an isomorphism M ' H(M) in D(R) for

every R-complex M. Hint: E 5.2.3.


6.5 Triangulation of D 241

E 6.4.11 Let M be an R-complex. Show that if H(M) is a complex of projective R-modules, thenthere is a quasi-isomorphism H(M)→M.

6.5 Triangulation of DDD

SYNOPSIS. Distinguished triangle; triangulated functor; universal property; distinguished trianglefrom short exact sequence; homology; Five Lemma, Horseshoe Lemma.

The homotopy category K(R) is triangulated but rarely Abelian, and same holdsfor the derived category D(R). While the triangulated structure on D(R) is inducedfrom K(R), it is closer to the Abelian structure on C(R) in the crucial sense thatevery short exact sequence in C(R) yields a distinguished triangle in D(R).

An important proof-technical tool in our approach to the triangulation on D(R)is the semi-projective resolution functor on K(R) and the one it induces on D(R).

6.5.1. By 6.3.10 the resolution functor P: K(R)→K(R) maps quasi-isomorphismsto isomorphisms, so it induces by 6.4.27 a functor P : D(R)→K(R) with PV = P.As P is k-linear and preserves coproducts, the induced functor P has the same prop-erties. When there is no risk of ambiguity, we write P for this induced functor.

6.5.2 Lemma. Let α/ϕ : M→ N be a morphism in D(R). The morphism P(α/ϕ) inK(R) fits into the following commutative diagram in D(R),

P(M)πM/1'//

P(α/ϕ)/1

M

α/ϕ

P(N)πN/1'// N .

PROOF. By 6.4.27 and 6.4.4 one has P(α/ϕ)/1 = (P(α)P(ϕ)−1)/1 = P(α)/P(ϕ). Now(6.4.9.2) and the fact that π : P→ IdK(R) is a natural transformation yields(

πN/1)(

P(α/ϕ)/1)=(πN/1

)(P(α)/P(ϕ)

)= (πN P(α))/P(ϕ) = (απU )/P(ϕ) ,

where U is the common domain of α and ϕ. Using again that π is a natural transfor-mation, one gets a commutative diagram in K(R),

P(U)P(ϕ)

uπU

P(M)1

=

πM

Uϕ

'

α

P(M) M N ,

which shows that (α/ϕ)(πM/1) = (απU )/P(ϕ) holds. Consequently, the composites(πN/1)(P(α/ϕ)/1) and (α/ϕ)(π

M/1) are identical.



6.5.3. By 4.2.8 and 6.4.31 there is a unique k-linear endofunctor ´Σ on D(R) thatmakes the following diagram commutative,

K(R)

Σ

V// D(R)

´Σ

K(R) V// D(R) .

By 6.4.36 it is an isomorphism with inverse induced by Σ−1 : K(R)→K(R). Bythe usual abuse of notation, the functor ´Σ is written Σ or, occasionally, ΣD. For amorphism α/ϕ in D(R) one has ΣD(α/ϕ) = (Σα)/(Σϕ).

6.5.4. By 6.3.10 there is a natural isomorphism φ : PΣ→ ΣP of endofunctors onK(R) that makes the functor P triangulated. With the notation from 6.5.1 one has

(PΣ)V = PVΣ = PΣ and (Σ P)V = ΣP ,

and hence the functors D(R)→K(R) induced by PΣ and ΣP are, respectively, PΣand Σ P. It follows from 6.4.28 that φ induces a natural isomorphism PΣ→ Σ P.

Consider the k-linear category D(R), see 6.4.26, equipped with the k-linear au-tomorphism Σ= ΣD. One may now speak of candidate triangles in D(R); cf. D.1.

6.5.5 Definition. A candidate triangle in D(R) is called a distinguished triangle ifit is isomorphic to the image of a distinguished triangle in K(R) under the canonicalfunctor V: K(R)→D(R).

6.5.6. For every morphism α : M→ N in C(R) it follows from 6.2.2, 6.2.3, and 6.5.5that M α−→ N −→ Coneα−→ ΣM is a distinguished triangle in D(R).

Next we show that the semi-projective resolution functor D(R)→ K(R) pre-serves distinguished triangles. Together with the already established fact that K(R)is triangulated, this yield a short pass to the fact that D(R) is triangulated.

6.5.7 Proposition. Consider the induced resolution functor P: D(R)→K(R) from6.5.1 and the induced natural isomorphism φ : PΣ→ ΣP from 6.5.4. For every dis-tinguished triangle in D(R),

Mα/ϕ−−→ N

β/ψ−−→ Xγ/χ−−→ ΣM ,

the candidate triangle

P(M)P(α/ϕ)−−−−→ P(N)

P(β/ψ)−−−−→ P(X)φM P(γ/χ)−−−−−−→ ΣP(M)

is distinguished in K(R).

PROOF. As in 6.5.1, write P for the resolution functor on K(R) and P for the inducedfunctor on D(R). As φ and the induced φ are natural isomorphisms, and since everydistinguished triangle in D(R) is isomorphic to one of the form



MV(α)−−−→ N

V(β)−−−→ XV(γ)−−−→ ΣM ,

where M α−→ Nβ−→ X

γ−→ ΣM is a distinguished triangle in K(R), it suffices to arguethat the candidate triangle

P(M)PV(α)−−−−→ P(N)

PV(β)−−−−→ P(X)φM P(γ)−−−−→ Σ PV(M) ,

is distinguished. However, this candidate triangle is nothing but

P(M)P(α)−−−→ P(N)

P(β)−−−→ P(X)φM P(γ)−−−−→ ΣP(M) ,

which is distinguished by 6.3.10.

6.5.8 Theorem. The derived category D(R), equipped with the automorphism Σ andthe collection of distinguished triangles defined in 6.5.5, is triangulated. Moreover,the canonical functor V: K(R)→D(R) is triangulated.

PROOF. By 6.5.3 one has VΣ = ΣV. Thus, once it has been established that thecategory D(R) is triangulated, the functor V is triangulated by 6.5.5.

(TR0): Follows immediately from the definition of distinguished triangles inD(R) and the fact that (K(R),Σ) satisfies (TR0).

(TR1): Let α/ϕ : M→ N be a morphism in D(R) and denote by U the commondomain of α and ϕ. As (K(R),Σ) satisfies (TR1), the morphism α fits in a distin-guished triangle in K(R), say,

U α−−→ Nβ−−→ X

γ−−→ ΣU .

Applying V to this diagram, one gets by 6.5.5 a distinguished triangle in D(R),namely the upper row in the following commutative diagram,

U

V(ϕ)

V(α)// N

V(β)// X

V(γ)// ΣU

ΣV(ϕ)

Mα/ϕ

// NV(β)

// X(ΣV(ϕ))V(γ)

// ΣM .

As V(ϕ) is an isomorphism by 6.4.18, the lower row is a distinguished triangle.(TR2’): Follows immediately since (K(R),Σ) satisfies (TR2’).(TR4’): Consider a commutative diagram in D(R),

(?)

M

ϕ

α// N

ψ

β// X

γ// ΣM

Σϕ

M′ α′// N′

β′// X ′

γ′// ΣM′ ,

where the rows are distinguished triangles. Note that in the diagram above, Greekletters denote morphisms in D(R), that is, left fractions. We shall construct a mor-phism χ : X → X ′ in D(R) that makes (?) commutative, and such that the mappingcone candidate triangle of (ϕ,ψ,χ) in D(R) is distinguished.



Consider the induced resolution functor P: D(R)→K(R) from 6.5.1 and the nat-ural isomorphism φ : PΣ→ ΣP from 6.5.4. As P is a functor, the following diagramin K(R) is commutative, and the rows are distinguished triangles by 6.5.7,

(‡)

P(M)

P(ϕ)

P(α)// P(N)

P(ψ)

P(β)// P(X)

φM P(γ)// ΣP(M)

ΣP(ϕ)

P(M′)P(α′)

// P(N′)P(β′)

// P(X ′)φM′ P(γ′)

// ΣP(M′) .

As the category K(R) is triangulated, there exists a morphism ϑ : P(X)→ P(X ′)that makes (‡) commutative and such that the mapping cone candidate triangle of(P(ϕ),P(ψ),ϑ) in K(R) is distinguished. The “front” in the next diagram in D(R) iscommutative, indeed, it is the image of (‡) under the functor V: K(R)→D(R),

()

M

ϕ

α// N

ψ

β// X

γ// ΣM

Σϕ

P(M)

πM/1'

??

P(ϕ)/1

P(α)/1// P(N)

πN/1'

??

P(ψ)/1

P(β)/1// P(X)

πX/1'

??

ϑ/1

(φM P(γ))/1// ΣP(M)

Σ (πM/1)'

??

Σ (P(ϕ)/1)

M′ α′// N′

β′// X ′

γ′// ΣM′ .

P(M′)πM′/1

'??

P(α′)/1// P(N′)

πN′/1

'??

P(β′)/1// P(X ′)

πX ′/1

'??

(φM′ P(γ′))/1

// ΣP(M′)Σ (πM′/1)

'??

The “back” in () is the commutative diagram (?). The three complete “walls” arecommutative by 6.5.2, and so are the left-hand and middle in both the “top” and the“bottom” The right-hand square in the “top” is also commutative; indeed, one has(

Σ(πM/1))(

(φM P(γ))/1)=((ΣπM)φM P(γ)

)/1

=(πΣM P(γ)

)/1

= (πΣM/1)(P(γ)/1)

= γ(πX/1) ,

where the first and third equalities are by (6.4.9.2), the second equality follows fromthe definition 6.3.9 of φM , and the fourth equality is by 6.5.2. A similar computationshows that the right-hand square in the “bottom” of () is commutative as well, andhence the diagram () is commutative.

Now define χ : X → X ′ to be the composite of morphisms

X(πX/1)−1

−−−−−→ P(X)ϑ/1−−→ P(X ′)

πX ′/1−−−→ X ′ .

A straightforward diagram chase shows that χmakes the “back” in () commutative,and hence (ϕ,ψ,χ) is a morphism of distinguished triangles. It remains to see thatthe mapping cone candidate triangle of (ϕ,ψ,χ) is distinguished. Since by D.18 it



is isomorphic to the mapping cone candidate triangle, ∆, of (P(ϕ)/1,P(ψ)/1,ϑ/1), itsuffices to argue that ∆ is distinguished. Evidently, ∆ is isomorphic to the image ofthe mapping cone candidate triangle, ∆0, of (P(ϕ),P(ψ),ϑ) under the functor V. Byassumption, ∆0 is distinguished in K(R), and hence its image ∆ is distinguished inD(R) by definition; see 6.5.5.

REMARK. Let Kprj(R) denote the full subcategory of K(R) whose objects are the semi-projectiveR-complexes. The proof of 6.2.4 shows that (Kprj(R),Σ) is a triangulated category, albeit not atriangulated subcategory of K(R); see E 6.2.10. The composite functor Kprj(R)→K(R)→D(R)is a triangulated equivalence; see E 6.5.4.

TRIANGULATED FUNCTORS

We revisit the derived category’s universal property and show that triangulated func-tors on K(R) induce triangulated functors on D(R).

6.5.9. Let U be a triangulated category and let F : K(R)→ U be a functor that mapsquasi-isomorphisms to isomorphisms. It follows from 6.4.35 and 6.5.3 that the func-tors D(R)→ U induced by FΣ and ΣU F are FΣD and ΣU F.

6.5.10 Theorem. Let U be a triangulated category and let F: K(R)→ U be a functorthat maps quasi-isomorphisms to isomorphisms. If F is triangulated with naturalisomorphism φ : FΣ→ ΣU F, then the induced functor F : D(R)→ U, see 6.4.27, istriangulated with natural isomorphism φ : FΣ→ ΣU F.

PROOF. For every distinguished triangle

(?) M α−−→ Nβ−−→ X

γ−−→ ΣM

in K(R), the diagram

(‡) F(M)F(α)−−−→ F(N)

F(β)−−→ F(X)φM F(γ)−−−−→ ΣU F(M)

is a distinguished triangle in U. By definition, every distinguished triangle in D(R)has, up to isomorphism, the form

MV(α)−−−→ N

V(β)−−−→ XV(γ)−−−→ ΣM

for some distinguished triangle (?) in K(R). Hence, it must be verified that

() F(M)FV(α)−−−−→ F(N)

FV(β)−−−−→ F(X)φM FV(γ)−−−−−−→ ΣU F(M)

is a distinguished triangle, which is evident since () is nothing but (‡).

6.5.11 Proposition. Let U be a triangulated category, let E,F: K(R)→ U be trian-gulated functors that map quasi-isomorphisms to isomorphisms, and let τ : E→ Fbe a natural transformation. If τ is triangulated, then the induced natural transfor-mation τ : E→ F of triangulated functors, see 6.4.28 and 6.5.10, is triangulated.



PROOF. By assumption the functors E and F are triangulated; denote the associatednatural isomorphisms by φ and ψ, respectively. By 6.5.10 the induced functors Eand F are triangulated with associated natural isomorphisms φ and ψ. For and R-complex M, the equality ψM τΣM = (Σ τM)φM holds as the left-hand side is ψMτΣM ,the right-hand side (ΣτM)φM , and those two composites agree by assumption.

6.5.12 Theorem. Let V be a triangulated category and let G: K(R)op→ V be a func-tor that maps quasi-isomorphisms to isomorphisms. If G is triangulated with naturalisomorphism ψ : Σ−1

V G→ GΣop, then the functor G : D(R)op→ V, see 6.4.29, is tri-angulated with natural isomorphism ψ : Σ−1

V G→ GΣop.

PROOF. Apply 6.5.10 to the functor Gop : K(R)→ Vop.

6.5.13 Proposition. Let V be a triangulated category, let G,J : C(R)op→ V be trian-gulated functors that map quasi-isomorphisms to isomorphisms, and let τ : G→ J bea natural transformation. If τ is triangulated, then the induced natural transformationτ : G→ J of triangulated functors, see 6.4.30 and 6.5.12, is triangulated.

PROOF. Apply 6.5.11 to the natural transformation τop : Jop→ Gop of functors fromK(R) to Vop.

THE UNIVERSAL PROPERTY REVISITED

6.5.14. By 6.5.8 the functor VS : K(S)→D(S) is triangulated with natural isomor-phism the identity transformation VSΣ= ΣVS; cf. 6.5.3. Thus, if F : K(R)→K(S)is triangulated functor with natural isomorphism φ, then the composite functor VS Fis triangulated with natural isomorphism VS φ.

6.5.15 Theorem. Let F: K(R)→K(S) be a functor that preserves quasi-isomor-phisms. If F is triangulated with associated natural isomorphism φ : FΣ→ ΣF, thenthe induced functor ´F : D(R)→D(S), see 6.4.31, is triangulated with associatednatural isomorphism ´φ : ´FΣ→ Σ ´F.

PROOF. By 6.5.14 the functor VS F is triangulated with natural isomorphism VS φ.Now 6.5.10 implies that the functor ´F = (VS F)´ is triangulated with natural isomor-phism ´φ= (VS φ)´; the equalities follow from 6.4.31 and 6.4.32.

6.5.16 Proposition. Let E,F: K(R)→K(S) be functors that preserve quasi-isomor-phisms and let τ : E→ F be a natural transformation. If E and F are triangulated andτ is triangulated, then the induced natural transformation ´τ : ´E→ ´F of triangulatedfunctors, see 6.4.32 and 6.5.15, is triangulated.

PROOF. Let E and F be triangulated with associated natural isomorphisms φ and ψ.By 6.5.14 the functors VS E,VS F: K(R)→D(S) are triangulated with natural iso-morphisms VS φ and VSψ. As the natural transformation τ : E→ F is triangu-lated and VSΣ = ΣVS, the natural transformation VS τ : VS E→ VS F is triangu-lated as well. By 6.5.11 the natural transformation ´τ = (VS τ)´ from ´E = (VS E)´to ´F = (VS F)´ is triangulated; the equalities follow from 6.4.31 and 6.4.32.



6.5.17 Theorem. Let G: K(R)op→K(S) be a functor that preserves quasi-isomor-phisms. If G is triangulated with natural isomorphism ψ : Σ−1 G→ GΣop, then thefunctor ´G, see 6.5.17, is triangulated with natural isomorphism ´ψ : Σ−1 ´G→ ´GΣop.


6.5.18 Proposition. Let G,J : K(R)op→K(S) be functors that preserve quasi-isomorphisms and let τ : G→ J be a natural transformation. If G and J are triangu-lated and τ is triangulated, then the induced transformation ´τ : ´G→ ´J of triangulatedfunctors, see 6.4.34 and 6.5.17, is triangulated.

PROOF. Analogous to the proof of 6.4.32, only apply 6.4.29 and 6.4.30 in place of6.4.27 and 6.4.28.

TRIANGLES FROM SEQUENCES

Every short exact sequence of complexes can be completed to a distinguished trian-gle in the derived category.

6.5.19 Theorem. For every commutative diagram in C(R),

0 // M α//

ϕ

Nβ//

ψ

X //

χ

0

0 // M′ α′// N′

β′// X ′ // 0 ,

with exact rows, there is a morphism of distinguished triangles in D(R),

M[α]/1

//

[ϕ]/1

N[β]/1

//

[ψ]/1

Xγ//

[χ]/1

ΣM

Σ([ϕ]/1)

M′[α′]/1

// N′[β′]/1

// X ′γ′// ΣM′ ,

with γ= [$]/[β] and γ′= [$′]/[β′] where β, β′ are the quasi-isomorphisms from 4.3.10and $, $′ are the canonical morphisms Coneα ΣM and Coneα′ ΣM′.

PROOF. It follows from 4.3.11 that there is a commutative diagram in D(R),

MVQ(ι)

//

VQ(ϕ)

CylαVQ(π)

//

VQ(α)

VQ(α)'

ConeαVQ($)

//

VQ(β)

VQ(β)'

ΣMΣVQ(ϕ)

M′VQ(ι′)

// Cylα′VQ(π′)

//

VQ(α′)'

Coneα′VQ($′)

//

VQ(β′)'

ΣM′

MVQ(α)

//

VQ(ϕ)

NVQ(β)

//

VQ(ψ)

XVQ($)VQ(β)−1

//

VQ(χ)

ΣM .

ΣVQ(ϕ)

M′VQ(α′)

// N′VQ(β′)

// X ′VQ($′)VQ(β′)−1

// ΣM′



The upper rows are by 6.2.5 images under V of distinguished triangles in K(R)and hence distinguished in D(R); thus the “top” is a morphism of distinguishedtriangles in D(R). As the vertical morphisms are isomorphisms, the “bottom” is amorphism of distinguished triangles as well. Finally, by 6.4.18 and (6.4.9.2) one hasVQ($)VQ(β)−1 = [$]/[β] and VQ($′)VQ(β′)−1 = [$′]/[β′].

REMARK. A general short exact sequence 0→M→ N → X → 0 of complexes can not be com-pleted to a distinguished triangle M→ N→ X → ΣM in the homotopy category; see E 6.2.2.

HOMOLOGY

Homology induces a homological functor on the derived category.

6.5.20 Proposition. The homology functor H: K(R)→ C(R) from 6.2.21 mapsquasi-isomorphisms to isomorphisms. The induced functor H: D(R)→ C(R) from6.4.27 is k-linear, it preserves products and coproducts, and one has HΣ= ΣH.

A morphism ξ in D(R) is an isomorphism if and only if H(ξ) is an isomorphism.

PROOF. The first assertion follows from the definition of quasi-isomorphisms, andthus H: K(R)→ C(R) induces by 6.4.27 a functor H : D(R)→ C(R), which out-side of this proof is denoted H. By 6.2.22, the functor H is k-linear and preservesproducts and coproducts. It follows from 6.4.27 that the induced functor H hasthe same properties. As one has HΣ = ΣH, the definition of H yields equalitiesHΣV = HVΣ= HΣ= ΣH = Σ HV, where V: K(R)→D(R) is the canonical func-tor. It now follows from the uniqueness assertion in 6.4.27 that one has HΣ= Σ H.

For a morphism α/ϕ in D(R) one has H(α/ϕ) = H(α)H(ϕ)−1. It follows from6.4.17 that α/ϕ is an isomorphism if and only if H(α/ϕ) is an isomorphism.

6.5.21. Notice from 6.5.20 that the supremum, infimum, and amplitude of com-plexes as defined in 2.5.4 are invariants of objects in the derived category. That is, forcomplexes M 'M′ one has supM = supM′, infM = infM′, and ampM = ampM′.

REMARK. While zero objects and isomorphisms in the derived category can be recognized inhomology, see 6.4.25 and 6.5.20, zero morphisms can not; see E 6.4.8.

The last assertion in the next theorem is the Five Lemma in D(R); cf. D.15.

6.5.22 Theorem. For every morphism of distinguished triangles in D(R),

M

ϕ

α// N

ψ

β// X

χ

γ// ΣM

Σϕ

M′ α′// N′

β′// X ′

γ′// ΣM′ ,

there is a commutative diagram in C(R),



(6.5.22.1)

H(M)

H(ϕ)

H(α)// H(N)

H(ψ)

H(β)// H(X)

H(χ)

H(γ)// ΣH(M)

ΣH(ϕ)

ΣH(α)// ΣH(N)

ΣH(ψ)

H(M′)H(α′)

// H(N′)H(β′)

// H(X ′)H(γ′)

// ΣH(M′)ΣH(α′)

// ΣH(N′) ,

with exact rows. In particular, if two of the morphisms ϕ, ψ, and χ are isomorphisms,then so is the third.

PROOF. By 6.5.20 the functor H: D(R)→ C(R) satisfies the identity HΣ = ΣH; itfollows that (6.5.22.1) is commutative. We argue that the upper row is exact; a par-allel argument shows that the lower row is exact. By the definition of distinguishedtriangles 6.5.5, there exists an isomorphism of candidate triangles in D(R),

M

'

α// N

'

β// X

'

γ// ΣM

'

MV(α)

// NV(β)

// XV(γ)

// ΣM ,

where M α−→ Nβ−→ X

γ−→ ΣM is a distinguished triangle in K(R). It follows thatthe upper row in (6.5.22.1) is isomorphic to the sequence

H(M)HV(α)−−−−→ H(N)

HV(β)−−−−→ H(X)HV(γ)−−−−→ ΣH(M)

ΣHV(α)−−−−−→ ΣH(N) .

This sequence is nothing but

H(M)H(α)−−−→ H(N)

H(β)−−−→ H(X)H(γ)−−−→ ΣH(M)

ΣH(α)−−−−→ ΣH(N) ,

which is exact by 6.2.22.The last assertion follows, in view of the final assertion in 6.5.20, from the Five

Lemma 2.1.40 applied to the diagram (6.5.22.1).

6.5.23 Corollary. Let M α−→ Nβ−→ X

γ−→ ΣM be a distinguished triangle in D(R).There are inequalities

supM 6 maxsupN,supX−1 , infM > mininfN, infX−1 ,supN 6 maxsupM,supX , infN > mininfM, infX ,supX 6 maxsupM+1,supN , and infX > mininfM+1, infN .

In particular, if two of the complexes M, N, and X are acyclic, then so is the third.Furthermore, the following assertions hold.

(a) M is acyclic if and only if β is an isomorphism.(b) N is acyclic if and only if γ is an isomorphism.(c) X is acyclic if and only if α is an isomorphism.

PROOF. Immediate from 6.5.22 in view of 6.4.25 and 6.5.20.



6.5.24 Corollary. Let F: D(R)→D(S) be a triangulated functor with natural iso-morphism φ : FΣ→ ΣF. For every commutative diagram in C(R),

0 // M α//

ϕ

Nβ//

ψ

X //

χ

0

0 // M′ α′// N′

β′// X ′ // 0 ,

with exact rows, there is a commutative diagram in C(S) with exact rows,

HF(M)

HF([ϕ]/1)

HF([α]/1)// HF(N)

HF([ψ]/1)

HF([β]/1)// HF(X)

HF([χ]/1)

H(δ)// ΣHF(M)

ΣHF([ϕ]/1)

ΣHF([α]/1)// ΣHF(N)

ΣHF([ψ]/1)

HF(M′)HF([α′]/1)

// HF(N′)HF([β′]/1)

// HF(X ′)H(δ′)

// ΣHF(M′)ΣHF([α′]/1)

// ΣHF(N′) .

Here δ and δ′ are the composites φM F(γ) and φM′ F(γ′), where γ : X → ΣM andγ′ : X ′→ ΣM′ are the morphisms in D(R) from 6.5.19.

PROOF. By 6.5.19 the commutative diagram in C(R) yields a morphism of distin-guished triangles in D(R), which the triangulated functor F maps to a morphism ofdistinguished triangles in D(S). Now 6.5.22 yields the desired diagram in C(S).

6.5.25 Corollary. Let G: D(R)op→D(S) be a triangulated functor with naturalisomorphism φ : GΣ−1→ ΣG. For every commutative diagram in C(R)op,

(6.5.25.1)

0 Moo Nαoo X

βoo 0oo

0 M′oo

ϕ

OO

N′α′oo

ψ

OO

X ′β′oo

χ

OO

0 ,oo

with exact rows, there is a commutative diagram in C(S) with exact rows,

HG(X ′)

HG([χ]/1)

HG([β′]/1)// HG(N′)

HG([ψ]/1)

HG([α′]/1)// HG(M′)

HG([ϕ]/1)

H(δ′)// ΣHG(X ′)

ΣHG([χ]/1)

ΣHG([β′]/1)// ΣHG(N′)

ΣHG([ψ]/1)

HG(X)HG([β]/1)

// HG(N)HG([α]/1)

// HG(M)H(δ)

// ΣHG(X)ΣHG([β]/1)

// ΣHG(N) .

Here δ and δ′ are the composites φX G(Σ−1 γ) and φX ′G(Σ−1 γ′), where γ : ΣM→ Xand γ′ : ΣM′→ X ′ are the morphisms in D(R)op obtained by applying 6.5.19 to(6.5.25.1) considered as a diagram in C(R).

PROOF. The diagram (6.5.25.1) corresponds to a commutative diagram in C(R),which by 6.5.19 yields a morphism of distinguished triangles in D(R); in view ofD.6 that morphism can be interpreted as a morphism of distinguished triangles inD(R)op, and it follows from the axioms for triangulated categories that the diagram



X N[β]/1

oo M[α]/1

oo Σ−1 XΣ−1 γoo

X ′

[ϕ]/1

OO

N′[β′]/1oo

[ψ]/1

OO

M′[α′]/1oo

[χ]/1

OO

Σ−1 X ′Σ−1 γ′oo

Σ−1 ([ϕ]/1)

OO

is a morphism of distinguished triangles in D(R)op. The triangulated functor G mapsit to a morphism of distinguished triangles in D(S), and 6.5.22 applies to yield thedesired diagram in C(S).

HORSESHOE LEMMAS

The next two results are known as the Horseshoe Lemmas.

6.5.26 Lemma. Let M′ → M → M′′ → ΣM′ be a distinguished triangle in D(R).If P′ and P′′ are semi-projective replacements of M′ and M′′, then there is an exactsequence 0→P′→P→P′′→ 0 in C(R) with P a semi-projective replacement of M.

PROOF. Rotation, see D.2, of the given triangle yields another distinguished trian-gle, Σ−1 M′′ → M′ → M → M′′. Since M′ ' P′ and M′′ ' P′′ hold in D(R) thereis also a distinguished triangle Σ−1 P′′

γ−→ P′→M→ P′′. As the complexes P′ andP′′ are semi-projective, the morphism γ has the form γ = [α]/1 for some morphismα : Σ−1 P′′→ P′ in C(R); see 6.4.6. The diagram Σ−1 P′′ α−→ P′→ Coneα→ P′′ inC(R) is a distinguished triangle when viewed as a diagram in D(R), see 6.5.6. AsD(R) is triangulated, there exists a morphism χ : M→ Coneα such that the diagram

Σ−1 P′′

Σ−1 1P′′

γ// P′

1P′

// M

χ

// P′′

1P′′

Σ−1 P′′γ// P′ // Coneα // ΣP′′ ,

is commutative. It follows from 6.5.22 that χ is an isomorphism. Set P = Coneα;by 4.1.5 there is a short exact sequence 0→ P′→ P→ P′′→ 0, whence it followsfrom 5.2.16 that the complex P is semi-projective.

6.5.27 Lemma. Let M′ → M → M′′ → ΣM′ be a distinguished triangle in D(R).If I′ and I′′ are semi-injective replacements of M′ and M′′, then there is an exactsequence 0→ I′→ I→ I′′→ 0 in C(R) with I a semi-injective replacement of M.


EXERCISES

E 6.5.1 Let R be semi-simple. Show that the categories D(R) and Mgr(R) are equivalent andconclude that D(R) is Abelian.

E 6.5.2 Show that D(R) and K(PrjR), see E 6.1.9, are equivalent categories if R is left hereditary.E 6.5.3 Show that D(R) and K(InjR), see E 6.1.10, are equivalent categories if R is left heredi-

tary.



E 6.5.4 Show that Kprj(R) and D(R) are equivalent as triangulated categories; see E 6.2.10.E 6.5.5 Show that Kinj(R) and D(R) are equivalent as triangulated categories; see E 6.2.12.E 6.5.6 Let (T,Σ) be a triangulated category. A commutative square in T,

Uϕ

α// N

ψ

Mβ// V

is called homotopy Cartesian if there exists a distinguished triangle of the form

U

(ϕ−α)//

M⊕N

(β ψ)// V

γ// ΣU .

The pair (ϕ,α) is called a homotopy pullback of (β,ψ), and (β,ψ) is called a homotopypushout of (ϕ,α). Show that homotopy pushouts and homotopy pullbacks always exist.

E 6.5.7 Let S be a triangulated subcategory of a triangulated category (T,Σ), and consider thehomotopy Cartesian square in T from E 6.5.6. Show that the morphism α is S-trivial ifand only if β is S-trivial in the sense of E 6.2.8. Hint: See [63, lem. 1.5.8].

E 6.5.8 (Cf. 6.5.23) Let M→ N→ X→ ΣM be a distinguished triangle in D(R). Show that thereinequalities

supM 6 maxsupN,supX−1 , infM > mininfN, infX−1 ,supN 6 maxsupM,supX , infN > mininfM, infX ,supX 6 maxsupM+1,supN , and infX > mininfM+1, infN .

E 6.5.9 Show that the full subcategories of D(R) defined by specifying their objects as follows:D@(R) = M ∈D(R) | there is a bounded above complex M′ with M 'M′ ,D@A(R) = M ∈D(R) | there is a bounded complex M′ with M 'M′ , and

DA(R) = M ∈D(R) | there is a bounded below complex M′ with M 'M′are triangulated subcategories of D(R).


Part II

Tools


Chapter 7Derived Functors

Exact functors are the structure preserving mappings between Abelian categories.In the world of triangulated categories, they are replaced by triangulated functors. Inthe context of this book, the utility of derived categories stems predominantly fromthe fact that well-behaved, though not necessarily exact, functors C(R)→ C(S) yieldtriangulated functors D(R)→D(S) through a process known as derivation.

7.1 Induced Functors on the Homotopy Category

SYNOPSIS. Functor that preserves homotopy; Hom functor; tensor product functor; unitor; coun-itor; commutativity; associativity; swap; adjunction; biduality; tensor evaluation; homomorphismevaluation.

We start by merging and expanding some key results from Chap. 6.

7.1.1 Definition. Set C = C(R1)op× ·· ·×C(Rm)

op×C(S1)× ·· ·×C(Sn). Parallelmorphisms (α1, . . . ,αm,β1, . . . ,βn) and (α′1, . . . ,α

′m,β

′1, . . . ,β

′n) in C are said to be

homotopic if one has αi ∼ α′i in C(Ri)op and β j ∼ β′j in C(S j) for all i = 1, . . . ,m and

j = 1, . . . ,n; see 4.3.14.A functor C→ C(T ) is said to preserve homotopy if it maps homotopic mor-

phisms in C to homotopic morphisms in C(T ); cf. 4.3.12 and 4.3.14.

7.1.2 Theorem. Let F: C(R1)op×·· ·×C(Rm)

op×C(S1)×·· ·×C(Sn)→ C(T ) be afunctor. If F preserves homotopy, then there exists a unique functor F that makes thefollowing diagram commutative,

C(R1)op×·· ·×C(Rm)

op×C(S1)×·· ·×C(Sn)F//

QopR1×···×Qop

Rm ×QS1 ×···×QSn

C(T )

QT

K(R1)op×·· ·×K(Rm)

op×K(S1)×·· ·×K(Sn)F// K(T ) .

The functor F acts on objects and morphisms as follows:

255

256 7 Derived Functors

F(M1, . . . ,Mm,N1, . . . ,Nn) = F(M1, . . . ,Mm,N1, . . . ,Nn) and

F([α1], . . . , [αm], [β1], . . . , [βn]) = [F(α1, . . . ,αm,β1, . . . ,βn)] .

Furthermore, the following assertions hold.(a) If F is k-linear in one of the variables, then F is k-linear in the corresponding

variable; in particular, if F is k-multilinear, then F is k-multilinear.(b) If F preserves products/coproducts in one of the variables, then F preserves

products/coproducts in the corresponding variable.(c) If F is a Σ-functor in one of the variables, then F is triangulated in the corre-

sponding variable.

PROOF. The arguments in the proofs of 6.1.20/6.1.22 and 6.2.17/6.2.19 apply mu-tatis mutandis to establish the claims.

REMARK. For every Abelian category A, one can construct the category C(A) of complexes andthe homotopy category K(A), and there is a canonical functor QA : C(A)→K(A) with theexpected universal properties. In Chaps. 2 and 6 we carried out these constructions explicitly forthe category A=M(R). The general constructions of C(A) and K(A) respect the formation ofproducts and opposites of categories; for example, one has

C(A1×A2) ∼= C(A1)×C(A2) and C(Aop) ∼= C(A)op .

In particular, for the Abelian category A=M(R)op×M(R) there are identifications,C(A) ∼= C(R)op×C(R) and K(A) ∼= K(R)op×K(R) .

Thus, the universal property of the canonical functor QA applies to show that the functorHomR(–,–) : C(R)op×C(R) −→ C(k) ,

induces a functor K(R)op×K(R)→K(k); this indicates another way to prove 7.1.2.

7.1.3 Proposition. Let

F,G : C(R1)op×·· ·×C(Rm)

op×C(S1)×·· ·×C(Sn) −→ C(T )

be functors that preserve homotopy. Consider the induced functors from 7.1.2,

F, G : K(R1)op×·· ·×K(Rm)

op×K(S1)×·· ·×K(Sn) −→ K(T ) .

Every natural transformation τ : F→ G induces a natural transformation τ : F→ Ggiven by τX = [τX ] for every object X = (M1, . . . ,Mm,N1, . . . ,Nn). Moreover, if Fand G are Σ-functors in the same variable and if τ is a Σ-transformation in that vari-able, then the natural transformation τ is triangulated in the corresponding variable.


THE HOM FUNCTOR

7.1.4 Example. By 2.3.9 the functor HomR(–,–) : C(R)op×C(R)→ C(k) preserveshomotopy. Thus, by 7.1.2 there exists a unique functor ¨HomR(–,–) that makes thefollowing diagram commutative,


7.1 Induced Functors on the Homotopy Category 257

C(R)op×C(R)HomR(–,–)

//

QopR ×QR

C(k)

Qk

K(R)op×K(R)¨HomR(–,–)

// K(k) .

It acts as follows on objects and morphisms:

¨HomR(M,N) = HomR(M,N) and ¨HomR([α], [β]) = [HomR(α,β)] .

Moreover, ¨HomR(–,–) is k-bilinear, see 2.3.10, it preserves products in both vari-ables, see 3.1.23 and 3.1.25, and it is triangulated in both variables, see 4.1.15 and4.1.16. By usual abuse of notation we denote the functor ¨HomR(–,–) by HomR(–,–).

7.1.5 Definition. Let K(R–So) denote the category K(R⊗k So).

7.1.6. In view of 2.1.37 the category K(R–So) is naturally identified with the cate-gory whose objects are complexes of R–So-bimodules, and whose hom-sets consistsof homotopy classes of R- and So-linear chain maps of degree 0. The homotopy re-lation used here is defined as in 2.2.23, only now homotopies are required to be bothR- and So-linear.

In particular, K(R–Ro) is identified with the category of complexes of R–Ro-bimodules, and the category K(Re) is naturally identified with the full subcategoryof K(R–Ro) whose objects are complexes of symmetric R–Ro-bimodules.

REMARK. Per the remark above, one can construct the homotopy category K(A) over anyAbelian category A. The content of 7.1.6 is that K(R–So) is naturally equivalent to K(M(R–So)).

7.1.7 Addendum to 7.1.4. By 2.3.11 there is a functor HomR(–,–) from the cate-gory C(R–Qo)op×C(R–So) to C(Q–So), and by 7.1.2 it induces a k-bilinear functor,

HomR(–,–) : K(R–Qo)op×K(R–So) −→ K(Q–So) ,

which preserves products and is triangulated in both variables.

THE TENSOR PRODUCT FUNCTOR

7.1.8 Example. By 2.4.8 the functor –⊗R –: C(Ro)×C(R)→ C(k) preserves homo-topy. Thus, 7.1.2 yields a unique functor – ⊗R – that makes the following diagramcommutative,

C(Ro)×C(R)–⊗R–

//

QRo×QR

C(k)

Qk

K(Ro)×K(R)–⊗R–

// K(k) .

It acts as follows on objects and morphisms:

M ⊗R N = M⊗R N and [α] ⊗R [β] = [α⊗R β] .



Moreover, – ⊗R – is k-bilinear, see 2.4.9, it preserves coproducts in both variables,see 3.1.11 and 3.1.12, and it is triangulated in both variables, see 4.1.17 and 4.1.18.By habitual abuse of notation we denote the functor – ⊗R – by –⊗R –.

7.1.9 Addendum. By 2.4.10 there is a functor –⊗R – from C(Q–Ro)×C(R–So) toC(Q–So), and by 7.1.2 it induces a k-bilinear functor,

–⊗R – : K(Q–Ro)×K(R–So) −→ K(Q–So) ,

which preserves coproducts and is triangulated in both variables.

STANDARD ISOMORPHISMS IN THE HOMOTOPY CATEGORY

Having established Hom and ⊗ as functors on the level of homotopy categories,our next objective is to extend the natural transformations and isomorphism fromSects. 4.4 and 4.5 to that setting.

7.1.10. Consider the functors F,G : C(Q–Ro)×C(R–So)×C(S–T o) −→ C(Q–T o)given by

F(M,X ,N) = (M⊗R X)⊗S N and G(M,X ,N) = M⊗R (X⊗S N)

and the associativity isomorphism ω : F→ G from 4.4.6. By 2.4.8 the functors Fand G preserve homotopy, so from 7.1.2 one gets functors F and G from K(Q–Ro)×K(R–So)×K(S–T o) to K(Q–T o) given by

F(M,X ,N) = (M ⊗R X) ⊗S N and G(M,X ,N) = M ⊗R (X ⊗S N) .

Moreover, by 7.1.3 there is an induced natural isomorphism ω = [ω] : F→ G, andit is triangulated in each variable, as ω is a Σ-transformation in each variable. Com-bined with 2.4.10 these arguments prove 7.1.13 below. The other results 7.1.11–7.1.18 are proved similarly.

7.1.11. For M in K(R–So) there are isomorphisms in K(R–So) induced by the unitorand counitor from 4.4.1,

µMR : R⊗R M −→ M and(7.1.11.1)

εMR : M −→ HomR(R,M) ,(7.1.11.2)

and they are natural in M. Moreover, as natural transformations of functors, µR andεR are triangulated.

7.1.12 Proposition. For M in K(Q–Ro) and N in K(R–So) there is an isomorphismin K(Q–So) induced by commutativity 4.4.3,

υMN : M⊗R N −→ N⊗Ro M ,

and it is natural in M and N. Moreover, as a natural transformation of functors, υ istriangulated in each variable.


7.1 Induced Functors on the Homotopy Category 259

7.1.13 Proposition. For M in K(Q–Ro), X in K(R–So), and N in K(S–T o) there isan isomorphism in K(Q–T o) induced by associativity 4.4.6,


and it is natural in M, X , and N. Moreover, as a natural transformation of functors,ω is triangulated in each variable.

7.1.14 Proposition. For M in K(R–Qo)op, X in K(R–So), and N in K(T –So)op thereis an isomorphism in K(Q–T o) induced by swap 4.4.9,


and it is natural in M, X , and N. Moreover, as a natural transformation of functors,ζ is triangulated in each variable.

7.1.15 Proposition. For M in K(R–T o), X in K(R–So)op, and N in K(S–Qo)op thereis an isomorphism in K(Q–T o) induced by adjunction 4.4.11,


and it is natural in M, X , and N. Moreover, as a natural transformation of functors,ρ is triangulated in each variable.

EVALUATION MORPHISMS IN THE HOMOTOPY CATEGORY

7.1.16 Proposition. Let X be in K(R–So). For M in K(R–T o) there is an morphismin K(R–T o) induced by biduality 4.5.2,


and it is natural in M. As a natural transformation of functors, δX is triangulated.

7.1.17 Proposition. For M in K(R–Qo)op, X in K(R–So), and N in K(S–T o) thereis an morphism in K(Q–T o) induced by tensor evaluation 4.5.6,


and it is natural in M, X , and N. Moreover, as a natural transformation of functors,θ is triangulated in each variable.

7.1.18 Proposition. For M in K(R–T o), X in K(R–So)op, and N in K(Q–So) thereis an morphism in K(Q–T o) induced by homomorphism evaluation 4.5.9,


and it is natural in M, X , and N. Moreover, as a natural transformation of functors,η is triangulated in each variable.

7.1.19. The evaluation morphisms in the homotopy category, 7.1.16–7.1.18, areisomorphisms under the hypotheses in 4.5.4, 4.5.7, and 4.5.10.



7.2 Induced Functors on the Derived Category

SYNOPSIS. Functor that preserves quasi-isomorphisms; universal property of left derived functor;universal property of right derived functor.

In the balance of this chapter, we use Greek letters for morphisms in K(R). Unlessotherwise specified α,β,γ, . . . denote homotopy classes of morphisms in C(R).

The next definition is in line with 7.1.1.

7.2.1 Definition. A morphism (α1, . . . ,αm,β1, . . . ,βn) in the product categoryK(R1)

op× ·· · ×K(Rm)op×K(S1)× ·· · ×K(Sn) is called a quasi-isomorphism if

each component αi and β j is a quasi-isomorphism as defined in 6.1.12.

7.2.2 Theorem. Let F: K(R1)op×·· ·×K(Rm)

op×K(S1)×·· ·×K(Sn)→K(T )be a functor. If F preserves quasi-isomorphisms, then there exists a unique func-tor ´F that makes the following diagram commutative,

K(R1)op×·· ·×K(Rm)

op×K(S1)×·· ·×K(Sn)F//

VopR1×···×Vop

Rm ×VS1 ×···×VSn

K(T )

VT

D(R1)op×·· ·×D(Rm)

op×D(S1)×·· ·×D(Sn)´F// D(T ) .

The functor ´F acts as follows on objects and morphisms:

´F(M1, . . . ,Mm,N1, . . . ,Nn) = F(M1, . . . ,Mm,N1, . . . ,Nn) and´F(α1/ϕ1, . . . ,αm/ϕm,β1/ψ1, . . . ,βn/ψn) =

F(1M′1 , . . . ,1M′m ,β1, . . . ,βn)

F(ϕ1, . . . ,ϕm,ψ1, . . . ,ψn) F(α1, . . . ,αm,1N1 , . . . ,1Nn)

1F(M1,...,Mm,N1,...,Nn),

where αi/ϕi : Mi→M′i in D(Ri)op and β j/ψ j : N j→ N′j in D(S j). Furthermore, the

following assertions hold.(a) If F is k-linear in one of the variables, then ´F is k-linear in the corresponding

variable; in particular, if F is k-multilinear, then ´F is k-multilinear.(b) If F preserves products/coproducts in one of the variables, then ´F preserves

products/coproducts in the corresponding variable.(c) If F is triangulated in one of the variables, then ´F is triangulated in the corre-

sponding variable.


7.2.3 Proposition. Let

F,G : K(R1)op×·· ·×K(Rm)

op×K(S1)×·· ·×K(Sn) −→ K(T )

be functors that preserve quasi-isomorphisms, and consider the induced functorsfrom 7.2.2,


7.2 Induced Functors on the Derived Category 261

´F, ´G : D(R1)op×·· ·×D(Rm)

op×D(S1)×·· ·×D(Sn) −→ D(T ) .

Every natural transformation τ : F→ G induces a natural transformation ´τ : ´F→ ´Ggiven by ´τX = τX/1 for every object X = (M1, . . . ,Mm,N1, . . . ,Nn). Moreover, if thefunctors F and G are triangulated in the same variable and if the transformation τ istriangulated in that variable, then ´τ is triangulated in the corresponding variable.


To facilitate further discussion, we introduce some shorthand notation.

7.2.4 Notation. For rings R1, . . . ,Rm and S1, . . . ,Sn set

K = K(R1)op×·· ·×K(Rm)

op×K(S1)×·· ·×K(Sn) ,

D = D(R1)op×·· ·×D(Rm)

op×D(S1)×·· ·×D(Sn) , and

V = VopR1×·· ·×Vop

Rm×VS1×·· ·×VSn .

DERIVED FUNCTORS

7.2.5 Construction. Adopt the notation from 7.2.4. Let τ : F→ G be a natural trans-formation of functors K→ K(T ). We use 6.3.10 and 6.3.14 to construct functorsand natural transformations Lτ : LF→ LG and Rτ : RF→ RG.

Consider the endofunctor A = IopR1×·· · × Iop

Rm×PS1×·· · × PSn on K. Since A

maps quasi-isomorphisms to isomorphisms, the same is true for the composite func-tor FA. Therefore 7.2.2 implies the existence of a unique functor LF = (FA) ´ thatmakes the following diagram commutative,

KFA//

V

K(T )

VT

DLF// D(T ) .

Further, 7.2.3 yields a unique transformation Lτ= (τA) ´: LF→ LG.Similarly, we consider the functor B = Pop

R1×·· ·×Pop

Rm× IS1×·· ·× ISn on K. As

the composite FB maps quasi-isomorphisms to isomorphisms, there is by 7.2.2 aunique functor RF = (FB) ´ that makes the following diagram commutative,

KFB//

V

K(T )

VT

DRF// D(T ) .

Further, 7.2.3 yields a unique transformation Rτ= (τB) ´: RF→ RG.

7.2.6 Definition. Let F: K(R1)op×·· ·×K(Rm)

op×K(S1)×·· ·×K(Sn)→K(T )be a functor. Every functor that is naturally isomorphic to LF from 7.2.5 is denoted



LF and called the left derived functor of F. Similarly, every functor that is naturallyisomorphic to RF is denoted RF and called the right derived functor of F.

Let G be another functor with the same domain and codomain as F and letτ : F→ G be a natural transformation. Every natural transformation that is naturallyisomorphic to Lτ from 7.2.5 is denoted Lτ and called the left derived transformationof τ. Similarly, every transformation that is naturally isomorphic to Rτ is denotedRτ and called the right derived transformation of τ.

At first sight, the definitions of derived functors in 7.2.6 may seem kind of ran-dom. They do, however, have universal properties as explained in 7.2.16 and 7.2.17.

7.2.7 Example. Consider for n> 1 the endofunctor F = HomZ(Z/nZ,–) on K(Z).For m> 1 the complex P = 0−→ Z m−→ Z−→ 0, concentrated in degrees 1 and 0, isa semi-projective resolution of Z/mZ, and hence there are isomorphisms in D(Z),

LF(Z/mZ) ' HomZ(Z/nZ,P) ' 0 .

In comparison, the module F(Z/mZ) is not necessarily zero; cf. 1.1.6.The complex I = 0→ Q→ Q/Z→ 0, concentrated in degrees 0 and −1, is a

semi-injective resolution of Z, and hence there are isomorphisms in D(Z),

RF(Z) ' HomZ(Z/nZ, I) ' Σ−1 HomZ(Z/nZ,Q/Z) ' Σ

−1 (Z/nZ) .

In comparison, the module F(Z) is zero.

7.2.8. Adopt the notation from 7.2.4. Consider the endofunctors A and B on K from7.2.5. It follows from 6.3.10 and 6.3.14 that there are natural transformations

Aλ0−→ IdK

ρ0−→ B

with λ0 = ιopR1×·· ·× ιop

Rm×πS1 ×·· ·×πSn and ρ0 = π

opR1×·· ·×πop

Rm× ιS1 ×·· ·× ιSn .

For every X in K the morphisms λX0 and ρX

0 are quasi-isomorphisms.From the natural transformation λ0 : A→ IdK one gets two natural transforma-

tions A2 → A, namely Aλ0 and λ0 A. However, it follows from 6.3.11 and 6.3.15that Aλ0 = λ0 A holds, and by 6.3.10 this natural transformation is a natural isomor-phism. Similarly, Bρ0 = ρ0 B is a natural isomorphism B→ B2.

A functor that preserves quasi-isomorphisms is its own derived functor.

7.2.9 Example. Adopt the notation from 7.2.4 and let F : K→K(T ) be a functorthat preserves quasi-isomorphisms. The functor ´F from 7.2.2 is both a left derivedand a right derived functor of F, that is, LF = ´F = RF. Indeed, application of Fto λ0 : A→ IdK from 7.2.8 yields a natural transformation Fλ0 : FA→ F. Since(Fλ0)

X = F(λX0 ) is a quasi-isomorphism for every X ∈K, there is by 7.2.3 a natu-

ral isomorphism (Fλ0) ´: LF = (FA) ´→ ´F. Similarly, one gets that RF is naturallyisomorphic to ´F.

Specific examples of such functors include the identity functor IdK(R), the re-solution functors PR and IR, the functors HomR(P,–) and HomR(–, I) where P is asemi-projective R-complex and I is a semi-injective R-complex, and F⊗R – whereF is a semi-flat Ro-complex; see 6.3.10, 6.3.10, 5.2.10, 5.3.16, and 5.4.9.



Quasi-isomorphic functors have isomorphic derived functors.

7.2.10 Example. Adopt the notation from 7.2.4. Let F,G: K→K(T ) be functorsand let τ : F→ G be a natural transformation. If τX is a quasi-isomorphism for everyX ∈K, then LF and RF are naturally isomorphic to LG and RG, respectively. Indeed,in this case Lτ= (τA) ´ and Rτ= (τB) ´ are natural isomorphisms.

In particular, the natural transformation π : P→ IdK(R) from 6.3.10 shows thatthe derived functors LP and RP are both naturally isomorphic to IdD(R). Similarly,the derived functors LI and RI of the semi-injective resolution functor, see 6.3.14,are both naturally isomorphic to IdD(R).

Before we move on to more substantial examples of derived functors in Sects. 7.3and 7.4, we explore some general properties of such functors.

7.2.11 Lemma. Let F,G: U→ V be naturally isomorphic functors.(a) Assume that U and V are k-prelinear categories. If F is k-linear, then so is G.(b) Assume that the categories U and V have products/coproducts. If F preserves

products/coproducts, then so does G.(c) Assume that U and V are triangulated and let τ : F→ G be a natural isomor-

phism. If F is triangulated with natural isomorphism φ : FΣU→ ΣV F, thenthe natural isomorphism ψ : GΣU→ ΣV G, given by ψM =(ΣτM)φM(τΣM)−1,makes the functor G and the natural transformation τ triangulated.

PROOF. All three assertions follow directly form the relavant definitions.

7.2.12 Theorem. Let F: K(R1)op×·· ·×K(Rm)

op×K(S1)×·· ·×K(Sn)→K(T )be a functor. The following assertions hold.

(a) If F is k-linear in one of its variables, then its derived functors LF and RF arek-linear in the corresponding variable. In particular, if F is k-multilinear, thenLF and RF are k-multilinear.

(b) If F preserves coproducts in one of its variables, then LF preserves coproductsin the corresponding variable.

(c) If F preserves products in one of its variables, then RF preserves products inthe corresponding variable.

(d) If F is triangulated in one of its variables, then LF and RF are triangulated inthe corresponding variable.

PROOF. By 7.2.6 and 7.2.11 it is sufficient to verify the assertions for the functorsLF = (FA) ´ and RF = (FB) ´ from 7.2.5. By 6.3.10 and 6.3.14 the resolution func-tors P and I are k-linear. Thus, if F is k-linear in one of its variables, then FA andFB are k-linear in the corresponding variable. Part (a) now follows from 7.2.2(a).

The remaining three parts follow from 7.2.2 by similar arguments. To this endrecall that P is triangulated and preserves coproducts, whence Pop is triangulatedand preserves products, and that I is triangulated and preserves products, whenceIop is triangulated and preserves coproducts; see 6.3.10, 6.3.14, and D.10.



UNIVERSAL PROPERTIES OF DERIVED FUNCTORS

Adopt the notation from 7.2.4. Let F: K→K(T ) be a functor. One can, in general,not expect to find a functor F′ that makes the diagram

KF//

V

K(T )

VT

DF′// D(T )

commutative. Indeed, it follows from 6.4.17 that a necessary condition for the ex-istence of such a functor F′ is that F preserves. On the other hand, 7.2.2 shows thatthis condition is sufficient to guarantee the existence of F′.

7.2.13 Definition. Adopt the notation from 7.2.4. Let F: K→K(T ) be a functor.To F we associate two categories LF and RF defined as follows.

The objects in LF are pairs (F′,ξ′) where F′ : D→D(T ) is a functor andξ′ : F′V→ VT F is a natural transformation. A morphism ζ : (F′,ξ′)→ (F′′,ξ′′) in LFis a natural transformation ζ : F′→ F′′ that makes the next diagram commutative,

F′V

ξ′

ζV

F′′Vξ′′// VT F .

The objects in RF are pairs (F′,ξ′) where F′ : D→D(T ) is a functor andξ′ : VT F→ F′V is a natural transformation. A morphism ζ : (F′,ξ′)→ (F′′,ξ′′) in RFis a natural transformation ζ : F′→ F′′ that makes the next diagram commutative,

VT Fξ′//

ξ′′

F′V

ζV||

F′′V .

The category LF has an initial object, namely the pair (O,0) where O: D→D(T )is the zero functor and 0: OV→ VT F is the zero transformation. Similarly, the pair(O,0), where this time 0 denotes the zero transformation VT F→ OV, is a terminalobject in RF. A less trivial fact is that LF has a terminal object and that RF has aninitial object; this is proved in 7.2.16 below.

7.2.14 Construction. Adopt the notation from 7.2.4. Let F : K→K(T ) be a functor.We construct objects (LF,λ) in LF and (RF,ρ) in RF, where the functors LF andRF are as in 7.2.5.

Recall the natural transformation λ0 : A→ IdK from 7.2.8. The natural transfor-mation λ is defined as follows:

(LF)V = VT FAVT Fλ0

// VT FIdK = VT F .



Similarly, the natural transformation ρ is given by

VT F = VT FIdKVT Fρ0

// VT FB = (RF)V ,

where ρ0 : IdK→ B is as in 7.2.8.

7.2.15 Lemma. Adopt the notation from 7.2.4. Let F,G: K→D(T ) be functorsand let τ1,τ2 : F→ G be natural transformations.

(a) Suppose that F maps quasi-isomorphisms in K to isomorphisms in D(T ). Ifτ1 A = τ2 A, then one has τ1 = τ2.

(b) Suppose that G maps quasi-isomorphisms in K to isomorphisms in D(T ). Ifτ1 B = τ2 B, then one has τ1 = τ2.

PROOF. (a): Consider the natural transformation λ0 : A→ IdK from 7.2.8. As τi isa natural transformation, there is a commutative diagram of natural transformations,

FA

τi A

Fλ0// F

τi

GAGλ0// G .

As λX0 is a quasi-isomorphism for every X in K, the assumption on F yields that Fλ0

is a natural isomorphism. It follows that τi = (Gλ0)(τi A)(Fλ0)−1. Consequently, if

one has τ1 A = τ2 A, then τ1 = τ2.(b): Similar to the proof of (a).

7.2.16 Theorem. Adopt the notation from 7.2.4. Let F: K→K(T ) be a functor.(a) The object (LF,λ) constructed in 7.2.14 is terminal in the category LF.(b) The object (RF,ρ) constructed in 7.2.14 is initial in the category RF.

PROOF. (a): It must be argued that for every object (F′,ξ′) in LF there exists aunique morphism (F′,ξ′)→ (LF,λ). Let ζ1 and ζ2 be any two such morphisms, thatis, each ζi : F′→ LF is a natural transformation such that the diagram

F′V

ξ′

ζi V

(LF)V λ// VT F

is commutative. As the functor V: K→D is the identity on objects, one has ζ1 = ζ2if and only if ζ1 V = ζ2 V, and by 7.2.15 this holds if and only if ζ1 VA = ζ2 VA. Toverify the latter, note that the diagram above yields the commutative diagram

F′VA

ξ′A

ζi VA

yy

(LF)VA λA// VT FA .



By definition, one has λA = VT Fλ0 A, which is a natural isomorphism by 7.2.8. Itfollows that ζi VA = (λA)−1(ξ′A) for i = 1,2, and hence ζ1 VA = ζ2 VA.

It remains to prove the existence of a morphism (F′,ξ′)→ (LF,λ) in LF. Thereare natural transformations

F′V F′VAF′Vλ0

'oo

ξ′A// VT FA = (LF)V ,

and hence a natural transformation ζ : F′V→ (LF)V given by ζ=(ξ′A)(F′Vλ0)−1.

As V is the identity on objects, this yields a natural transformation ζ : F′→ LF withζV= ζ; cf. 6.4.28. To verify the equality λ(ζV)= ξ′, that is, λζ = ξ′, amounts by thedefinitions of λ and ζ to showing that (VT Fλ0)(ξ

′A) = ξ′(F′Vλ0) holds; however,this follows from the naturality of ξ′ : F′V→ VT F applied to λ0 : A→ IdK.

(b): Similar to the proof of (a).

7.2.17. All terminal objects in a given category are isomorphic. In particular, everyterminal object in LF is isomorphic to (LF,λ). If (F′,ξ′) and (LF,λ) are isomorphicin LF then, in particular, F′ and LF are naturally isomorphic functors, i.e. F′ is LF.On the other hand, if ϕ : F′→ LF is a natural isomorphism—that is, F′ is LF—thenthere is a unique natural transformation ξ′ : F′V→ VT F, namely ξ′= λ(ϕV), suchthat (F′,ξ′) is isomorphic to (LF,λ) in LF. Thus, for most purposes one can suppressthe natural transformation and identify a terminal object in LF with a left derivedfunctor of F in the sense of 7.2.6.

Similarly, an initial object in RF is nothing but a right derived functor of F in thesense of 7.2.6.

7.2.18 Proposition. Adopt the notation from 7.2.4. Let F,G: K→K(T ) be func-tors and let τ : F→ G be a natural transformation.

(a) Let (F′,ξ′)∈LF and (G′,η′)∈LG. If (G′,η′) is terminal, then there is a uniquenatural transformation τ′ : F′→ G′ that makes the next diagram commutative,

F′V

ξ′

τ′V// G′V

η′

VT FVT τ// VT G .

(b) Let (F′,ξ′) ∈ RF and (G′,η′) ∈ RG. If (F′,ξ′) is initial, then there is a uniquenatural transformation τ′ : F′→ G′ that makes the next diagram commutative,

VT F

ξ′

VT τ// VT G

η′

F′V τ′V// G′V .

PROOF. (a): Note that (F′,(VT τ) ξ′) is an object in LG. A natural transformationτ′ that makes the diagram commutative is a morphism τ′ : (F′,(VT τ) ξ′)→ (G′,η′)in LG, and there exists a unique such morphism as (G′,η′) is terminal in LG.


7.3 Right Derived Hom Functor 267

(b): Similar to the proof of (a).

EXERCISES

E 7.2.1 (Cf. 7.2.7) Show that there is an isomorphism HomZ(Z/nZ,Q/Z)∼= Z/nZ in M(Z).E 7.2.2 Let R be left hereditary and let F : C(R)→C(S) be a functor that preserves homotopy.

Show that for every R-module M one has Hv(LF(M)) = 0 = H−v(RF(M)) for v 6= 0,1.Hint: E 1.3.17 and E 1.4.8.

E 7.2.3 Let N be an Ro-module and set F = –⊗R N. Show that for every R-module M there is anisomorphism H0(LF(M))∼= M⊗R N. Are H0(RF(M)) and M⊗R N isomorphic?

E 7.2.4 Let n > 0 be an integer and set F = –⊗Z Z/nZ. Compute the groups Hv(LF(M)) andH−v(RF(M)) for v = 0,1 and M = Z,Z/mZ,Q,Q/Z.

E 7.2.5 Let R be left hereditary and let G: C(R)op→C(S) be a functor that preserves homotopy.Show that for every R-module M one has Hv(LG(M)) = 0 = H−v(RG(M)) for v 6= 0,1.Hint: E 1.3.17 and E 1.4.8.

E 7.2.6 Let N be an R-module and set G = HomR(–,N). Show that for every R-module M onehas H0(RG(M))∼= HomR(M,N). Are H0(LG(M)) and HomR(M,N) isomorphic?

E 7.2.7 Let n > 0 be an integer and set G = HomZ(–,Z/nZ). Compute the groups Hv(LG(M))and H−v(RG(M)) for v = 0,1 and M = Z,Z/mZ,Q,Q/Z.

E 7.2.8 (a) Give an example of a functor F, such that F preserves products but LF does not.(b) Give an example of a functor F, such that LF preserves products but F does not.

E 7.2.9 (a) Give an example of a functor F, such that F preserves coproducts but RF does not.(b) Give an example of a functor F, such that RF preserves coproducts but F does not.

7.3 Right Derived Hom Functor

SYNOPSIS. RHom; Ext; exact Ext sequence; complexes of bimodules.

The next result shows that one can compute the right derived functor of HomR(–,–)by resolving appropriately over R in either or both variables.

7.3.1 Theorem. Each of the functors

HomR(Pop(–),–) , HomR(Pop(–), I(–)) , and HomR(–, I(–))

from K(R)op×K(R) to K(k) preserves quasi-isomorphisms and the induced func-tors HomR(Pop(–),–) , HomR(Pop(–), I(–)) , and HomR(–, I(–)) ´ from 7.2.2 are all

RHomR(–,–) : D(R)op×D(R) −→ D(k) .

This functor is k-bilinear, it preserves products in both variables, and it is triangu-lated in both variables.

PROOF. The natural transformations π and ι from 6.3.10 and 6.3.14 yield naturaltransformations of functors from K(R)op×K(R) to K(k),

HomR(Pop(–),–)τ=Hom(Pop(–),ι)

// HomR(Pop(–), I(–)) HomR(–, I(–)) .σ=Hom(πop,I(–))oo



By 6.3.10, 6.3.14, 5.2.10, and 5.3.16 all three functors preserve quasi-isomorphisms,and τMN and σMN are quasi-isomorphisms for all R-complexes M and N. Thus 7.2.3yields natural isomorphisms,

HomR(Pop(–),–) ´ ´τ'// HomR(Pop(–), I(–)) ´ HomR(–, I(–)) ´ ,´σ

'oo

of functors from D(R)op ×D(R) to D(k). The middle functor is RHomR(–,–),see 7.2.5, and all three functors are therefore RHomR(–,–); see 7.2.6.

It follows from 7.1.4 and 7.2.12 that the functor RHomR(–,–) is k-bilinear, pre-serves products in both variables, and is triangulated in both variables.

7.3.2. One way to compute the functor RHomR(–,–) : D(R)op×D(R)→D(k) is

RHomR(M,N) = HomR(M, I(N)) andRHomR(α/ϕ,β/ψ) =

(Hom(M′, I(β))/Hom(ϕ, I(ψ))

)(Hom(α, I(N))/1Hom(M,I(N))

)for morphisms α/ϕ : M→M′ in D(R)op and β/ψ : N→ N′ in D(R); see 7.3.1 and7.2.2. It also follows that another way to compute the functor is

RHomR(M,N) = HomR(P(M),N) andRHomR(α/ϕ,β/ψ) =

(Hom(P(M′),β)/Hom(P(ϕ),ψ)

)(Hom(P(α),N)/1Hom(P(M),N)

).

Note that the last composite in D(k) is equal to

Hom(P(α)P(ϕ)−1,β)/Hom(P(M),ψ) .

7.3.3 Definition. Let D(R–So) denote the category D(R⊗k So).

7.3.4. In view of 7.1.6 the category D(R–So) is naturally identified with the categorywhose objects are complexes of R–So-bimodules, and whose hom-sets consist of leftfractions, i.e. equivalence classes of diagrams

Mϕ←−−'

U α−−→ N

where α and ϕ are homotopy classes of R- and So-linear chain maps of degree 0, andϕ is a quasi-isomorphism. The equivalence relation used here is defined as in 6.4.1,only the morphisms µ1 and µ2 are now required to be homotopy classes of R- andSo-linear chain maps of degree 0.

In particular, D(R–Ro) is identified with the category whose objects are com-plexes of R–Ro-bimodules, and the category D(Re) is naturally identified with thesubcategory whose objects are complexes of symmetric R–Ro-bimodules.

7.3.5 Addendum to 7.3.1. Let ring homomorphisms

α : R⊗kQo −→ A and β : R⊗k So −→ B ,

be given. By restriction of scalars K(A)→K(R–Qo) and K(B)→K(R–So) it fol-lows from 7.1.7 that there is a functor HomR(–,–) from K(A)op×K(B) to K(Q–So).It has a right derived functor, which we temporarily denote



F : D(A)op×D(B) −→ D(Q–So) .

Per 7.2.5 and 7.2.6 it is given by

(7.3.5.1) F = HomR(PopA (–), IB(–)) ´ .

One can compare F to the functor RHom from 7.3.1 via the canonical restrictionfunctors; that is, one can consider the diagram,

(7.3.5.2)

D(A)op×D(B) F//

D(Q–So)

D(R)op×D(R)RHomR(–,–)

// D(k) .

Without extra assumptions, the diagram is not commutative, not even up to naturalisomorphism; see 7.3.6. If Q = k, A = R, and α is the canonical map R⊗k k→ R,then F is naturally isomorphic to F1 =HomR(P

opR (–),–) ´ and the diagram is commu-

tative up to natural isomorphism. Similarly, if S = k, B = R, and β is the canonicalmap, then the functor F is naturally isomorphic to F2 = HomR(–, IR(–)) ´ and thediagram is commutative up to natural isomorphism. In these situations one can thuschoose the right derived functor F such that it restricts to RHom, and hence we usethat symbol for F. That is, there are functors

RHomR(–,–) = F' F1 : D(R)op×D(R–So)−→D(So) and(7.3.5.3)RHomR(–,–) = F' F2 : D(R–Qo)op×D(R)−→D(Q) .(7.3.5.4)

Though, according to 7.2.6, the symbol RHomR(–,–) would be the correct nota-tion for the functor (7.3.5.1), we only use that symbol for this derived functor insituations where the diagram (7.3.5.2) is commutative up to natural isomorphism.

7.3.6 Example. Set k= Z= R and A = Q = Z/2Z= S = B; one then has R⊗kQo =A = B = R⊗k So. Set M = Z/2Z= N; with the notation from (7.3.5.1) one has

F(M,N) = HomZ(Z/2Z,Z/2Z) = Z/2Z .

One can compute RHomZ(M,N) from 7.3.1 as HomZ(P,Z/2Z), where P is the Z-complex 0−→ Z 2−→ Z−→ 0 concentrated in degrees 1 and 0; whence

RHomZ(M,N) = 0−→ Z/2Z 0−−→ Z/2Z−→ 0

has homology in degrees 0 and −1 and is not isomorphic to F(M,N) in D(Z).

EXT

For the next definition, recall from 6.5.20 that complexes that are isomorphic in thederived category have isomorphic homology.

7.3.7 Definition. For every m in Z denote by ExtmR (–,–) the functor

H−m(RHomR(–,–)) : D(R)op×D(R) −→ M(k) .



It follows from 7.3.1 and 6.5.20 that the functors Extm are k-bilinear.

7.3.8 Addendum. In view of 7.3.5 they extend to functors,

ExtmR (–,–) : D(R–Qo)op×D(R) −→ M(Q) andExtmR (–,–) : D(R)op×D(R–So) −→ M(So) .

7.3.9. Let M and N be R-complexes. Per 6.5.21 there are, with the usual conventionsinf∅= ∞ and sup∅=−∞, equalities,

(7.3.9.1)infm ∈ Z | ExtmR (M,N) 6= 0 = −supRHomR(M,N) and

supm ∈ Z | ExtmR (M,N) 6= 0 = − infRHomR(M,N) .

The Ext functors are closely related to hom-sets in the derived category.

7.3.10 Proposition. Let M and N be R-complexes. For every integer m there is anisomorphism of k-modules,

ExtmR (M,N) ∼= D(R)(M,Σm N) ,


PROOF. Let P be the semi-projective resolution functor from 6.3.10. There are iso-morphisms of k-modules, which are all natural in M and N,

ExtmR (M,N) = H0(Σm RHomR(M,N))

∼= H0(RHomR(M,Σm N))∼= H0(HomR(P(M),P(Σm N)))∼= K(R)(P(M),P(Σm N))∼= D(R)(M,Σm N) .

The equality holds by 6.5.20 and the definition of Ext. The first isomorphism followsas RHomR(M,–) is triangulated; see 7.3.1. The second isomorphism follows fromthe definition of RHom, the third holds by 6.1.2, and the fourth holds by 6.4.7.

For modules there is an even simpler description of Ext0.

7.3.11 Corollary. For R-modules M and N there is an isomorphism of k-modules,

Ext0R(M,N) ∼= HomR(M,N) ,

and it is natural in M and N. Furthermore, one has ExtmR (M,N) = 0 for all m < 0.

PROOF. The isomorphism follows from 7.3.10 and 6.4.14. By 5.2.14 there is a semi-projective resolution P '−→M with Pm = 0 for m < 0. Thus HomR(P,N)−m = 0 holdsfor all m < 0, and as the complex HomR(P,N) is RHomR(M,N), it follows thatExtmR (M,N) = 0 holds for all m < 0.

7.3.12. Let M be an R-complex and let n be an integer. If one has Hv(M) = 0 for allv 6= n, then by 4.2.4 there are quasi-isomorphisms,



MτM

Ěn←−−MĚnτ

MĚnĎn−−−→ (MĚn)Ďn = Σ

n Hn(M) .

In particular, there is an isomorphism M ' Σn Hn(M) in D(R).

7.3.13 Proposition. Let M and N be R-modules. If one has ExtmR (M,N) = 0 for allm > 0, then there is an isomorphism RHomR(M,N)' HomR(M,N) in D(k).


REMARK. Let M and N be R-modules. An extension of M by N (some authors say “N by M”) isan exact sequence of R-modules of the form 0→ N→ X →M→ 0. Two such extensions,

Ξ = 0−→ N −→ X −→M −→ 0 and Ξ′ = 0−→ N −→ X ′ −→M −→ 0 ,

are equivalent, in symbols Ξ∼ Ξ′, if there is a homomorphism X → X ′ that makes the diagram

0 // N

=

// X

// M

=

// 0

0 // N // X ′ // M // 0

commutative. It follows from the Five Lemma 1.1.2 that the map X → X ′ is an isomorphism and,consequently, ∼ is an equivalence relation. Because it was first studied by Yoneda [90], the set ofequivalence classes of extensions of M by N is called Yoneda Ext and denoted by YExt1R(M,N).It can be shown that YExt1R(M,N) is an Abelian group under the so-called Baer sum, definedvia pullbacks. It follows from the proof of 2.1.44 that the set of all split extensions of M by Nconstitutes an equivalence class under∼; this is the zero element in the group YExt1R(M,N). It canalso be shown that YExt1R yields a functor from M(R)op×M(R) to M(Z), and that there is anisomorphism of Abelian groups,

YExt1R(M,N) ∼= Ext1R(M,N) ,

which is natural in M and N. One consequence of this isomorphism is that Ext1R(M,N) = 0 holdsif and only if every short exact sequence of the form 0→ N→ X →M→ 0 is split. For a proof ofthis last assertion that works directly with the definition of Ext, see 7.3.15 and E 7.3.2.

In view of 6.5.19 the following result applies, in particular, to exact sequences0→M′→M→M′′→ 0 and 0→ N′→ N→ N′′→ 0 of complexes.

7.3.14 Theorem. Let M′ α′−→M α−→M′′−→ ΣM′ be a distinguished triangle in D(R).

For every R-complex N there is an exact sequence of k-modules,

· · · −−−→ ExtmR (M′′,N)

Extm(α,N)// ExtmR (M,N)

Extm(α′,N)// ExtmR (M

′,N)

−−−→ Extm+1R (M′′,N)

Extm+1(α,N)// Extm+1

R (M,N) −−−→ ·· · .

Let N′β′−→ N

β−→ N′′ −→ ΣN′ be a distinguished triangle in D(R). For every R-complex M there is an exact sequence of k-modules,

· · · −−−→ ExtmR (M,N′)Extm(M,β′)

// ExtmR (M,N)Extm(M,β)

// ExtmR (M,N′′)

−−−→ Extm+1R (M,N′)

Extm+1(M,β′)// Extm+1

R (M,N)−−−→ ·· · .

PROOF. The functor RHomR(–,N) is triangulated, see 7.3.1, so applied to the distin-guished triangle M′→M→M′′→ ΣM′ in D(R), viewed as a distinguished triangle



M′′→M→M′→ Σ−1 M′′ in D(R)op, it yields a distinguished triangle in D(k). Anapplication of 6.5.22 to this triangle and the definition, 7.3.7, of Ext gives the firstexact sequence. The second exact sequence is obtained similarly.

7.3.15 Proposition. Let M and N be R-modules. If one has Ext1R(M,N) = 0, thenevery short exact sequence of the form 0→ N→ X →M→ 0 is split.

PROOF. Write π for the homomorphism X →M. By 7.3.14 and 7.3.11, there is anexact sequence of k-modules,

HomR(M,X)Hom(M,π)−−−−−−→ HomR(M,M)−→ Ext1R(M,N) .

Thus if one has Ext1R(M,N) = 0, then HomR(M,π) is surjective and there exists ahomomorphism γ : M→ X with 1M = HomR(M,π)(γ) = πγ.

7.3.16 Proposition. Let M be an R-complex and 0→ N′ → N → N′′ → 0 be anexact sequence in C(R). If Ext1R(Mv,N′i ) = 0 holds for all v, i ∈ Z, then the sequenceof complexes, 0→ HomR(M,N′)→ HomR(M,N)→ HomR(M,N′′)→ 0, is exact.

PROOF. For every i ∈ Z there is an exact sequence 0→ N′i → Ni → N′′i → 0 ofR-modules. By 7.3.14 and 7.3.11 it yields for every v ∈ Z an exact sequence

0−→ HomR(Mv,N′i )−→ HomR(Mv,Ni)−→ HomR(Mv,N′′i )−→ Ext1R(Mv,N′i ) .

The assertion now follows from 2.3.16.

7.3.17 Proposition. Let N be an R-complex and 0→ M′ → M → M′′ → 0 be anexact sequence in C(R). If Ext1R(M

′′v ,Ni) = 0 holds for all v, i ∈ Z, then the sequence

of complexes, 0→ HomR(M′′,N)→ HomR(M,N)→ HomR(M′,N)→ 0, is exact.

PROOF. The proof is parallel to that of 7.3.16, using 2.3.18 instead of 2.3.16.

RESOLUTIONS OF COMPLEXES OF BIMODULES

For future reference we record a few special cases of results from Chap. 5.

7.3.18. (a) If S is projective as a k-module, then R⊗k So is projective as an R-moduleby 5.2.21. Thus every semi-projective R⊗k So-complex is semi-projective as an R-complex by 5.2.22.

(b) If S is flat as a k-module, then R⊗k So is flat as an R-module by 5.4.17. Thusevery semi-flat R⊗k So-complex is semi-flat as an R-complex by 5.4.18.

(c) If S is flat as a k-module, then R⊗k So is flat as an Ro-module by 5.4.17. Thusevery semi-injective R⊗k So-complex is semi-injective as an R-complex by 5.4.20.

7.3.19 Addendum to 7.3.1. Adopt the notation from 7.3.5. It follows from 5.2.22that if the next condition is satisfied, then every semi-projective A complex is semi-projective over R;

(7.3.19.1) A is projective as an R-module via R−→ R⊗kQo α−−→ A .



Thus, the condition ensures that PA(M) yields a semi-projective resolution of Mas an R-complex, whence the functor F from (7.3.5.1) makes the diagram (7.3.5.2)commutative up to natural isomorphism. Moreover, this functor is naturally isomor-phic to F1 = HomR(P

opA (–),–) ´ and per convention we set

(7.3.19.2) RHomR(–,–) = F' F1 : D(A)op×D(R–So)−→D(Q–So) .

Note that this use of the symbol RHomR(–,–) is an extension of the usage introducedin (7.3.5.3), in as much as (7.3.19.1) is trivially satisfied for Q = k and A = R.

It follows from 5.4.20 that if the next condition is satisfied, then every semi-injective B-complex is semi-injective over R;

(7.3.19.3) B is flat as an Ro-module via R−→ R⊗k So β−−→ B .

Under this condition, the functor F from (7.3.5.1) also makes the diagram (7.3.5.2)commutative up to natural isomorphism. Moreover, this functor is naturally isomor-phic to F2 = HomR(–, IB(–)) , and per convention we set

(7.3.19.4) RHomR(–,–) = F' F2 : D(R–Qo)op×D(B)−→D(Q–So) .

With S = k and B = R this recovers (7.3.5.4).

7.3.20 Example. If Q is projective as a k-module, then α= 1R⊗kQosatisfies condi-

tion (7.3.19.1); see 7.3.18(a). Similarly (7.3.19.3) is satisfied by β= 1R⊗kSoif S is flat

as a k-module; see 7.3.18(c). Examples of k-algebras that are projective, even free,as k-modules are matrix algebras Mn×n(k) and polynomial algebras k[x1, . . . ,xn].

If k is a field—more generally a semi-simple ring or even a von Neuman regularring—then every k-algebra is flat so RHomR(–,–) is per (7.3.19.4) a functor fromD(R–Qo)op×D(R–So) to D(Q–So).

7.3.21 Example. Recall from 1.1.25 that symmetric R–Ro-bimodules are modulesover the quotient Re = Re/C of the enveloping algebra. If R is commutative, thenevery R-module is by default considered to be a symmetric bimodule. Moreover, itis elementary to verify that the map R→ Re given by x 7→ [x⊗1]C is a ring homo-morphism with inverse given by [x⊗y]C 7→ xy. Thus the composite α : R⊗k Ro→ Rof this isomorphism and the canonical map Re→ Re satisfies condition (7.3.19.1),and RHomR(–,–) is per (7.3.19.2) a functor from D(R)op×D(R–So) to D(R–So).

Similarly, as R coincides with Ro it follows from (7.3.19.4) that RHomR(–,–) isa functor from D(R–Qo)op×D(R) to D(Q–R).

7.3.22 Proposition. Let α : R⊗kQo→ A be a ring homomorphism such that condi-tion (7.3.19.1) is satisfied and consider the functor

RHomR(–,–) = HomR(PopA (–),–) ´ : D(A)op×D(R–So) −→ D(Q–So)

from (7.3.19.2). If M and L are complexes of A-modules such that M ' L in D(A)and L is semi-projective as an R-complex, then there is an equality,

RHomR(M,–) = HomR(L,–) ´ : D(R–So) −→ D(Q–So) .



PROOF. The A-complex P = PA(M) is semi-projective. As P ' M ' L in D(A)there exists by 6.4.20 a quasi-isomorphism ϕ : P→ L in K(A). By assumption, L issemi-projective as an R-complex, and so is P by (7.3.19.1). Thus it follows from5.2.20 that ϕ : P→ L is an isomorphism in K(R). Consequently, for every com-plex N of R–So-bimodules, HomR(ϕ,N) is a morphism in K(Q–So) and a quasi-isomorphism, even an isomorphism, in K(k); thus it is a quasi-isomorphism inK(Q–So). Now the natural transformation HomR(ϕ,–) : HomR(L,–)→ HomR(P,–)of functors from K(R–So) to K(Q–So) induces by 6.4.32 an isomorphism betweenHomR(L,–) ´ and RHomR(M,–) = HomR(P,–) .

7.3.23 Proposition. Let β : R⊗k So→ B be a ring homomorphism such that condi-tion (7.3.19.3) is satisfied and consider the functor

RHomR(–,–) = HomR(–, IB(–)) ´ : D(R–Qo)op×D(B) −→ D(Q–So)

from (7.3.19.4). If N and J are complexes of B-modules such that N ' J in D(B)and J is semi-injective as an R-complex, then there is an equality,

RHomR(–,N) = HomR(–,J) ´ : D(R–Qo)op −→ D(Q–So) .

PROOF. The B-complex I = IB(N) is semi-injective. As one has J ' N ' I in D(B)there exists by 6.4.21 a quasi-isomorphism ϕ : J→ I in K(B). By assumption, J issemi-injective as an R-complex, and so is I by (7.3.19.3). Thus it follows from 5.3.24that ϕ : J→ I is an isomorphism in K(R). Consequently, for every complex M ofR–Qo-bimodules, HomR(M,ϕ) is a morphism in K(Q–So) and a quasi-isomorphism,even an isomorphism, in K(k); thus it is a quasi-isomorphism in K(Q–So). Nowthe natural transformation HomR(–,ϕ) : HomR(–,J)→ HomR(–, I) of functors fromK(R–Qo)op to K(Q–So) induces by 6.4.32 an isomorphism between HomR(–,J) ´and RHomR(–,N) = HomR(–, I) .

EXERCISES

E 7.3.1 Let M be an R-module and let P '−→M be a projective resolution. Set Z =Z0(P) and let ιdenote the embedding Z P0. Show that for every R-module N there is an isomorphismExt1R(M,N) ∼= HomR(Z,N)/ImHomR(ι,N). Conclude that M is projective if and onlyif Ext1R(M,–) = 0 if and only if Ext1R(M,Z) = 0.

E 7.3.2 Let M and N be R-modules such that every exact sequence 0→ N → X → M → 0 issplit. Show that Ext1R(M,N) = 0 holds. Hint: Adopt the notation from E 7.3.1 and givenα : Z→ N consider the pushout X of (α, ι).

E 7.3.3 Use Ext to solve E 1.4.7.E 7.3.4 Let R be a principal left ideal domain. Show that every finitely generated R-module M

with Ext1R(M,R) = 0 is free.E 7.3.5 Show that every finite Z-module has the form Ext1Z(M,N) for suitable Z-modules M and

N. Hint: One has Ext1Z(Z/nZ,Z)∼= Z/nZ for n > 1.E 7.3.6 Let M and N be Z-modules. Show that if the Z-module Ext1Z(M,N) is free, then it is

zero.E 7.3.7 An R-module M is called cotorsion if Ext1R(F,M) = 0 holds for every flat R-module F .

Show that for every Ro-module N the R-module Homk(N,E) is cotorsion.


7.4 Left Derived Tensor Product Functor 275

E 7.3.8 An R-module E is called fp-injective if Ext1R(M,E) = 0 holds for every finitely presentedR-module M. Show that for an Ro-module F the next conditions are equivalent: (i) F isflat; (ii) Homk(F,E) is injective; (iii) Homk(F,E) is fp-injective.

E 7.3.9 Let 0→ E ′→ E→ E ′′→ 0 be an exact sequence of R-modules. Show that if E ′ and E ′′

are fp-injective, then E is fp-injective.E 7.3.10 Show that there is a bijective correspondence between ring homomorphisms R⊗k S→ T

and pairs of ring homomorphisms R→ T and S→ T with commuting images.E 7.3.11 Show that S is a projective k-module if and only if R⊗k So is projective as an R-module

via the canonical map R→ R⊗k So.E 7.3.12 Let R→ S be a ring homomorphism. Show that S is projective as an R-module if and

only if every semi-projective S-complex is semi-projective over R.E 7.3.13 Show that S is a flat k-module if and only if R⊗k So is flat as an R-module (equivalently

as an Ro-module) via the canonical map R→ R⊗k So.E 7.3.14 Let R→ S be a ring homomorphism. Show that S is flat as an Ro-module if and only if

every semi-injective S-complex is semi-injective over R.E 7.3.15 Let M and N be R-complexes and let X be a K-projective complex that is isomorphic to

M in D(R). Show that there is an isomorphism HomR(X ,N)' RHomR(M,N) in D(k).E 7.3.16 Let M and N be R-complexes and let Y be a K-injective complex that is isomorphic to N

in D(R). Show that there is an isomorphism HomR(M,Y )' RHomR(M,N) in D(k).

7.4 Left Derived Tensor Product Functor

SYNOPSIS. ⊗L; Tor; exact Tor sequence; complexes of bimodules.

7.4.1 Definition. For the functor –⊗R –: K(Ro)×K(R)→K(k) we use the nota-tions –⊗L

R – and –⊗LR – instead of L(–⊗R –) and L(–⊗R –); see 7.2.5 and 7.2.6.

The next result shows that one can compute the left derived functor of –⊗R – byresolving appropriately in either or both variables.

7.4.2 Theorem. Each of the functors

PRo(–)⊗R – , PRo(–)⊗R PR(–) , and –⊗R PR(–)

from K(Ro)×K(R) to K(k) preserves quasi-isomorphisms and the induced functors(PRo(–)⊗R –) , (PRo(–)⊗R PR(–)) , and (–⊗R PR(–)) ´ from 7.2.2 are all

–⊗LR – : D(Ro)×D(R) −→ D(k) .

This functor is k-bilinear, it preserves coproducts in both variables, and it is trian-gulated in both variables.

PROOF. The natural transformation π from 6.3.10 yields natural transformations offunctors from K(Ro)×K(R) to K(k),

PRo(–)⊗R – PRo(–)⊗R PR(–)τ=P(–)⊗πoo

σ=π⊗P(–)// –⊗R PR(–) .



By 6.3.10, 5.4.10, and 5.4.9 all three functors preserve quasi-isomorphisms, andτMN and σMN are quasi-isomorphisms for all Ro-complexes M and R-complexes N.Thus 7.2.3 yields natural isomorphisms,

(PRo(–)⊗R –) ´ (PRo(–)⊗R PR(–)) ´´τ'oo

´σ'// (–⊗R PR(–)) ´ ,

of functors from D(Ro)×D(R) to D(k). The middle functor is –⊗LR –, see 7.2.5,

and all three functors are therefore –⊗LR –; see 7.2.6.

It follows from 7.1.8 and 7.2.12 that the functor –⊗LR – is k-bilinear, preserves

coproducts in both variables, and is triangulated in both variables.

7.4.3. By 7.4.2 and 7.2.2, one way to compute –⊗LR –: D(Ro)×D(R)→D(k) is

M⊗LR N = PRo(M)⊗R N and α/ϕ⊗L

R β/ψ = (P(α)⊗ β)/(P(ϕ)⊗ ψ)

for morphisms α/ϕ in D(Ro) and β/ψ in D(R). It also follows that another way tocompute the functor is

M⊗LR N = M⊗R PR(N) and α/ϕ⊗L

R β/ψ = (α⊗ P(β))/(ϕ⊗ P(ψ)) .

7.4.4 Definition. Let M be an R-complex. A semi-flat R-complex F that is isomor-phism to M in D(R) is called a semi-flat replacement of M.

7.4.5 Proposition. The following assertions hold.(a) Let M be an Ro-complex with a semi-flat replacement G. As functors from

D(R) to D(k) one hasM⊗L

R – = (G⊗R –) ´ .

(b) Let G be an endofunctor on K(Ro) and let ϕ : G→ IdK(Ro) be a natural trans-formation such that G(M) is semi-flat and ϕM is a quasi-isomorphism forevery Ro-complex M. As functors from D(Ro)×D(R) to D(k) one has

–⊗LR – = (G(–)⊗R –) ´ .

PROOF. (a): By 6.4.20 there is a quasi-isomorphism ψ : PRo(M)→ G in K(Ro), andhence a natural transformation τ= ψ⊗R –: PRo(M)⊗R –→ G⊗R – of functors fromK(R) to K(k). From 5.4.9 and 5.4.10 it follows that both functors preserve quasi-isomorphisms, and by 5.4.16 the morphism τN = ψ⊗R N is a quasi-isomorphism forevery R-complex N. Thus ´τ : (PRo(M)⊗R –) ´→ (G⊗R –) ´ is a natural isomorphismof functors from D(R) to D(k), see 7.2.3, and the assertion follows from 7.4.2.

(b): By 6.3.1 there is for every M a unique morphism ψM : PRo(M)→ G(M) inK(Ro) with ϕMψM = πM where π : PRo→ IdK(Ro) is the natural transformation from6.3.10. It follows that each ψM is a quasi-isomorphism and arguments parallel tothose in 6.3.12 show that ψ : PRo→ G is a natural transformation. It follows thatτ= ψ⊗R –: PRo(–)⊗R –→ G(–)⊗R – is a natural transformation of functors fromK(Ro)×K(R) to K(k) such that τMN = ψM⊗R N is a quasi-isomorphism for all Mand N. Thus ´τ : (PRo(–)⊗R –) ´→ (G(–)⊗R –) ´ is a natural isomorphism of functorsfrom D(Ro)×D(R) to D(k), see 7.2.3, and the assertion follows from 7.4.2.



7.4.6 Proposition. The following assertions hold.(a) Let N be an R-complex with a semi-flat replacement G. As functors from

D(Ro) to D(k) one has

–⊗LR N = (–⊗R G) ´ .

(b) Let G be an endofunctor on K(R) and let ϕ : G→ IdK(R) be a natural transfor-mation such that G(N) is semi-flat and ϕN is a quasi-isomorphism for everyR-complex N. As functors from D(Ro)×D(R) to D(k) one has

–⊗LR – = (–⊗R G(–)) ´ .


7.4.7 Addendum to 7.4.2. Let ring homomorphisms

α : Q⊗k Ro −→ A and β : R⊗k So −→ B ,

be given. By restriction of scalars K(A) → K(Q–Ro) and K(B)→K(R–So) itfollows from 7.1.9 there is a functor –⊗R – from the category K(A)×K(B) toK(Q–So). It has a left derived functor, which we temporarily denote

F: D(A)×D(B)→D(Q–So) .

Per 7.2.5 and 7.2.6 it is given by

(7.4.7.1) F = (PA(–)⊗R PB(–)) ´ .

One can compare F to the functor ⊗L from 7.4.2 via the canonical restriction func-tors; that is, one can consider the diagram,

(7.4.7.2)

D(A)×D(B) F//

D(Q–So)

D(Ro)×D(R)–⊗L

R–// D(k) .

Without extra assumptions, the diagram is not commutative, not even up to naturalisomorphism. If Q = k, A = Ro, and α is the canonical map k⊗k Ro→ Ro, thenF is naturally isomorphic to F1 = (PRo(–)⊗R –) , and the diagram is commutativeup to natural isomorphism. Similarly, if S = k, B = R, and β is the canonical map,then the functor F is naturally isomorphic to F2 = (–⊗R PR(–)) ´ and the diagram iscommutative up to natural isomorphism. In these situations one can thus choose theleft derived functor F such that it restricts to ⊗L, and hence we use that symbol forF. That is, there are functors

–⊗LR – = F' F1 : D(Ro)×D(R–So)−→D(So) and(7.4.7.3)

–⊗LR – = F' F2 : D(Q–Ro)×D(R)−→D(Q) .(7.4.7.4)

Though, according to 7.2.6, the symbol –⊗LR – would be the correct notation for the

functor (7.4.7.1), we only use that symbol in situations where the diagram (7.4.7.2)is commutative up to natural isomorphism.



TOR

For the next definition, recall from 6.5.20 that complexes that are isomorphic in thederived category have isomorphic homology.

7.4.8 Definition. For every m in Z denote by TorRm(–,–) the functor

Hm(–⊗LR –) : D(Ro)×D(R) −→ M(k) .

It follows from 7.4.2 and 6.5.20 that the functors Torm are k-bilinear.

REMARK. The name “Tor” for the functors defined in 7.4.8 stems from the fact that these functorsmeasure torsion: If k is an integral domain with field of fractions Q, then for every k-module Mthere is an isomorphism Tork1(M,Q/k)∼= MT, where MT is the torsion submodule of M.

7.4.9 Addendum. In view of 7.4.7 they extend to functors,

TorRm(–,–) : D(Ro)×D(R–So) −→ M(So) and

TorRm(–,–) : D(Q–Ro)×D(R) −→ M(Q) .

7.4.10. Let M be an Ro-complex and let N be an R-complex. Per 6.5.21 there are,with the usual conventions inf∅= ∞ and sup∅=−∞, equalities,

(7.4.10.1)infm ∈ Z | TorR

m(M,N) 6= 0 = inf(M⊗LR N) and

supm ∈ Z | TorRm(M,N) 6= 0 = sup(M⊗L

R N) .

For modules there is a simple description of Tor0.

7.4.11 Proposition. For every Ro-module M and every R-module N there is anisomorphism of k-modules,

TorR0 (M,N) ∼= M⊗R N ,

and it is natural in M and N. Furthermore, one has TorRm(M,N) = 0 for all m < 0.

PROOF. The natural transformation π : P→ IdK(R) from 6.3.10 induces a naturaltransformation of functors from M(Ro)×M(R) to M(k),

TorR0 (M,N) = H0(M⊗R P(N))

H0(M⊗πN)−−−−−−−→ H0(M⊗R N) = M⊗R N .

The equality on the left-hand side of the arrow follows from 7.4.2 and the definition7.4.8 of Tor; the equality on the right-hand side of the arrow holds as M⊗R N isa module. To see that the displayed map is an isomorphism, choose by 5.2.14 asurjective semi-projective resolution π : P '−→ N in C(R) with Pm = 0 for m < 0.Note that the homomorphism H0(M⊗ π) is the map

(M⊗R P0)/Im(M⊗R ∂P1 )−→M⊗R N given by [x]Im(M⊗∂P

1 )7−→ (M⊗R π0)(x) .

Application of the right exact functor M⊗R – to the exact sequence

P1∂P

1−−→ P0π0−−→ N −→ 0



shows that M⊗R π0 is surjective with kernel Im(M⊗R ∂P1 ); hence H0(M⊗R π) is an

isomorphism. With P as above one has (M⊗R P)m = M⊗R Pm = 0 for m < 0, andhence also TorR

m(M,N) = Hm(M⊗R P) = 0 for m < 0, as claimed.

7.4.12 Proposition. Let M be an Ro-module and N be an R-module. If TorRm(M,N)=

0 holds for all m > 0, then there is an isomorphism M⊗LR N 'M⊗R N in D(k).


In view of 6.5.19 the following result applies, in particular, to exact sequences0→M′→M→M′′→ 0 and 0→ N′→ N→ N′′→ 0 of complexes.

7.4.13 Theorem. Let M′ α′−→M α−→M′′ −→ ΣM′ be a distinguished triangle inD(Ro). For every R-complex N there is an exact sequence of k-modules,

· · · −−−→ TorRm(M

′,N)Torm(α

′,N)// TorR

m(M,N)Torm(α,N)

// TorRm(M

′′,N)

−−−→ TorRm−1(M

′,N)Torm−1(α

′,N)// TorR

m−1(M,N)−−−→ ·· · .

Let N′β′−→ N

β−→ N′′ −→ ΣN′ be a distinguished triangle in D(R). For every Ro-complex M there is an exact sequence of k-modules,

· · · −−−→ TorRm(M,N′)

Torm(M,β′)// TorR

m(M,N)Torm(M,β)

// TorRm(M,N′′)

−−−→ TorRm−1(M,N′)

Torm−1(M,β′)// TorR

m−1(M,N)−−−→ ·· · .

PROOF. The functor –⊗LR N is triangulated by 7.4.2, so applied to the distinguished

triangle M′→M→M′′→ ΣM′ in D(R), it yields a distinguished triangle in D(k).An application of 6.5.22 to this distinguished triangle and the definition 7.4.8 of Torgives the first exact sequence. The second exact sequence is obtained similarly.

7.4.14 Proposition. Let M be an Ro-complex and 0→ N′ → N → N′′ → 0 be anexact sequence in C(R). If TorR

1 (Mv,N′′i ) = 0 holds for all v, i ∈ Z, then the sequenceof complexes 0→M⊗R N′→M⊗R N→M⊗R N′′→ 0 is exact.

PROOF. For every i ∈ Z there is an exact sequence 0→ N′i → Ni → N′′i → 0 ofR-modules. By 7.4.13 and 7.4.11 it yields for every v ∈ Z an exact sequence

TorR1 (Mv,N′′i )−→Mv⊗R N′i −→Mv⊗R Ni −→Mv⊗R N′′i −→ 0 .

The assertion now follows from 2.4.14.

7.4.15 Proposition. Let N be an R-complex and 0→ M′ → M → M′′ → 0 be anexact sequence in C(Ro). If TorR

1 (M′′v ,Ni) = 0 holds for all v, i∈ Z, then the sequence

of complexes 0→M′⊗R N→M⊗R N→M′′⊗R N→ 0 is exact.

PROOF. The proof is parallel to that of 7.4.14, using 2.4.15 instead of 2.4.14.



RESOLUTIONS OF COMPLEXES OF BIMODULES

7.4.16 Addendum to 7.4.2. Adopt the notation from 7.4.7. It follows from 5.4.18that if the next condition is satisfied, then every semi-flat A-complex, in particularevery semi-projective A-complex, is semi-flat over Ro;

(7.4.16.1) A is flat as an Ro-module via Ro−→ Q⊗k Ro α−−→ A .

Thus, the condition ensures that PA(M) yields a semi-flat replacement of M as anRo-complex, whence the functor F from (7.4.7.1) makes the diagram (7.4.7.2) com-mutative up to natural isomorphism. Moreover, this functor is naturally isomorphicto F1 = (PA(–)⊗R –) , and per convention we set

(7.4.16.2) –⊗LR – = F' F1 : D(A)×D(R–So)−→D(Q–So) .

Notice that this use of the symbol –⊗LR – is an extension of the usage introduced in

(7.4.7.3), in as much as (7.4.16.1) is trivially satisfied for Q = k and A = Ro.Similarly, every semi-flat B-complex, in particular every semi-projective B-

complex, is semiflat over R if the next condition is satisfied;

(7.4.16.3) B is flat as an R-module via R−→ R⊗k So β−−→ B .

Under this condition, the functor F from (7.4.7.1) also makes the diagram (7.4.7.2)commutative up to natural isomorphism. Moreover, this functor is naturally isomor-phic to F2 = (–⊗R PB(–)) , and per convention we set

(7.4.16.4) –⊗LR – = F' F2 : D(Q–Ro)×D(B)−→D(Q–So) .

With S = k and B = R this recovers (7.4.7.4).

7.4.17 Example. If Q is flat as a k-module, then α = 1Q⊗kRosatisfies condition

(7.4.16.1); see 7.3.18(b). Similarly condition (7.4.16.3) is satisfied by β= 1R⊗kSoif

S is flat as a k-module. Examples of k-algebras that are flat, even free, as k-modulesare matrix algebras Mn×n(k) and polynomial algebras k[x1, . . . ,xn].

If k is a field—more generally a von Neuman regular ring—then every k-algebrais flat so –⊗L

R – is per (7.4.16.1) a functor from D(Q–Ro)×D(R–So) to D(Q–So).

7.4.18 Example. If R is commutative, then the composite R⊗k Ro−→ Re ∼=−→ R, see7.3.21, satisfies condition (7.4.16.1) and it follows from (7.4.16.2) that –⊗L

R – is afunctor from D(R)×D(R–So) to D(R–So). The same composite satisfies condition(7.4.16.3), and as Ro coincides with R it follows from (7.4.16.4) that –⊗L

R – is afunctor from D(Q–R)×D(R) to D(Q–R).

7.4.19 Proposition. Let α : R⊗kQo→ A be a ring homomorphism such that condi-tion (7.4.16.1) holds and consider the functor

–⊗LR – = (PA(–)⊗R –) ´ : D(A)×D(R–So) −→ D(Q–So)

from (7.4.16.2). If M and G are complexes of A-modules such that M ' G in D(A)and G is semi-flat as an Ro-complex, then there is an equality,


7.5 Standard Isomorphisms in the Derived Category 281

M⊗LR – = (G⊗L

R –) ´ : D(R–So) −→ D(Q–So) .

PROOF. The A-complex P = PA(M) is semi-projective. As one has P ' M ' Gin D(A) there exists by 6.4.20 a quasi-isomorphism ϕ : P→ G in K(A). By as-sumption, G is semi-flat as an Ro-complex, and so is P by (7.4.16.1). It followsfrom 5.4.16 and commutativity 7.1.12 that for every complex N of R–So-bimodules,ϕ⊗R N is a quasi-isomorphism in K(Q–So). By 6.4.32 the natural transformationϕ⊗R –: P⊗R –→ G⊗R – of functors from K(R–So) to K(Q–So) induces an iso-morphism between M⊗L

R – = (P⊗R –) ´ and (G⊗R –) .

7.4.20 Proposition. Let β : R⊗k So→ B be a ring homomorphism such that condi-tion (7.4.16.3) holds and consider the functor

–⊗LR – = (–⊗R PB(–)) ´ : D(Q–Ro)×D(B) −→ D(Q–So)

from (7.4.16.4). If N and G are complexes of B-modules such that N ' G in D(B)and G is semi-flat as an R-complex, then there is an equality,

–⊗LR N = (–⊗L

R G) ´ : D(Q–Ro) −→ D(Q–So) .


EXERCISES

E 7.4.1 Note thatQ is a semi-flat replacement of the Z-complex L from 5.4.15 and show that theredoes not exist a quasi-isomorphism Q→ L in C(Z).

E 7.4.2 Let M be an Ro-complex, let N be an R-complex, and let Z be a K-flat complex that isisomorphic to N in D(R). Show that there is an isomorphism M⊗R Z 'M⊗L

R N in D(k).E 7.4.3 Let R→ S be a ring homomorphism. Show that S is flat as an R-module if and only if

every semi-flat S-complex is semi-flat over R.

7.5 Standard Isomorphisms in the Derived Category


UNITOR AND COUNITOR

7.5.1. Let F,G: K(R–So)→K(R–So) be the functors F=R⊗R – and G= IdK(R–So)

and consider the unitor µR : F→ G from (7.1.11.1); it is a triangulated natural iso-morphism. Both functors preserve quasi-isomorphisms and hence 7.2.3 yields a tri-angulated natural isomorphism ´µR : ´F→ ´G. It follows from 7.2.9 that ´F is a left de-rived functor of F, and one has ´F = R⊗L

R –, as 1R is a semi-projective resolution ofR. It is evident that ´G = IdD(R–So). These arguments establish the unitor µR = ´µR inD(R–So); see (7.5.1.1). Similar arguments yields the counitor (7.5.1.2) in D(R–So).

For every complex M of R–So-bimodules there are isomorphisms in D(R–So),

µMR : R⊗L

R M −→ M and(7.5.1.1)



εMR : M −→ RHomR(R,M) ,(7.5.1.2)

which are natural in M. As natural transformations, µR and εR are triangulated.

COMMUTATIVITY

7.5.2 Construction. LetR⊗k So−→ B

be a ring homomorphism such that B is flat over R; cf. (7.4.16.3). Restriction of sca-lars K(B)→K(R–So) yields by 7.1.9 functors F,G: K(Q–Ro)×K(B)→K(Q–So)given by

F(M,N) = M⊗R N and G(M,N) = N⊗Ro M .

Now consider the commutativity isomorphism υ : F→ G obtained from 7.1.12. Pre-composing F, G, and υ with the functor IdK(Q–Ro)×PB, one gets the natural isomor-phism given by

M⊗R PB(N)υM P(N)

−−−−→ PB(N)⊗Ro M ,

which we abbreviate υ1 : F1→ G1. Both F1 and G1 preserve quasi-isomorphisms,see 7.4.16, and hence 7.2.2 and 7.2.3 yield a natural isomorphism ´υ1 : ´F1→ ´G1 offunctors from D(Q–Ro)×D(B) to D(Q–So). By definition, one has ´F1 = –⊗L

R –and ´G1 = –⊗L

Ro –.

7.5.3 Proposition. Adopt the setup from 7.5.2. Let M be a complex of Q–Ro-bimodules and N be a B-complex. Commutativity 7.1.12 induces an isomorphismD(Q–So),

υMN : M⊗LR N −→ N⊗L

Ro M ,

which is natural in M and N. Moreover, as a natural transformation of functors, υ istriangulated in each variable.

PROOF. The desired natural isomorphism υ is ´υ1 constructed in 7.5.2. The naturaltransformation υ is triangulated in each variable by 7.1.12, and the functor PB istriangulated by 6.3.10. It follows that the natural transformation υ1 : F1→ G1 in7.5.2 is triangulated in each variable, whence so is υ = ´υ1 by 7.2.3.

7.5.4 Example. As in 7.5.3 adopt the setup from 7.5.2, let M be a complex ofQ–Ro-bimodules and N be a B-complex. Let F be a semi-flat replacement of N overB. By 6.4.20 there is a quasi-isomorphism π : PB(N)→ F in K(B) which yields acommutative diagram in K(Q–So),

M⊗R PB(N)

'M⊗π

υM P(N)// PB(N)⊗Ro M

' π⊗M

M⊗R F υMF// F⊗Ro M ;



the vertical morphisms are quasi-isomorphisms by 5.4.16. Application of the functorVQ⊗kSo to this diagram yields υMF/1 = υMN ; cf. 7.5.3.

7.5.5 Corollary. Let M be an Ro-complex and N be an R-complex. Commutativ-ity 7.1.12 induces an isomorphism in D(k),

υMN : M⊗LR N −→ N⊗L

Ro M ,

which is natural in M and N. As a natural transformation of functors, υ is triangu-lated in each variable. Moreover, the following assertions hold.

(a) If M is a complex of Q–Ro-bimodules, then υMN is an isomorphism in D(Q).(b) If N is a complex of R–So-bimodules, then υMN is an isomorphism in D(So).

PROOF. Part (a) is immediate from commutativity 7.5.3 with S = k and B = R, andthe statement about the isomorphism in D(k) is the special case Q = k. Part (b) alsofollows from an application of commutativity 7.5.3, now with the roles of M and N,the roles of R and Ro interchanged, and Q replaced by So. Finally, of both (a) andyields the isomorphism υMN in D(Q).

ASSOCIATIVITY

7.5.6 Construction. Let

Q⊗k Ro−→ A and S⊗k T o−→ B

be ring homomorphisms such that A is flat over Ro and B is flat over S; cf. (7.4.16.1)and (7.4.16.3). Restriction of scalars K(A) → K(Q–Ro) and K(B) → K(S–T o)yields by 7.1.9 functors F,G: K(A)×K(R–So)×K(B)→K(Q–T o) given by

F(M,X ,N) = (M⊗R X)⊗S N and G(M,X ,N) = M⊗R (X⊗S N) .

Now consider the associativity isomorphism ω : F→ G obtained from 7.1.13. Pre-composing F, G, and ω with the functor PA× IdK(R–So)×PB, one gets the naturalisomorphism given by

(PA(M)⊗R X)⊗S PB(N)ωP(M)X P(N)

−−−−−−−→ PA(M)⊗R (X⊗S PB(N)) ,

which we abbreviate ω1 : F1→ G1. Both F1 and G1 preserve quasi-isomorphisms,see 7.3.19, and hence 7.2.2 and 7.2.3 yield a natural isomorphism ´ω1 : ´F1→ ´G1 offunctors from D(A)×D(R–So)×D(B) to D(Q–T o).

Consider the diagram below; by definition the squares are commutative. As thefunctor F1 is the composite of the functors in the top row, it follows from the unique-ness statement in 7.2.2 that ´F1 is the composite of the functors in the bottom row.



K(A)×K(R–So)×K(B)

VA×VR⊗So×VB

F1

''(PA(–)⊗R–)×IdK(B)// K(Q–So)×K(B)

VA⊗So×VB

–⊗SPB(–)// K(Q–T o)

VQ⊗T o

D(A)×D(R–So)×D(B)

´F1

77

(PA(–)⊗R–) ´×IdD(B)// D(Q–So)×D(B)

(–⊗SPB(–)) ´// D(Q–T o) .

By 7.4.16 the functor (PA(–)⊗R –) ´ is –⊗LR –: D(A)×D(R–So)→D(Q–So), and

(–⊗S PB(–)) ´ is –⊗LS –: D(Q–So)×D(B)→D(Q–T o) by 7.4.2. Thus, one has

´F1(M,X ,N) = (M⊗LR X)⊗L

S N .

Analogously one gets,

´G1(M,X ,N) = M⊗LR (X⊗L

S N) .

7.5.7 Proposition. Adopt the setup from 7.5.6. Let M be an A-complex, X be acomplex of R–So-bimodules, and N be a B-complex. Associativity 7.1.13 inducesan isomorphism in D(Q–T o),

ωMXN : (M⊗LR X)⊗L

S N −→ M⊗LR (X⊗L

S N) ,

which is natural in M, X , and N. Moreover, as a natural transformation of functors,ω is triangulated in each variable.

PROOF. The desired natural isomorphism ω is ´ω1 constructed in 7.5.6. The naturaltransformation ω is triangulated in each variable by 7.1.13, and the functors PA andPB are triangulated by 6.3.10. It follows that the natural transformationω1 : F1→ G1in 7.5.6 is triangulated in each variable, whence so is ω = ´ω1 by 7.2.3.

7.5.8 Corollary. Let M be an Ro-complex, X be a complex of R–So-bimodules, andN be an S-complex. Associativity 7.1.13 induces an isomorphism in D(k),

ωMXN : (M⊗LR X)⊗L

S N −→ M⊗LR (X⊗L

S N) ,

which is natural in M, X , and N. Moreover, as a natural transformation of functors,ω is triangulated in each variable.

PROOF. Apply 7.5.7 with Q = k= T , A = Ro, and B = S.

7.5.9 Example. Adopt the setup from 7.5.6 and let M, X , and N be as in 7.5.7. Let Fbe a semi-flat replacement of the A-complex M and G be a semi-flat replacement ofthe B-complex N. By 6.4.20 there are a quasi-isomorphisms π : PA(M)→ F in K(A)and π′ : PB(N)→ G in K(B) which yield a commutative diagram in K(Q–T o),



(PA(M)⊗R X)⊗S PB(N)

'(π⊗X)⊗π′

ωP(M)X P(N)// PA(M)⊗R (X⊗S PB(N))

' π⊗(X⊗π′)

(F⊗R X)⊗S G ωFXG// F⊗R (X⊗S G) ;

the vertical morphisms are quasi-isomorphisms by 5.4.16. Application of the functorVQ⊗kT o to this diagram yields ωFXG/1 = ωMXN ; cf. 7.5.7.

REMARK. There is an abstract approach to associativity ω that makes no reference to resolutions;it is based on the universal property of derived functors 7.2.16. Consider again the natural transfor-mation ω : F→ G from 7.5.6. For simplicity set Q = k= T , A = Ro, and B = S. The functor F is acomposite F = F2 (F1× IdK(S)) with

F1 = –⊗R –: K(Ro)×K(R–So)−→K(So) and

F2 = –⊗S –: K(So)×K(S)−→K(k) .

For i = 1,2 let (LFi,λi) be a terminal object in the category LFi , that is, LFi is a left derived functorof Fi. The composite

λ = λ2(F1× IdK(S))LF2(λ1×1VS )

is a natural transformation of functors

LF2 (LF1× IdD(S)) (VRo×VR⊗kSo×VS)λ−−→ Vk F2 (F1× IdK(S)) = Vk F ,

and thus (LF2 (LF1× IdD(S)),λ) is an object in LF. A straightforward formal argument showsthat this is a terminal object in LF, and hence LF2 (LF1× IdD(S)) is a left derived functor of F.Thus we may write(7.5.9.1) LF = LF2 (LF1× IdD(S)) = (–⊗L

R –)⊗LS – ,

and this expresses, loosely speaking, that the composite of left derived functors is a left derivedfunctor of the composite. Similarly, the functor G is a composite G = G2 (IdK(Ro)×G1) with

G1 = –⊗S –: K(R–So)×K(S)−→K(R) and

G2 = –⊗R –: K(Ro)×K(R)−→K(k) .

And for any choice of left derived functors LG1 and LG2, the composite LG2 (IdD(Ro)×LG1) isa left derived functor of G, that is, one has(7.5.9.2) LG = LG2 (IdD(Ro)×LG1) = –⊗L

R (–⊗LS –) .

With these choices (7.5.9.1) and (7.5.9.2) of left derived functors, it follows from 7.2.18 thatω : F→ G induces a natural transformation ω′ : LF→ LG which is uniquely determined by com-mutativity of a certain diagram. A formal argument shows that ω′ is isomorphic to ω from 7.5.7.

SWAP


R⊗kQo−→ A and T ⊗k So−→ B

be ring homomorphisms such that A is projective over R and B is projective overSo; cf. (7.3.19.1). Restriction of scalars K(A)→ K(R–Qo) and K(B)→ K(T –So)yields by 7.1.7 functors F,G: K(A)op×K(R–So)×K(B)op→K(Q–T o) given by

F(M,X ,N) = HomR(M,HomSo(N,X)) andG(M,X ,N) = HomSo(N,HomR(M,X)) .



Now consider the swap isomorphism ζ : F→ G obtained from 7.1.14. Precompos-ing F, G, and ζ with the functor Pop

A × IdK(R–So)×PopB , one gets the natural isomor-

phism given by

HomR(PA(M),HomSo(PB(N),X))ζP(M)X P(N)

−−−−−−→ HomSo(PB(N),HomR(PA(M),X)) ,

which we abbreviate ζ1 : F1→ G1. Both F1 and G1 preserve quasi-isomorphisms,see 7.3.19, and hence 7.2.2 and 7.2.3 yield a natural isomorphism ´ζ1 : ´F1→ ´G1 offunctors from D(A)op×D(R–So)×D(B)op to D(Q–T o). Proceeding as in 7.5.6,one obtains

´F1(M,X ,N) = RHomR(M,RHomSo(N,X)) and´G1(M,X ,N) = RHomSo(N,RHomR(M,X)) .

7.5.11 Proposition. Adopt the setup from 7.5.10. Let M be an A-complex, let X bea complex of R–So-bimodules, and let N be a B-complex. Swap 7.1.14 induces anisomorphism in D(Q–T o),

ζMXN : RHomR(M,RHomSo(N,X)) −→ RHomSo(N,RHomR(M,X)) ,

which is natural in M, X , and N. Moreover, as a natural transformation of functors,ζ is triangulated in each variable.

PROOF. The desired natural isomorphism ζ is ´ζ1 constructed in 7.5.10. The naturaltransformation ζ is triangulated in each variable by 7.1.14, and the functors Pop

A andPop

B are triangulated by 6.3.10 and D.10. It follows that the natural transformationζ1 : F1→ G1 in 7.5.10 is triangulated in each variable, and so is ζ = ´ζ1 by 7.2.3.

7.5.12 Corollary. Let M be an R-complex, let X be a complex of R–So-bimodules,and let N be an So-complex. Swap 7.1.14 induces an isomorphism in D(k),

ζMXN : RHomR(M,RHomSo(N,X)) −→ RHomSo(N,RHomR(M,X)) ,

which is natural in M, X , and N. Moreover, as a natural transformation of functors,ζ is triangulated in each variable.

PROOF. Apply 7.5.11 with Q = k= T , A = R, and B = So.

7.5.13 Example. Let M, X , and N be k-complexes. Let a choice of resolution func-tors P and I on K(k) be given, see 6.3.10 and 6.3.14, and consider resolutions

πM : P(M)'−−→M , ιX : X '−−→ I(X) , and πN : P(N)

'−−→ N .

Set π = πM , P = P(M), ι = ιX , I = I(X), π′ = πN , and P′ = P(N). As ζ is a naturaltransformation, see 7.1.14, there is a commutative diagram in K(k),



Homk(P,Homk(N, I))

'ϕ′=Hom(P,Hom(π′,I))

ζPIN// Homk(N,Homk(P, I))

' ψ′=Hom(π′,Hom(P,I))

Homk(P,Homk(P′, I))ζPIP′

// Homk(P′,Homk(P, I))

Homk(P,Homk(P′,X))

'ϕ=Hom(P,Hom(P′,ι))

OO

ζPXP′// Homk(P′,Homk(P,X)) ,

' ψ=Hom(P′,Hom(P,ι))

OO

where the vertical morphisms are quasi-isomorphisms by 5.2.10 and 5.3.16. It in-duces by application of Vk a commutative diagram in D(k),

Homk(P,Homk(N, I))

'(1/ϕ)(ϕ′/1)

ζPIN/1// Homk(N,Homk(P, I))

' (1/ψ)(ψ′/1)

RHomk(M,RHomk(N,X))ζMXN

// RHomk(N,RHomk(M,X)) .

The diagram shows that also ζPIN/1 is swap ζMXN .Even without this diagram, it is clear that ζPIN/1 is a morphism in D(k) from

RHomk(M,Homk(N,X)) to RHomk(N,RHomk(M,X)). Indeed, HomR(P, I) is asemi-injective k-complex by 5.4.10 and 5.4.19. Now 6.3.5 and 6.3.6 yield an isomor-phism Homk(P, I)u I(HomR(P, I)) in K(k), so Homk(N,Homk(P, I)) is isomorphicto RHomk(N,RHomk(M,X)) in D(k).

ADJUNCTION


R⊗k T o−→ A and S⊗kQo−→ B

be ring homomorphisms such that A is flat over Ro and B is projective over S;cf. 7.3.19. Restriction of scalars K(A)→K(R–T o) and K(B)→K(S–Qo) yields by7.1.7 and 7.1.9 functors F,G: K(A)×K(R–So)op×K(B)op→K(Q–T o) given by

F(M,X ,N) = HomR(X⊗S N,M) andG(M,X ,N) = HomS(N,HomR(X ,M)) .

Now consider the adjunction isomorphism ρ : F→ G obtained from 7.1.15. Pre-composing F, G, and ω with the functor IA× IdK(R–So)op×Pop

B , one gets the naturalisomorphism given by

HomR(X⊗S PB(N), IA(M))ρI(M)X P(N)

−−−−−−→ HomR(PB(N),HomS(X , IA(M))) ,

which we abbreviate ρ1 : F1→ G1. Both F1 and G1 preserve quasi-isomorphisms,see 7.3.19, and hence 7.2.2 and 7.2.3 yield a natural isomorphism ´ρ1 : ´F1→ ´G1 of



functors from D(A)×D(R–So)op×D(B)op to D(Q–T o). Proceeding as in 7.5.6,one obtains

´F1(M,X ,N) = RHomR(X⊗LS N,M) and

´G1(M,X ,N) = RHomS(N,RHomR(X ,M)) .

7.5.15 Proposition. Adopt the setup from 7.5.14. Let M be an A-complex, X be acomplex of R–So-bimodules, and N be a B-complex. Adjunction 7.1.15 induces anisomorphism in D(Q–T o),

ρMXN : RHomR(X⊗LS N,M) −→ RHomS(N,RHomR(X ,M)) ,

which is natural in M, X , and N. Moreover, as a natural transformation of functors,ρ is triangulated in each variable.

PROOF. The desired natural isomorphism ρ is ´ρ1 constructed in 7.5.14. The naturaltransformation ρ is triangulated in each variable by 7.1.15, and the functors Pop

B andIA are triangulated by 6.3.10, 6.3.14, and D.10. Therefore, the natural transformationρ1 : F1→ G1 in 7.5.14 is triangulated in each variable, and so is ρ= ´ρ1 by 7.2.3.

7.5.16 Corollary. Let M be an R-complex, X be a complex of R–So-bimodules, andN be an S-complex. Adjunction 7.1.15 induces an isomorphism in D(k),

ρMXN : RHomR(X⊗LS N,M) −→ RHomS(N,RHomR(X ,M)) ,

which is natural in M, X , and N. Moreover, as a natural transformation of functors,ρ is triangulated in each variable.

PROOF. Apply 7.5.15 with Q = k= T , A = R, and B = S.

EXERCISES

E 7.5.1 Let M be an Ro-module and let N be an R-module. Show that there is an isomorphism,TorR

m(M,N)∼= TorRom (N,M).

E 7.5.2 Let M be an Ro-module, let X be an R–So-bimodule, and let F be a flat S-module. Showthat there is an isomorphism, TorR

m(M,X)⊗S F ∼= TorRm(M,X⊗S F).

E 7.5.3 Let P be a projective R-module, let X be an R–So-bimodule, and let N be an So-module.Show that there is an isomorphism, HomR(P,ExtmSo(N,X))∼= ExtmSo(N,HomR(P,X)).

E 7.5.4 Let I be an injective R-module, let X be an R–So-bimodule, and let N be an S-module.Show that there is an isomorphism, HomR(TorS

m(X ,N), I)∼= ExtmS (N,HomR(X , I)).

7.6 Boundedness and Finiteness

SYNOPSIS. Bounded derived category; preservation of boundedness and finiteness by RHom and⊗L.

Conditions of vanishing and finiteness of homology modules identify useful subcat-egories of D(R).



BOUNDED SUBCATEGORIES OF THE DERIVED CATEGORY

Recall from 6.5.20 that homology is a functor on D(R).

7.6.1 Definition. The full subcategories D@(R), DA(R), and D@A(R) of D(R) aredefined by specifying their objects as follows,

D@(R) = M ∈D(R) | H(M) is bounded above ,DA(R) = M ∈D(R) | H(M) is bounded below , andD@A(R) = M ∈D(R) | H(M) is bounded .

7.6.2. Note that D@A(R) is the intersection of the subcategories D@(R) and DA(R).In terms of the invariants from 2.5.4 these categories can be described as follows,

D@(R) = M ∈D(R) | supM < ∞ ,DA(R) = M ∈D(R) | infM >−∞ , andD@A(R) = M ∈D(R) | ampM < ∞ .

7.6.3. Recall from 2.5.19 that soft truncation above (–)Ďn and soft truncation below(–)Ěn are k-linear endofunctors on C(R). It follows from 4.3.21 and 4.2.9 that (–)Ďnand (–)Ěn preserve homotopy and quasi-isomorphisms, so by 6.1.20 and 6.4.31 theyyield well-defined k-linear endofunctors on D(R), also denoted (–)Ďn and (–)Ěn.

For an R-complex M it follows from 4.2.4 and 6.4.18 that there are isomorphismsM 'MĎn for every n> supM and M 'MĚn for every n6 infM.

7.6.4 Proposition. The categories D@(R), DA(R), and D@A(R) are triangulatedsubcategories of D(R), and they are closed under soft truncations.

PROOF. We verify the axioms from D.14. By definition, all three subcategories ofD(R) are full; evidently they are k-linear and closed under isomorphisms and Σ.It follows from 6.5.22 that they are closed under distinguished triangles. Closureunder soft truncations is evident.

7.6.5 Proposition. There are equalities of full subcategories of D(R),

D@(R) = M ∈D(R) |M is isomorphic in D(R) to a bounded above complex ,DA(R) = M ∈D(R) |M is isomorphic in D(R) to a bounded below complex ,D@A(R) = M ∈D(R) |M is isomorphic in D(R) to a bounded complex .PROOF. The inclusions “⊇” are trivial; the inclusions “⊆” follow from 7.6.3.

7.6.6 Proposition. For every R-complex M and every integer n, the following aredistinguished triangles in D(R),

Σn Hn(M)−→MĎn −→MĎn−1 −→ Σ(Σn Hn(M)) ,(7.6.6.1)

MĚn+1 −→MĚn −→ Σn Hn(M)−→ Σ(MĚn+1) , and(7.6.6.2)

MĚn −→M −→MĎn−1 −→ Σ(MĚn) .(7.6.6.3)



PROOF. To construct (7.6.6.1), note that there is a short exact sequence in C(R),

0−→ H −→MĎn −→MĎn−1 −→ 0 ,

and hence by 6.5.19 a distinguished triangle H →MĎn →MĎn−1 → ΣH in D(R).The complex

H = 0−→ Cn(M)∂M

n−−→ Bn−1(M)−→ 0

is concentrated in degrees n and n− 1, and the embedding Σn Hn(M)→ H is evi-dently a quasi-isomorphism; in particular it yields an isomorphism in D(R).

A dual argument establishes (7.6.6.2). To construct (7.6.6.3), consider the shortexact sequence in C(R),

0−→MĚn −→M −→C −→ 0 ,

where the cokernel C is the complex 0→Bn−1(M)→Mn−1→Mn−2→··· . The sur-jection C→MĎn−1 is a quasi-isomorphism; in particular it yields an isomorphismin D(R), and the existence of the desired triangle follows from 6.5.19.

REMARK. A t-structure on a triangulated category (T,Σ) is a pair (A,B) of full subcategoriesthat are closed under isomorphisms, direct sums, and direct summands such that (1) ΣA⊆A andΣ−1B⊆B; (2) T(A,B) = 0 for all A∈A and B∈B; (3) for every M ∈T there is a distinguishedtriangle A→M→ B→ ΣA with A ∈A and B ∈B. The triangle (7.6.6.3) together with the factthat there are no non-zero morphisms MĚ1 → MĎ0 in D(R), see E 7.6.5, show that truncationsgive rise to a t-structure on D(R).

7.6.7 Proposition. Let M be a complex in DA(R) and N be a complex in D@(R).For integers u6 infM and w> supN one has,

(a) supRHomR(M,N)6 w−u.(b) Hw−u(RHomR(M,N))∼= HomR(Hu(M),Hw(N)).

In particular, the complex RHomR(M,N) belongs to D@(k).

PROOF. By 5.2.15 there exists a semi-projective resolution P '−→ M with Pv = 0for all v < u. By 7.6.3 one has RHomR(M,N)' RHomR(M,NĎw)'HomR(P,NĎw)in D(k), and the assertions follow from 2.5.9.

7.6.8 Proposition. Let M be a complex in DA(Ro) and N be a complex in DA(R).For integers u6 infM and w6 infN one has,

(a) inf(M⊗LR N)> u+w.

(b) Hu+w(M⊗LR N)∼= Hu(M)⊗R Hw(N).

In particular, the complex M⊗LR N belongs to DA(k).

PROOF. By 5.2.15 there exists a semi-projective resolution P '−→ M with Pv = 0for all v < u. By 7.6.3 one has M⊗L

R N 'M⊗LR NĚw ' P⊗R NĚw in D(k), and the

assertions follow from 2.5.13.



FINITE HOMOLOGY AND NOETHERIAN RINGS

7.6.9 Definition. The full subcategory Df(R) of D(R) is defined by specifying itsobjects as follows,

Df(R) = M ∈D(R) | H(M) is degreewise finitely generated .

The full subcategory Df(R)∩D@(R) is denoted by Df@(R). Similarly, one defines

the subcategories DfA(R) and Df

@A(R).

7.6.10 Proposition. If R is left Noetherian, then Df(R) is a triangulated subcategoryof D(R), and it is closed under soft truncations.

PROOF. We verify the axioms from D.14. By definition Df(R) is full, and evidentlyit is additive and closed under isomorphisms and Σ; see 6.5.20. Since R is left Noe-therian, it follows from 6.5.22 that Df(R) is closed under distinguished triangles.Closure under soft truncations is evident.

7.6.11 Proposition. If R is left Noetherian, then the following hold:

DfA(R) =

M ∈D(R)

∣∣∣∣ M is isomorphic in D(R) to a bounded belowand degreewise finitely generated complex

and

Df@A(R) =

M ∈D(R)

∣∣∣∣ M is isomorphic in D(R) to a boundedand degreewise finitely generated complex

.

PROOF. The inclusions “⊇” are trivial. For a complex M in DfA(R), choose u ∈ Z

with u6 infM. By 5.1.14 there is a semi-free resolution L '−→M with L degreewisefinitely generated and Lv = 0 for all v < u. This proves the first equality.

For M in Df@A(R) there exists by the argument above a bounded below and de-

greewise finitely generated complex L with M ' L in D(R). Choose w ∈ Z withw > supM, by 7.6.3 one has M ' MĎw ' LĎw in D(R), and LĎw is bounded anddegreewise finitely generated.

REMARK. Under extra assumptions on R one can prove that a complex in Df@(R) is isomorphic

in D(R) to a bounded above complex of finitely generated R-modules; see E 9.1.5.

7.6.12 Proposition. Let R be left Noetherian and let S be right Noetherian. Let M bea complex in Df

A(R) and let X be a complex in D@(R–So). If H(X) is degreewisefinitely generated over So, then the complex RHomR(M,X) belongs to Df

@(So).

PROOF. Since M is in DA(R) the functor RHomR(M,–) : D(R–So)→D(So) mapsD@(R–So) to D@(So) and it is way-out below; see (7.3.5.3) and 7.6.7. To show thatRHomR(M,X) belongs to Df(So), apply A.28(c) to the functor F = RHomR(M,–)and the category U = X ′ ∈ D(R–So) | H(X ′) ∈ Df(So). It suffices to argue thatRHomR(M,X ′) is in Df(So) for every R–So-bimodule X ′ that is finitely generatedover So. To see this choose by 5.1.14 a semi-free resolution L '−→M with L de-greewise finitely generated and Lv = 0 for v < infM. By 2.5.10 the So-complexRHomR(M,X ′)'HomR(L,X ′) consists of finitely generated modules, and hence itshomology is degreewise finitely generated, again because S is right Noetherian.



7.6.13 Proposition. Let R and S be left Noetherian. Let N be a complex in DfA(S)

and let X be a complex in DA(R–So). If H(X) is degreewise finitely generated overR, then the complex X⊗L

S N belongs to DfA(R).

PROOF. As N is in DA(R) the functor –⊗LS N : D(R–So)→D(R) maps DA(R–So)

to DA(R) and it is way-out above; see (7.4.7.4) and 7.6.8. To show that X⊗LS N is in

Df(R), apply A.28(b) to F = –⊗LR N and U= X ′ ∈D(R–So) | H(X ′) ∈Df(R). It

suffices to argue that X ′⊗LS N is in Df(R) for every R–So-bimodule X ′ that is finitely

generated over R. To see this choose by 5.1.14 a semi-free resolution L '−→ N withL degreewise finitely generated and Lv = 0 for v < infN. By 2.5.14 the R-complexX ′⊗L

S N ' X ′⊗S L consists of finitely generated modules, and hence its homologyis degreewise finitely generated as R is left Noetherian.

EXERCISES

E 7.6.1 Let M be an R-complex with Hv(M) = 0 for v 6= 0. (a) Show that if Mv = 0 holds forv < 0, then there is a quasi-isomorphism M '−→ H(M) in C(R). (b) Show that if Mv = 0holds for v > 0, then there is a quasi-isomorphism H(M)

'−→ M in C(R). (c) Concludethat for a complex M′ with ampM′ = 0 there is an isomorphism M′ ' H(M′) in D(R).

E 7.6.2 Show that the functors (–)Ďn,(–)Ěn : K(R)→K(R) and (–)Ďn,(–)Ěn : D(R)→D(R)are not triangulated.

E 7.6.3 Show that D@A(R) is the smallest triangulated subcategory of D(R) that contains the fullsubcategory M(R).

E 7.6.4 Let U be a triangulated subcategory of D(R) that is closed under soft truncations. Showthat for every complex M in U and every v ∈ Z the module Hv(M) is in U.

E 7.6.5 Let M be an R-complex; show that there is no non-zero morphism MĚ1→MĎ0 in D(R).E 7.6.6 Give examples of complexes M ∈D@(Z) and N ∈DA(Z) such that supRHomZ(M,N)

is strictly less than supN− infM.E 7.6.7 Give examples of complexes M ∈DA(Z) and N ∈DA(Z) such that inf(M⊗L

ZN) isstrictly greater than infM+ infN.


Chapter 8Homological Dimensions

The construction of a semi-free resolution, or a semi-injective resolution, of an R-complex yields a complex that, though it has remarkable properties, is isomorphicin the derived category to the original complex. What one may have to concede inthis exchange is boundedness: Even if one starts with a module, the resolutions maynot be bounded. Homological dimensions control the size of such resolution.

8.1 Projective Dimension

SYNOPSIS. Vanishing of Ext; semi-projective replacement; projective dimension; Schanuel’slemma; projective dimension over Noetherian ring; ∼ over semi-perfect ring; ∼ over perfect ring;∼ of module.

First we connect projectivity of modules to vanishing of Ext.

8.1.1 Lemma. For an R-module P, the following conditions are equivalent.(i) P is projective.

(ii) infRHomR(P,N)> infN holds for every R-complex N.(iii) There exists a surjective homomorphism π : L→ P with L projective and with

Ext1R(P,Kerπ) = 0.

PROOF. If P is projective, then 1P is a semi-projective resolution of P, and one hasRHomR(P,N)'HomR(P,N) in D(k). Now 2.5.6(a) yields infRHomR(P,N)> infN.Thus, (i) implies (ii) which, in particular, implies that the functor Ext1R(P,–) is zeroon modules, whence (iii) follows. To see that (iii) implies (i), consider the short exactsequence 0→Kerπ→ L→ P→ 0. It is split by 7.3.15, so P is a direct summand ofa projective module and hence projective by 1.3.20.

293

294 8 Homological Dimensions

SEMI-PROJECTIVE REPLACEMENTS AND PROJECTIVE DIMENSION

Recall from 5.2.14 that every complex is isomorphic in the derived category to asemi-projective complex. A complex of finite projective dimension is one that isisomorphic to a bounded above semi-projective complex.

8.1.2 Definition. Let M be an R-complex. A semi-projective R-complex that is iso-morphic to M in D(R) is called a semi-projective replacement of M.

The projective dimension of M, written pdR M, is defined as

pdR M = inf

n ∈ Z∣∣∣∣ There is a semi-projective replacement

P of M with Pv = 0 for all v > n

,

with the convention inf∅= ∞. One says that pdR M is finite if pdR M < ∞ holds.

8.1.3. Let M be an R-complex. For every semi-projective replacement P of M onehas H(P)∼= H(M); the next (in)equalities are hence immediate from the definition,

pdR M > supM and pdRΣs M = pdR M+ s for every s ∈ Z .

Moreover, one has pdR M =−∞ if and only if M is acyclic.Note that the definition of projective dimension could also be written

(8.1.3.1) pdR M = infsupP\ | P is a semi-projective replacement of M .

8.1.4. Let M be an R-complex. For every semi-projective replacement P of M itfollows from 6.4.20 that there is semi-projective resolution P '−→M.

8.1.5 Proposition. Let R→ S be a ring homomorphism. For every R-complex Mthere is an inequality,

pdS(S⊗LR M)6 pdR M .

PROOF. For every semi-projective replacement P of the R-complex M, the S-complex S⊗R P is a semi-projective replacement of S⊗L

R M by 5.2.22. As one hassup(S⊗R P)\ 6 supP\, the desired inequality follows.

The inequality in 8.1.5 may be strict.

8.1.6 Example. It follows from 5.4.15 that one has pdZQ 6 1. On the other hand,if there were a semi-projective replacement P of Q in D(Z) with P1 = 0, then onewould have Q ∼= H0(P) ∼= Z0(P), but Z0(P) would be a submodule of a free Z-module and hence free by 1.3.10; a contradiction. Thus pdZQ= 1 holds, and for thebase changed module Q⊗ZQ∼= Q one gets pdQ(Q⊗ZQ) = 0 from 8.1.2 and 8.1.3.

We now aim for a theorem that asserts, in summary form, that complexes offinite projective dimension are characterized by vanishing of Ext and that any semi-projective replacement of such a complex can be truncated to yield one of minimallength. We need the next two lemmas to prove it.


8.1 Projective Dimension 295

8.1.7 Lemma. Let M be an R-complex and let P be a semi-projective replacementof M. For every R-module N and for all integers m > 0 and n > supM there is anisomorphism

Extn+mR (M,N) ∼= ExtmR (Cn(P),N) .

PROOF. Recall from 5.2.8 that Pěn is a semi-projective R-complex. As one hasn > supM = supP, the canonical morphism Σ−n Pěn Cn(P) is a semi-projectiveresolution. In the next computation, the final and the first two identities follow fromthe definitions of RHom and Hom; the fourth follows from 2.3.13.

H−(n+m)(RHomR(M,N)) ∼= H−(n+m)(HomR(P,N))

= H−(n+m)(HomR(Pěn,N))

= H−m(Σn HomR(Pěn,N))

∼= H−m(HomR(Σ−n Pěn,N))

∼= H−m(RHomR(Cn(P),N)) ;

the definition of Ext, 7.3.7, now yields the asserted isomorphism.

8.1.8 Lemma. Let P be a semi-projective R-complex and let n be an integer. Thecomplex PĎn is semi-projective if and only if the module Cn(P) is projective.

PROOF. If the complex PĎn is semi-projective, then the module Cn(P) = (PĎn)n isprojective. If Cn(P) is a projective module, then it is semi-projective as a complex by5.2.12. There is an exact sequence 0→ Pďn−1→ PĎn→ Σn Cn(P)→ 0, so by 5.2.16the complex PĎn is semi-projective if and only if Pďn−1 is semi-projective, which itis by 5.2.8 and 5.2.16 applied to the exact sequence 0→Pďn−1→P→Pěn→ 0.

8.1.9 Theorem. Let M be an R-complex and let n be an integer. The followingconditions are equivalent.

(i) pdR M 6 n.(ii) − infRHomR(M,N)6 n− infN holds for every R-complex N.

(iii) n> supM and Extn+1R (M,N) = 0 holds for every R-module N.

(iv) n> supM and Ext1R(Cn(P),Cn+1(P)) = 0 holds for some, equivalently every,semi-projective replacement P of M.

(v) n> supM and for some, equivalently every, semi-projective replacement P ofM, the module Cn(P) is projective.

(vi) n> supM and for every semi-projective replacement P of M, there is a semi-projective resolution PĎn

'−→M.(vii) There is a semi-projective resolution P '−→ M with Pv = 0 for all v > n and

for all v < infM.In particular, there are equalities

pdR M = sup− infRHomR(M,N) | N is an R-module= supm ∈ Z | ExtmR (M,N) 6= 0 for some R-module N .

PROOF. The implication (vii)=⇒ (i) is trivial.



(i)=⇒ (ii): One can assume that N is in DA(R) and not acyclic; otherwise theinequality is trivial. Set u = infN; it is an integer and there is an isomorphismRHomR(M,N) ' RHomR(M,NĚu); see 4.2.4. By assumption there exists a semi-projective replacement P of M with Pv = 0 for all v> n, and by 8.1.4 it yields a semi-projective resolution P '−→M. Thus, one has infRHomR(M,N) = infHomR(P,NĚu).For v < u−n and p ∈ Z, one of the inequalities p > n or p+v6 n+v < u holds, sothe module

HomR(P,NĚu)v =∏

p∈ZHomR(Pp,(NĚu)p+v)

is zero. In particular, one has Hv(HomR(P,NĚu)) = 0 for v < u−n, so the inequalityinfRHomR(M,N)> u−n holds as desired.

(ii)=⇒ (iii): The second assertion is immediate. By way of 2.5.6(b) one getsn>− infRHomR(M,ER) =− infHomR(M,ER) = supM.

(iii)=⇒ (iv): Let P be a semi-projective replacement of M. It follows from 8.1.7and the assumptions that one has Ext1R(Cn(P),Cn+1(P)) = 0.

(iv)=⇒ (v): Let P be a semi-projective replacement of M. As n> supM = supPholds, the sequence 0→ Cn+1(P)→ Pn → Cn(P)→ 0 is exact, and the assertionfollows from 8.1.1.

(v)=⇒ (i) and (v)=⇒ (vi): Let P be a semi-projective replacement of M. As onehas n> supM = supP, the morphism τP

Ďn : P PĎn is a quasi-isomorphism. Sincethe module Cn(P) is projective, the complex PĎn is semi-projective by 8.1.8. Finally,it follows from the isomorphisms PĎn ' P ' M in D(R) and 8.1.4 that there is aquasi-isomorphism PĎn→M.

This argument shows that the “some” part of (v) implies (i), and thus (i)–(v) areequivalent. The argument also shows that the “every” part of (v) implies (vi).

(vi)=⇒ (vii): Choose by 5.2.15 a semi-projective resolution P '−→M with Pv = 0for all v< infM. By (vi) there is a quasi-isomorphism PĎn→M, which is the desiredresolution.

The asserted equalities follow from the equivalence of (i), (ii), and (iii).

The next corollary implies, in view of 6.5.19, that if two out of three complexesin a short exact sequence have finite projective dimension, then so has the third.

8.1.10 Corollary. Let M′→M→M′′→ ΣM′ be a distinguished triangle in D(R).With p′ = pdR M′, p = pdR M, and p′′ = pdR M′′ there are inequalities,

p′ 6maxp, p′′−1 , p6maxp′, p′′ , and p′′ 6maxp′+1, p .

In particular, if two of the complexes in the triangle have finite projective dimension,then so has the third.


Every vector space has projective dimension 0 as a module over the ground field.It follows from 1.3.10 and 1.3.17 that a module over a left principal ideal domainhas projective dimension at most 1. It is not, though, hard to showcase a ring withmodules of infinite projective dimension.



8.1.11 Example. The complex P = · · · −→ Z/4Z 2−→ Z/4Z 2−→ Z/4Z −→ 0, con-centrated in non-negative degrees, is a semi-projective replacement of the Z/4Z-module 2Z/4Z. Since Z/4Z is indecomposable as a Z/4Z-module, no cokernelCn(P) = 2Z/4Z for n> 0 is projective, so one has pdZ/4Z(2Z/4Z) = ∞.

8.1.12 Proposition. For every family of R-complexes Muu∈U there is an equality,

pdR(∐

u∈U

Mu) = supu∈UpdR Mu .

PROOF. Let N be an R-module. The functor RHomR(–,N) preserves products by3.1.25 and 7.2.12(c), and so does homology by 3.1.22, whence one has

− inf(RHomR(∐

u∈UMu,N)) = − inf(

∏u∈U

RHomR(Mu,N))

= supu∈U− infRHomR(Mu,N) .

The desired equality now follows from 8.1.9.

The following result, which is known as Schanuel’s lemma (for semi-projectivecomplexes), can be seen as a refinement of the statement 8.1.9(v).

8.1.13 Proposition. Let M be an R-complex and let P and P′ be semi-projectivereplacements of M. For every v ∈ Z there exist projective R-modules L and L′ withCv(P)⊕L∼= Cv(P′)⊕L′.

PROOF. By 6.4.20 and 5.2.20 there is a homotopy equivalence α : P→ P′, and by4.3.30 the complex Coneα is contractible. Furthermore, it consists of projectivemodules, so 4.3.32 shows that every module Zn(Coneα) is projective. By 4.3.21 themorphism αĎv : PĎv→ P′Ďv is also a homotopy equivalence so Cone(αĎv), that is,

0−→ Cv(P)−→ Cv(P′)⊕Pv−1 −→ P′v−1⊕Pv−2∂Coneα

v−1−−→ P′v−2⊕Pv−3 −→ ·· · ,

is contractible. Hence one has Cv(P)⊕Zv−1(Coneα)∼= Cv(P′)⊕Pv−1.

NOETHERIAN RINGS

Over a Noetherian ring, the projective dimension of a finitely generated module canbe detected by vanishing of homology with coefficients in cyclic modules.

8.1.14 Theorem. Let R be left Noetherian. Let M be an R-complex in DfA(R) and

let n be an integer. The following conditions are equivalent.(i) pdR M 6 n.

(ii) n> supM and Extn+1R (M,R/a) = 0 for every left ideal a in R.

(iii) There is a semi-projective resolution P '−→M with P degreewise finitely gen-erated and Pv = 0 for all v > n and for all v < infM.



In particular, there is an equality

pdR M = sup− infRHomR(M,R/a) | a is a left ideal in R= supm ∈ Z | ExtmR (M,R/a) 6= 0 for some left ideal a in R .

PROOF. It follows from the definition of projective dimension that (iii) implies (i),and the implication (i)=⇒ (ii) holds by 8.1.9. To see that (ii) implies (iii), noticefirst that by (ii) and A.1 one has Extn+1

R (M,N) = 0 for every finitely generated R-module N. Choose by 5.1.14 a semi-free resolution L '−→ M with L degreewisefinitely generated and Lv = 0 for all v < infM. As one has n > supM = supL, thesequence 0→ Cn+1(L)→ Ln→ Cn(L)→ 0 is exact; moreover, the modules Cv(L)are finitely generated as R is left Noetherian. By 8.1.7 there is an isomorphism

Ext1R(Cn(L),Cn+1(L)) ∼= Extn+1R (M,Cn+1(L)) = 0 ,

whence Cn(L) is projective by 8.1.1. Now 8.1.9 finishes the proof.

PERFECT AND SEMI-PERFECT RINGS

Existence of minimal projective resolutions is treated in B.58 and B.59.

8.1.15 Proposition. Let M be an R-complex and let n be an integer. If P '−→M is aminimal semi-projective resolution, then the following conditions are equivalent.

(i) pdR M 6 n.(ii) n> supM and Pn+1 = 0.

(iii) Pv = 0 for all v > n.In particular, one has pdR M = supP\.

PROOF. If one has pdR M 6 n, then there is a semi-projective replacement P′ of Mwith P′v = 0 for all v > n. It follows from 6.4.20 that there is a quasi-isomorphismP′→ P, and by B.54 it has a right inverse, whence one has Pv = 0 for v > n. Thus(i) implies (iii) which, in turn, implies (ii). If one has n> supM and Pn+1 = 0, thenthe module Cn(P) = Pn is projective, whence pdR M 6 n holds by 8.1.9.

Every complex over a left perfect ring has a minimal projective resolution.

8.1.16 Theorem. Let R be left perfect with Jacobson radical J and set k = R/J. LetM be an R-complex and let n be an integer. The following conditions are equivalent.

(i) pdR M 6 n.(ii) n> supM and Extn+1

R (M,k) = 0.(iii) n> supM and TorR

n+1(k,M) = 0.In particular, one has

pdR M =− infRHomR(M,k) = sup(k⊗LR M) .

PROOF. Choose by B.58 a semi-projective resolution P '−→M with P minimal. Thegraded module P\ is semi-perfect by B.51, so one has ∂P(P)⊆ JP by B.53. It followsthat the complex k⊗R P has zero differential, so one has



TorRv (k,M) = Hv(k⊗L

R M) = (k⊗R P)v = k⊗R Pv .

By B.38(a) the module k⊗R Pv ∼= Pv/JPv is non-zero for every Pv 6= 0, so one hasTorR

v (k,M) = 0 if and only if Pv = 0. The equivalence of conditions (i) and (iii) nowfollows from 8.1.15.

It also follows from the inclusion ∂P(P) ⊆ JP that the complex HomR(P,k) haszero-differential, whence one has

ExtvR(M,k) = H−v(RHomR(M,k)) = HomR(P,k)−v = HomR(Pv,k) .

For every v ∈ Z with Pv 6= 0 the k-module Pv/JPv is non-zero. As k is semi-simple, itfollows that there is a direct summand of Pv/JPv which is isomorphic to a non-zeroideal in k. Thus, there is a non-zero homomorphism Pv Pv/JPv→ k. In particular,one has ExtvR(M,k) = 0 if and only if Pv = 0. The equivalence of (i) and (ii) nowfollows from 8.1.15. The final equalities follow from the equivalence of (i)–(iii).

8.1.17 Theorem. Let R be left Noetherian and semi-perfect with Jacobson radicalJ, and set k = R/J. Let M be an R-complex in Df

A(R) and let n be an integer. Thefollowing conditions are equivalent.

(i) pdR M 6 n.(ii) n> supM and Extn+1

R (M,k) = 0.(iii) n> supM and TorR

n+1(k,M) = 0.In particular one has,

pdR M =− infRHomR(M,k) = sup(k⊗LR M) .

PROOF. Choose by B.59 a semi-projective resolution P '−→M with P minimal anddegreewise finitely generated. The graded module P\ is semi-perfect by B.44, andfrom this point the proof of 8.1.16 applies verbatim.

THE CASE OF MODULES

8.1.18. Notice from 8.1.9 that a non-zero R-module is projective if and only if it hasprojective dimension 0 as an R-complex.

8.1.19 Theorem. Let M be an R-module and let n> 0 be an integer. The followingconditions are equivalent.

(i) pdR M 6 n.(ii) One has ExtmR (M,N) = 0 for every R-module N and every integer m > n.

(iii) One has Extn+1R (M,N) = 0 for every R-module N.

(iv) There is a projective resolution 0→ Pn→ Pn−1→ ·· · → P0→M→ 0.(v) In every projective resolution · · · → Pv → Pv−1 → ··· → P0 → M → 0 the

module Coker(Pn+1→ Pn) is projective.In particular, there is an equality

pdR M = supm ∈ N0 | ExtmR (M,N) 6= 0 for some R-module N .



PROOF. In a projective resolution · · · → Pv → Pv−1 → ·· · → P0 → M → 0, themorphism P0 M yields a semi-projective resolution of M, considered as an R-complex; cf. 5.2.27. In particular, the complex · · · → Pv→ Pv−1→ ···→ P0→ 0 is asemi-projective replacement of M. Conversely, a semi-projective resolution P '−→Mwith Pv = 0 for all v < 0 yields a projective resolution of M. The equivalence of thefive conditions now follows from the equivalence of (i)–(iii), (v), and (vii) in 8.1.9.The final equality follows from the equivalence of (i) and (ii).

REMARK. Let A 6= 0 be a Z-module; it follows from 1.3.10 that pdZA is 0 or 1. If pdZA = 0,which by 1.3.17 means that A is free, then Ext1Z(A,Z) = 0 holds. By E 7.3.4 the converse is true ifA is finitely generated and, more generally, Stein [81] has shown that every countably generated Z-module A with Ext1Z(A,Z) = 0 is free. The question whether every Z-module A with Ext1Z(A,Z) = 0is free is known as Whitehead’s problem; Shelah [77] has shown that it is undecidable within ZFC.

8.1.20 Theorem. Let R be left Noetherian. Let M be a finitely generated R-moduleand let n> 0 be an integer. The following conditions are equivalent.

(i) pdR M 6 n.(ii) One has Extn+1

R (M,R/a) = 0 for every left ideal a in R.(iii) There is a projective resolution 0→ Pn → Pn−1 → ··· → P0 → M→ 0 with

each module Pv finitely generated.In particular, there is an equality

pdR M = supm ∈ N0 | ExtmR (M,R/a) 6= 0 for some left ideal a in R .PROOF. An argument parallel to the proof of 8.1.19 shows that the equivalence ofthe three conditions follows from 8.1.14 and that the final equality holds.

EXERCISES

E 8.1.1 Let R→ S be a ring homomorphism. Show that if S is a projective R-module, then onehas pdR N 6 pdS N for every S-complex N.

E 8.1.2 Let R be semi-simple. Show that one has pdR M = supM for every R-complex M.E 8.1.3 Let M 6' 0 be a complex in D@(R) and set w = supM. Show that for every semi-

projective replacement P of M one has pdR M = w+pdR Cw(P).E 8.1.4 Let M be an R-complex. Show that pdR M is finite if and only if RHomR(M,N) is a

complex with bounded below homology for every R-module N.E 8.1.5 Show that the R-complexes of finite projective dimension form a triangulated subcate-

gory of D(R).E 8.1.6 Let M be an R-complex and assume that it is isomorphic in D(R) to a K-projective

complex X with Xv = 0 for all v > n. Show that pdR M is at most n, and conclude thatone could use K-projective replacements in 8.1.2.

E 8.1.7 Let R be left hereditary. Show that one has pdR M 6 supM+1 for every R-complex M.E 8.1.8 Show that every finitely generated R-module has finite projective dimension if and only

if every left ideal in R has finite projective dimension. Hint: Proof of A.1.E 8.1.9 Let M be an R-module of finite projective dimension n. (a) Show that there exists

a projective R-module P with ExtnR(M,P) 6= 0. (b) Show that if R is left Noether-ian and M is finitely generated, then one has ExtnR(M,R) 6= 0. (c) Show that one hasExtnZ/4Z(2Z/4Z,Z/4Z) = 0 for all n> 1 and compare to 8.1.11.


8.2 Injective Dimension 301

E 8.1.10 Show that a finitely generated Z/4Z-module M is projective if and only if one hasExt1Z/4Z(M,Z/2Z) = 0.

E 8.1.11 Show that the full subcategory of R-complexes of finite projective dimension is a trian-gulated subcategory of D(R).

8.2 Injective Dimension

SYNOPSIS. Vanishing of Ext; semi-injective replacement; injective dimension; Schanuel’s lemma;minimal semi-injective resolution; injective dimension over Artinian ring;∼ of module; coproductof injective modules.

This sesstion develops, initially, in close parallel to the previous one.

8.2.1 Lemma. For an R-module I the following conditions are equivalent.(i) I is injective.

(ii) infRHomR(M, I)>−supM holds for every R-complex M.(iii) One has Ext1R(R/a, I) = 0 for every left ideal a in R.(iv) There exists an injective homomorphism ι : I→ E with E injective and with

Ext1R(Coker ι, I) = 0.

PROOF. If I is injective, then 1I is a semi-injective resolution of I, and one hasRHomR(M, I) ' HomR(M, I) in D(k). Now 2.5.6(b) yields infRHomR(M, I) >−supM. Thus, (i) implies (ii) which, in particular, implies that the functor Ext1R(–, I)is zero on modules. Hence (ii) implies (iii) and, in view of 5.3.4, also (iv).

(iii)=⇒ (i): Consider the short exact sequence 0→ a→ R→ R/a→ 0. It yieldsby 7.3.14 an exact sequence HomR(R, I)→ HomR(a, I)→ Ext1R(R/a, I) = 0. Thus,every homomorphism a→ I is the restriction of a homomorphism R→ I, whence Iis injective by Baer’s criterion 1.3.26.

(iv)=⇒ (i): The exact sequence 0→ I→ E→ Coker ι→ 0 is split by 7.3.15, so Iis a direct summand of an injective module and hence injective by 1.3.23.

SEMI-INJECTIVE REPLACEMENTS AND INJECTIVE DIMENSION

8.2.2 Definition. Let M be an R-complex. A semi-injective R-complex I that isisomorphic to M in D(R) is called a semi-injective replacement of M.

The injective dimension of M, written idR M, is defined as

idR M = inf

n ∈ Z∣∣∣∣ There is a semi-injective replacement

I of M with Iv = 0 for all v <−n

,

with the convention inf∅= ∞. One says that idR M is finite if idR M < ∞ holds.

REMARK. The minus sign in the definition above appears because we use homological notation;if one is concerned, primarily, with semi-injective replacements, then cohomological notation is amore natural choice.



8.2.3. Let M be an R-complex. For every semi-injective replacement I of M one hasH(I)∼= H(M); the next (in)equalities are hence immediate from the definition,

idR M > − infM and idRΣs M = idR M− s for every s ∈ Z .

Moreover, one has idR M =−∞ if and only if M is acyclic.Note that the definition of injective dimension could also be written

(8.2.3.1) idR M = inf− inf I\ | I is a semi-injective replacement of M .

8.2.4. Let M be an R-complex. For every semi-injective replacement I of M it fol-lows from 6.4.21 that there is semi-injective resolution M '−→ I.


idS RHomR(S,M)6 idR M .

PROOF. For every semi-injective replacement I of the R-complex M, the S-complexHomR(S, I) is a semi-injective replacement of RHomR(S,M) by 5.4.20. As one hasinfHomR(S, I)\ > inf I\, the desired inequality follows.

8.2.6 Lemma. Let M be an R-complex and let I be a semi-injective replacementof M. For every R-module N and for all integers m > 0 and n > − infM there is anisomorphism

Extn+mR (N,M) ∼= ExtmR (N,Z−n(I)) .

PROOF. Recall from 5.3.12 that Iď−n is a semi-injective R-complex. As one hasn>− infM =− inf I, the canonical morphism Z−n(I) Σn Iď−n is a semi-injectiveresolution. In the next computation, the final and the first two identities follow fromthe definitions of RHom and Hom; the fourth follows from 2.3.15.

H−(n+m)(RHomR(N,M)) ∼= H−(n+m)(HomR(N, I))

= H−(n+m)(HomR(N, Iď−n))

= H−m(Σn HomR(N, Iď−n))

∼= H−m(HomR(N,Σn Iď−n))∼= H−m(RHomR(N,Z−n(I))) ;

the definition of Ext, 7.3.7, now yields the asserted isomorphism.

8.2.7 Lemma. Let I be a semi-injective R-complex and let n be an integer. Thecomplex IĚn is semi-injective if and only if the module Zn(I) is injective.

PROOF. If the complex IĚn is semi-injective, then the module Zn(I) = (IĚn)n isinjective. If Zn(I) is an injective module, then it is semi-injective as a complex by5.3.18. There is an exact sequence 0→ Σn Zn(I)→ IĚn→ Iěn+1→ 0, so by 5.3.20the complex IĚn is semi-injective if and only if Iěn+1 is semi-injective, which it isby 5.3.12 and 5.3.20 applied to the exact sequence 0→ Iďn→ I→ Iěn+1→ 0.




(i) idR M 6 n.(ii) − infRHomR(N,M)6 n+ supN holds for every R-complex N.

(iii) n>− infM and Extn+1R (R/a,M) = 0 holds for every left ideal a in R.

(iv) n>− infM and one has Ext1R(Z−(n+1)(I),Z−n(I)) = 0 for some, equivalentlyevery, semi-injective replacement I of M.

(v) n>− infM and for some, equivalently every, semi-injective replacement I ofM, the module Z−n(I) is injective.

(vi) n>− infM and for every semi-injective replacement I of M, there is a semi-injective resolution M '−→ IĚ−n.

(vii) There is a semi-injective resolution M '−→ I with Iv = 0 for all v <−n and forall v > supM.

In particular, there are equalities

idR M = sup− infRHomR(R/a,M) | a is a left ideal in R= supm ∈ Z | ExtmR (R/a,M) 6= 0 for some left ideal a in R .

PROOF. The implication (vii)=⇒ (i) is trivial.(i)=⇒ (ii): One can assume that N is in D@(R) and not acyclic; otherwise the

inequality is trivial. Set w = supN; it is an integer and there is an isomorphismRHomR(N,M)' RHomR(NĎw,M); see 4.2.4. Choose a semi-injective replacementI of M with Iv = 0 for all v < −n; by 8.2.4 it yields a semi-injective resolutionM '−→ I. Thus, one has infRHomR(N,M) = infHomR(NĎw, I). For v <−n−w andp ∈ Z, one of the inequalities p > w or p+ v6 w+ v <−n holds, so the module

HomR(NĎw, I)v =∏

p∈ZHomR((NĎw)p, Ip+v)

is zero. In particular, one has Hv(HomR(NĎw, I)) = 0 for v <−n−w, so the inequal-ity infRHomR(M,N)>−n−w holds as desired.

(ii)=⇒ (iii): The second assertion is immediate, and an application of (ii) withN = R yields − infM =− infRHomR(R,M)6 n.

(iii)=⇒ (v): Let I be a semi-injective replacement of M. It follows from 8.2.6that one has Ext1R(R/a,Z−n(I)) = 0 for every left ideal a in R, whence Z−n(I) is aninjective module by 8.2.1.

(v)=⇒ (iv): Immediate from 8.2.1.(iv)=⇒ (i) and (iv)=⇒ (vi): Let I be a semi-injective replacement of M. As one

has −n6 infM = inf I, the sequence 0→ Z−n(I)→ I−n→ Z−(n+1)(I)→ 0 is exact,and it follows from 8.2.1 that the module Z−n(I) is injective. Further, the canonicalmorphism τI

Ě−n : IĚ−n I is a quasi-isomorphism, so in D(R) there is an isomor-phism M ' IĚ−n, and the right-hand complex is semi-injective by 8.2.7. Now itfollows from 8.2.4 that there is quasi-isomorphism M→ IĚ−n.

This argument shows that the “some” part of (v) implies (i), and thus (i)–(v) areequivalent. The argument also shows that the “every” part of (v) implies (vi).

(vi)=⇒ (vii): Choose by 5.3.26 a semi-injective resolution M '−→ I, wit Iv = 0 forall v > supM. In particular, I is a semi-injective replacement of M, so by (vi) thereis a quasi-isomorphism M→ IĚ−n, which is the desired resolution.




The next corollary implies, in view of 6.5.19, that if two out of three complexesin a short exact sequence have finite injective dimension, then so has the third.

8.2.9 Corollary. Let M′ → M→ M′′ → ΣM′ be a distinguished triangle in D(R).With ı′ = idR M′, ı= idR M, and ı′′ = idR M′′ there are inequalities,

ı′ 6maxı, ı′′+1 , ı6maxı′, ı′′ , and ı′′ 6maxı′−1, ı .

In particular, if two of the complexes in the triangle have finite injective dimension,then so has the third.


Over a semi-simple ring every module is injective, see 1.3.24; in particular such aring is self-injective. As the example below illustrates, modules over a self-injectivering can have infinite injective dimension.

8.2.10 Proposition. If k is a principal ideal domain and a is a proper non-zero idealin k, then the quotient ring k/a is self-injective.

PROOF. We use Baer’s criterion 1.3.26. Choose an element a 6= 0 in k with a= (a).Every ideal in k/(a) has the form (b)/(a) with a = bc for some c ∈ k. Given ahomomorphism ϕ : (b)/(a)→ k/(a) of k/(a)-modules, one has ϕ([b](a)) = [x](a)for some x ∈ k. As bc = a we get cx ∈ (a), that is, cx = ay for some y ∈ k. Henceax = bcx = bay and thus x = by. So ϕ′ : k/(a)→ k/(a) given by [r](a) 7→ [ry](a) is ahomomorphism of k/(a)-modules whose restriction to (b)/(a) is ϕ.

8.2.11 Example. The ring Z/4Z is self-injective by 8.2.10. Hence the complex

I = 0−→ Z/4Z 2−−→ Z/4Z 2−−→ Z/4Z−→ ·· · ,

concentrated in non-positive degrees, is a semi-injective replacement of the Z/4Z-module 2Z/4Z. As Z/4Z is indecomposable as a Z/4Z-module, no kernel Z−n(I) =2Z/4Z for n> 0 is injective, so one has idZ/4Z(2Z/4Z) = ∞.


idR(∏

u∈U

Mu) = supu∈UidR Mu .

PROOF. Let N be an R-module. The functor RHomR(N,–) preserves products by3.1.23 and 7.2.12, and so does homology by 3.1.22, whence one has

− inf(RHomR(N,∏

u∈UMu)) = − inf(

∏u∈U

RHomR(N,Mu))

= supu∈U− infRHomR(N,Mu) .


The following result—often called Schanuel’s lemma (for semi-injective com-plexes)—can be seen as a refinement of the statement 8.2.8(v).



8.2.13 Proposition. Let M be an R-complex and let I and I′ be semi-injective re-placements of M. For every v ∈ Z there exist injective R-modules E and E ′ withZv(I)⊕E ∼= Zv(I′)⊕E ′.

PROOF. By 6.4.21 and 5.3.24 there is a homotopy equivalence α : I→ I′, and by4.3.30 the complex Coneα is contractible. Furthermore, it consists of injective mod-ules, so 4.3.32 shows that every module Cn(Coneα) is injective. By 4.3.21 the mor-phism αĚv : IĚv→ I′Ěv is also a homotopy equivalence so Cone(αĚv), that is,

· · · −→ I′v+3⊕ Iv+2∂Coneα

v+3−−−−→ I′v+2⊕ Iv+1 −→ I′v+1⊕Zv(I)−→ Zv(I′)−→ 0 ,

is contractible. Hence one has Zv(I′)⊕Cv+2(Coneα)∼= I′v+1⊕Zv(I).

MINIMALITY AND ARTINIAN RINGS

By B.24 every complex has a minimal semi-injective resolution.

8.2.14 Proposition. Let M be an R-complex and let n be an integer. If M '−→ I is aminimal semi-injective resolution, then the following conditions are equivalent.

(i) idR M 6 n.(ii) n>− infM and I−(n+1) = 0.

(iii) Iv = 0 for all v <−n.In particular, one has idR M =− inf I\.

PROOF. If one has idR M 6 n, then there is a semi-injective replacement I′ of Mwith I′v = 0 for all v <−n. It follows from 6.4.21 that there is a quasi-isomorphismI→ I′, and by B.21 it has a left inverse, whence one has Iv = 0 for v <−n. Thus (i)implies (iii) which, in turn, implies (ii). If one has n>− infM and I−(n+1) = 0, thenthe module Z−n(I) = I−n is injective, whence idR M 6 n holds by 8.2.8.

Existence of minimal resolutions and structure theory for injective modules overArtinian rings yields a result akin to 8.1.16.

8.2.15 Theorem. Let R be left Artinian with Jacobson radical J and set k = R/J.Let M be an R-complex and let n be an integer. The next conditions are equivalent.

(i) idR M 6 n.(ii) n>− infM and Extn+1

R (k,M) 6= 0.In particular, there is an equality

idR M =− infRHomR(k,M) .

PROOF. Choose by B.24 a semi-injective resolution M '−→ I with I minimal. Firstwe argue that the complex HomR(k, I) has zero differential. As k is semi-simple, itsuffices to argue that HomR(N, I) has zero differential for every simple module N.Suppose ∂HomR(N,I) is non-zero for some simple R-module N. For some n ∈ Z thereis then a homomorphism α : N→ In with ∂I

nα 6= 0. In particular, α is non-zero, andsince N is simple it follows that α and ∂I

nα are injective. Let A be the complex



0 −→ N =−→ N −→ 0 concentrated in degrees n and n− 1; there is now an exactsequence of R-complexes, 0→ A→ I → I/A→ 0, induced by α. The complex Ais acyclic, so the morphism π : I→ I/A is a quasi-isomorphism by 4.2.6. Since I isminimal and semi-injective, it follows from B.21 that π has a left inverse, which isabsurd in view of B.19. Thus, HomR(k, I) has zero differential; and hence

Hv(RHomR(k,M)) = HomR(k, I)v = HomR(k, Iv)

holds for every v ∈ Z. For every maximal left ideal m in R there is a non-zero homo-morphism k R/m, so it follows from F.4 and F.6 that HomR(k, Iv) is non-zero forevery Iv 6= 0. In particular, one has Hv(RHomR(k,M)) = 0 if and only if Iv = 0. Now8.2.14 yields the equivalence of (i) and (ii), and from here the displayed equality isimmediate.

THE CASE OF MODULES

8.2.16. Notice from 8.2.8 that a non-zero R-module is injective if and only if it hasinjective dimension 0 as an R-complex.


(i) idR M 6 n.(ii) One has ExtmR (N,M) = 0 for every R-module N and every integer m > n.

(iii) One has Extn+1R (R/a,M) = 0 for every left ideal a in R.

(iv) There is an injective resolution 0→M→ I0→ ··· → I−(n−1)→ I−n→ 0.(v) In every injective resolution 0→ M → I0 → ··· → I−(v−1) → I−v → ··· the

module Ker(I−n→ I−(n+1)) is injective.In particular, there is an equality

idR M = supm ∈ Z | ExtmR (R/a,M) 6= 0 for some left ideal a in R .PROOF. In an injective resolution 0→ M → I0 → ··· → I−(v−1) → I−v → ··· , themorphism M I0 yields a semi-injective resolution of M, considered as an R-complex; in particular, the complex 0→ I0→ ·· · → I−(v−1)→ I−v→ ·· · is a semi-injective replacement of M. Conversely, a semi-injective resolution M '−→ I withIv = 0 for all v > 0 yields an injective resolution of M. The equivalence of the fiveconditions now follows from the equivalence of (i)–(iii), (v), and (vii) in 8.2.8. Thefinal equality follows from the equivalence of (i) and (ii).

COPRODUCTS OF INJECTIVE MODULES

8.2.18 Theorem. The following conditions are equivalent.(i) R is left Noetherian.

(ii) For every U-direct system µvu : Iu→ Ivu∈U of injective R-modules with Ufiltered, the colimit colimu∈U Iu is injective.



(iii) For every countable family Iuu∈N of injective R-modules the coproduct∐u∈U Iu is injective.

PROOF. The implication (ii) =⇒ (iii) is evident; cf. 3.2.31.(i)=⇒ (ii): Let a be a left ideal in R, and let ι denote the embedding a R. There

is a commutative diagram in M(k),

colimu∈U

HomR(R, Iu)colimu∈U Hom(ι,Iu)

//

∼=

colimu∈U

HomR(a, Iu)

∼=

HomR(R,colimu∈U

Iu)Hom(ι,colimu∈U Iu)

// HomR(a,colimu∈U

Iu) ,

where the vertical maps are the canonical homomorphisms (3.2.36.1). As R is leftNoetherian, a is a finitely generated R-module, whence the vertical maps are iso-morphisms by 3.2.36. For every u in U the homomorphism HomR(ι, Iu) is surjec-tive, see 1.3.21, so the upper horizontal map is surjective by 3.2.11, and hence so isHomR(ι,colimu∈U Iu). Now it follows from Baer’s criterion 1.3.26 that the R-modulecolimu∈U Iu is injective.

(iii)=⇒ (i): Let a1 ⊆ a2 ⊆ ·· · be a chain of left ideals in R and set a=⋃

u∈N au.For every u ∈ N choose by 5.3.4 an injective homomorphism R/au → Iu with Iu

injective and identify R/au with its image in Iu. Set I =∐

u∈N Iu; the assignmenta 7→ ([a]au)u∈N defines a homomorphism α : a→ I, as one has a ∈ au for u 0. By(iii) the module I is injective, so α lifts to a homomorphism R→ I; in particular,there is an element i = (iu)u∈N in I such that α(a) = ai holds for every a ∈ R. Onehas iu = 0 for u 0 and, therefore, au = a for u 0.

8.2.19 Corollary. Let R be left Noetherian. For every family of R-modules Muu∈Uthere is an equality,

idR(∐

u∈U

Mu) = supu∈UidR Mu .

PROOF. For every u ∈U the module Mu is a direct summand of∐

u∈U Mu, so theinequality “>” holds by 8.2.12. To prove the opposite inequality, consider for everyu in U the minimal injective resolution Mu '−→ Iu. There is a quasi-isomorphism∐

u∈U Mu '−→∐

u∈U Iu, and the complex∐

u∈U Iu is semi-injective by 8.2.18 and5.3.12. The inequality now follows from 8.2.14 and (8.2.3.1).

REMARK. The proof of 8.2.19 applies to a family Muu∈U of complexes that are uniformlybounded above, see E 8.2.10, and that is the end of its range, since a coproduct of semi-injectivecomplexes need not be semi-injective. Indeed, Iacob and Iyengar [38] show that all coproducts ofsemi-injective R-complexes are semi-injective only if R is left Noetherian and regular in the sensethat every finitely generated R-module has finite projective dimension.

EXERCISES

E 8.2.1 Show that an R-module M is injective if Ext1R(R/a,M) = 0 holds for every essential leftideal a⊂ R. ([83])



E 8.2.2 Let M be an R-module with pdR M > n > 0. Show that ExtnR(M,–) : M(R)→M(k) ishalf exact but neither left nor right exact.

E 8.2.3 Let R→ S be a ring homomorphism. Show that if S is a flat Ro-module, then the inequal-ity idR N 6 idS N holds for every S-complex N.

E 8.2.4 Let R be semi-simple. Show that idR M =− infM holds for every R-complex M.E 8.2.5 Let M 6' 0 be a complex in DA(R) and set u = infM. Show that for every semi-injective

replacement I of M one has idR M = idR Zu(I)−u.E 8.2.6 Let M be an R-complex. Show that idR M is finite if and only if the homology of the

complex RHomR(N,M) is bounded below for every R-module N.E 8.2.7 Show that the R-complexes of finite injective dimension form a triangulated subcategory

of D(R).E 8.2.8 Let M be an R-complex and assume that it is isomorphic in D(R) to a K-injective com-

plex Y with Yv = 0 for all v < −n. Show that idR M is at most n, and conclude that onecould use K-injective replacements in 8.2.2.

E 8.2.9 Let R be left hereditary. Show that the inequality idR M 6 − infM + 1 holds for everyR-complex M.

E 8.2.10 Let R be left Noetherian and let Muu∈U be a family of R-complexes. Show that ifsupu∈UsupMu< ∞ holds, then one has idR

(∐u∈U Mu)= supu∈UidR Mu.

E 8.2.11 Let R be left Noetherian and assume that every left ideal in R has finite projective di-mension. Show that for every family Muu∈U of R-complexes the following equalityholds, idR

(∐u∈U Mu)= supu∈UidR Mu.

E 8.2.12 Let M be an R-module of finite injective dimension n. (a) Show that there exists an injec-tive R-module I with ExtnR(I,M) 6= 0. (b) Show that if R is commutative and Noetherian,then there is a prime ideal p in R with ExtnR(ER(R/p),M) 6= 0.

E 8.2.13 Let R be the ring of 2×2 matrices over a field. (a) Show that R is the direct sum of thetwo maximal left ideals m and n made up of matrices with zeroes in the first and secondcolumn, respectively. (b) Show that the simple R-modules R/m and R/n are isomorphic.

E 8.2.14 Show that the full subcategory of R-complexes of finite injective dimension is a triangu-lated subcategory of D(R).

8.3 Flat Dimension

SYNOPSIS. Vanishing of Tor; semi-flat replacement, flat dimension; ∼ vs. injective dimension;Schanuel’s lemma; flat dimension over Noetherian ring;∼ over perfect ring;∼ of module; productof flat modules.

8.3.1 Lemma. Let M be an R-complex, N be an Ro-complex, and m be an integer.There are isomorphisms of k-modules,

ExtmR (M,Homk(N,E)) ∼= Homk(TorRm(N,M),E) ∼= ExtmRo(N,Homk(M,E)) .

PROOF. As E is an injective k-module one has RHomk(–,E) = Homk(–,E). Thefirst and last isomorphisms below follow from the definitions, 7.3.7 and 7.4.8; thesecond isomorphism is adjunction 7.5.15, and the third one holds by 2.2.18.

ExtmR (M,Homk(N,E)) ∼= H−m(RHomR(M,Homk(N,E)))

∼= H−m(Homk(N⊗LR M,E))

∼= Homk(Hm(N⊗LR M),E)


8.3 Flat Dimension 309

∼= Homk(TorRm(N,M),E) .

This proves the first of the asserted isomorphisms; the other is proved similarly.

Recall from 7.4.4 that a semi-flat replacement of an R-complex M is a semi-flatR-complex F that is isomorphic to M in D(R).

8.3.2 Lemma. For an R-module F the following conditions are equivalent.(i) F is flat.

(ii) sup(N⊗LR F)6 supN holds for every Ro-complex N.

(iii) One has TorR1 (R/b,F) = 0 for every finitely generated right ideal b in R.

(iv) There exists a surjective homomorphism π : L→ F with L flat and such thatTorR

1 (Homk(Kerπ,E),F) = 0 holds.

PROOF. If F is flat, then F is a semi-flat replacement of itself, and by 7.4.5 one hasN⊗L

R F ' N⊗R F in D(k). Now 2.5.6(c) yields sup(N⊗LR F) 6 supN, whence (i)

implies (ii). From (ii) it follows that the functor TorR1 (–,F) is zero on modules, and

therefore (ii) implies (iii) and, in view of 1.3.11, (ii) implies (iv) as well.(iii)=⇒ (i): Let b be a right ideal in R; by 3.2.27 it is the filtered colimit of its

finitely generated right subideals. For each such subideal b′ the embedding b′ Ryields an exact sequence TorR

1 (R/b′,F)→ b′⊗R F→ R⊗R F ; see 7.4.13. From (iii)

it follows that the morphism b′⊗R F → R⊗R F is injective, so by exactness of fil-tered colimits 3.2.32, and the fact that –⊗R F preserves colimits 3.2.16, the mor-phism b⊗R F → R⊗R F is injective. Thus, F is flat by 1.3.43.

(iv)=⇒ (i): Set K = Kerπ; there is an exact sequence of Ro-modules,

0−→ Homk(F,E)−→ Homk(L,E)−→ Homk(K,E)−→ 0 ,

where Homk(L,E) is injective by 1.3.43. By 8.3.1 one has

Ext1Ro(Homk(K,E),Homk(F,E)) ∼= Homk(TorR1 (Homk(K,E),F),E) .

It now follows from (iv) and 8.2.1 that the Ro-module Homk(F,E) is injective,whence F is flat by 1.3.43.

SEMI-FLAT REPLACEMENTS AND FLAT DIMENSION

8.3.3 Definition. Let M be an R-complex. The flat dimension of M, written fdR M,is defined as

fdR M = inf

n ∈ Z

∣∣∣∣∣ There is a semi-flat replacementF of M with Fv = 0 for all v > n

,

with the convention inf∅= ∞. One says that fdR M is finite if fdR M < ∞ holds.

8.3.4. Let M be an R-complex. For every semi-flat replacement F of M one hasH(F)∼= H(M); the next (in)equalities are hence immediate from the definition,

fdR M > supM and fdRΣs M = fdR M+ s for every s ∈ Z .



Moreover, one has fdR M =−∞ if and only if M is acyclic.Note that the definition of flat dimension could also be written

(8.3.4.1) fdR M = infsupF\ | F is a semi-flat replacement of M .

While projective and injective dimension could be defined using resolutions—by 6.4.20 and 6.4.21replacements and resolutions are one and the same thing—this is not the case for the flat dimension. The

next example illustrates this and shows, in the process, that semi-flat complexes may not allow comparison

maps. This, again, contrasts sharply with 6.4.20 and 6.4.21.

8.3.5 Example. As Q is a flat Z-module, see 1.3.38, one has fdZQ= 0 by 8.3.4.Let L '−→ Q be a semi-free resolution with Lv = 0 for v 6= 0,1; cf. 5.4.15. One

has fdZL = fdZQ = 0, but there does not exist a quasi-isomorphism F → L, whereF is a (semi-flat) complex with Fv = 0 for v > 0. Indeed, such a quasi-isomorphismwould induce an injective homomorphism Z0(F)→Z0(L), but that would contradict1.3.10 as Z0(L) = L0 is free while Z0(F)∼= Q is not.

8.3.6. Let M be an R-complex. Every semi-projective replacement P of M is a semi-flat replacement by 5.4.10, so there is an inequality

fdR M 6 pdR M .

8.3.7 Example. The inequality in 8.3.6 may be strict. Indeed, fdZQ = 0 holds by8.3.5, while 8.1.6 yields pdZQ= 1.

8.3.8 Proposition. Assume that every flat R-module has finite projective dimensionand let M be an R-complex. If fdR M is finite, then pdR M is finite.

PROOF. One can assume that M is not acyclic. Set n = fdR M and w = supM; bothare integers by assumption and 8.3.4. Let P be a semi-projective replacement of M.Note that Σ−w Pěw is a semi-projective replacement of the module Cw(P) and thatpdR M = w+ pdR Cw(P) holds by 8.1.9. The module Cn(P) is flat by 8.3.12, so ithas finite projective dimension by assumption. It follows that also Cw(P) has finiteprojective dimension, whence pdR M = w+pdR Cw(P) is finite.


fdS(S⊗LR M)6 fdR M .

PROOF. For every semi-flat replacement F of the R-complex M, the S-complexS⊗R F is a semi-flat replacement of S⊗L

R M by 5.4.18. As sup(S⊗R F)\ 6 supF\

holds, the desired inequality follows.

8.3.10 Lemma. Let M be an R-complex and let F be a semi-flat replacement ofM. For every Ro-module N and for all integers m > 0 and n > supM there is anisomorphism,

TorRn+m(N,M) ∼= TorR

m(N,Cn(F)) .



PROOF. Recall from 5.4.8 that Fěn is a semi-flat R-complex. As one has n >supM = supF , the canonical morphism Σ−n Fěn Cn(F) is a quasi-isomorphism;in particular, Σ−n Fěn is a semi-flat replacement of Cn(F). In the next computation,the final and the first two identities follow from the definitions of (derived) tensorproducts; the fourth follows from 2.4.12.

Hn+m(N⊗LR M) ∼= Hn+m(N⊗R F)

= Hn+m(N⊗R Fěn)

= Hm(Σ−n (N⊗R Fěn))

∼= Hm(N⊗R Σ−n Fěn)

∼= Hm(N⊗LR Cn(F)) ;

the definition of Tor, 7.4.8, now yields the asserted isomorphism.

8.3.11 Lemma. Let F be a semi-flat R-complex and let n be an integer. The complexFĎn is semi-flat if and only if the module Cn(F) is flat.

PROOF. If the complex FĎn is semi-flat, then the module Cn(F) = (FĎn)n is flat. Ifthe module Cn(F) is flat, then it is semi-flat as a complex by 5.4.11. There is anexact sequence 0→ Fďn−1→ FĎn→ Σn Cn(F)→ 0, so by 5.4.12 the complex FĎnis semi-flat if and only if Fďn−1 is semi-flat, which it is by 5.4.8 and 5.4.12 appliedto the exact sequence 0→ Fďn−1→ F → Fěn→ 0.


(i) fdR M 6 n.(ii) sup(N⊗L

R M)6 n+ supN holds for every Ro-complex N.(iii) n > supM and TorR

n+1(R/b,M) = 0 holds for every finitely generated rightideal b in R.

(iv) n> supM and one has TorR1 (Homk(Cn+1(F),E),Cn(F)) = 0 for some, equiv-

alently every, semi-flat replacement F of M.(v) n> supM and for some, equivalently every, semi-flat replacement F of M the

module Cn(F) is flat.(vi) n > supM and for every semi-flat replacement F of M the complex FĎn is a

semi-flat replacement of M.(vii) There exists a semi-flat replacement F of M with Fv = 0 for all v > n and for

all v < infM.In particular, there are equalities

fdR M = supsup(R/b⊗LR M) | b is a finitely generated right ideal in R

= supm ∈ Z | TorRm(R/b,M) 6= 0 for a finitely generated right ideal b in R .

PROOF. The implication (vii)=⇒ (i) is trivial.(i)=⇒ (ii): One can assume that N is in D@(Ro) and not acyclic; otherwise the

inequality is trivial. Set w = supN; it is an integer and there is an isomorphismN⊗L

R M 'NĎw⊗LR M; see 4.2.4. Choose a semi-flat replacement F of M with Fv = 0



for all v> n. Now one has sup(N⊗LR M) = sup(NĎw⊗R F). For v> n+w and p∈Z,

one of the inequalities p > w or v− p> v−w > n holds, so the module

(NĎw⊗R F)v =∐

p∈Z(NĎw)p⊗R Fv−p

is zero. In particular, one has Hv(NĎw⊗R F) = 0 for v > n+w, so the desired in-equality sup(N⊗L

R M)6 n+w holds.(ii)=⇒ (iii): Apply (ii) to the module N = R to get supM = sup(R⊗R M) 6 n;

the second assertion is immediate.(iii)=⇒ (v): Let F be a semi-flat replacement of M. It follows from 8.3.10 that

one has TorR1 (R/b,Cn(F)) = 0 for every finitely generated right ideal b in R, whence

Cn(F) is flat by 8.3.2.(v)=⇒ (iv): Immediate from 8.3.2.(iv)=⇒ (i) and (iv)=⇒ (vi): Let F be a semi-flat replacement of M. As one has

n> supM = supF , the sequence 0→ Cn+1(F)→ Fn→ Cn(F)→ 0 is exact, and itfollows from 8.3.2 that the module Cn(F) is flat. Further, the canonical morphismτF

Ďn : F FĎn is a quasi-isomorphism, so M is isomorphic to FĎn in D(R), and thelatter complex is semi-flat by 8.3.11.

This argument shows that the “some” part of (iv) implies (i), and thus (i)–(v) areequivalent. The argument also shows that the “every” part of (iv) implies (vi).

(vi)=⇒ (vii): Choose by 5.1.12 a semi-free resolution L of M with Lv = 0 forv < infM. By 5.4.10 the complex L is semi-flat, so LĎn is the desired replacement.


The next corollary implies, in view of 6.5.19, that if two out of three complexesin a short exact sequence have finite flat dimension, then so has the third.

8.3.13 Corollary. Let M′→M→M′′→ ΣM′ be a distinguished triangle in D(R).With f ′ = fdR M′, f = fdR M, and f ′′ = fdR M′′ there are inequalities,

f ′ 6max f , f ′′−1 , f 6max f ′, f ′′ , and f ′′ 6max f ′+1, f .

In particular, if two of the complexes M′, M, and M′′ in the triangle have finite flatdimension, then so has the third.



fdR(∐

u∈U

Mu) = supu∈UfdR Mu .

PROOF. Let N be an Ro-module. The functor N⊗LR – preserves coproducts by 3.1.12

and 7.2.12, and so does homology by 3.1.10, whence one has

sup(N⊗LR∐

u∈UMu) = sup(

∐u∈U

(N⊗LR Mu)) = sup

u∈Usup(N⊗L

R Mu) .




Notice that 8.1.5, 8.2.5, and 8.3.9 are special cases of the next result. The assump-tion about non-acyclicity is only there to avoid potentially meaningless expressionslike ∞−∞ on the right-hand sides of the displayed inequalities.

8.3.15 Proposition. Let M be an R-complex, let X be a complex of R–So-bimodules,and let N be an S-complex. If none of the complexes M, X , and N are acyclic, thenthe following inequalities hold.

idS RHomR(X ,M) 6 fdSo X + idR M .(8.3.15.1)idSo RHomR(M,X) 6 idSo X +pdR M .(8.3.15.2)

fdR(X⊗LS N) 6 fdR X + fdS N .(8.3.15.3)

pdR(X⊗LS N) 6 pdR X +pdS N .(8.3.15.4)

PROOF. For every left ideal a in S, adjunction 7.5.15, 8.2.8, and 8.3.12 yield:

− infRHomS(S/a,RHomR(X ,M)) = − infRHomR(X⊗LS S/a,M)

6 idR M+ sup(X⊗LS S/a)

6 idR M+ fdSo X .

The inequality (8.3.15.1) now follows from another application of 8.2.8.For every right ideal b in S, swap 7.5.11, 8.1.9, and 8.2.8 yield:

− infRHomSo(S/b,RHomR(M,X)) = − infRHomR(M,RHomSo(S/b,X))

6 pdR M− infRHomSo(S/b,X)

6 pdR M+ idSo X .

The inequality (8.3.15.2) now follows from another application of 8.2.8.For every right ideal b in R, associativity 7.5.7 and 8.3.12 yield:

sup(R/b⊗LR (X⊗L

S N)) = sup((R/b⊗LR X)⊗L

S N)

6 fdS N + sup(R/b⊗LR X)

6 fdS N + fdR X .

The inequality (8.3.15.3) now follows from another application of 8.3.12.For every R-module C, adjunction 7.5.15 and 8.1.9 yield:

− infRHomR(X⊗LS N,C) = − infRHomS(N,RHomR(X ,C))

6 pdS N− infRHomR(X ,C)

6 pdS N +pdR X .

The inequality (8.3.15.4) now follows from another application of 8.1.9.

The next two results exhibit the flat–injective duality.

8.3.16 Proposition. For every R-complex M one has

fdR M = idRo Homk(M,E) .



PROOF. Let K be an Ro-module; by 2.5.6(b), commutativity 7.5.5, adjunction 7.5.15,and 7.2.9 one has,

sup(K⊗LR M) = − infRHomk(K⊗L

R M,E)

= − infRHomRo(K,RHomk(M,E))

= − infRHomRo(K,Homk(M,E)) ,

and the desired equality now follows from 8.2.8 and 8.3.12.

The next result, which could be called Schanuel’s lemma (for semi-flat com-plexes), can be seen as a refinement of the statement 8.3.12(v).

8.3.17 Proposition. Let M be an R-complex with a semi-flat replacement F anda semi-projective replacement P. For every v ∈ Z there is an exact sequence of R-modules 0→ Cv(P)→ Cv(F)⊕L→ G→ 0 where L is projective and G is flat.

PROOF. By 6.4.20 there is a quasi-isomorphism α : P→ F . Both complexes P andF are semi-flat, see 5.4.10, so Coneα is semi-flat and acylic by 4.1.5, 5.4.12, and4.2.15. By C.23 each module Zn(Coneα) is flat. The morphism αĎv : PĎv→ FĎv isa quasi-isomorphism by 4.2.9, and acyclicity of the complex Cone(αĎv):

0−→ Cv(P)−→ Cv(F)⊕Pv−1 −→ Fv−1⊕Pv−2∂Coneα

v−1−−−−→ Fv−2⊕Pv−3 −→ ·· · ,

yields the desired sequence 0→ Cv(P)→ Cv(F)⊕Pv−1→ Zv−1(Coneα)→ 0.

NOETHERIAN RINGS

8.3.18 Theorem. Let R be left Noetherian. For every complex M in D@(R) one has

fdRo Homk(M,E) = idR M .

PROOF. The equality is trivial if M is acyclic, so we assume that it is not and sets = supM. Let K be a cyclic R-module and choose by 5.1.14 a semi-projectiveresolution P '−→ K with P degreewise finitely generated and Pv = 0 for all v < 0. Inthe following chain of isomorphisms in D(k) the first holds by definition, the secondand fourth hold by 4.2.4, and the third is homomorphism evaluation 4.5.10(1,a),

K⊗LRo Homk(M,E) ' P⊗Ro Homk(M,E)

' P⊗Ro Homk(MĎs,E)∼= Homk(HomR(P,MĎs),E)

' Homk(HomR(P,M),E) .

Now 7.6.7 yields

sup(K⊗LRo Homk(M,E)) = − infHomR(P,M) = − infRHomR(K,M) ,

and the desired inequality follows from 8.2.8 and 8.3.12.



8.3.19 Theorem. Let R be left Noetherian. For every R-complex M in DfA(R) there

is an equality,fdR M = pdR M .

PROOF. The inequality fdR M 6 pdR M holds by 8.3.6. To prove the opposite in-equality, it is sufficient to consider the case where n = fdR M is an integer. Chooseby 5.1.14 a semi-free resolution L '−→M with L degreewise finitely generated. Thecomplex L is a semi-flat replacement of M by 5.4.10, so the module Cn(L) is flat by8.3.12 and hence projective by 1.3.42, as Cn(L) is finitely presented. Thus LĎn is asemi-projective replacement of M, whence one has n> pdR M.

PERFECT RINGS

For perfect rings, the equality in 8.3.19 extends to all complexes.

8.3.20 Theorem. Let R be left perfect with Jacobson radical J and set k = R/J. Forevery R-complex M there are equalities,

fdR M = sup(k⊗LR M) = pdR M .

PROOF. The inequality fdR M 6 pdR M holds by 8.3.6. From 8.1.16 and 8.3.12 onegets pdR M = sup(k⊗L

R M)6 fdR M.

THE CASE OF MODULES

8.3.21. Notice from 8.3.12 that a non-zero R-module is flat if and only if it has flatdimension 0 as an R-complex.

Semi-flat replacements subsume the classic notion of flat resolutions of modules.

8.3.22 Definition. Let M be an R-module. A flat resolution of M consists of a com-plex · · · → Fv→ Fv−1→ ·· · → F0→ 0 of flat R-modules and a surjective homomor-phism F0→M, such that the following sequence of R-modules is exact,

· · · −→ Fv −→ Fv−1 −→ ·· · −→ F0 −→M −→ 0 .

8.3.23. By 5.1.16 every R-module has a flat resolution. In every flat resolution of anR-module M, the complex · · · → Fv→ Fv−1→ ··· → F0→ 0 is a semi-flat replace-ment of M as an R-complex.

Conversely, let M be and R-module and let F be a semi-flat replacement of Mwith Fv = 0 for all v < 0. One then has Hv(F) = 0 for all v 6= 0, and there is anisomorphism H0(F)

∼=−→M. The complex F together with the composite homomor-phism F0 = Z0(F) H0(F)

∼=−→M now yields a flat resolution of M.


(i) fdR M 6 n.



(ii) One has TorRm(N,M) = 0 for every Ro-module N and every integer m > n.

(iii) One has TorRn+1(R/b,M) = 0 for every finitely generated right ideal b in R.

(iv) There is a flat resolution 0→ Fn→ Fn−1→ ··· → F0→M→ 0.(v) In every flat resolution · · · → Fv → Fv−1 → ·· · → F0 → M → 0 the module

Coker(Fn+1→ Fn) is flat.In particular, there is an equality

fdR M = supm∈N0 |TorRm(R/b,M) 6= 0 for a finitely generated right ideal b in R .

PROOF. In view of 8.3.23 the equivalence of the five conditions follows from theequivalence of (i)–(iii), (v), and (vii) in 8.3.12. The final equality follows from theequivalence of (i) and (ii).

PRODUCTS OF FLAT MODULES

8.3.25 Proposition. Let R be right Noetherian. For every family Fuu∈U of flatR-modules, the product

∏u∈U Fu is flat.

PROOF. Let b be a right ideal in R, and let ι denote the embedding b R. There isa commutative diagram in M(k),

b⊗R∏

u∈UFu ι⊗

∏u∈U Fu

//

∼=

R⊗R∏

u∈UFu

∼=∏

u∈U(b⊗R Fu)

∏u∈U (ι⊗Fu)

//∏

u∈U(R⊗R Fu) ,

where the vertical maps are the canonical homomorphisms (3.1.28.1). As R is rightNoetherian, b is a finitely presented Ro-module, whence the vertical maps are iso-morphisms by 3.1.28. For every u in U the homomorphism ι⊗R Fu is injective, see1.3.43, so the lower horizontal map is injective and hence so is β = ι⊗R

∏u∈U Fu.

By 7.4.13 there is an exact sequence,

0 = TorR1 (R,

∏u∈U

Fu)−→ TorR1 (R/b,

∏u∈U

Fu)−→ b⊗R∏

u∈UFu β−−→ R⊗R

∏u∈U

Fu ,

which shows that TorR1 (R/b,

∏u∈U Fu) is zero. Thus

∏u∈U Fu is flat by 8.3.2.

REMARK. The proof of 8.3.25 applies under the weaker assumption that R is right coherent. Infact, the property that products of flat modules are flat characterizes right coherent rings; this is aresult of Chase [17].

8.3.26 Corollary. Let R be right Noetherian. For every family Muu∈U of R-modules there is an equality,

fdR(∏

u∈U

Mu) = supu∈UfdR Mu .



PROOF. For every u ∈U the module Mu is a direct summand of∏

u∈U Mu, so theinequality “>” holds by 8.3.14. The opposite inequality is trivial if the supremumsupu∈UfdR Mu is infinite, so assume that it is not and call it s. By 8.3.12 eachmodule Mu has a semi-flat replacement Fu with Fu

v = 0 for all v > s and for allv < 0. There is an isomorphism

∏u∈U Fu ∼=

∏u∈U Mu in D(R), and the complex∏

u∈U Fu is semi-flat by 8.3.25 and 5.4.8 and concentrated in degrees v6 s.

REMARK. The proof of 8.3.26 also applies to a family of complexes that are uniformly boundedbelow, see E 8.3.9, but it reaches no further, as a product of semi-flat complexes need not be semi-flat. Indeed, Iacob and Iyengar [38] show that all products of semi-flat R-complexes are semi-flatonly if R is right coherent and every finitely presented Ro-module has finite projective dimension.

EXERCISES

E 8.3.1 Show that every R-module is flat if and only if R/b is flat for every finitely generatedright ideal b in R.

E 8.3.2 Let R→ S be a ring homomorphism. Show that if S is flat as an R-module, then one hasfdR N 6 fdS N for every S-complex N.

E 8.3.3 Let R be von Neumann regular. Show that fdR M = supM holds for every R-complex M.E 8.3.4 Let M 6' 0 be a complex in D@(R) and set w = supM. Show that for every semi-flat

replacement F of M one has fdR M = w+ fdR Cw(F).E 8.3.5 Let M be an R-complex. Show that fdR M is finite if and only if the complex N⊗L

R Mhas bounded above homology for every Ro-module N.

E 8.3.6 Show that the R-complexes of finite flat dimension form a triangulated subcategory ofD(R).

E 8.3.7 Let M be an R-complex and assume that it is isomorphic in D(R) to a K-flat complex Zwith Zv = 0 for all v > n. Show that fdR M is at most n, and conclude, that one could useK-flat replacements in 8.3.3.

E 8.3.8 Let µvu : Mu→Mvu6v be a U-direct system of R-complexes. Show that if U is fil-tered, then there is an inequality fdR(colimu∈U Mu)6 supu∈UfdR Mu, and show thatequality may not hold.

E 8.3.9 Let R be right coherent and let Muu∈U be a family of R-complexes. Show that ifinfu∈UinfMu>−∞ holds, then one has fdR

(∏u∈U Mu)= supu∈UfdR Mu.

E 8.3.10 Let R be right coherent and let Muu∈U be a family of R-complexes. Show that ifevery finitely generated right ideal in R has finite projective dimension, then one hasfdR(∏

u∈U Mu)= supu∈UfdR Mu. Hint: 3.1.28E 8.3.11 Let M be an R-module of finite flat dimension n. (a) Show that there exists an injective

Ro-module I such that TorRn (I,M) 6= 0. (b) Show that if R is commutative and Noetherian,

then one has TorRn (ER(R/p),M) 6= 0 for some prime ideal p in R.

E 8.3.12 Show that the full subcategory of R-complexes of finite flat dimension is a triangulatedsubcategory of D(R).



8.4 Evaluation Morphisms in the Derived Category


BIDUALITY

8.4.1 Construction. LetR⊗k So−→ B

be a ring homomorphism such that B is flat over Ro and over S; cf. (7.3.19.3). Let Xbe a B-complex and X '−→ I be a semi-injective resolution in K(B). By restrictionof scalars K(B)→K(T –So), 7.1.7, and 7.3.5 one has the functors

RHomR(–,X) = (HomR(–, I)) ´: D(R–Qo)op −→ D(Q–So) and

RHomSo(–,X) = (HomSo(–, I)) ´: D(Q–So)op −→ D(R–Qo) .

Let F,G: K(R–Qo)→K(R–Qo) be the functors given by

F = IdK(R–Qo) and G = HomSo(HomR(–, I), I)

and consider the biduality morphism δI : F→ G from 7.1.16. Both functors preservequasi-isomorphisms, so 7.2.3 yields a natural transformation ´δI : ´F→ ´G. One has

´F = IdD(R–Qo) and ´G = RHomSo(RHomR(–,X),X) ,

where the right-hand identity follows from 6.4.36.

8.4.2 Proposition. Adopt the setup from 8.4.1. Let X be a B-complex and M be acomplex of R–Qo-bimodules. Biduality 7.1.16 induces a morphism in D(R–Qo),

δMX : M −→ RHomSo(RHomR(M,X),X) ,

which is natural in M. As a natural transformation of functors, δX is triangulated.

PROOF. The desired natural transformation δX is ´δI constructed in 8.4.1. As thenatural transformation δI is triangulated by 7.1.16, it follows from 7.2.3 that δX = ´δIis triangulated as well.

TENSOR EVALUATION


R⊗kQo−→ A and S⊗k T o−→ B

be ring homomorphisms such that A is projective over R and B is flat over S;cf. (7.3.19.1) and (7.4.16.3). From 7.1.7 and 7.1.9 one gets, by restriction of sca-lars K(A)→K(R–Qo) and K(B)→K(S–T o), functors

F,G : K(A)op×K(R–So)×K(B) −→ K(Q–T o)

given by


8.4 Evaluation Morphisms in the Derived Category 319

F(M,X ,N) = HomR(M,X)⊗S N andG(M,X ,N) = HomR(M,X⊗S N) .

Now consider tensor evaluation θ : F→ G obtained from 7.1.17. Precomposing F,G, and θ with the functor Pop

A × IdK(R–So)×PB, one gets the natural transformationgiven by

HomR(PA(M),X)⊗S PB(N)θP(M)X P(N)

−−−−−−→ HomR(PA(M),X⊗B PS(N)) ,

which we abbreviate θ1 : F1→ G1. The functors F1 and G1 map every triple ofquasi-isomorphisms to a quasi-isomorphism, and hence 7.2.2 and 7.2.3 yield a nat-ural morphism ´θ1 : ´F1→ ´G1 of functors from D(A)op×D(R–So)×D(B) to D(k).Proceeding as in 7.5.6, one obtains

´F1(M,X ,N) = RHomR(M,X)⊗LS N and

´G1(M,X ,N) = RHomR(M,X⊗LS N) .

8.4.4 Proposition. Adopt the setup from 8.4.3. Let M be an A-complex, X be a com-plex of R–So-bimodules, and N be a B-complex. Tensor evaluation 7.1.17 induces amorphism in D(Q–T o),

θMXN : RHomR(M,X)⊗LS N −→ RHomR(M,X⊗L

S N) ,

which is natural in M, X , and N. Moreover, as a transformation of functors, θ istriangulated in each variable.

PROOF. The desired natural transformation θ is ´θ1 constructed in 8.4.3. The naturaltransformation θ is triangulated in each variable by 7.1.17, and the functors Pop

Aand PB are triangulated by 6.3.10 and D.10. It follows that θ1 : F1→ G1 in 8.4.3 istriangulated in each variable; hence so is θ = ´θ1 by 7.2.3.

8.4.5 Corollary. Let M be an R-complex, X be a complex of R–So-bimodules, andN be an S-complex. Tensor evaluation 7.1.17 induces a morphism in D(k),


S N) ,

which is natural in M, X , and N. Moreover, as a transformation of functors, θ istriangulated in each variable.

PROOF. Apply 8.4.4 with Q = k= T , A = R and B = S.

8.4.6 Example. Adopt the setup from 8.4.3, and let M, X , and N be as in 8.4.4. LetP '−→ M be a semi-projective resolution over A, and let F be a semi-flat replace-ment of N over B. By 6.3.2 and 6.4.20 there is an isomorphism π : P→ PA(M) inK(A) and a quasi-isomorphism π′ : PB(N)→ F in K(B) which yield a commutativediagram in K(Q–T o),



HomR(PA(M),X)⊗S PB(N)

'Hom(π,X)⊗π′

θP(M)X P(N)// HomR(PA(M),X⊗S PB(N))

' Hom(π,X⊗π′)

HomR(P,X)⊗S F θPXF// HomR(P,X⊗S F) ;

the vertical morphisms are quasi-isomorphisms by 5.2.10, 5.4.10, and 5.4.16. Ap-plication of the functor VQ⊗kT o to this diagram yields θPXF/1 = θMXN ; cf. 8.4.4.

8.4.7 Theorem. Let M be an R-complex, X be a complex of R–So-bimodules, andN be an S-complex. If R is left Noetherian, then tensor evaluation 8.4.5,


S N) ,

is an isomorphism under each of the following conditions.(a) M is in Df

A(R) and has finite projective dimension.(b) M is in Df

A(R), X is in D@(R–So), and N has finite flat dimension.If S is left Noetherian, then θMXN is an isomorphism under each of the following

conditions.(c) N is in Df

A(S) and has finite projective dimension.(d) M has finite projective dimension, X is in DA(R–So), and N is in Df

A(S).

PROOF. It suffices by 8.4.6 to prove that there exist a semi-projective resolutionP '−→M over R and a semi-flat replacement F of N over S such that tensor evaluationθPXF is an isomorphism in K(k).

Assume first that R is left Noetherian and that M is in DfA(R). By 5.1.14 there is

a semi-projective resolution P '−→M with P bounded below and degreewise finitelygenerated. Let F be a semi-flat replacement of N over S. Under the assumptions in(a), one can take P to be bounded, see 8.1.14, and then θPXF is an isomorphism by4.5.7(d). Under the conditions in (b), one can assume that F is bounded above, see8.3.12, and that X is bounded above. Now θPXF is an isomorphism by 4.5.7(1,c).

Assume now that S is left Noetherian and that N is in DfA(S). There is a semi-

projective resolution F '−→ N over S with F bounded below and degreewise finitelygenerated. Let P '−→M be a semi-projective resolution over R. Under the assump-tions in (c), one can take F to be bounded, and then θPXF is an isomorphism by4.5.7(e). Under the conditions in (d), one can assume that P is bounded above, see8.1.9 and that X is bounded below. Now θPXF is an isomorphism by 4.5.7(2,a).



R⊗k T o−→ A and Q⊗k So−→ B

be ring homomorphisms such that A is flat over Ro and B is projective over So;cf. 7.3.19. Restriction of scalars K(A)→K(R–T o) and K(B)→K(Q–So) yields by7.1.7 and 7.1.9 functors Let F,G: K(A)×K(R–So)op×K(B)→K(Q–T o) given by


8.4 Evaluation Morphisms in the Derived Category 321

F(M,X ,N) = N⊗S HomR(X ,M) andG(M,X ,N) = HomR(HomSo(N,X),M) .

Now consider homomorphism evaluation η : F→ G from 7.1.18. Precomposing F,G, and η with the functor IA× IdK(R–So)op×PB, one gets the natural morphismgiven by

PB(N)⊗S HomR(X , IA(M))ηI(M)X P(N)

−−−−−−→ HomR(HomSo(PB(N),X), IA(M)) ,

which we abbreviate η1 : F1→ G1. The functors F1 and G1 map every triple ofquasi-isomorphisms to a quasi-isomorphism, and hence 7.2.2 and 7.2.3 yield a nat-ural transformation ´η1 : ´F1→ ´G1 of functors from D(A)×D(R–So)op ×D(B) toD(Q–T o). Proceeding as in 7.5.6, one obtains

´F1(M,X ,N) = N⊗LS RHomR(X ,M) and

´G1(M,X ,N) = RHomR(RHomSo(N,X),M) .

8.4.9 Proposition. Adopt the setup from 8.4.8. Let M be an A-complex, X be a com-plex of R–So-bimodules, and N be a B-complex. Homomorphism evaluation 7.1.18induces a morphism in D(Q–T o),

ηMXN : N⊗LS RHomR(X ,M) −→ RHomR(RHomSo(N,X),M) ,

which is natural in M, X , and N. Moreover, as a transformation of functors, η istriangulated in each variable.

PROOF. The desired natural transformation η is ´η1 constructed in 8.4.8. The naturaltransformation η is triangulated in each variable by 7.1.18, and the functors IA andPB are triangulated by 6.3.14 and 6.3.10. It follows that η1 : F1→ G1 in 8.4.8 istriangulated in each variable; hence so is η = ´η1 by 7.2.3.

8.4.10 Corollary. Let M be an R-complex, X be a complex of R–So-bimodules, andN be an S-complex. Homomorphism evaluation 7.1.18 induces a morphism in D(k),


which is natural in M, X , and N. Moreover, as a transformation of functors, η istriangulated in each variable.

PROOF. Apply 8.4.9 to Q = k= T , A = R, and B = So.

8.4.11 Theorem. Let M be an R-complex, X be a complex of R–So-bimodules, andN be an So-complex. If S is right Noetherian, then homomorphism evaluation 8.4.10,


is an isomorphism under each of the following conditions.(a) N is in Df

A(So) and has finite projective dimension, or

(b) N is in DfA(S

o), X is in D@(R–So), and M has finite injective dimension.



PROOF. It suffices by to prove that there is a semi-projective resolution P '−→N overSo and a semi-injective resolution M '−→ I over R such that homomorphism evalu-ation ηIXP is an isomorphism in K(k). Let M '−→ I be a semi-injective resolution.By assumption, S is right Noetherian, so under the conditions in (a) as well as in (b)there exists by 5.1.14 a semi-projective resolution P '−→ N with P bounded belowand degreewise finitely generated. Under the conditions in (a) one can assume thatP is bounded, see 8.1.14, and then ηIXP is an isomorphism by 4.5.10(c). Under theconditions in (b), one can assume that I is bounded below, see 8.2.8, and that X isbounded above. Now ηIXP is an isomorphism by 4.5.10(1,a).

EXERCISES

E 8.4.1 Let M be an R-complex and X be k-complex. Show that biduality 4.5.2 yields a morphism,δM

X : M→ RHomk(RHomk(M,X),X), in D(R) which is natural in M and X .

E 8.4.2 Let R be left Noetherian. Show that if M ∈Df@A(R) has finite projective dimension, then

so has the Ro-complex RHomR(M,R).E 8.4.3 Let R be left Noetherian. Show that if M ∈Df

@A(R) has finite projective dimension, thenthere is an isomorphism M ' RHomRo(RHomR(M,R),R) in D(R).

E 8.4.4 Let S be right Noetherian. Show that for every M ∈D(R) with idR M finite and everyX ∈D@(R–So) one has fdS RHomR(X ,M)6 idSo X + supM.

E 8.4.5 Let S be left Noetherian. Show that for every M ∈D(R) with pdR M finite and everyX ∈DA(R–So) one has fdSo RHomR(M,X)6 fdSo X− infM.

E 8.4.6 Let R be left Noetherian. Show that for every M ∈DfA(R) with pdR M finite and every

X ∈D(R–So) one has pdSo RHomR(M,X)6 pdSo X− infM.E 8.4.7 Let R be left Noetherian. Show that for every N ∈D(S) with fdS N finite and every X ∈

D@(R–So) one has idR(X⊗LS N)6 idR X− infN.

E 8.4.8 Assume that k is Noetherian. Use way-out techniques to show that tensor evaluationRHomk(M,X)⊗L

k N→ RHomk(M,X⊗Lk N) is an isomorphism in D(k) under each of the

following conditions: (1) M,X ∈D@A(k) and N ∈Df(k) with pdkM < ∞ and fdkX < ∞;(2) M ∈Df(k) and X ,N ∈D@A(k) with idkX < ∞ and fdkN < ∞.

E 8.4.9 Assume that k is Noetherian. Use way-out techniques to show that homomorphism eval-uation N⊗L

k RHomk(X ,M)→ RHomk(RHomk(N,X),M) is an isomorphism in D(k) forcomplexes M,X ∈D@A(k) and N ∈Df(k) with idkM < ∞ and idkX < ∞.

8.5 Global and Finitistic Dimensions

SYNOPSIS. (Weak) global dimension, ∼ of Noetherian ring, finitistic projective/injective/flat di-mension, ∼ of perfect ring, ∼ of Noetherian ring.

GLOBAL DIMENSION

8.5.1 Definition. The global dimension of R, written gldim(R), is defined as

gldim(R) = suppdR M |M is an R-module .


8.5 Global and Finitistic Dimensions 323

8.5.2. It follows from 1.3.24 that the rings of global dimension 0 are precisely thesemi-simple rings.

REMARK. Since semi-simplicity is a left/right symmetric property, one has gldim(R) = 0 if andonly if gldim(Ro) = 0. We prove in 8.5.12 that the global dimensions of R and Ro agree if R isNoetherian; in general, though, they need not agree. For example, Small [79] shows that the ring

R =(Z Q0 Q

)=(

x y0 z

)∣∣∣ x ∈ Z and y,z ∈ Q

has gldim(R) = 2 and gldim(Ro) = 1. In fact, the value of gldim(R)−gldim(Ro) can be any integer;examples are supplied by Fossum, Griffiths and Reiten [28].

8.5.3 Example. One has gldim(Z)= 1 and gldim(Z/4Z)=∞; see 1.3.10 and 8.1.11.

8.5.4 Theorem. There are equalities,

gldim(R) = suppdR M− supM |M 6' 0 is a complex in D@(R)= suppdR(R/a) | a is a left ideal in R= supidR M+ infM |M 6' 0 is a complex in DA(R)= supidR M |M is an R-module .

PROOF. Denote the suprema in the display by s1, . . . ,s4, in the order they appear.The inequalities s1 > gldim(R) > s2 and s3 > s4 are evident. Let n be an integer.If one has s2 6 n, then 8.1.9 yields − infRHomR(R/a,M) 6 n− infM for everyR-complex M and every left ideal a, so 8.2.8 yields idR M 6 n− infM for everyR-complex M, whence s3 6 n holds and one has s3 6 s2. A parallel argument thatapplies 8.1.9 and 8.2.8 in opposite order yields s1 6 s4.

WEAK GLOBAL DIMENSION

8.5.5 Definition. The weak global dimension of R, written w.gldim(R), is given by

w.gldim(R) = supfdR M |M is an R-module .

8.5.6. By 8.3.6 there is an inequality

w.gldim(R)6 gldim(R) ,

and by 8.3.20 equality holds if R is left perfect.

8.5.7 Example. By 1.3.42 and 8.5.3 one has w.gldim(Z) = 1 and w.gldim(Z/4Z) =∞.

8.5.8 Theorem. One has w.gldim(R) = 0 if and only if R is von Neumann regular.

PROOF. If w.gldim(R) = 0 holds, then every R-module is flat and, therefore, R isvon Neumann regular by 1.3.41. Conversely, if R is von Neumann regular, thenevery finitely generated right ideal b in R is generated by an idempotent and hencea direct summand of RR; thus the Ro-module R/b is projective. It follows from 8.3.2that TorR

1 (R/b,M) = 0 holds for every R-module M and for every finitely generatedright ideal b and, consequently, every R-module is flat.



8.5.9 Example. The commutative ring QN is von Neumann regular by 1.3.40, soby 8.5.8 one has w.gldim(QN) = 0. As QN is not Noetherian it is, in particular, notsemi-simple, so one has gldim(QN)> 0 by 8.5.2.

REMARK. It is a result of Osofsky [67] that one has gldim(QN) = n+1 if and only if 2ℵ0 = ℵn,where n is a natural number. Thus, gldim(QN) is at least 2, but the exact value can not be decidedwithin ZFC. Moreover if U is a set of cardinality > ℵω, where ω denotes the first infinite ordinal,then one has gldim(QU ) = ∞. In contrast, one has w.gldim(QU ) = 0 by 1.3.40 and 8.5.8.

The weak global dimension is a left/right symmetric invariant.

8.5.10 Theorem. There is an equality w.gldim(R) = w.gldim(Ro), and one has

w.gldim(R) = supfdR M− supM |M 6' 0 is a complex in D@(R)= supfdR(R/a) | a is a finitely generated left ideal in R .

PROOF. Let n be an integer. If the inequality w.gldim(R) 6 n holds, then 8.3.12yields TorR

n+1(N,M) = 0 for all R-modules M and all Ro-modules N. By commuta-tivity 4.4.3 and the definition of Tor, 7.4.8, one has TorR

n+1(N,M) = TorRo

n+1(M,N),so another application of 8.3.12 yields fdRo N 6 n for all Ro-modules N, whencew.gldim(Ro)6 n holds. This proves the inequality w.gldim(Ro)6 w.gldim(R), andthe opposite inequality holds by symmetry.

To prove the equalities in the display, denote the suprema by s1 and s2 in the orderthey appear. The inequalities s1 > w.gldim(R)> s2 are evident. Let n be an integer.If one has w.gldim(R) 6 n, then w.gldim(Ro) 6 n holds as well, so for every Ro-module N one has sup(N⊗L

R M) = sup(M⊗LRo N)6 n+ supM by 8.3.12. It follows

that fdR M6 n+supM holds, so one has s16 n and hence s16w.gldim(R). If s26 nholds, then one has 0 = TorR

n+1(N,R/a) = TorRo

n+1(R/a,N) for every Ro-module Nand every finitely generated left ideal a in R. Thus, n> w.gldim(Ro) holds, and onehas s2 > w.gldim(Ro) = w.gldim(R).

GLOBAL DIMENSIONS OF NOETHERIAN RINGS

8.5.11 Proposition. If R is left Noetherian, then there is an equality

gldim(R) = supidR(R/a) | a is a left ideal in R .PROOF. In view of 8.5.4 one need only prove the inequality “6”. Let n be an integerand assume that idR(R/a)6 n holds for every left ideal a in R. It suffices, again by8.5.4, to prove that pdR M 6 n holds for every finitely generated R-module M. By8.1.14 that follows as one has Extn+1

R (M,R/a) = 0 for every left ideal a in R.

REMARK. Osofsky [65] shows that for every n> 1 there is a (valuation) ring R with gldim(R) = nand supidR(R/a) | a is a left ideal in R= 1.

8.5.12 Theorem. If R is left Noetherian, then one has w.gldim(R) = gldim(R).

PROOF. By 8.3.19 one has pdR M = fdR M for every finitely generated R-module M.The equality gldim(R) = w.gldim(R) now follows from 8.5.10 and 8.5.4.



8.5.13 Corollary. If R is left Noetherian and von Neumann regular, then R is semi-simple; in particular, R is Artinian.

PROOF. Immediate from 8.5.12, 8.5.8, and 8.5.2.

8.5.14 Corollary. If R is Noetherian, then one has gldim(R) = gldim(Ro).

PROOF. When R is both left and right Noetherian, 8.5.12 and 8.5.10 yield equalitiesgldim(R) = w.gldim(R) = w.gldim(Ro) = gldim(Ro).

FINITISTIC DIMENSIONS

8.5.15 Definition. The finitistic projective dimension of R, written FPD(R), the fini-tistic injective dimension of R, written FID(R), and the finitistic flat dimension of R,written FFD(R), are defined as follows,

FPD(R) = suppdR M |M is an R-module with pdR M < ∞ ,FID(R) = supidR M |M is an R-module with idR M < ∞ , andFFD(R) = supfdR M |M is an R-module with fdR M < ∞ .

REMARK. The finitistic dimensions are not left/right symmetric invariants. The following exampledue to Small is published in [42]. Let k be a field and denote by n the maximal ideal of the localring Q = k[x]/(x2); the ring

R =(k Q/n0 Q

)=(

x y0 z

)∣∣∣ x ∈ k, y ∈ Q/n, and z ∈ Q

has FPD(R) = 1 and FPD(Ro) = 0. More generally, Green, Kirkman, and Kuzmanovich [32] showthat for every integer n > 0 there exists a ring R with FPD(R) = n and FPD(Ro) = 0.

It would be too much of a detour to include the proof of the next theorem here, soit has been relegated to an appendix. It is known as Jensen’s theorem (on projectivedimension of flat modules).

8.5.16 Theorem. For every flat R-module F one has pdR F 6 FPD(R).

PROOF. See Appn. E.

8.5.17 Corollary. For every R-complex M with fdR M finite there is an inequality,

pdR M 6 supM+FPD(R) .

PROOF. One can assume that M is not acyclic and FPD(R) is finite. Set n = fdR Mand w= supM; both are integers by assumption and 8.3.4. Let P be a semi-projectivereplacement of M. Note that Σ−w Pěw is a semi-projective replacement of the moduleCw(P) and that pdR M = w+pdR Cw(P) holds by 8.1.9. The module Cn(P) is flat by8.3.12 so it has finite projective dimension by 8.5.16. It follows that Cw(P) has finiteprojective dimension, so pdR Cw(P)6 FPD(R) holds by definition.

8.5.18 Corollary. If FPD(R) is finite, then an R-complex M has finite flat dimen-sion, i.e. fdR M < ∞, if and only if it has finite projective dimension, i.e. pdR M < ∞.




REMARK. Without recourse to 8.5.16, we show in 9.3.11 that the flat and projective dimensions aresimultaneously finite for complexes over any ring that admits a so-called dualizing complex. This,however, may not imply that such rings have finite finitistic projective dimension. Indeed, everyfinite dimensional algebra over a field has a dualizing complex, but it remains an open problem ifthe finitistic projective dimension is finite for every such algebra.

Per the Remark after 8.5.9 there exist rings with flat modules of infinite projective dimension.

8.5.19 Proposition. The following inequalities hold,

gldim(R) > w.gldim(R) 6 gldim(Ro)

6 6 6

FPD(R) > FFD(R) 6 FID(Ro) .

If one of the quantities in the first row is finite, then it equals the quantity below it.

PROOF. The inequalities in the first row follow from 8.5.6 and 8.5.10. The “vertical”inequalities and the last assertion are immediate from the definitions and 8.5.4. Thelast inequality in the second row follows from 8.3.16. The first one follows from8.5.17 and 8.3.6.

By 8.5.19 one has FPD(R) = gldim(R) provided that gldim(R) is finite. As aconsequence of 8.5.16, the following stronger result holds.

8.5.20 Proposition. If w.gldim(R) is finite, then one has FPD(R) = gldim(R).

PROOF. It is sufficient to prove the inequality FPD(R)> gldim(R). Let M be an R-module; one has fdR M < ∞ by assumption, so 8.5.16 yields FPD(R)> pdR M.

REMARK. By 8.5.20 and the Remark after 8.5.9 and there exist rings with finitistic flat dimensionzero and infinite finitistic projective dimension.

8.5.21 Theorem. If R is left perfect, then there are (in)equalities

FFD(R) = FPD(R)6 idR R .

PROOF. The equality is immediate from 8.3.20. To prove the inequality, let M be anR-module with pdR M = n finite. Let J be the Jacobson radical of R and set k = R/J.By 8.1.16 one has ExtnR(M,k) 6= 0, and by 7.3.14 there is an exact sequence,

ExtnR(M,R)−→ ExtnR(M,k)−→ Extn+1R (M,J) = 0 ,

whence one has ExtnR(M,R) 6= 0 and, thus, n6 idR R by 8.2.8.

FINITISTIC DIMENSIONS OF NOETHERIAN RINGS

Over Noetherian rings, there are further relations among the quantities in 8.5.19.

8.5.22 Theorem. Let R be left Noetherian. There are (in)equalities

FPD(R) 6 idR R and FID(R) = FFD(Ro) .

Moreover, if both quantities idR R and FID(R) are finite, then they are equal.



PROOF. To prove the equality, it suffices by 8.5.19 to establish the inequality “6”,which follows immediately from 8.3.18.

To prove the inequality, let M be an R-module with pdR M = n finite. Choose anR-module N with ExtnR(M,N) 6= 0 and an exact sequence 0→K→ L→N→ 0 withL a free R-module; cf. 1.3.11. By 7.3.14 there is an exact sequence

ExtnR(M,L)−→ ExtnR(M,N)−→ Extn+1R (M,K) = 0 ,

whence one has ExtnR(M,L) 6= 0 and, thus, n6 idR L = idR R by 8.2.8 and 8.2.19.Assume that idR R and FID(R) are finite; evidently one has idR R 6 FID(R). To

prove the opposite inequality, set n = FID(R) and note that Extn+1R (−,K) = 0 holds

for every R-module K with idR K finite. Furthermore, there exists an R-module Mwith idR M = n. Consider an exact sequence 0→ K→ L→M→ 0 with L free. AsidR R is finite, it follows from 8.2.19 and 8.2.9 that L and K have finite injectivedimension. Let N be an R-module with ExtnR(N,M) 6= 0. The exact sequence

ExtnR(N,L)−→ ExtnR(N,M)−→ Extn+1R (N,K) = 0 ,

shows that ExtnR(N,L) is non-zero, whence n6 idR L = idR R holds as desired.

8.5.23 Corollary. If R is Noetherian, then there are (in)equalities

FID(Ro) = FFD(R) 6 FPD(R) 6 idR R 6 gldim(R) .

PROOF. The equality holds by 8.5.22 as Ro is left Noetherian. The first inequalityis part of 8.5.19. The second inequality holds by 8.5.22 as R is left Noetherian. Thethird and final inequality holds by 8.5.4.

EXERCISES

E 8.5.1 Show that if R is left Noetherian, then one has gldim(R)6 gldim(Ro).E 8.5.2 Show that there are equalities

FPD(R) = suppdR M− supM |M 6' 0 is an R-complex with pdR M < ∞ ,FID(R) = supidR M+ infM |M 6' 0 is an R-complex with idR M < ∞ , and

FFD(R) = supfdR M− supM |M 6' 0 is an R-complex with fdR M < ∞ .E 8.5.3 Show it if k is Noetherian and idk k is finite, then FPD(k) = FFD(k) = FID(k) = idk k.E 8.5.4 Show that for silpR = supidR P | P is a projective R-module one has FPD(R)6 silpR.E 8.5.5 Let R be left Noetherian; show that idR R = silpR holds for silpR as defined in E 8.5.4.E 8.5.6 Set silfR = supidR F | F is a flat R-module, define silpR as in E 8.5.4, and show that

the equality silfR = silpR holds. Hint: 8.5.16E 8.5.7 Set sfliR = supfdR I | I is an injective R-module. (a) Show that there is an inequality

FFD(R) 6 sfliRo. (b) Show that sfliR 6 FFD(R) holds if sfliR is finite. (c) Show thatFFD(R) = FFD(Ro) = sfliR = sfliRo holds if sfliR and sfliRo are finite.

E 8.5.8 Let silpR and sfliR be as in E 8.5.4 and E 8.5.7 and define spliR and splfR analogously.Show that there are inequalities spliR6 sfliR+ splfR and splfR6 FPD(R).

E 8.5.9 Show that if gldim(R) is finite, then every complex of projective R-modules is semi-projective, and every complex of injective R-modules is semi-injective.

E 8.5.10 Let P be a bounded above R-complex of projective modules and assume that FPD(R) isfinite. Show that if pdR P is finite, then P is semi-projective.



E 8.5.11 Let I be a bounded below R-complex of injective modules and assume that FID(R) isfinite. Show that if idR I is finite, then I is semi-injective.


Chapter 9Dualizing Complexes

9.1 Grothendieck Duality

SYNOPSIS. Homothety formation; dualizing complex; Grothendieck Duality.

9.1.1 Definition. The full subcategories P(R), I(R), and F(R) of D@A(R) are definedby specifying their objects as follows

P(R) = M ∈D@A(R) | pdR M < ∞ ,I(R) = M ∈D@A(R) | idR M < ∞ , andF(R) = M ∈D@A(R) | fdR M < ∞ .

The full subcategories Pf(R), If(R), and Ff(R) of Df@A(R) are defined similarly.

9.1.2 Proposition. The categories P(R), I(R), and F(R) are triangulated subcate-gories of D@A(R). Further, if R is left Noetherian, then Pf(R), If(R), and Ff(R) aretriangulated subcategories of Df

@A(R) and one has Pf(R) = Ff(R).

PROOF. The full subcategory P(R) of D@A(R) is triangulated by 8.1.3 and 8.1.10.Similarly, I(R) is triangulated by 8.2.3 and 8.2.9, and F(R) is triangulated by 8.3.4and 8.3.13. The remaining assertions follow from 7.6.10 and 8.3.19.

HOMOTHETY FORMATION

Let s be an element in S and let X be a complex of R–So-bimodules. It is straight-forward to verify that the homothety sX , see 2.1.9, is R-linear. We now consider themap that to each element of S assigns the corresponding homothety.

9.1.3 Proposition. For a complex X of R–So-bimodules the homothety formationmap,

χXSoR : S −→ HomR(X ,X) given by χX

SoR(s) = sX ,

is a morphism in C(S–So).

329

330 9 Dualizing Complexes

PROOF. Both S and HomR(X ,X) are complexes of S–So-bimodules; see 2.3.11. Thecomputations below, where s,u,v ∈ S and x ∈ X , show that the map χ= χX

SoR is bothS- and So-linear.

χ(us)(x) = x(us) = (xu)s = χ(s)(xu) = (uχ(s))(x) andχ(sv)(x) = x(sv) = (xs)v = (χ(s)(x))v = (χ(s)v)(x) .

As χ is graded of degree 0, it remains to show that it commutes with the differentials.On S the differential is zero, and one has

∂Hom(X ,X)χ(s) = ∂X sX − sX∂X = 0

by So-linearity of the differential on X .

REMARK. Homothety formation is, in fact, a morphism of k-algebras; cf. E 9.1.1.

9.1.4 Example. Homothety formation χRRoR : R→ HomR(R,R) is an isomorphism

of R–Ro-bimodules. Indeed, it is the counitor εRR from (4.4.1.2).

9.1.5. Let X be a complex of R–So-bimodules. Homothety formation in C(S–So),

χXSoR : S −→ HomR(X ,X) ,

from 9.1.3 induces a morphism in K(S–So) which we denote by the same symbol.

9.1.6. Since a complex of R–So-bimodules is a complex of So–R-bimodules, andsince R = Ro as R–Ro-bimodules, it follows from 9.1.3 and 9.1.5 that homothetyformation χX

RSo : R→ HomSo(X ,X) is a morphism in C(R–Ro) and in K(R–Ro).If R is commutative, then an R-complex X is tacitly considered to be a complex

of symmetric R–Ro-bimodules, and the morphisms χXRoR and χX

RRo coincide. Thismorphism is written χX

R .

DEFINITION AND EXAMPLES OF DUALIZING COMPLEXES

9.1.7 Definition. Let R be left Noetherian and S be right Noetherian. A complex Dof R–So-bimodules is called dualizing for (R,So) if it meets the requirements:

(1) H(D) is bounded and degreewise finitely generated over R and over So.(2) D has finite injective dimension over R and over So.(3) There exists a complex J of R–So-bimodules such that

− there is an isomorphism D' J in D(R–So),− J is semi-injective over R and over So,− homothety formation χJ

RSo : R→ HomSo(J,J) in K(R–Ro) and homothetyformation χJ

SoR : S→ HomR(J,J) in K(S–So) are quasi-isomorphisms.

9.1.8. The definition of a dualizing complex for (R,So) is symmetric in R and So.Indeed, a complex of R–So-bimodules is the same as a complex of So–R-bimodules,and a ring is left Noetherian if and only if the opposite ring is right Noetherian.


9.1 Grothendieck Duality 331

9.1.9 Definition. Let R be Noetherian. A complex of R–Ro-bimodules is called du-alizing for R if it is dualizing for (R,Ro) as defined in 9.1.7.

9.1.10 Example. Let k be a field and let R be a finite dimensional k-algebra; i.e. Rhas finite rank as a k-vector space. Consider the R–Ro-bimodule D = Homk(R,k).As a k-vector space D is finitely generated, and hence it is finitely generated as amodule over R and over Ro. Moreover, D is injective over R and over Ro by 5.4.20.The isomorphisms in the commutative diagram

RχD

//

χR ∼=

HomRo(Homk(R,k),Homk(R,k))

HomRo(R,R)Hom(R,δR)

∼=// HomRo(R,Homk(Homk(R,k),k))

∼= ζRDk

OO

come from 9.1.4, 4.5.4, and 4.4.9; it shows that χDRRo is an isomorphism in C(R–Ro).

A a similar diagram shows that χDRoR is an isomorphism as well. Thus, with J = D

and S = R the conditions 9.1.7 are met, so D is dualizing for R per 9.1.9.

9.1.11 Definition. Let R be commutative Noetherian. An R-complex D is calleddualizing for R if it is dualizing as defined in 9.1.9 when considered as a complexof symmetric R–Ro-bimodules.

9.1.12 Example. Let R be commutative Artinian with Jacobson radical J=⋂n

u=1muwhere m1, . . . ,mn are the finitely many maximal ideals of R. The direct sum ofindecomposable injective R-modules D =

⊕nu=1 ER(R/mu) is the injective envelope

of the module⊕n

u=1 R/mu, which by the Chinese Remainder Theorem is isomorphicto R/J. By F.23 the module D is finitely generated. By F.13, F.15, and F.18 there areisomorphisms of rings,

HomR(D,D) ∼=×nu=1 HomR(ER(R/mu),ER(R/mu))

∼=×nu=1 HomRmu (ERmu (Rmu/muRmu),ERmu (Rmu/muRmu))

∼=×nu=1 Rmu

∼= R .

It is straightforward to verify that the composite isomorphism is homothety forma-tion χD

R . By injectivity of D it follows that D is dualizing for R per 9.1.11.

9.1.13 Example. If R is commutative, Noetherian, and self-injective, then R is adualizing complex for R, as the conditions in 9.1.7 are met by D = R = J; see 9.1.4.Specific examples are Z/(n) and k[x]/(xn) for every n> 1 and any field k; see 8.2.10.

HOMOTHETY FORMATION IN THE DERIVED CATEGORY

Recall from 7.3.18 that R⊗k So is flat as an Ro-module if S is flat as a k-module.



9.1.14 Construction. Assume that R⊗k So is flat as an Ro-module and recall from(7.3.19.4) that that RHomR(–,–) is a functor D(R–So)op ×D(R–So)→ D(S–So)and for X ∈D(R–So) one has RHomR(X ,X) = HomR(I, I) for every semi-injectiveresolution in X '−→ I in K(R–So). Thus, homothety formation χI

SoR : S→ HomR(I, I)in K(S–So) from 9.1.5 yields a morphism in D(S–So),

χXSoR : S −→ RHomR(X ,X) .

9.1.15. If R⊗k So is flat as an S-module, then 9.1.14 applies to the morphism χXRSo

in K(R–Ro), see 9.1.6, and yields homothety formation in D(R–Ro),

χXRSo : R −→ RHomSo(X ,X) .

9.1.16 Lemma. Assume that R⊗k So is flat as an Ro-module. Let X and J be com-plexes of R–So-bimodules such that X ' J in D(R–So) and J is semi-injective overR. The morphisms χX

SoR from 9.1.14 and χJSoR/1 are isomorphic in D(S–So).

PROOF. First recall from 7.3.23 that HomR(–,J) is RHomR(–,X). Let X '−→ I be asemi-injective resolution in K(R–So). As in the proof of 7.3.23 there exists a quasi-isomorphism ϕ : J→ I in K(R–So) for which HomR(M,ϕ) is a quasi-isomorphismin K(S–So) for every complex M of R–So-bimodules. It follows from 7.3.19 that Iis semi-injective over R and hence HomR(ϕ, I) is a quasi-isomorphism in K(S–So).The assertion now follows from the commutative diagram

SχJ

//

χI

HomR(J,J)

Hom(J,ϕ)'

HomR(I, I)Hom(ϕ,I)

'// HomR(J, I) .

9.1.17 Proposition. Assume that R and S are flat as k-modules. Let R be left Noe-therian and S be right Noetherian. A complex D of R–So-bimodules is dualizing for(R,So) if and only if it meets the requirements:

(1) H(D) is bounded and degreewise finitely generated over R and over So.(2) D has finite injective dimension over R and over So.(3) Homothety formation χD

RSo : R→ RHomSo(D,D) in D(R–Ro) and homothetyformation χD

SoR : S→ RHomR(D,D) in D(S–So) are isomorphisms.

PROOF. The assumptions on R and S imply that R⊗k So is flat as an Ro-module andas an S-module; cf. 7.3.18. “Only if” now follows from 9.1.16. For “if” let D '−→ Jbe a semi-injective resolution in K(R–So). One has D ' J in D(R–So), and thecomplex J is semi-injective over R and over So; see 7.3.19. By assumption χD

RSo =χJ

RSo/1 and χDSoR = χJ

SoR/1 are isomorphisms so χJRSo and χJ

SoR are quasi-isomorphismsby 6.4.17.

REMARK. There exists a left Noetherian (even a left Artinian local, and even a commutative Noe-therian local) algebra R over a field k such that there is no dualizing complex for (R,So) for anyright Noetherian k-algebra S. See Yekutieli and Zhang [89, sec. 1] and Wu and Zhang [87, sec. 7].In [87] there is also an example of an artinian local k-algebra R that does not have a dualizing com-plex; however, for some right artinian k-algebra S the pair (R,So) does have a dualizing complex.



9.1.18. Let R, S, and D be as in 9.1.17 and let D '−→ I be a semi-injective resolutionin K(R–So) with Iv = 0 for v > supD; see 5.3.26. Recall from 7.3.18(c) that I issemi-injective over R and over So. Thus, for every integer n > maxidR D, idSo Dthe bounded complex J = IĚ−n is semi-injective over R and over So by 8.2.8, andone has D' J in D(R–So).

9.1.19 Definition. If R is Noetherian with idR R and idRo R finite, then R is calledIwanaga–Gorenstein.

See 8.2.10 for examples of commutative Iwanaga–Gorenstein rings.

REMARK. For an Iwanaga–Gorenstein ring R one has idR R = idRo R by a result of Zaks [91].

9.1.20 Example. If R is Iwanaga–Gorenstein and flat as a k-module, then R is adualizing complex for R. Indeed, let ι : R '−→ I be a semi-injective resolution inK(R–Ro). Per 9.1.4 the homothety morphism χR

RRo is an isomorphism, so it followsfrom the commutative diagram,

RχR

∼=//

χI

HomRo(R,R)

Hom(R,ι)'

HomRo(I, I)Hom(ι,I)

'// HomRo(R, I) ,

that χIRRo is a quasi-isomorphism; i.e. χR

RRo = χIRRo/1 is an isomorphism in D(R–Ro).

The same diagram, with R and Ro interchanged, shows that χRRoR is an isomorphism.

Now apply 9.1.17 and 9.1.9.

EXISTENCE OF DUALIZING COMPLEXES

To parse the next result, notice that if k is Noetherian and R is finitely generated as ak-module, then R is Noetherian as well. Indeed, every finitely generated R-moduleand every finitely generated Ro-module is finitely generated over k.

9.1.21 Theorem. Let k be Noetherian and R be finitely generated as a k-module. IfD is a dualizing complex for k then RHomk(R,D) is a dualizing complex for R.

PROOF. As a k-complex, the homology of E = RHomk(R,D) is bounded and de-greewise finitely generated by 7.6.12 and 8.2.8; in particular, it is degreewise finitelygenerated over R and over Ro. Moreover, idR E and idRo E are finite by 8.2.5.

It remains to show that E satisfies 9.1.7(3). Let D '−→ I be a bounded semi-injective resolution over k. The complex J = Homk(R, I) in C(R–Ro) is by 5.4.20semi-injective over R and over Ro, and one has E ' J in D(R–Ro). There is a com-mutative diagram in C(R–Ro),



RχJ

// HomR(J,J)

R⊗k k

∼=µRυRk

OO

'R⊗χI

Homk(R⊗R J, I)

∼= ρJRI

OO

R⊗kHomk(I, I) ∼=ηIIR

// Homk(J, I) ,

∼= Hom(µJ ,I)

OO

where adjunction ρJRI and homomorphism evaluation ηIIR are isomorphisms by4.4.11 and 4.5.9(3,b). The modules in the bounded k-complex Homk(I, I) are flat by8.3.18, so it is semi-flat by 5.4.8. The homothety morphism χI

k is by 9.1.17 a quasi-isomorphism and by 5.4.16 also R⊗k χI

k is a quasi-isomorphism. Now it followsfrom the diagram that χJ

RoR is a quasi-isomorphism. With R and Ro interchanged thediagram shows that χJ

RRo is a quasi-isomorphism.

9.1.22 Definition. Let k be Artinian. A k-algebra that is finitely generated as a k-module is called an Artin algebra or, more elaborately, an Artin k-algebra.

9.1.23 Example. If k is Artinian with Jacobson radical J, then the injective envelopeD of k/J is a dualizing complex for k; see 9.1.10. By 9.1.21 every Artin k-algebraR now has a dualizing complex, namely RHomk(R,D)' Homk(R,D).

GROTHENDIECK DUALITY

The next example motivates the developments in the rest of this section.

9.1.24 Example. Let R be an Artin algebra, by 9.1.23 it has a dualizing complex Dwhich is, in fact, an R–Ro-bimodule that is injective over R and over Ro. For everydegreewise finitely generated R-complex M there is a commutative diagram in C(R),

M⊗Ro R ∼=

M⊗χD//

∼=µMυMR

M⊗Ro HomRo(D,D)

∼=ηDDM

MδM

// HomRo(HomR(M,D),D) ,

where the evaluation morphism ηDDM is an isomorphism by 4.5.10(3,b). The dia-gram shows that the biduality morphism δM

D is an isomorphism of R-complexes.

9.1.25 Theorem. Assume that R and S are flat as k-modules. Let R be left Noether-ian, let S be right Noetherian, and let D be a dualizing complex for (R,So). For everycomplex M in Df(R) the biduality morphism

δMD : M −→ RHomSo(RHomR(M,D),D)

is an isomorphism in D(R).



PROOF. From 8.2.8 and 7.6.7 one gets

− infRHomSo(RHomR(M,D),D)6 idSo D+ supD− infM andsupRHomSo(RHomR(M,D),D)6 idR D+ supD+ supM ;

in particular, the functor RHomSo(RHomR(–,D),D) is way-out; see A.24. So is theidentity functor on D(R), and the transformation δD is triangulated by 8.4.2. Thus,by A.27(d) it suffices to show that δM

D is an isomorphism for every finitely generatedR-module M. For such a module the evaluation morphism ηDDM is an isomorphismby 8.4.11, whence the commutative diagram,

M⊗LRo R

µMυMR '

'M⊗LχD

// M⊗LRo HomSo(D,D)

' ηDDM

MδM

// RHomSo(RHomR(M,D),D) ,

shows that δMD is an isomorphism.

9.1.26 Proposition. Assume that R and S are flat as k-modules. For every complexX of R–So-bimodules there is an adjunction,

D(So)RHomSo(–,X)op

//D(R)op ,,

RHomR(–,X)oo

whose unit is biduality δX : IdD(So)→ RHomR(RHomSo(–,X)op,X), and whose co-unit is δX : IdD(R)→ RHomSo(RHomR(–,X)op,X) when viewed as transformationof endofunctors on D(R).

PROOF. For every R-complex M and every So-complex N one gets from 7.3.10 andswap 7.5.11 the following natural isomorphisms

D(R)op(RHomSo(N,X),M)∼= H0(RHomR(M,RHomSo(N,X)))∼= H0(RHomSo(N,HomR(M,X)))∼=D(So)(N,RHomR(M,X)) .

This establishes the asserted adjunction and it is elementary to verify the claimsabout the unit and counit.

Grothendieck Duality is the phenomenon that for a commutative Noetherian ringk with a dualizing complex D the functor RHomk(–,D) yields a duality on Df(k).This is a special case of the next result.

9.1.27 Theorem. Assume that R and S are flat as k-modules. Let R be left Noether-ian, let S be right Noetherian, and let D be a dualizing complex for (R,So). There isan adjoint equivalence of k-linear triangulated categories,

Df(So)RHomSo(–,D)op

//Df(R)op ,,

RHomR(–,D)oo



and it restricts to adjoint equivalences of triangulated subcategories

Df@(S

o)DfA(R)

op , DfA(S

o)Df@(R)

op , Df@A(S

o)Df@A(R)

op, and

If(So) Pf(R)op .

PROOF. The functors RHomSo(–,D)op : D(So)D(R)op : RHomR(–,D) are ad-joint by 9.1.26 and they are k-linear and triangulated by 7.3.1. By A.30 the func-tor RHomR(–,D) is way-out, so it restricts to a functor Df(So) ← Df(R)op by7.6.12 and A.32. Similarly, RHomSo(−,D)op is way-out and restricts to a functorDf(So)→Df(R)op. By 9.1.25 these restrictions yield an adjoint equivalence.

By 7.6.7 the complex RHomR(M,D) is in D@(So) for every M ∈DfA(R), and by

8.2.8 the complex RHomSo(N,D) is in DA(R) for every N ∈D@(So). This provesthe first restriction; a parallel argument yields the second, and the third restrictionfollows from the two first.

To establish the final restriction it suffices to argue that (1) idSo RHomSo(M,D)is finite for every M ∈ Pf(R) and that (2) pdR RHomSo(N,D) is finite for every N ∈If(So). Here (1) follows from 8.3.15. To prove (2), let N ∈ If(So) and let a be a leftideal in R. Now biduality 9.1.25 and swap 7.5.11 yield,

RHomR(RHomSo(N,D),R/a)

' RHomR(RHomSo(N,D),RHomSo(RHomR(R/a,D),D))

' RHomSo(RHomR(R/a,D),RHomR(RHomSo(N,D),D))

' RHomSo(RHomR(R/a,D),N) .

By 8.2.8 the complex RHomSo(RHomR(R/a,D),N) has bounded homology withnegative infimum at most idR D+ idSo N and thus 8.1.14 yields pdR RHomSo(N,D)6idR D+ idSo N.

EXERCISES

E 9.1.1 Let X be a complex of R–So-bimodules. Show that the maps χXRSo : R→ HomSo(X ,X) and

χXSoR : So→ HomR(X ,X) are morphisms of k-algebras.

E 9.1.2 Show that every quasi-Frobenius ring has a dualizing complex.E 9.1.3 Let R be Noetherian and assume that it is flat as a k-module. A complex C of R–Ro-

bimodules is called semi-dualizing for R if the homothety morphisms χCRRo and χC

RoR inD(R–Ro) are isomorphisms. Show that R is semi-dualizing for R.

E 9.1.4 Let R be flat as a k-module and Noetherian. Let C be a semi-dualizing complex for R;see E 9.1.3. Show that the biduality morphism M → RHomRo(RHomR(M,C),C) is anisomorphism in D(R) for every complex M ∈Df

@A(R) of finite projective dimension.E 9.1.5 Let k be Noetherian. Show that if there is a dualizing complex for k, then every complex

in Df@(k) is isomorphic in D(k) to a bounded above complex of finitely generated k-

modules.E 9.1.6 Let R be flat as a k-module and Noetherian. Let D be a dualizing complex for R and

D '−→ I be a semi-injective resolution in K(R–Ro). Show that for a complex P of finitelygenerated projective R-modules the complex HomR(P,R) is acyclic if and only if I⊗R Pis acyclic.


9.2 Morita Equivalence 337

9.2 Morita Equivalence

SYNOPSIS. Invertible complex; (co)unit of adjunction; Morita Equivalence; inverse complex.

Recall from 7.3.18(a) that if S is projective as a k-module then R⊗k So is projectiveas an R-module.

9.2.1 Lemma. Assume that R⊗k So is projective as an R-module. Let X and L becomplexes of R–So-bimodules such that X ' L in D(R–So) and L is semi-projectiveover R. The morphisms χX

SoR from 9.1.14 and χLSoR/1 are isomorphic in D(S–So).

PROOF. First recall from 7.3.22 that HomR(L,–) is RHomR(X ,–). Let X '−→ I be asemi-injective resolution in K(R–So). By 6.4.21 there exists a quasi-isomorphismψ : L→ I in K(R–So). As L is semi-projective over R, the morphism HomR(L,ψ) isa quasi-isomorphism in K(S–So). It follows from 7.3.18(c) that I is semi-injectiveover R and hence HomR(ψ, I) is a quasi-isomorphism. The assertion now followsfrom the commutative diagram in K(S–So) below,

SχL

//

χI

HomR(L,L)

Hom(L,ψ)'

HomR(I, I)Hom(ψ,I)

'// HomR(L, I) .

INVERTIBLE COMPLEXES

9.2.2 Definition. Assume that R and S are projective as k-modules. Let R be leftNoetherian and let S be right Noetherian. A complex U of R–So-bimodules is calledinvertible for (R,So) if it satisfies the conditions:

(1) H(U) is bounded and degreewise finitely generated over R and over So.(2) U has finite projective dimension over R and over So.(3) The homothety morphism χU

RSo : R→ RHomSo(U,U) in D(R–Ro) and the ho-mothety morphism χU

SoR : S→ RHomR(U,U) in D(S–So) are isomorphisms.

9.2.3. The definition of an invertible complex is symmetric in R and So; cf. 9.1.8.

9.2.4. Let R, S, and U be as in 9.2.2 and let P '−→U be a semi-projective resolutionin K(R–So) with Pv = 0 for v < infU ; see 5.2.15. Recall from 7.3.18(a) that P issemi-projective over R and over So. Thus, for every integer n>maxpdR U,pdSo Uthe bounded complex PĎn is semi-projective over R and over So by 8.1.9, and onehas U ' PĎn in D(R–So).

REMARK. An R-complex that is isomorphic in D(R) to a bounded complex of finitely generatedprojective R-modules is known as a perfect R-complex. By 8.1.14 an invertible complex for (R,So)is thus both a perfect R-complex and a perfect So-complex.

9.2.5 Example. If R is Noetherian and projective as a k-module, then R is invertiblefor (R,Ro); cf. 9.1.4 and 9.2.1.



MORITA EQUIVALENCE

9.2.6 Proposition. Assume that R and S are projective as k-modules. For everycomplex X of R–So-bimodules there is an adjunction,

D(S)X⊗L

S–//D(R) .

RHomR(X ,–)oo

For an S-complex N the unit αNX is the unique morphism in D(S) that makes the

following diagram commutative,

(9.2.6.1)

S⊗LS N

'µN

χX⊗LN// RHomR(X ,X)⊗L

S N

θXXN

NαN

X// RHomR(X ,X⊗L

S N) .

For an R-complex M the counit βMX is the unique morphism in D(R) that makes the

following diagram commutative,

(9.2.6.2)

X⊗LS RHomR(X ,M)

ηXMX

βMX

// M

εM '

RHomR(RHomSo(X ,X),M)RHom(χX ,M)

// RHomR(R,M) .

PROOF. For every R-complex M and every S-complex N one gets from 7.3.10 andadjunction 7.5.15 the following natural isomorphisms

D(R)(X⊗LS N,M)∼= H0(RHomR(X⊗L

S N,M))∼= H0(RHomS(N,RHomR(X ,M)))∼=D(S)(N,RHomR(X ,M)) .

This establishes the asserted adjunction and it is elementary to verify the claimsabout the unit and counit.

9.2.7 Addendum. For a complex N of S–T o-bimodules it follows from (9.2.6.1),7.5.1, and tensor evaluation 8.4.4 that αN

X is a morphism in D(S–T o). Similarly,(9.2.6.2), 7.5.1, and homomorphism evaluation 8.4.9 show that for a complex M ofR–Qo-bimodules, βM

X is a morphism in D(R–Qo).

9.2.8 Lemma. Assume that R and S are projective as k-modules. Let R be leftNoetherian, let S be right Noetherian, and let U be invertible for (R,So). There is anisomorphism in D(S–Ro),

RHomR(U,R) ' RHomSo(U,S) .

Denoting this complex U∗ there are natural isomorphisms of functors,



U⊗LS – ' RHomS(U∗,–) : D(S)−→D(R) and

RHomR(U,–) ' U∗⊗LR – : D(R)−→D(S) .

Finally, U∗ has bounded homology and it has finite projective dimension over S andover Ro.

PROOF. The isomorphism of complexes follows from the definition of an invertiblecomplex and swap 7.5.11,

RHomR(U,R)' RHomR(U,RHomSo(U,U))

' RHomSo(U,RHomR(U,U))

' RHomSo(U,S) .

The first natural isomorphism of functors follows from homomorphism evalua-tion 8.4.11(a) and an application of the counitor (7.5.1.2); the second follows fromtensor evaluation 8.4.7(a) and the unitor (7.5.1.1).

It is evident from 9.2.4 that U∗ has bounded homology. In view of the first naturalisomorphism, it is immediate from 8.1.9 and 8.3.12 that pdS U∗ is finite. That alsopdRo U∗ is finite now follows from 9.2.3.

9.2.9 Theorem. Assume that R and S are projective as k-modules. Let R be leftNoetherian, let S be right Noetherian, and let U be invertible for (R,So). There is anadjoint equivalence of k-linear triangulated categories,

D(S)U⊗L

S–//D(R) .

RHomR(U,–)oo

It restricts to adjoint equivalences of triangulated subcategories,

DA(S)DA(R) , D@(S)D@(R) , and D@A(S)D@A(R) ,

and further to equivalences

P(S) P(R) , I(S) I(R) , and F(S) F(R) .

PROOF. The displayed functors are adjoints by 9.2.6 and they are k-linear and tri-angulated by 7.3.1 and 7.4.2. By assumption, the homothety morphism χU

SoR is anisomorphism in D(S–So), and it follows from 8.4.4 and 8.4.7(a) that the evaluationmorphism θUUN is an isomorphism in D(S) for every S-complex N. Thus, the com-mutative diagram (9.2.6.1) shows that the unit of the adjunction αU is an isomor-phism. Similarly, it follows from 8.4.9 and 8.4.11(a) and the commutative diagram(9.2.6.2) that the counit βU is an isomorphism.

It follows from 7.6.8 that the functor U⊗LR – maps DA(S) to DA(R), and as

fdSo U is finite the functor maps D@(S) to D@(R) by 8.3.12. Likewise it followsfrom 7.6.7 and 8.1.9 that the functor RHomR(U,–) maps D@(R) to D@(S) andDA(R) to DA(S). This establishes the first two restrictions of the equivalence, andthe third one follows by combining them.

The final three equivalences are immediate by 8.3.15 and 9.2.8.



REMARK. Rings are said to be derived Morita equivalent if their (bounded) derived categories areequivalent as triangulated categories. Thus, with T = k the theorem shows that R and S are derivedMorita equivalent if there exists an invertible complex for (R,So). Such a complex is a specialinstance of a tilting complex in D(R–So), and by a theorem of Rickard [71], R and S are derivedMorita equivalent if and only if there exists a tilting complex in D(R–So).

INVERSE COMPLEX

9.2.10 Lemma. Assume that R and S are projective as k-modules. Let R be leftNoetherian, let S be right Noetherian, and let U be invertible for (R,So). There isan isomorphism U⊗L

S RHomR(U,R)' R in D(R–Ro). Moreover, any complex V ∈D(S–Ro) that satisfies U⊗L

S V ' R in D(R–Ro) is isomorphic to RHomR(U,R).


Let U be an invertible complex for (R,So); with the notation from 9.2.8 there areby 9.2.10 and 9.2.3, and by commutativity 7.5.5, isomorphisms,

U⊗LS U∗ ' R and U∗⊗L

R U ' S ,

in D(R–Ro) and D(S–So). These isomorphisms suggest why U is called invertible.To fully justify the terminology, it remains to show that U∗ is invertible.

9.2.11 Lemma. Let R be Noetherian. For every complex M in DfA(R) the complex

RHomR(M,R) is in Df@(R

o), and if pdR M is finite, then pdRo RHomR(M,R) is finite.

PROOF. The first assertion is immediate from 7.6.12. If pdR M is finite, then by8.1.14 one can choose a semi-projective resolution P '−→ M with P bounded anddegreewise finitely generated. The Ro-complex RHomR(M,R) ' HomR(P,R) isbounded, and by 4.5.4 it is a complex of finitely generated projective modules, soalso pdRo RHomR(M,R) is finite.

9.2.12 Proposition. Assume that R is flat as a k-module and left Noetherian. For ev-ery R-complex M in Df

@A(R) of finite projective dimension, the biduality morphismδM

R : M→ RHomRo(RHomR(M,R),R) is an isomorphism in D(R).

PROOF. The vertical isomorphisms in the commutative diagram,

M⊗LRo R

µMυMR '

M⊗LχR

'// M⊗L

Ro RHomRo(R,R)

ηRRM'

MδM

// RHomRo(RHomR(M,R),R) ,

come from (7.5.1.1) and 8.4.9 and 8.4.11(a); it shows that δMR is an isomorphism.

9.2.13 Theorem. Assume that R and S are projective as k-modules and Noetherian.If U is an invertible complex for (R,So), then the complex U∗ from 9.2.8 is aninvertible complex for (S,Ro).



PROOF. By 9.2.8 the complex U∗ belongs to D@A(S–Ro) and it has finite projectivedimension over S and over Ro. Further, it has degreewise finitely generated homo-logy over either ring by 9.2.11. It remains to see that the homothety morphisms areisomorphisms. There is a commutative diagram in D(S–So),

SχU

'//

χU∗

RHomR(U,U)

' RHom(U,δU )

RHomRo(U∗,U∗)ζU∗RU

'// RHomR(U,RHomRo(U∗,R)) ,

where the isomorphisms come from biduality 9.2.12, swap 7.5.11, and the assump-tions on U . It shows that χU∗

SRo is an isomorphism in D(S–So); a similar diagram withthe rings interchanged shows that χU∗

RoS is an isomorphism in D(R–Ro).

9.2.14 Definition. Assume that R and S are projective as k-modules and Noetherian.For an invertible complex U for (R,So), the complex RHomR(U,R)'RHomSo(U,S)in D(S–Ro) is called the inverse of U and denoted U∗.

For rings that are Noetherian on both sides, 9.2.9 restricts to an equivalence ofthe subcategories of complexes with degreewise finitely generated homology.

9.2.15 Theorem. Assume that R and S are projective as k-modules and Noether-ian, and let U be invertible for (R,So). There is an adjoint equivalence of k-lineartriangulated categories,

Df(S)U⊗L

S–//Df(R) .

RHomR(U,–)oo

It restricts to adjoint equivalences of triangulated subcategories,

DfA(S)Df

A(R) , Df@(S)Df

@(R) , and Df@A(S)Df

@A(R) ,

and further to equivalences

Pf(S) Pf(R) , and If(S) If(R) .

PROOF. It is sufficient to prove that the equivalence D(S) D(R) established in9.2.9 restricts to an equivalence of triangulated subcategories Df(S)Df(R). As Uhas finite flat dimension, see 8.3.6, the functor U⊗L

S – is way-out by A.26. By A.28 itis thus enough to show that U⊗L

S N belongs to Df(R) for every finitely generated S-module N, and that follows from 7.6.13. By 9.2.8 one has RHomR(U,–)'U∗⊗L

R –,and the complex U∗ is invertible for (S,Ro) by 9.2.13, so it follows from what hasalready been proved that the functor RHomR(U,–) maps Df(R) to Df(S).

EXERCISES

E 9.2.1 Let R, S, and T be Noetherian. Show that if U is invertible for (R,So) and V is invertiblefor (S,T o), then U⊗L

S V is invertible for (R,T o).



E 9.2.2 Let R be projective as a k-module and Noetherian. Show that the invertible complexes forfor (R,Ro) form a group.

E 9.2.3 Let R be projective as a k-module and Noetherian. Show that if D and D′ are dualizingcomplexes for R, then RHomR(D,D′) is invertible for (R,Ro).

E 9.2.4 Let R be projective as a k-module and Noetherian. Show that if D is dualizing complexfor R and U is invertible for (R,Ro), then U⊗L

R D is dualizing for R.

9.3 Foxby Equivalence

SYNOPSIS. Auslander Category; Bass Category; Foxby Equivalence; projective dimension of flatmodules; parametrization of dualizing complexes; invertible complex.

Assume that R and S are projective as k-modules. Let R be left Noetherian, let S beright Noetherian, and let D be a dualizing complexe for (R,So). In this section, westudy the adjunction from 9.2.6:

D(S)D⊗L

S–//D(R) .

RHomR(D,–)oo

9.3.1 Lemma. Assume that R and S are projective as k-modules. Let R be leftNoetherian, let S be right Noetherian, and let D be a dualizing complex for (R,So).

(a) For N in D(S) with fdS N finite the unit αND : N→ RHomR(D,D⊗L

S N) is anisomorphism in D(S).

(b) For M in D(R) with idR M finite the counit βMD : D⊗L

S RHomR(D,M)→M isan isomorphism in D(R).

PROOF. Part (a) follows from (9.2.6.1) and tensor evaluation 8.4.7(b) and part (b)from (9.2.6.2) and homomorphism evaluation 8.4.11(b).

AUSLANDER AND BASS CATEGORIES

9.3.2 Lemma. Assume that R and S are projective as k-modules. Let R be left Noe-therian, S be right Noetherian, and D and D′ be a dualizing complexes for (R,So).

(a) For an S-complex N the unit αND is an isomorphism in D(S) if and only if αN

D′is an isomorphism. Furthermore, D⊗L

S N has bounded homology if and onlyif D′⊗L

S N has bounded homology.(b) For an R-complex M the counit βM

D is an isomorphism in D(R) if and only ifβM

D′ is an isomorphism. Furthermore, RHomR(D,M) has bounded homologyif and only if RHomR(D′,M) has bounded homology.

PROOF. Set U = RHomSo(D,D′); it follows from 9.1.8, 9.3.1, and 9.2.7 that the co-unit β= βD′

D : D⊗LRo U → D′ is an isomorphism in D(R–So). The diagram in D(S),


9.3 Foxby Equivalence 343

NαN

//

αN

RHomR(D′,D′⊗LS N)

' RHom(βυUD,D′⊗LN)

RHomR(D,D⊗LS N)

'RHom(D,δD⊗LN)

RHomR(U⊗LR D,D′⊗L

S N)

' ρDU(D′⊗LN)

RHomR(D,RHomR(U,D′)⊗LS N)

RHom(D,θUD′N)

'// RHomR(D,RHomR(U,D′⊗L

S N)) ,

is commutative. The vertical isomorphism on the left follows from biduality 9.1.25and the lower horizontal isomorphism is induced by tensor evaluation 8.4.4/8.4.7(a),as U is a complex in Df

@A(R) of finite projective dimension; see 9.1.27. Thus αND

is an isomorphism if and only if αND′ is an isomorphism. Furthermore, as one has

D′⊗LS N 'U⊗L

R (D⊗LS N) it follows from 7.6.8, 8.3.6, and 8.3.12 that if D⊗L

S N hasbounded homology, then so has D′⊗L

S N. This proves part (a).To prove part (b) one establishes a similar commutative diagram in D(R) from

the following sequences of isomorphisms induced by biduality 9.1.25 and homo-morphism evaluation 8.4.9/8.4.11(a),

D′⊗LS RHomR(D′,M)' D′⊗L

S RHomR(RHomSo(RHomR(D′,D),D),M)

' D′⊗LS (RHomR(D′,D)⊗L

S RHomR(D,M))

' D⊗LS RHomR(D,M) .

The lemma above ensures that the next definitions are unambiguous.

9.3.3 Definition. Assume that R and S are projective as k-modules. Let R be leftNoetherian, let S be right Noetherian, and let D be a dualizing complex for (R,So).

The gross Auslander Category of S, denoted A(S), is the full subcategory of D(S)defined by specifying its objects as follows,

A(S) = N ∈D(S) | the unit morphism αND is an isomorphism .

The Auslander Category of S is the following full subcategory of A(S),

A(S) = N ∈ A(S) | N and D⊗LS N have bounded homology ;

we also consider the full subcategory Af(S) =A(S)∩Df(S).The gross Bass Category of R, denoted B(R), is the full subcategory of D(R)

defined by specifying its objects as follows,

B(R) = M ∈D(R) | the counit morphism βMD is an isomorphism .

The Bass Category of R is the following full subcategory of B(R)

B(R) = M ∈ B(R) |M and RHomR(D,M) have bounded homology ;

we also considered the full subcategory Bf(R) =B(R)∩Df(R).



9.3.4. As the functors D⊗LS – and RHomR(D,−) are triangulated, it follows from

D.15 that A(S) and B(R) are triangulated subcategories of D(S) and D(R), andhence so are A(S) and B(R) in view of 7.6.4. If R and S are left Noetherian, thenAf(S) and Bf(R) are triangulated by 7.6.10.

The (gross) Auslander Category and Bass Category are by construction equiva-lent; a precise statement is made in 9.3.8.

9.3.5 Example. By 9.3.1 every S-complex of finite flat dimension belongs to A(S),and every R-complex of finite injective dimension belongs to B(R).

It follows from 8.3.12 and 7.6.8 that every complex in D@A(S) of finite flat dimen-sion belongs to A(S), in particular, S ∈A(S). And by 8.2.8 and 7.6.7 every complexin D@A(R) of finite injective dimension belongs to B(R), in particular, D ∈B(R).

9.3.6 Example. Recall from 9.1.20 that if R is Iwanaga–Gorenstein and projectiveas a k-module, then R is a dualizing complex for R. For such a ring one evidently has

A(R) = D(R) = B(R) and A(R) = D@A(R) = B(R) .

In particular, every R-module belongs to the Auslander/Bass Category but need nothave finite flat/injective dimension; see 8.1.11/8.2.11 and 8.3.19.

9.3.7 Proposition. Let X be a complex in D@(R–So). The following hold.(a) If R is left Noetherian, idR X is finite, and N ∈DA(S) is a complex of finite

flat dimension, then one has

idR(X⊗LS N) 6 idR X− infN .

In particular, if I is an R–So-bimodule injective over R and F is a flat S-module, then I⊗S F is an injective R-module.

(b) If S is right Noetherian, idSo X is finite, and M ∈D@(R) is a complex of finiteinjective dimension, then one has

fdS RHomR(X ,M) 6 supM+ idSo X .

In particular, if I is an R–So-bimodule injective over So and E is an injectiveR-module, then HomR(I,E) is a flat S-module.

PROOF. (a): For every left ideal a in R one has

RHomR(R/a,X⊗LS N) ' RHomR(R/a,X)⊗L

S N

by tensor evaluation 8.4.7(b). By 7.6.8 and 8.2.8 the right-hand complex is boundedbelow with minus infimum at most idR X− infN. The claim follows from 8.2.8.

(b): For every right ideal b in S one has

S/b⊗LS RHomR(X ,M) ' RHomR(RHomSo(S/b,X),M)

by commutativity 7.5.5 and homomorphism evaluation 8.4.11(b). By 7.6.7 and 8.2.8the right-hand complex is homologically bounded above with supremum at mostsupM+ idSo X , so the desired conclusion follows from 8.3.12.



FOXBY EQUIVALENCE

9.3.8 Theorem. Assume that R and S are projective as k-modules. Let R be leftNoetherian, S be right Noetherian, and D be a dualizing complex for (R,So). Thereis an adjoint equivalence of k-linear triangulated categories,

A(S)D⊗L

S–//B(R) .

RHomR(D,–)oo

It restricts to adjoint equivalences of triangulated categories

A(S)B(R) and further to F(S) I(R) .

PROOF. For every S-complex N and every R-complex M one has:

βD⊗L

SND D(D⊗L

S αND) = 1D⊗L

SN and RHomR(D,βMD )αRHomR(D,M)

D = 1RHomR(D,M) .

This follows as the functor D⊗LS – is left adjoint of RHomR(D,–). It is now imme-

diate from the definitions in 9.3.3 that the functors D⊗LS – and RHomR(D,–) yield

an adjoint equivalence A(S) B(R). It is evident that this restricts to an equiva-lence A(S) B(R). It follows from 9.3.5 that F(S) and I(R) are subcategories ofA(S) and B(R), respectively. That the already established equivalence restricts toan equivalence F(S) I(R) now follows from 9.3.7.

For rings that are Noetherian on both sides, 9.3.8 restricts to an equivalence ofthe subcategories of complexes with degreewise finitely generated homology.

9.3.9 Theorem. Assume that R and S are projective as k-modules and Noetherian,and let D be a dualizing complex for (R,So). There is an adjoint equivalence ofk-linear triangulated categories,

Af(S)D⊗L

S–//Bf(R) ,

RHomR(D,–)oo

and it restricts to an adjoint equivalence Pf(S) If(R).

PROOF. For an S-complex N in Af(S) it follows from 7.6.13 that the R-complexD⊗L

S N has degreewise finitely generated homology, so it is in Bf(R) by 9.3.8.For an R-complex M in Bf(R) one has

RHomR(D,M) ' RHomR(D,RHomSo(RHomR(M,D),D))

' RHomSo(RHomR(M,D),RHomR(D,D))

' RHomSo(RHomR(M,D),S)

by biduality 9.1.25, swap 7.5.11, and 9.1.17. The complex RHomR(M,D) belongsby 9.1.27 to Df

@A(So), so RHomSo(RHomR(M,D),S) has degreewise finitely gener-

ated homology over S by 7.6.12. Hence, RHomR(D,M) belongs to Af(S) by 9.3.8.By 9.3.8 and 9.1.2 the equivalence restricts to Pf(S) If(R).



9.3.10 Proposition. Assume that R and S are projective as k-modules. Let R be leftNoetherian, let S be right Noetherian, and let D be a dualizing complexes for (R,So).

(a) Let M be a complex of S–Ro-bimodules and let E be an S-complex of finiteinjective dimension. If M is in A(Ro), then RHomS(M,E) belongs to B(R),and the converse holds if E is a faithfully injective S-module.

(b) Let N be a complex of R–So-bimodules and let E be an R-complex of finiteinjective dimension. If N is in B(So), then RHomR(N,E) belongs to A(S),and the converse holds if E is a faithfully injective R-module.

PROOF. (a): There is a commutative diagram in D(R),

D⊗LS RHomR(D,RHomS(M,E))

βRHom(M,E)//

'D⊗LρDME

RHomS(M,E)

D⊗LS RHomS(M⊗L

R D,E) // RHomS(RHomSo(D,D⊗LRo M),E) ,

RHom(αM ,E)

OO

where the lower horizontal morphism comes from commutativity 7.5.5 and homo-morphism evaluation 8.4.9. If M is in A(Ro), then the homology of M is bounded,and hence so is the homology of RHomS(M,E) by 8.2.8 and 7.6.7. Furthermore,the homology of D⊗L

Ro M is bounded and hence so is the homology of the complexRHomR(D,RHomS(M,E)), as adjunction 7.5.15 and commutativity 7.5.5 yield

(?) RHomR(D,RHomS(M,E))' RHomS(D⊗LRo M,E) .

The morphism RHom(αMD ,E) is an isomorphism, and the assumption on D⊗L

Ro Mensures that lower horizontal morphism is an isomorphism; cf. 8.4.11(b). It followsthat βRHomS(M,E)

D is an isomorphism and hence RHomS(M,E) is in B(R).If E is a faithfully injective S-module and RHomS(M,E) is in B(R), then M

and D⊗LRo M have bounded homology by 2.5.6(b) and (?). Again, the lower vertical

morphism is an isomorphism, so it follows that RHom(αMD ,E) is an isomorphism,

and hence so is αMD .

(b): Similar to the proof of part (b). One expands the following chain of isomor-phisms into a commutative diagram:

RHomR(D,D⊗LS RHomR(N,E)) ' RHomR(D,RHomR(RHomSo(D,N),E))

' RHomR(D⊗LRo RHomSo(D,N),E)

' RHomR(N,E) .

PROJECTIVE DIMENSION OF FLAT MODULES

The utility of dualizing complexes is illustrated by the next results. The theoremcompares to Jensen’s theorem 8.5.16 and the corollary 9.3.12 provides the sameconclusion as Corollary 8.5.18, but under a different assumption.



9.3.11 Theorem. Assume that R and S are projective as k-modules. Let R be leftNoetherian, let S be right Noetherian, and let D be a dualizing complex for (R,So).For every S-complex N of finite flat dimension one has,

pdS N 6 maxidR D+ sup(D⊗LS N) , supN ;

in particular, N has finite projective dimension.

PROOF. The quantity n = maxidR D+ sup(D⊗LS N),supN is finite by 8.3.12. Let

P '−→ N be a semi-projective resolution over S. It suffices by 8.1.9 to show thatExt1S(Cn(P),Cn+1(P)) is zero. The S-module C = Cn+1(P) has finite flat dimension;indeed, the sequence 0→ Cu(P)→ Pu−1→ ·· · → Pn+2→C→ 0 is exact for everyu > n+ 2, and by 8.3.12 the module Cu(P) is flat for u 0. By 8.1.7, 9.3.5, andadjunction 7.5.15 there are isomorphisms of k-modules,

Ext1S(Cn(P),C) ∼= H−(n+1)(RHomS(N,C))

∼= H−(n+1)(RHomS(N,RHomR(D,D⊗LS C)))

∼= H−(n+1)(RHomR(D⊗LS N,D⊗L

S C)) .

It is now sufficient to show that infRHomR(D⊗LS N,D⊗L

S C) is at least −n, and thatfollows as 8.2.8 and 9.3.7 yield,

− infRHomR(D⊗LS N,D⊗L

S C) 6 idR(D⊗LS C)+ sup(D⊗L

S N)

6 idR D− infC+ sup(D⊗LS N)

6 n .

9.3.12 Corollary. Let R be projective as a k-module and Noetherian with a dualizingcomplex. For every R-complex M the quantities fdR M and pdR M are simultaneouslyfinite.


9.3.13 Corollary. Let R be projective as a k-module and Iwanaga–Gorenstein. Forevery M ∈D@A(R) the quantities fdR M, pdR M, and idR M are simultaneously finite.

PROOF. Recall from 9.1.20 that R is a dualizing complex for R. The assertions aretherefore immediate from 9.3.12 and 9.3.8.

9.3.14 Lemma. Assume that R and S are projective as k-modules. Let R be leftNoetherian, S be right Noetherian, and D be a dualizing complex for (R,So). Let Mbe an R-complex and N an S-complex; the following assertions hold.

(a) For every R-complex I of finite injective dimension and with H(I) non-zeroand bounded, one has− infRHomR(I,M)6 idR D+sup I− infRHomR(D,M).

(b) For every S-complex F of finite flat dimension and with H(F) non-zero andbounded, one has − infRHomS(N,F)6 idR D− infF + sup(D⊗L

S N).(c) For every So-complex E of finite injective dimension and with H(E) non-zero

and bounded, one has sup(E⊗LS N)6 idR D+ supE + sup(D⊗L

S N).



PROOF. (a) By 9.3.5 the R-complex I belongs to the Bass category, so one has

(?)RHomR(I,M) ' RHomR(D⊗L

S RHomR(D, I),M)

' RHomS(RHomR(D, I),RHomR(D,M))

where the second isomorphism is adjunction 7.5.15. The S-complex RHomR(D, I)has finite flat dimension by 9.3.8, so 9.3.11 yields the first inequality in the chain

pdS RHomR(D, I) 6 maxidR D+ sup(D⊗S RHomR(D, I)),supRHomR(D, I)6 maxidR D+ sup I,sup I− infD= idR D+ sup I .

The second inequality follows from 9.3.8 and 7.6.7, and finally the equality followsin view of 8.2.3. The desired inequality now follows from (?) and 8.1.9.

(b): By 9.3.5 the complex F belongs to the Auslander Category, so one has

RHomS(N,F) ' RHomS(N,RHomR(D,D⊗LS F)) ' RHomR(D⊗L

S N,D⊗LS F)

where the second isomorphism is adjunction 7.5.15. Now 8.2.8 and 9.3.7 yield

− infRHomS(N,F) 6 idR(D⊗LS F)+ sup(D⊗L

S N)

6 idR D− infF + sup(D⊗LS N) .

(c): The S-complex RHomk(E,E) ' Homk(E,E) has finite flat dimension by8.3.18, so by 2.5.6(b), adjunction 7.5.15, and part (b) one has

sup(E⊗LS N) = − infRHomk(E⊗L

S N,E)

= − infRHomS(N,RHomk(E,E))

6 idR D− infRHomk(E,E)+ sup(D⊗LS N)

= idR D+ supE + sup(D⊗LS N) .

PARAMETRIZATION OF DUALIZING COMPLEXES

We close this chapter with a theorem, due to Yekutieli, that says that dualizing com-plexes are parametrized by invertible complexes.

9.3.15 Lemma. Assume that R and S are projective as k-modules and Noetherian.If D is a dualizing complex for (R,So) and U is an invertible complex for (S,So),then the complex D⊗L

S U is dualizing for (R,So).

PROOF. By 9.2.15 and 9.3.9 the complex D⊗LS U belongs to If(So) and to If(R).

Furthermore, there is a commutative diagram in D(S–So),

SχD⊗LU

//

'χU

RHomR(D⊗LS U ,D⊗L

S U)

ρUD(D⊗LU)'

RHomS(U,U)RHom(U,αU )

'// RHomS(U,RHomR(D,D⊗L

S U)) ,



where the lower horizontal isomorphism comes from 9.3.5 and 9.2.7. The diagram

shows that χD⊗LUSoR is an isomorphism. To see that χD⊗LU

RSo is an isomorphism inD(R–Ro), expand the chain of isomorphisms,

RHomSo(D⊗LS U ,D⊗L

S U) ' RHomSo(D,RHomSo(U,U⊗LSo D)) ' RHomSo(D,D) ,

into a commutative diagram. The first isomorphism is adjunction 7.5.15, and thesecond is induced by the unit αD

U , which is an isomorphism by 9.2.9. It now followsfrom 9.1.17 that D⊗L

S U is a dualizing complex for (R,So).

9.3.16 Lemma. Assume that R and S are projective as k-modules and Noetherian.If D and D′ are dualizing complexes for (R,So), then the complex RHomR(D,D′) isinvertible for (S,So) with inverse RHomR(D′,D).

PROOF. It follows from 9.3.9 and from 9.1.27 in view of 9.1.8 that the complexU = RHomR(D,D′) belongs to Pf(S) and Pf(So). To prove that it is invertible,it remains to see that the homothety morphisms are isomorphisms; cf. 9.2.2. It isstraightforward to verify that there is a commutative diagram in D(S–So),

SχU

//

'χD′

RHomS(RHomR(D,D′),RHomR(D,D′))

RHomR(D′,D′) 'RHom(βD′ ,D′)

// RHomR(D⊗LS RHomR(D,D′),D′) ,

' ρRHom(D,D′)DD′OO

where the lower horizontal isomorphism comes from 9.3.5 and 9.2.7. It follows thatχU

SoS is an isomorphism. To see that also χUSSo is an isomorphism in D(S–So), expand

the following chain of isomorphisms into a commutative diagram,

RHomSo(RHomR(D,D′),RHomR(D,D′))

' RHomR(D,RHomSo(RHomR(D,D′),D′))

' RHomR(D,D)

' S .

The first isomorphism is swap 7.5.11 and the second follows from biduality 9.1.25.Finally, biduality 9.1.25 and swap 7.5.11 yield,

(?)

RHomR(D′,D)' RHomR(D′,RHomSo(RHomR(D,D′),D′))

' RHomSo(RHomR(D,D′),RHomR(D′,D′))

' RHomSo(U,S) ,

whence the inverse of U is RHomR(D′,D); see 9.2.14.

9.3.17 Theorem. Assume that R and S are projective as k-modules and Noetherian.If D is a dualizing complex for (R,So), then there is a bijective correspondence ofisomorphism classes

invertible complexes for (S,So) ←→ dualizing complexes for (R,So) .



For every invertible complex U for (S,So) and for every dualizing complex D′ for(R,So) the correspondance is given by

U 7−→ D⊗LS U and RHomR(D,D′) 7−→D′ .

PROOF. For every invertible complex U for (S,So) the complex D⊗LS U is dualizing

for (R,So) by 9.3.15. By 9.3.5 the complex U belongs to A(S), and hence there isan isomorphism U ' RHomR(D,D⊗L

S U) in D(S–So), see 9.2.7.For every dualizing complex D′ for (R,So) the complex RHomR(D,D′) is invert-

ible for (S,So) by 9.3.16. By 9.3.5 the complex D′ belongs to B(R), so there is anisomorphism D′ ' D⊗L

S RHomR(D,D′) in D(R–So), see 9.2.7.

9.3.18 Corollary. Assume that R and S are projective as k-modules and Noetherian.If D is a dualizing complex for (R,So), then there is a bijective correspondence ofisomorphism classes

invertible complexes for (S,So) ←→ invertible complexes for (R,Ro) .

For every invertible complex U for (S,So) and for every invertible complex V for(R,Ro) the correspondance is given by

U 7−→ RHomSo(D,D⊗LS U) and RHomR(D,D⊗L

Ro V ) 7−→V .

PROOF. Recall from 9.1.8 that a dualizing complex for (R,So) is also dualizing for(So,R). The assertions are now immediate from 9.3.17.

EXERCISES

E 9.3.1 Let R and S be projective as k-modules and Noetherian. Let D be a dualizing complex for(R,So), and let U be an invertible complex for (S,So). Show directly, i.e. without using9.3.18, that RHomSo(D,D⊗L

S U) is invertible for (R,Ro).


Chapter 10Gorenstein Homological Dimensions

10.1 Gorenstein Projective Dimension

SYNOPSIS. Totally acyclic complex of (finitely generated) projective modules; Gorenstein projec-tive module; Gorenstein projective dimension; ∼ is refinement of projective dimension; ∼ overNoetherian ring; ∼ of module.

GORENSTEIN PROJECTIVE MODULES

10.1.1 Definition. A complex P of projective R-modules is called totally acyclic ifit is acyclic and HomR(P,L) is acyclic for every projective R-module L.

An R-module G is called Gorenstein projective if one has G ∼= C0(P) for sometotally acyclic complex P of projective R-modules.

Notice that if P is a totally acyclic complex of projective R-modules, then themodule Cv(P) is Gorenstein projective for every v ∈ Z.

10.1.2 Example. Every projective module P is Gorenstein projective as the disccomplex D0(P) = 0→ P =−→ P→ 0 is totally acyclic.

10.1.3 Lemma. Let P be an acyclic complex of finitely generated projective R-modules. The following conditions are equivalent.

(i) HomR(P,R) is acyclic.(ii) P is totally acyclic.

(iii) HomR(P,F) is acyclic for every flat R-module F .(iv) E⊗R P is acyclic for every injective Ro-module E.

PROOF. The implications (iii)=⇒(ii)=⇒(i) are evident. Considering R as an R–Ro-bimodule one has HomR(P,R)⊗R F ∼= HomR(P,F) for every R-module F by tensorevaluation 1.4.6 and the unitor (1.2.1.1). This shows that (i) implies (iii). One alsohas E⊗R P ∼= HomRo(R,E)⊗R P ∼= HomRo(HomR(P,R),E) for every Ro-module Eby the counitor (1.2.1.2), commutativity 1.2.2, and homomorphism evaluation 1.4.9.From this isomorphism it follows that (i) and (iv) are equivalent.

351

352 10 Gorenstein Homological Dimensions

10.1.4 Example. The Dold complex D from 2.1.22 is a totally acyclic complex ofprojective Z/4Z-modules; this follows from 10.1.3 as HomZ/4Z(D,Z/4Z) ∼= D. Inparticular, the module C0(D) = (Z/4Z)/(2Z/4Z)∼= Z/2Z is Gorenstein projective.

The proof of the next result applies the argument known as Eilenberg’s swindle.

10.1.5 Proposition. Let X be a class of R-modules. Assume that X is closed un-der countable coproducts or countable products, and furthermore that X is closedunder kernels of surjective homomorphisms or under cokernels of injective homo-morphisms. The class X is closed under direct summands.

PROOF. Let X ∈ X and assume that X = M⊕N holds in M(R). Recall from 1.1.17that M⊕N is both a coproduct and a product. If X is closed under countable co-products, set Y = X (N), and if X is closed under countable products, set Y = XN. Ineither case, Y belongs to X and there is an isomorphism Y ∼= M⊕Y . Thus there areshort exact sequences 0→M→ Y → Y → 0 and 0→ Y → Y →M→ 0. Since X isclosed under kernels of surjective homomorphisms or under cokernels of injectivehomomorphisms, it follows that M belongs to X.

10.1.6 Proposition. The following assertions hold.(a) Let 0→ G′ → G→ G′′ → 0 be an exact sequence of R-modules. If G′′ is

Gorenstein projective, then G is Gorenstein projective if and only if G′ is so.If G′ and G are Gorenstein projective, then G′′ is Gorenstein projective if andonly if one has Ext1R(G

′′,L) = 0 for every projective R-module L.(b) Let Guu∈U be a family of R-modules. The module

∐u∈U Gu is Gorenstein

projective if and only if each Gu is Gorenstein projective. In particular, a directsummand of a Gorenstein projective module is Gorenstein projective.

PROOF. First note that an R-module G is Gorenstein projective if and only if thefollowing two conditions hold:

(1) One has ExtmR (G,L) = 0 for every m > 0 and every projective R-module L.(2) There is a complex P = 0→ P0 → P−1 → ··· of projective R-modules and

a quasi-isomorphism ε : G→ P such that HomR(ε,L) is a quasi-isomorphismfor every projective R-module L.

Now, let 0→ G′ α′−→ G α−→ G′′→ 0 be an exact sequence of R-modules where Gand G′′ are Gorenstein projective. As G and G′′ satisfy condition (1), so does G′

by 7.3.14. As G and G′′ satisfy (2) there exist complexes P = 0→ P0→ P−1→ ···and P′′ = 0→ P′′0 → P′′−1 → ··· of projective R-modules and quasi-isomorphismsε : G→ P and ε′′ : G′′→ P′′ such that HomR(ε,L) and HomR(ε

′′,L) are quasi-iso-morphisms for every projective R-module L. In view of 4.1.16 and 4.2.15 the com-plex HomR(Coneε,L) is acyclic for every projective R-module L, so A.12 im-plies that HomR(ε,L) is a quasi-isomorphism for every complex L of projective R-modules. There is an exact sequence of R-complexes 0→ G ε−→ P→ Cokerε→ 0where Cokerε= 0→Cokerε0→ P−1→ P−2→ ··· . The module Cokerε0 is Goren-stein projective, so it follows from 7.3.17 that HomR(ε,L) is surjective for every


10.1 Gorenstein Projective Dimension 353

complex L of projective R-modules. In particular, HomR(ε,P′′) is a surjective quasi-isomorphism, and hence the map C(R)(ε,P′′) : C(R)(P,P′′)→ C(R)(G,P′′) is sur-jective by 4.2.7 and 2.3.10. Thus there is a morphism of R-complexes β : P→ P′′

with βε = ε′′α. Let $ : Cone(1P′′) ΣP′′ be the canonical morphism 4.1.5. WithP′ = Ker(β Σ−1$) there is a commutative diagram in C(R) with exact rows,

(?)

0 // G′

ε′

α′// G

ε=(ε0)

α// G′′

ε′′

// 0

0 // P′ //

P⊕

Σ−1 Cone(1P′′)

(β Σ−1$)// P′′ // 0 ,

where ε′ is the induced morphism. As the complex Cone(1P′′) is contractible, see4.3.31, the morphism ε is a quasi-isomorphism and so is HomR(ε,L) for every pro-jective R-module L. The complex P′ consists of projective modules and is concen-trated in degrees 6 0 as this is the case for P, P′′ and Σ−1 Cone(1P′′); see 5.2.3.

As ε and ε′′ are quasi-isomorphisms, so is ε′ by 4.2.5 applied to the diagram (?).If L is a projective R-module, then application of the functor HomR(–,L) to (?)yields a commutative diagram with exact rows. This is because Ext1R(G

′′,L) = 0holds and because the bottom row in (?) is degreewise split exact. As HomR(ε,L)and HomR(ε

′′,L) are quasi-isomorphisms, so is HomR(ε′,L) by another application

of 4.2.5. This shows that the module G′ satisfies condition (2), and hence we haveproved that the class of Gorenstein projective modules is closed under kernels ofsurjective homomorphisms.

Next we show that the class of Gorenstein projective modules is closed undercoproducts. Let Guu∈U be a family of Gorenstein projective R-modules. By def-inition there exists for each u ∈ U a totally acyclic complex Pu of projective R-modules with Gu ∼= C0(Pu). The complex P =

∐u∈U Pu consists of projective mod-

ules by 1.3.20 and it is acyclic by 3.1.10. For every R-module L there is by 3.1.25an isomorphism HomR(P,L)∼=

∏u∈U HomR(Pu,L), and this complex is acyclic if L

is projective. Thus, P is a totally acyclic complex of projective R-modules. Sinceone has C0(P) ∼=

∐u∈U C0(Pu) ∼=

∐u∈U Gu, cf. 3.1.8, we conclude that

∐u∈U Gu is

Gorenstein projective.Having established these properties, it now follows from 10.1.5 that the class of

Gorenstein projective modules is closed under direct summands. This proves (b).We now finish the proof of (a). To show that the class of Gorenstein projective

modules is closed under extensions, let 0→G′→G→G′′→ 0 be an exact sequencewhere G′ and G′′ are Gorenstein projective. By the definition 10.1.1 of Gorensteinprojective modules, there exists an exact sequence 0→G′→ P′→G′′′→ 0 with P′

projective and G′′′ Gorenstein projective. By 3.2.23 there is a commutative diagramwith exact rows and columns,



(‡)

0

0

0 // G′

// G

// G′′ // 0

0 // P′

// P′tG′ G

// G′′ // 0

G′′′

G′′′

0 0 .

As G′′ is Gorenstein projective and P′ is projective one has Ext1R(G′′,P) = 0. Thus

the second row in (‡) splits by 7.3.15, so one has P′tG′ G ∼= P′⊕G′′. This moduleis Gorenstein projective by (b). Now the already established part of (a) applied tothe second column in (‡) shows that G is Gorenstein projective.

It remains to prove the final assertion in (a). The “only if” part is evident. For theconverse, consider again the diagram (‡). By assumption, P′ is projective and G′, G,and G′′′ are Gorenstein projective. The second column and the already establishedpart of (a) show that P′ tG′ G is Gorenstein projective. As Ext1R(G

′′,P′) = 0 holds,by assumption, we get from 7.3.15 that the second row splits. Hence part (b) showsthat G′′ is Gorenstein projective.

10.1.7 Lemma. Let N be the class of R-modules N with ExtiR(G,N) = 0 for everyGorenstein projective R-module G and all i > 0. Let 0→ N′→ N→ N′′→ 0 be anexact sequence of R-modules. If two of the modules N′, N, and N′′ are in N, then sois the third. In particular, every R-module N with pdR N or idR N finite belongs to N.

PROOF. It follows from 7.3.14 that N′,N′′ ∈ N implies N ∈ N, and that N′,N ∈ N

implies N′′ ∈ N. Now assume that N,N′′ ∈ N. Another application of 7.3.14 yieldsExtiR(G,N′) = 0 for every Gorenstein projective R-module G and all i > 1. To showthat also Ext1R(G,N′) = 0 holds, note that every Gorenstein projective module G bydefinition fits into an exact sequence 0→G→ P→G′→ 0 with P projective and G′

Gorenstein projective. Application of 7.3.14 to this sequence gives an isomorphismExt1R(G,N′)∼= Ext2R(G

′,N′), and the right-hand side is zero, as just shown. The lastassertion now follows as N contains every projective and every injective R-module,see 10.1.1 and 8.2.17.

GORENSTEIN PROJECTIVE DIMENSION

10.1.8 Definition. Let M be an R-complex. The Gorenstein projective dimension ofM, written GpdR M, is defined as

GpdR M = inf

n ∈ Z∣∣∣∣ There exists a semi-projective replacement P of M withHv(P) = 0 for all v > n and Cn(P) Gorenstein projective

with the convention inf∅= ∞. One says that GpdR M is finite if GpdR M < ∞ holds.



10.1.9. Let M be an R-complex. For every semi-projective replacement P of M onehas H(P)∼= H(M); the next (in)equalities are hence immediate from the definition,

GpdR M > supM and GpdRΣs M = GpdR M+ s for every s ∈ Z .

Moreover, one has GpdR M =−∞ if and only if M is acyclic.

10.1.10 Lemma. Let M be an R-complex. For every semi-projective replacement Pof M and every integer v> GpdR M the module Cv(P) is Gorenstein projective.

PROOF. By 8.1.13 and 10.1.6 it suffices to prove the assertion for some specificsemi-projective replacement P of M. We can assume that GpdR M is finite; otherwisethe statement is empty. We can also assume that M is not acyclic; otherwise P =0 is a semi-projective replacement of M. Thus g = GpdR M is an integer and bydefinition, 10.1.8, there is a semi-projective replacement P of M with Hv(P) = 0for all v > g and Cg(P) Gorenstein projective. Since there are short exact sequences0→ Cv+1(P)→ Pv→ Cv(P)→ 0 for all v> g, we conclude from 10.1.2 and 10.1.6that the modules Cg(P),Cg+1(P), . . . are Gorenstein projective.

The following result is sometimes expressed by saying that GpdR is a refinementof pdR. It follows, in particular, that a Gorenstein projective module is either pro-jective or has infinite projective dimension. The Gorenstein projective module from10.1.4 has infinite projective dimension; see 8.1.11.

10.1.11 Theorem. Let M be an R-complex. There is an inequality,

GpdR M 6 pdR M ,

and equality holds if M has finite projective dimension.

PROOF. The inequality is evident from the definitions of the dimensions, see 8.1.2and 10.1.8, and from the fact that every projective module is Gorenstein projec-tive, see 10.1.2. Now assume that p = pdR M is an integer. To prove GpdR M > pit must be shown that if P is a semi-projective replacement of M with Hv(P) = 0for all v > n and Cn(P) Gorenstein projective, then n > p holds. Suppose one hasn < p. There are exact sequences 0→Cp(P)→ Pp−1→···→ Pn+1→Cn+1(P)→ 0and 0→ Cn+1(P)→ Pn→ Cn(P)→ 0. The first sequence shows that the moduleCn+1(P) has finite projective dimension, as Cp(P) is projective by 8.1.9, and hence7.3.15 and 10.1.7 imply that the latter sequence splits. Thus, Cn(P) is a direct sum-mand of Pn, in particular, Cn(P) is projective. Now another application of 8.1.9yields p6 n, which is a contradiction. We conclude that n> p, as desired.

Equality also holds in 10.1.11 if M has finite injective dimension; see 10.1.18.

10.1.12 Proposition. Let M′ → M → M′′ → ΣM′ be a distinguished triangle inD(R). With g′ = GpdR M′, g = GpdR M, and g′′ = GpdR M′′ there are inequalities,

g′ 6maxg,g′′−1 , g6maxg′,g′′ , and g′′ 6maxg′+1,g .

In particular, if two of the complexes M′, M, and M′′ have finite Gorenstein projec-tive dimension, then so has the third.



PROOF. It suffices to prove the second inequality since the first and third inequalitiesfollow by applying the second inequality and 10.1.9 to the distinguished trianglesΣ−1 M′′→M′→M→M′′ and M→M′′→ ΣM′→ ΣM; see (TR2) in D.2.

To prove the second inequality, apply 6.5.26 to get an exact sequence of com-plexes 0→ P′→ P→ P′′→ 0 where P′, P, and P′′ are semi-projective replacementsof M′, M, and M′′. Set s′ = supM′, s = supM, and s′′ = supM′′. We may assumethat g′ and g′′ are finite and that M is not acyclic; otherwise the inequality is trivial.It follows from 2.2.20 that M′ and M′′ can not both be acyclic, so g′ or g′′ is aninteger, and hence so is m = maxg′,g′′. Note that 10.1.9 and 6.5.23 yield m > s,so Hv(P) = 0 for v > m. As m+1 > g′′ > s′′ one has Hm+1(P′′) = 0, so the sequence0→ Cm(P′)→ Cm(P)→ Cm(P′′)→ 0 is exact by 2.2.13. As m> g′ and m> g′′ themodules Cm(P′) and Cm(P′′) are Gorenstein projective by 10.1.10, and hence so isCm(P) by 10.1.6. Thus one has g = GpdR M 6 m by 10.1.8.

10.1.13 Proposition. Let M be an R-complex of finite Gorenstein projective dimen-sion g = GpdR M. For every semi-projective replacement P of M and every integeru with g > u there is a distinguished triangle in D(R),

K −→M −→ N −→ ΣK ,

where the complexes K and N have the following properties:(a) There is a degreewise split exact sequence 0→ Pďu→K→ Σu G→ 0 in C(R)

where G is a Gorenstein projective R-module. Furthermore, one has

GpdR K 6 u and Hv(K)∼=

0 for v> u+1Hv(M) for v6 u−1 .

(b) The complex N satisfies

pdR N = g and Hv(N)∼=

Hv(M) for v> u+20 for v6 u .

Furthermore, there is an exact sequence of R-modules,

(10.1.13.1) 0−→ Hu+1(M)−→ Hu+1(N)−→ Hu(K)−→ Hu(M)−→ 0 .

PROOF. If M is acyclic the statement is void as no integer u satisfies −∞ = g > u.Now assume that M is not acyclic, in which case g is an integer. By 10.1.10 the mo-dule Cg(P) is Gorenstein projective and by 10.1.1 there exists an acyclic complexL = 0→ Cg(P)→ Lg−1 · · · → Lu→ G→ 0, concentrated in degrees g, . . . ,u− 1,where the modules Lg−1, . . . ,Lu are projective and the cokernels are Gorenstein pro-jective. Set Lg = Cg(P) and Lu−1 = G and notice that, in particular, the cokernelCu(L)∼= G is Gorenstein projective.

Consider the short exact sequence of R-complexes,

(?) 0−→ Σg−1 Cg(P)

α−−→ Lďg−1 −→ LĎg−1 −→ 0 ,

where α is a quasi-isomorphism as LĎg−1 is acyclic; see 4.2.6. Let F be any com-plex of projective R-modules. As the complex LĎg−1 consists of Gorenstein projec-tive modules, it follows from 7.3.17 and 10.1.7 that the functor HomR(–,F) leaves



the sequence (?) exact. Since LĎg−1 is an acyclic complex with Cv(LĎg−1) Goren-stein projective for every v, it follows from 10.1.7 and A.11 that HomR(LĎg−1,Fn)is acyclic for every n ∈ Z, and hence HomR(LĎg−1,F) is acyclic by A.12. It nowfollows from 4.2.6 that HomR(α,F) is a surjective quasi-isomorphism, whencethe morphism C(R)(α,F) is surjective as well by 4.2.7 and 2.3.10. Surjectivity ofC(R)(α,Pďg−1) yields a commutative diagram of R-complexes,

Σg−1 Cg(P)β//

'α

Pďg−1

Lďg−1γ// Pďg−1 ,

where β is induced by ∂Pg . This diagram—in conjunction with the definition of dis-

tinguished triangles in D(R), see 6.5.6, the axiom (TR3) in D.2, and 6.5.22—showsthat the complexes Coneβ and Coneγ are isomorphic in D(R). Evidently, one hasConeβ = PĎg, and this complex is isomorphic to P ' M in D(R). Consequently,Coneγ 'M holds in D(R).

Set C = Coneγ. By 2.5.17 and 6.5.19 there is a distinguished triangle in D(R),

Cďu −→C −→Cěu+1 −→ ΣCďu ,

which we argue is the desired one. As already noticed, C'M in D(R). The complex

K = Cďu = 0−→ Pu⊕G−→ Pu−1 −→ Pu−2 −→ ·· ·

fits into the degreewise split exact sequence in C(R),

(‡) 0−→ Pďu −→ K −→ Σu G−→ 0 .

The complex P is semi-projective, and so is Pěu+1 by 5.2.8. Hence 5.2.16 applied tothe exact sequence 0→ Pďu → P→ Pěu+1 → 0 shows that Pďu is semi-projective.It follows from 8.1.2 and 10.1.11 that GpdR(Pďu) = pdR(Pďu) 6 u holds. As G is aGorenstein projective module, one has GpdR(Σ

u G) 6 u, with equality if G is non-zero, so application of 6.5.19 and 10.1.12 to (‡) shows that GpdR K 6 u. The asser-tion about the homology of K =Cďu is immediate as C 'M in D(R). Note that

N = Cěu+1 = 0−→ Lg−1 −→ Pg−1⊕Lg−2 −→ ·· · −→ Pu+1⊕Lu −→ 0

is a complex of projective R-modules concentrated in degrees g, . . . ,u+ 1. In theextremal case u = g−1 one has N = Σg Lg−1. In particular, N is semi-projective by5.2.8 and one has pdR N 6 g. Note that 10.1.12 yields g 6 maxGpdR K,GpdR N.As one has GpdR K 6 u < g we get GpdR N > g, so it follows from 10.1.11 thatpdR N = g holds. The assertion about the homology of N = Cěu+1 is immediate asC 'M in D(R). The exact sequence (10.1.13.1) follows by applying 6.5.22 to theconstructed distinguished triangle.

10.1.14 Corollary. Let M be an R-module of finite Gorenstein projective dimensiong = GpdR M. The following assertions hold.



(a) There is an exact sequence of R-modules 0→M→ X → G→ 0 where G isGorenstein projective and pdR X = g.

(b) If g > 0, then there is an exact sequence of R-modules 0→ X →G→M→ 0where G is Gorenstein projective and pdR X = g−1.

PROOF. (a): For u =−1 the sequence (10.1.13.1) reads

0−→M −→ H0(N)−→ H−1(K)−→ 0 .

If follows from 10.1.13 that K is isomorphic to Σ−1 H−1(K) in D(R), whence onehas GpdR H−1(K)− 1 = GpdR K 6 u = −1. Consequently, the module H−1(K) isGorenstein projective. Similarly, one has N ' H0(N) and pdR H0(N) = pdR N = g.

(b): As g > 0 one can apply (10.1.13.1) with u = 0 to obtain the exact sequence

0−→ H1(N)−→ H0(K)−→M −→ 0 .

If follows from 10.1.13 that K is isomorphic to H0(K) in D(R), whence one hasGpdR H0(K) =GpdR K 6 u= 0. That is, the module H0(K) is Gorenstein projective.Similarly, one has N ' ΣH1(N) and pdR H1(N)+1 = pdR N = g.

10.1.15 Proposition. Let P and L be semi-projective R-complexes. If L has finiteprojective or finite injective dimension, then the morphism

HomR(τPĎn,L) : HomR(PĎn,L) −→ HomR(P,L)

is a quasi-isomorphism for every integer n> GpdR P.

PROOF. Given a semi-projective R-complex F that is isomorphic to L in the derivedcategory, there is by 6.4.20 and 5.2.20 a homotopy equivalence λ : L→ F . From10.1.9 one gets n > supP, so the map τP

Ďn : P→ PĎn is a quasi-isomorphism by4.2.4. Choose a semi-injective resolution ι : F '−→ I. In the commutative diagram,

HomR(PĎn,L)Hom(PĎn,λ)

u//

Hom(τPĎn,L)

HomR(PĎn,F)

Hom(τPĎn,F)

Hom(PĎn,ι)// HomR(PĎn, I)

Hom(τPĎn,I)'

HomR(P,L)Hom(P,λ)

u// HomR(P,F)

Hom(P,ι)

'// HomR(P, I) ,

the left-hand horizontal maps are homotopy equivalences by 4.3.19 while HomR(P, ι)and HomR(τ

PĎn, I) are quasi-isomorphisms by semi-projectivity of P and semi-

injectivity of I. The diagram shows that it suffices to verify that HomR(PĎn, ι) isa quasi-isomorphism. Set C = Cone ι and notice that it is an acyclic complex ofmodules that are direct sums of projective and injective modules. The goal is toshow that the complex HomR(PĎn,C)∼= ConeHomR(PĎn, ι) is acyclic; cf. 4.1.15.

By (2.5.20.3) there is an exact sequence 0→ PĚn+1→ P→ PĎn→ 0. By 10.1.2and 10.1.10 these are complexes of Gorenstein projective modules, and it followsfrom 7.3.17 and 10.1.7 that the sequence

0−→ HomR(PĎn,C)−→ HomR(P,C)−→ HomR(PĚn+1,C)−→ 0



is exact. The middle complex is acyclic as P is semi-projective and C is acyclic.To prove acyclicity of HomR(PĎn,C) it suffices to argue that HomR(PĚn+1,C) isacyclic; see 2.2.20. To that end, it suffices by A.10 to show that HomR(G,C) isacyclic for every Gorenstein projective module G. It is already known from 10.1.7that one has ExtiR(G,Cv) = 0 for all v ∈ Z and all i > 0, so it suffices by A.9 to arguethat ExtiR(G,Zv(C)) = 0 holds for v 0 and all i > 0.

If L has finite projective dimension, then one can assume that the complexes Fand I are bounded above; see 8.1.2 and 5.3.26. It follows that C is bounded above,in particular, Zv(C) = 0 holds for v 0.

If L has finite injective dimension, then one can assume that the complexes Iand F are bounded below; see 8.2.2 and 5.2.15. It follows that C is bounded be-low; in particular, Zv(C) = 0 holds for for v 0. Let G be a Gorenstein projectiveR-module. For every v ∈ Z one has ExtiR(G,Cv) = 0 for all i > 0, so in view of10.1.7, induction on the exact sequences 0→ Zv(C)→ Cv → Zv−1(C)→ 0 yieldsExtiR(G,Zv(C)) = 0 for all v ∈ Z and all i > 0.

10.1.16 Corollary. Let M be an R-complex of finite Gorenstein projective dimen-sion and N be an R-complex of finite projective or finite injective dimension. Forevery semi-projective replacement P of M, every semi-projective replacement L ofN, and every integer n> GpdR M there is an isomorphism in D(k),

RHomR(M,N) ' HomR(PĎn,L) .


10.1.17 Theorem. Let M be an R-complex of finite Gorenstein projective dimensionand let n be an integer. The following conditions are equivalent.

(i) GpdR M 6 n.(ii) − infRHomR(M,N) 6 n− infN holds for every R-complex N with pdR N fi-

nite or idR N finite.(iii) − infRHomR(M,N)6 n holds for every projective R-module N.(iv) n> supM and Extn+1

R (M,N) = 0 for every R-module N with pdR N finite.(v) n> supM and for some, equivalently every, semi-projective replacement P of

M, the module Cv(P) is Gorenstein projective for every v> n.(vi) There is a semi-projective resolution P '−→M with Pv = 0 for every v < infM,

Hv(P) = 0 for every v > n, and Cv(P) Gorenstein projective for every v> n.In particular, there are equalities

GpdR M = sup− infRHomR(M,N) | N is an R-module with pdR N finite= supm ∈ Z | ExtmR (M,N) 6= 0 for some projective R-module N .

PROOF. We start by establishing the equivalence of (i), (ii), and (iii).(i)=⇒ (ii): One can assume that N is in DA(R) and not acyclic; otherwise the

inequality is trivial. In this case, u = infN is an integer. By 5.2.15 there is a semi-projective resolution L '−→ N with Lv = 0 for v < u. If pdR N or idR N is finite, then10.1.16 implies that one has RHomR(M,N)' HomR(PĎn,L), where P is any semi-projective replacement of M. For every v < u−n and p ∈ Z one of the inequalities



p > n or p+ v6 n+ v < u holds, so the module

HomR(PĎn,L)v =∏

p∈ZHomR((PĎn)p,Lp+v)

is zero. In particular, one has Hv(RHomR(M,N)) = 0 for v < u−n = infN−n.(ii)=⇒ (iii): Trivial.(iii)=⇒ (i): By assumption, g = GpdR M is finite. We must show that (iii) implies

g6 n. We can assume that M is not acyclic as otherwise the inequality is trivial. Bydefinition there exists a semi-projective replacement P of M with Cg(P) Gorensteinprojective. By 10.1.16 there is an isomorphism,

(?) RHomR(M,L) ' HomR(PĎg,L) ,

for every projective R-module L. Recall from 10.1.9 that g > supM = supP. Weconsider two different cases:

First assume that g = supP holds; this implies Hg(P) 6= 0 so the homomorphismCg(P)→ Pg−1 is not injective. Since Cg(P) is Gorenstein projective there exists, inparticular, an embedding Cg(P) L where L is a projective module. This map doesnot admit a factorization Cg(P)→ Pg−1→ L, as this would force Cg(P)→ Pg−1 tobe injective. Thus the map HomR(Pg−1,L)→ HomR(Cg(P),L) is not surjective andhence infHomR(PĎg,L) =−g. Now (?) and (iii) yield g =− infRHomR(M,L)6 n.

Next assume that g > supP. In this case there is a short exact sequence of mod-ules, 0→Cg(P)→Pg−1→Cg−1(P)→ 0. As g> supP and g=GpdR M, the moduleCg−1(P) is not Gorenstein projective. Hence 10.1.6 yields Ext1R(Cg−1(P),L) 6= 0 forsome projective module L, and by 8.1.7 this means that H−g(RHomR(M,L)) 6= 0.Hence g6− infRHomR(M,L)6 n, where the last inequality holds by (iii).

To finish the proof we show (ii)=⇒(iv)=⇒ (v)=⇒(vi)=⇒(i).(ii)=⇒ (iv): The second assersion in (iv) is immediate from (ii). The inequality

n> supM follows, in view of 10.1.9, from (i), which is equivalent to (ii).(iv)=⇒ (v): First note that by 8.1.13 and 10.1.6 the “some” version and the “ev-

ery” version of condition (v) are equivalent. By assumption, g = GpdR M is finite, soin any semi-projective replacement P of M the module Cv(P) Gorenstein projectivefor every integer v > g; see 10.1.10. Thus, to show (v) it is enough to prove n> g.Assume towards a contradiction that n < g holds. Notice that by the assumptionn> supM, the module Cn(P) can not be Gorenstein projective. There is an exact se-quence 0→ Cg(P)→ Pg−1→ ··· → Pn→ Cn(P)→ 0, which shows that Cn(P) hasfinite Gorenstein projective dimension, as Pěn is a semi-projective replacement ofΣn Cn(P). Now 10.1.14 yields an exact sequence 0→ N→ G→ Cn(P)→ 0 whereG is Gorenstein projective and N has finite projective dimension. By 8.1.7 and (iv)one has Ext1R(Cn(P),N)∼= Extn+1

R (M,N) = 0, so the sequence splits by 7.3.15. Now10.1.6 implies that Cn(P) is Gorenstein projective, which is a contradiction.

(v)=⇒ (vi): Immediate in view of 5.2.15.(vi)=⇒ (i): Immediate from the definition, 10.1.8, of GpdR.

10.1.18 Theorem. Let M be an R-complex. If M has finite injective dimension, thenone has GpdR M = pdR M.



PROOF. The inequality GpdR M 6 pdR M holds by 10.1.11. To show the oppositeinequality, we may assume that g = GpdR M is an integer. Set u = infM−1, whichis an integer by 8.2.3, and note that one has u < infM 6 supM 6 g by 10.1.9. Thus10.1.13 yields a distinguished triangle in D(R),

(?) K −→M −→ N −→ ΣK ,

with GpdR K 6 u and pdR N = g. As idR M is finite, 10.1.17 shows that one has

− infRHomR(K,M)6 GpdR K− infM 6 u− infM = −1 ,

and hence H0(RHomR(K,M)) = 0. By 7.3.10 this means that D(R)(K,M) = 0, inparticular, the morphism K→M in the distinguished triangle above is zero. By D.17this means the (?) splits, and hence N 'M⊕ΣK holds in D(R). In particular onehas pdR M 6 pdR N = g, as claimed.

10.1.19 Proposition. For every family of R-complexes Muu∈U one has,

GpdR(∐

u∈U

Mu) = supu∈UGpdR Mu .

PROOF. To prove the inequality “6”, one can assume that the right-hand side isfinite, say, s ∈ Z. By the definition, 10.1.8, of Gorenstein projective dimension andby 10.1.10 every Mu admits a semi-projective replacement Pu with Hv(Pu) = 0 forall v > s and Cv(Pu) Gorenstein projective for all v> s. Now P =

∐u∈U Pu is a semi-

projective replacement of∐

u∈U Mu with Hv(P) = 0 for all v > s, see 5.2.17 and3.1.10; and Cv(P) =

∐u∈U Cv(Pu) Gorenstein projective for all v> s, see 10.1.6.

To prove the opposite inequality “>” it suffices, as each Mu is a direct summandin∐

u∈U Mu, to argue that if M′ is a direct summand in an R-complex M, then onehas GpdR M′ 6 GpdR M. To this end, we may assume that M is not acyclic and thatg = GpdR M is finite. Let M′′ be an R-complex with M = M′⊕M′′. Let P′ and P′′ besemi-projective replacements of M′ and M′′. Now P = P′⊕P′′ is a semi-projectivereplacement of M, see 5.2.17. As Hv(P) = 0 holds for every v > g, even for everyv > supM, one has Hv(P′) = 0 for every v > g by 3.1.10. One gets from 10.1.10 thatCv(P) = Cv(P′)⊕Cv(P′′) is Gorenstein projective for all v> g, and hence Cv(P′) isGorenstein projective for all v> g by 10.1.6. Thus GpdR M′ 6 g holds.

NOETHERIAN RINGS AND HOMOLOGICAL FINITENESS

10.1.20 Lemma. Let G be a finitely generated R-module. If G is Gorenstein projec-tive, then there exists an exact sequence 0→G→ F→G′→ 0 of finitely generatedR-modules with F free and G′ Gorenstein projective.

PROOF. By assumption there exists a totally acyclic complex P of projective R-modules with G ∼= C0(P). Without loss of generality, one can assume that the mo-dule P−1 is free. Indeed, there exists a projective module P′ such that P−1⊕P′ is free,see 1.3.15, and the complex P = P⊕D−1(P′) is totally acyclic with C0(P) = C0(P);cf. 10.1.2. As P is acyclic, there is an injective homomorphism ι : G→ P−1, and



since G is finitely generated, the image of ι is contained in a finitely generated freesubmodule F of P−1. It remains to show that G′ = F/Im ι is Gorenstein projective,and to this end it suffices by 10.1.6 to show that Ext1R(G

′,L) is zero for every pro-jective R-module L. The commutative diagram

0 // G // F //

G′ //

0

0 // G // P−1 // C−1(P) // 0

yields for every projective R-module L a commutative diagram with exact rows,

HomR(P−1,L) //

HomR(G,L) // Ext1R(C−1(P),L)

// 0

HomR(F,L) // HomR(G,L) // Ext1R(G′,L) // 0 ;

see 6.5.22, 7.3.7, and 7.3.14. As the complex HomR(P,L) is acyclic, one hasExt1R(C−1(P),L) = 0, and it follows that that Ext1R(G

′,L) vanishes as well.

10.1.21 Proposition. Assume that R is left Noetherian. A finitely generated R-module G is Gorenstein projective if and only if there exists a totally acyclic com-plex P of finitely generated free R-modules with G∼= C0(P).

PROOF. The “if” statement evident. To prove the converse, assume that G is Goren-stein projective. It follows from 10.1.20 that there is an injectiv quasi-isomorphismι : G→ F , where F is a complex of finitely generated free R-modules with Fv = 0for v > 0 and Cv(F) Gorenstein projective for every v ∈ Z. By 5.1.13 there is asurjective semi-free resolution π : L '−→ G where L is degreewise finitely gener-ated with Lv = 0 for v < 0. The complex P = Σ−1 Cone ιπ is a complex of finitelygenerated free R-modules with C0(P) = C0(L) ∼= G. Moreover, for v 0 the mo-dule Cv(P) = Cv+1(F) is Gorenstein projective, so by 10.1.7 and A.11 the complexHomR(P,L) is acyclic for every projective module L. That is, P is totally acyclic.

10.1.22 Lemma. Assume that R is left Noetherian. Let N be the class of R-modulesN with ExtiR(G,N) = 0 for every finitely generated Gorenstein projective R-moduleG and all i > 0. Let 0→ N′→ N→ N′′→ 0 be an exact sequence of R-modules. Iftwo of the modules N′, N, and N′′ are in N, then so is the third. In particular, everymodule N with fdR N or idR N finite is in N.

PROOF. It follows from 7.3.14 that N′,N′′ ∈ N implies N ∈ N, and that N′,N ∈ N

implies N′′ ∈ N. Now assume that N,N′′ ∈ N. Another application of 7.3.14 yieldsExtiR(G,N′) = 0 for every finitely generated Gorenstein projective R-module G andall i > 1. To show that also Ext1R(G,N′) = 0 holds, note that by 10.1.20 every finitelygenerated Gorenstein projective module G fits in an exact sequence of finitely gen-erated modules 0→ G→ F → G′ → 0 with F free and G′ Gorenstein projective.Application of 7.3.14 to this sequence yields Ext1R(G,N′) ∼= Ext2R(G

′,N′), and the



right-hand side is zero, as just shown. The last assertion now follows as N containsevery flat and every injective R-module; see 10.1.21, 10.1.3, and 8.2.17.

REMARK. Finitely generated Gorenstein projective modules are also known as totally reflexivemodules; see Avramov and Martsinkovsky [8]. See also 10.4.8.

The next result applies by 5.1.14 to complexes M in DfA(R) of finite Gorenstein

projective dimension.

10.1.23 Proposition. Assume that R is left Noetherian. Let M be a complex in Df(R)of finite Gorenstein projective dimension g = GpdR M. For every degreewise finitelygenerated semi-projective replacement P of M and every integer u with g > u thereis a distinguished triangle in Df(R),

K −→M −→ N −→ ΣK ,

where the complexes K and N have the properties listed in 10.1.13.

PROOF. The proof of 10.1.13 applies with one change: By 10.1.21 one can assumethat the acyclic complex L is degreewise finitely generated.

10.1.24 Corollary. Let R be left Noetherian and M be a finitely generated R-moduleof finite Gorenstein projective dimension g = GpdR M. The next assertions hold.

(a) There is an exact sequence 0→ M → X → G→ 0 of finitely generated R-modules where G is Gorenstein projective and pdR X = g.

(b) If g > 0, then there is an exact sequence 0→ X → G→ M → 0 of finitelygenerated R-modules where G is Gorenstein projective and pdR X = g−1.

PROOF. The proof of 10.1.14 applies when one replaces references to 10.1.13 withreferences to 10.1.23.

10.1.25 Proposition. Assume that R is left Noetherian. Let L be a bounded belowand degreewise finitely generated complex of projective R-moduls and F be a semi-flat R-complex. If F has finite flat or finite injective dimension, then the morphism

HomR(τLĎn,F) : HomR(LĎn,F) −→ HomR(L,F)

is a quasi-isomorphism for every integer n> GpdR L.

PROOF. Given a semi-projective R-complex P that is isomorphic to F in the derivedcategory, there is by 6.4.20 a quasi-isomorphism π : P→ F . From 10.1.9 one getsn > supL, so the map τL

Ďn : L→ LĎn is a quasi-isomorphism by 4.2.4. Choose asemi-injective resolution ι : P '−→ I. In the commutative diagram,

HomR(LĎn, I)

Hom(τLĎn,I) '

HomR(LĎn,P)Hom(LĎn,ι)oo

Hom(τLĎn,P)

Hom(LĎn,π)

'// HomR(LĎn,F)

Hom(τLĎn,F)

HomR(L, I) HomR(L,P)Hom(L,ι)

'oo

Hom(L,π)

'// HomR(L,F) ,



the right-hand horizontal maps are quasi-isomorphisms by C.24, while HomR(L, ι)and HomR(τ

LĎn, I) are quasi-isomorphisms by semi-projectivity of L, see 5.2.8, and

semi-injectivity of I. The diagram shows that it suffices to verify that HomR(LĎn, ι)is a quasi-isomorphism. Set C = Cone ι and notice that it is an acyclic complex ofmodules that are direct sums of projective and injective R-modules. The goal is toshow that the complex HomR(LĎn,C)∼= ConeHomR(LĎn, ι) is acyclic; cf. 4.1.15.

By (2.5.20.3) there is an exact sequence 0→ LĚn+1→ L→ LĎn→ 0. By 10.1.2and 10.1.10 these are complexes of finitely generated Gorenstein projective mod-ules, and it follows from 7.3.17 and 10.1.22 that the sequence

0−→ HomR(LĎn,C)−→ HomR(L,C)−→ HomR(LĚn+1,C)−→ 0

is exact. The middle complex is acyclic as L is semi-projective and C is acyclic.To prove acyclicity of HomR(LĎn,C) it suffices to argue that HomR(LĚn+1,C) isacyclic; see 2.2.20. To this end, it suffices by A.10 to show that HomR(G,C) isacyclic for every finitely generated Gorenstein projective module G. It is alreadyknown from 10.1.22 that one has ExtiR(G,Cv) = 0 for all v ∈ Z and all i > 0, so itsuffices by A.9 to argue that ExtiR(G,Zv(C)) = 0 holds for v 0 and all i > 0.

If F has finite flat dimension, then one can assume that I is bounded above; see8.3.3 and 5.3.26. Moreover, the modules Cv(P) are flat for v 0 by 5.4.10 and8.3.12. For v 0 one thus has Zv(C)∼= Cv+1(C) = Cv(P), and as these modules areflat, it follows from 10.1.22 that ExtiR(G,Zv(C)) = 0 holds for v 0 and all i > 0.

If F has finite injective dimension, then one can assume that the complexes Iand P are bounded below; see 8.2.2 and 5.2.15. It follows that C is bounded be-low; in particular, Zv(C) = 0 holds for for v 0. Let G be a Gorenstein projectiveR-module. For every v ∈ Z one has ExtiR(G,Cv) = 0 for all i > 0, so in view of10.1.22, induction on the exact sequences 0→ Zv(C)→Cv→ Zv−1(C)→ 0 yieldsExtiR(G,Zv(C)) = 0 for all v ∈ Z and all i > 0.

10.1.26 Corollary. Assume that R is left Noetherian. Let M be a complex in DfA(R)

of finite Gorenstein projective dimension and N an R-complex of finite flat or finiteinjective dimension. For every bounded below and degreewise finitely generatedsemi-projective replacement L of M, every semi-flat replacement F of N, and everyinteger n> GpdR M there is an isomorphism in D(k),

RHomR(M,N) ' HomR(LĎn,F) .


10.1.27 Theorem. Let R be left Noetherian, let M be a complex in DfA(R) of finite

Gorenstein projective dimension, and let n ∈ Z. The next conditions are equivalent.(i) GpdR M 6 n.

(ii) − infRHomR(M,N)6 n− infN holds for every R-complex N with fdR N finiteor idR N finite.

(iii) − infRHomR(M,R)6 n holds.(iv) n> supM and Extn+1

R (M,N) = 0 holds for every finitely generated R-moduleN with pdR N finite.



(v) There is a degreewise finitely generated semi-projective resolution P '−→ Mwith Pv = 0 for every v < infM, and Hv(P) = 0 for every v > n, and Cv(P)Gorenstein projective for every v> n.


GpdR M = − infRHomR(M,R) = supm ∈ Z | ExtmR (M,R) 6= 0 .PROOF. The implication (i)=⇒ (iv) follows from 10.1.17 and (v)=⇒ (i) is immedi-ate from the definition, 10.1.8. Moreover, (ii)=⇒ (iii) is trivial.

(i)=⇒ (ii): The complex M has by 5.1.14 a bounded below and degreewisefinitely generated semi-projective replacement. The argument in the proof of 10.1.17now applies when one refers to 10.1.26 in place of 10.1.16.

(iii)=⇒ (i): Let L '−→M be a semi-free resolution with L degreewise finitely gen-erated and bounded below; see 5.1.14. The functor HomR(L,–), which by 7.3.1 isRHomR(M,–), preserves coproducts of modules by 3.1.31 applied degreewise. Itfollows that − infRHomR(M,P) 6 n holds for every free, and hence every projec-tive, R-module P. Now invoke 10.1.17.

(iv)=⇒ (v): Let P '−→ M be a a semi-free resolution with P degreewise finitelygenerated and Pv = 0 for every v < infM; see 5.1.14. By assumption, g = GpdR Mis finite, so the module Cv(P) Gorenstein projective for every integer v > g; see10.1.10. It now suffices to show that n> g. Assume that n < g holds. Notice thatby the assumption n > supM, the module Cn(P) can not be Gorenstein projec-tive. There is an exact sequence 0→ Cg(P)→ Pg−1 → ··· → Pn → Cn(P)→ 0,which shows that Cn(P) has finite Gorenstein projective dimension, as Pěn is asemi-projective replacement of Σn Cn(P). Now 10.1.24 yields an exact sequence0 → N → G → Cn(P) → 0 of finitely generated R-modules where G is Goren-stein projective and N has finite projective dimension. By 8.1.7 and (iv) one hasExt1R(Cn(P),N) ∼= Extn+1

R (M,N) = 0, so the sequence splits by 7.3.15. Now 10.1.6implies that Cn(P) is Gorenstein projective, which is a contradiction.

THE CASE OF MODULES

10.1.28. Notice from 10.1.17 that a non-zero R-module is Gorenstein projective ifand only if it has Gorenstein projective dimension 0 as an R-complex.

10.1.29 Theorem. Let M be an R-module of finite Gorenstein projective dimensionand let n> 0 be an integer. The following conditions are equivalent.

(i) GpdR M 6 n.(ii) ExtmR (M,N) = 0 holds for every R-module N with pdR N finite or idR N finite

and all integers m > n.(iii) ExtmR (M,N) = 0 holds for every projective R-module N and all integers m> n.(iv) Extn+1

R (M,N) = 0 holds for every R-module N with pdR N finite.(v) In some/every projective resolution · · · → Pv → Pv−1 → ··· → P0 → M→ 0

the module Coker(Pv+1→ Pv) is Gorenstein projective for every v> n.



(vi) There is an exact sequence of R-modules 0→G→Pn−1→·· ·→P0→M→ 0with each Pi projective and G Gorenstein projective.


GpdR M = supm ∈ N0 | ExtmR (M,N) 6= 0 for some projective R-module N .PROOF. In a projective resolution · · · → Pv → Pv−1 → ·· · → P0 → M → 0, themorphism P0 M yields a semi-projective resolution of M, considered as an R-complex; cf. 5.2.27. In particular, the complex · · · → Pv→ Pv−1→ ···→ P0→ 0 is asemi-projective replacement of M. Conversely, a semi-projective resolution P '−→Mwith Pv = 0 for all v < 0 yields a projective resolution of M. The equivalence of thesix conditions is now immediate from 10.1.17.

10.1.30 Theorem. Assume that R is left Noetherian. Let M be a finitely generatedR-module of finite Gorenstein projective dimension and let n> 0 be an integer. Thefollowing conditions are equivalent.

(i) GpdR M 6 n.(ii) ExtmR (M,N) = 0 holds for every R-module N with fdR N finite or idR N finite

and all integers m > n.(iii) ExtmR (M,R) = 0 holds for all integers m > n.(iv) Extn+1

R (M,N) = 0 for every finitely generated R-module N with pdR N finite.(v) There is an exact sequence 0→ G→ Pn−1 → ··· → P0 →M→ 0 of finitely

generated R-modules with each Pi projective and G Gorenstein projective.In particular, there is an equality

GpdR M = supm ∈ N0 | ExtmR (M,R) 6= 0 .PROOF. An argument parallel to the proof of 10.1.29 shows that the statement fol-lows from 10.1.27.

REMARK. Examples by Jorgensen and Sega [44] show that in 10.1.30 one can not dispense withthe a priori assumption that GpdR M is finite.

EXERCISES

E 10.1.1 Let n ∈ Z. Show that an R-complex M has GpdR M 6 n if and only if there is a is adiagram P τ−→ P′ π−→ M in C(R) where P is a totally acyclic complex of projectiveR-modules, τv is an isomorphism for v 0 and π is a semi-projective resolution.

E 10.1.2 The invariant FGPD(R) = supGpdR M |M is an R-module with GpdR M < ∞ is cal-led the finitistic Gorenstein projective dimension of R. Show that FGPD(R) = FPD(R).

E 10.1.3 Let N be the class of Ro-modules N with TorRi (N,G) = 0 for every finitely generated

Gorenstein projective R-module G and all i > 0. Show that if in an exact sequence ofRo-modules 0→ N′ → N → N′′ → 0 any two of the modules N′, N, and N′′ are in N,then so is the third. Conclude that every Ro-module N with fdRo N is in N. Concludefurther that if R is left Noetherian, then every Ro-module N with idRo N finite is in N.

E 10.1.4 Assume that R is left Noetherian. Let L be a degreewise finitely generated semi-projective R-complex and E a semi-injective R-complex. Show that if E has finite flat orfinite injective dimension, then E⊗R τ

LĎn : E⊗R L→ E⊗R LĎn is a quasi-isomorphism

for every integer n> GpdR L.


10.2 Gorenstein Injective Dimension 367

10.2 Gorenstein Injective Dimension

SYNOPSIS. Totally acyclic complex of injective modules; Gorenstein injective module; Gorensteininjective dimension; ∼ is refinement of injective dimension; ∼ of module.

GORENSTEIN INJECTIVE MODULES

10.2.1 Definition. A complex I of injective R-modules is called totally acyclic if itis acyclic and HomR(E, I) is acyclic for every injective R-module E.

An R-module G is called Gorenstein injective if one has G ∼= Z0(I) for sometotally acyclic complex I of injective R-modules.

Notice that if I is a totally acyclic complex of injective R-modules, then the mo-dule Zv(I) is Gorenstein injective for every v ∈ Z.

10.2.2 Example. Every injective module I is Gorenstein injective as the disc com-plex D1(I) = 0→ I =−→ I→ 0 is totally acyclic.

10.2.3 Example. As the ring R = Z/4Z is self-injective, see 8.2.10, the Dold com-plex D from 2.1.22 is an acyclic complex of injective R-modules. To see that it istotally acyclic, note that p= 2Z/4Z is the only prime ideal in R. The endomorphismof R that multiplies with [2]4Z has kernel p, so there is an injective homomorphismR/p→ R whose image is clearly essential. Hence one has ER(R/p) ∼= R, see B.12and B.16. It now follows from F.4 and F.10 that every injective R-module E is free,that is, E ∼= R(U) for some set U . For any such module E there are isomorphisms

HomR(E,D) ∼= HomR(R(U),D) ∼= HomR(R,D)U ∼= DU

by 3.1.25 and 4.4.1; in particular, the complex HomR(E,D) is acyclic. Thus the mo-dule Z0(D) = Ker([2]4Z ·) = p (which is isomorphic to R/p) is Gorenstein injective.

10.2.4 Proposition. The following assertions hold.(a) Let 0→ G′ → G→ G′′ → 0 be an exact sequence of R-modules. If G′ is

Gorenstein injective, then G is Gorenstein injective if and only if G′′ is so. IfG and G′′ are Gorenstein injective, then G′ is Gorenstein injective if and onlyif one has Ext1R(E,G

′) = 0 for every injective R-module E.(b) Let Guu∈U be a family of R-modules. The module

∏u∈U Gu is Gorenstein

injective if and only if each Gu is Gorenstein injective. In particular, a directsummand of a Gorenstein injective module is Gorenstein injective.

PROOF. First note that an R-module G is Gorenstein injective if and only if thefollowing two conditions hold:

(1) There is a complex I = · · · → I1→ I0→ 0 of injective R-modules and a quasi-isomorphism π : I→ G such that HomR(E,π) is a quasi-isomorphism for ev-ery injective R-module E.

(2) One has ExtmR (E,G) = 0 for every m > 0 and every injective R-module E.



Now, let 0→ G′ α′−→ G α−→ G′′→ 0 be an exact sequence of R-modules where Gand G′ are Gorenstein injective. As G and G′ satisfy condition (2), so does G′′ by7.3.14. As G and G′ satisfy (1) there exist complexes I = · · · → I1→ I0→ 0 andI′ = · · · → I′1 → I′0 → 0 of injective R-modules and quasi-isomorphisms π : I→ Gand π′′ : I′→ G′ such that HomR(E,π) and HomR(E,π′) are quasi-isomorphisms forevery injective R-module E. By 4.1.15 and 4.2.15 the complex HomR(E,Coneπ) isacyclic for every injective R-module E, so A.10 implies that HomR(E,π) is a quasi-isomorphism for every complex E of injective R-modules. There is an exact se-quence 0→ Kerπ→ I π−→ G→ 0 where Kerπ= · · · → I2→ I1→ Kerπ0→ 0. Themodule Kerπ0 is Gorenstein injective, so it follows from 7.3.16 that HomR(E,π) issurjective for every complex E of injective R-modules. In particular, HomR(I′,π)is a surjective quasi-isomorphism, and hence C(R)(I′,π) : C(R)(I′, I)→ C(R)(I′,G)is surjective by 4.2.7 and 2.3.10. Thus there exists a morphism of R-complexesβ : I′→ I with πβ = α′π′. Let ι : I′ Cone(1I′) be the canonical morphism 4.1.5.With I′′ = Coker

(βι

)there is a commutative diagram in C(R) with exact rows,

(?)0 // I′

π′

(βι

)//

I⊕

Cone(1I′)

π=(π0)

// I′′

π′′

// 0

0 // G′ α′// G α

// G′′ // 0 ,

where π′′ is the induced morphism. As the complex Cone(1I′) is contractible, see4.3.31, the morphism π is a quasi-isomorphism and so is HomR(E, π) for everyinjective R-module E. The complex I′′ consists of injective modules and is concen-trated in degrees > 0 as this is the case for I, I′ and Cone(1I′); see 5.3.7.

As π and π′ are quasi-isomorphisms, so is π′′ by 4.2.5 applied to the diagram (?).If E is an injective R-module, then application of the functor HomR(E,–) to (?)yields a commutative diagram with exact rows. This is because Ext1R(E, I

′) = 0holds and because the top row in (?) is degreewise split exact. As HomR(E, π) andHomR(E,π′) are quasi-isomorphisms, so is HomR(E,π′′) by another application of4.2.5. This shows that the module G′′ satisfies condition (1), and hence we haveproved that the class of Gorenstein injective modules is closed under cokernels ofinjective homomorphisms.

Next we show that the class of Gorenstein injective modules is closed underproducts. Let Guu∈U be a family of Gorenstein injective R-modules. By defini-tion there exists for each u ∈U a totally acyclic complex Iu of injective R-moduleswith Gu ∼= Z0(Iu). The complex I =

∏u∈U Iu consists of injective modules by 1.3.23

and it is acyclic by 3.1.22. For every R-module E there is by 3.1.23 an isomor-phism HomR(E, I) ∼=

∏u∈U HomR(E, Iu), and this complex is acyclic if E is in-

jective. Thus, I is a totally acyclic complex of injective R-modules. Since one hasZ0(I) ∼=

∏u∈U Z0(Iu) ∼=

∏u∈U Gu, see 3.1.20, we conclude that

∏u∈U Gu is Goren-

stein injective.Having established these properties, it now follows from 10.1.5 that the class of

Gorenstein injective modules is closed under direct summands. This proves (b).



We now finish the proof of (a). To show that the class of Gorenstein injectivemodules is closed under extensions, let 0→G′→G→G′′→ 0 be an exact sequencewhere G′ and G′′ are Gorenstein injective. By the definition 10.2.1 of Gorensteininjective modules, there exists an exact sequence 0→ G′′′→ I′′→ G′′→ 0 with I′′

injective and G′′′ Gorenstein injective. By 3.3.27 there is a commutative diagramwith exact rows and columns,

(‡)

0

0

G′′′

G′′′

0 // G′ // GuG′′ I′′ //

I′′

// 0

0 // G′ // G

// G′′

// 0

0 0 .

As G′ is Gorenstein injective and I′′ is injective one has Ext1R(I′′,G′) = 0. Thus

the first row in (‡) splits by 7.3.15, so one has GuG′′ I′′ ∼= G′⊕ I′′. This module isGorenstein injective by (b). Now the already established part of (a) applied to thefirst column in (‡) shows that G is Gorenstein injective.

It remains to prove the final assertion in (a). The “only if” part is evident. Forthe converse, consider again the diagram (‡). By assumption, I′′ is injective and G,G′′, and G′′′ are Gorenstein injective. The first column and the already establishedpart of (a) show that GuG′′ I′′ is Gorenstein injective. As Ext1R(I

′′,G′) = 0 holds, byassumption, we get from 7.3.15 that the first row splits. Hence part (b) shows thatG′ is Gorenstein injective.

10.2.5 Lemma. Let N be the class of R-modules N with ExtiR(N,G) = 0 for everyGorenstein injective R-module G and all i > 0. Let 0→ N′→ N → N′′→ 0 be anexact sequence of R-modules. If two of the modules N′, N, and N′′ are in N, then sois the third. In particular every R-module N with pdR N or idR N finite belongs to N.

PROOF. It follows from 7.3.14 that N′,N′′ ∈ N implies N ∈ N, and that N,N′′ ∈N

implies N′ ∈ N. Now assume that N′,N ∈ N. Another application of 7.3.14 yieldsExtiR(N

′′,G) = 0 for every Gorenstein injective R-module G and all i > 1. To showthat also Ext1R(N

′′,G) = 0 holds, note that every Gorenstein injective module G bydefinition fits into an exact sequence 0→ G′→ I→ G→ 0 with I injective and G′

Gorenstein injective. Application of 7.3.14 to this sequence gives an isomorphismExt1R(N

′′,G)∼= Ext2R(N′′,G′), and the right-hand side is zero, as just shown. The last

assertion now follows as N contains every injective and every projective R-module,see 10.2.1 and 8.1.19.



GORENSTEIN INJECTIVE DIMENSION

10.2.6 Definition. Let M be an R-complex. The Gorenstein injective dimension ofM, written GidR M, is defined as

GidR M = inf

n ∈ Z∣∣∣∣ There exists a semi-injective replacement I of M withHv(I) = 0 for all v <−n and Z−n(I) Gorenstein injective

with the convention inf∅= ∞. One says that GidR M is finite if GidR M < ∞ holds.

10.2.7. Let M be an R-complex. For every semi-injective replacement I of M onehas H(I)∼= H(M); the next (in)equalities are hence immediate from the definition,

GidR M > − infM and GidRΣs M = GidR M− s for every s ∈ Z .

Moreover, one has GidR M =−∞ if and only if M is acyclic.

10.2.8 Lemma. Let M be an R-complex. For every semi-injective replacement I ofM and every integer v> GidR M the module Z−v(I) is Gorenstein injective.

PROOF. By 8.2.13 and 10.2.4 it suffices to prove the assertion for some specificsemi-injective replacement I of M. We can assume that GidR M is finite; otherwisethe statement is empty. We can also assume that M is not acyclic; otherwise I = 0is a semi-injective replacement of M. Thus g = GidR M is an integer and by def-inition, 10.2.6, there is a semi-injective replacement I of M with Hv(I) = 0 forall v < −g and Z−g(I) Gorenstein injective. Since there are short exact sequences0→ Z−v(I)→ I−v→ Z−(v+1)(I)→ 0 for all v > g, we conclude from 10.2.2 and10.2.4 that the modules Z−g(I),Z−(g+1)(I), . . . are Gorenstein injective.

The following result is sometimes expressed by saying that GidR is a refinementof idR. It follows, in particular, that a Gorenstein injective module is either injectiveor has infinite injective dimension. The Gorenstein injective module from 10.2.3 hasinfinite injective dimension; see 8.2.11.


GidR M 6 idR M ,

and equality holds if M has finite injective dimension.

PROOF. The inequality is evident from the definitions of the dimensions, see 8.2.2and 10.2.6, and from the fact that every injective module is Gorenstein injective,see 10.2.2. Now assume that ı = idR M is an integer. To prove GidR M > ı it mustbe shown that if I is a semi-injective replacement of M with Hv(I) = 0 for allv < −n and Z−n(I) Gorenstein injective, then n > ı holds. Suppose one has n < ı.There is an exact sequence 0→ Z−n(I)→ I−n→ Z−(n+1)(I)→ 0 and an exact se-quence 0→ Z−(n+1)(I)→ I−(n+1)→ ·· · → I−(ı−1)→ Z−ı(I)→ 0. The second se-quence shows that the module Z−(n+1)(I) has finite injective dimension, as Z−ı(I) isinjective by 8.2.8, and hence 7.3.15 and 10.2.5 imply that the first sequence splits.Thus, Z−n(I) is a direct summand of I−n, in particular, Z−n(I) is injective. Nowanother application of 8.2.8 yields ı6 n, which is a contradiction.



Equality also holds in 10.2.9 if M has finite projective dimension; see 10.2.16.

10.2.10 Proposition. Let M′ → M → M′′ → ΣM′ be a distinguished triangle inD(R). With g′ = GidR M′, g = GidR M, and g′′ = GidR M′′ there are inequalities,

g′ 6maxg,g′′+1 , g6maxg′,g′′ , and g′′ 6maxg′−1,g .

In particular, if two of the complexes M′, M, and M′′ have finite Gorenstein injectivedimension, then so has the third.

PROOF. It suffices to prove the second inequality since the first and third inequalitiesfollow by applying the second inequality and 10.2.7 to the distinguished trianglesΣ−1 M′′→M′→M→M′′ and M→M′′→ ΣM′→ ΣM; see (TR2) in D.2.

To prove the second inequality, apply 6.5.27 to get an exact sequence of com-plexes 0→ I′→ I→ I′′→ 0 where I′, I, and I′′ are semi-injective replacements ofM′, M, and M′′. Set u′ = infM′, u = infM, and u′′ = infM′′. We may assume thatg′ and g′′ are finite and that M is not acyclic; otherwise the inequality is trivial. Itfollows from 2.2.20 that M′ and M′′ can not both be acyclic, so g′ or g′′ is an inte-ger, and hence so is m = maxg′,g′′. Note that 10.2.7 and 6.5.23 yield m>−u, soHv(I) = 0 for v <−m. As m+1 > g′ >−u′ one has H−m−1(I′) = 0, so the sequence0→ Z−m(I′)→ Z−m(I)→ Z−m(I′′)→ 0 is exact by 2.2.13. As m > g′ and m > g′′

the modules Z−m(I′) and Z−m(I′′) are Gorenstein injective by 10.2.8, and hence sois Z−m(I) by 10.2.4. Thus one has g = GidR M 6 m by 10.2.6.

10.2.11 Proposition. Let M be an R-complex of finite Gorenstein injective dimen-sion g = GidR M. For every semi-injective replacement I of M and every integer wwith g > w there is a distinguished triangle in D(R),

Σ−1 K −→ N −→M −→ K ,

where the complexes K and N have the following properties:(a) There is a degreewise split exact sequence 0→ Σ−w G→ K → Iě−w → 0 in

C(R) where G is a Gorenstein injective R-module. Furthermore, one has

GidR K 6 w and Hv(K)∼=

Hv(M) for v>−w+10 for v6−w−1 .


idR N = g and Hv(N)∼=

0 for v>−wHv(M) for v6−w−2 .


(10.2.11.1) 0−→ H−w(M)−→ H−w(K)−→ H−w−1(N)−→ H−w−1(M)−→ 0 .

PROOF. If M is acyclic the statement is void as no integer w satisfies −∞ = g > w.Now assume that M is not acyclic, in which case g is an integer. By 10.2.8 themodule Z−g(I) is Gorenstein injective and by 10.2.1 there exists an acyclic R-complex E = 0→ G→ E−w · · · → E−g+1→ Z−g(I)→ 0, concentrated in degrees−w+1, . . . ,−g, where the modules E−w, . . . ,E−g+1 are injective and the kernels



are Gorenstein injective. Set E−w+1 = G and E−g = Z−g(I) and notice that, in par-ticular, the kernel Z−w(E)∼= G is Gorenstein injective.


(?) 0−→ EĚ−g+1 −→ Eě−g+1α−−→ Σ

−g+1 Z−g(I)−→ 0 ,

where α is a quasi-isomorphism as EĚ−g+1 is acyclic; see 4.2.6. Let J be any com-plex of injective R-modules. As the complex EĚ−g+1 consists of Gorenstein injectivemodules, it follows from 7.3.16 and 10.2.5 that the functor HomR(J,–) leaves the se-quence (?) exact. As EĚ−g+1 is an acyclic complex with Zv(EĚ−g+1) Gorenstein in-jective for every v, it follows from 10.2.5 and A.9 that HomR(Jn,EĚ−g+1) is acyclicfor every n ∈ Z, and hence HomR(J,EĚ−g+1) is acyclic by A.10. It now followsfrom 4.2.6 that HomR(J,α) is a surjective quasi-isomorphism, whence the morphismC(R)(J,α) is surjective as well by 4.2.7 and 2.3.10. Surjectivity of C(R)(Iě−g+1,α)yields a commutative diagram of R-complexes,

Iě−g+1γ// Eě−g+1

' α

Iě−g+1β// Σ−g+1 Z−g(I) ,

where β is induced by ∂I−g+1. This diagram—in conjunction with the definition of

distinguished triangles in D(R), see 6.5.5, the axiom (TR3) in D.2, and 6.5.22—shows that the complexes Coneβ and Coneγ are isomorphic in D(R). Evidently,one has Σ−1 Coneβ= IĚ−g, and this complex is isomorphic to I 'M in D(R). Con-sequently, Σ−1 Coneγ 'M holds in D(R).

Set C = Coneγ. By 2.5.17 and 6.5.19, and by the axiom (TR2) in D.2, there is adistinguished triangle in D(R),

Σ−1 (Σ−1 Cě−w+1)−→ Σ

−1 Cď−w −→ Σ−1 C −→ Σ

−1 Cě−w+1 ,

which we argue is the desired one. As noticed, Σ−1 C 'M in D(R). The complex

K = Σ−1 Cě−w+1 = · · · −→ I−w+2 −→ I−w+1 −→ I−w⊕G−→ 0

is concentrated in degrees >−w and fits into the degreewise split exact sequence

(‡) 0−→ Σ−w G−→ K −→ Iě−w −→ 0 .

The complex I is semi-injective, and so is Iď−w−1 by 5.3.12. Hence 5.3.20 applied tothe exact sequence 0→ Iď−w−1→ I→ Iě−w→ 0 shows that Iě−w is semi-injective.It follows from 8.2.2 and 10.2.9 that GidR(Iě−w) = idR(Iě−w) 6 w holds. As Gis a Gorenstein injective module, one has GidR(Σ

−w G) 6 w, with equality if G isnon-zero, so application of 6.5.19 and 10.2.10 to (‡) shows that GidR K 6 w. Theassertion about the homology of K = Σ−1 Cě−w+1 follows as Σ−1 C 'M in D(R).Notice that

N = Σ−1 Cď−w = 0−→ E−w⊕ I−w−1 −→ ·· · −→ E−g+2⊕ I−g+1 −→ E−g+1 −→ 0



is a complex of injective R-modules concentrated in degrees −w−1, . . . ,−g. In theextremal case w= g−1 one has N = Σ−g E−g+1. In particular, N is semi-injective by5.3.12 and one has idR N 6 g. Note that 10.2.10 yields g6maxGidR N,GidR K. Asone has GidR K 6 w < g we get GidR N > g, so it follows from 10.2.9 that idR N = gholds. The assertion about the homology of N = Σ−1 Cď−w follows as Σ−1 C'M inD(R). The exact sequence (10.1.13.1) follows by applying 6.5.22 to the constructeddistinguished triangle.

10.2.12 Corollary. Let M be an R-module of finite Gorenstein injective dimensiong = GidR M. The following assertions hold.

(a) There is an exact sequence of R-modules 0→ G→ X →M→ 0 where G isGorenstein injective and idR X = g.

(b) If g > 0, then there is an exact sequence of R-modules 0→M→G→ X → 0where G is Gorenstein injective and idR X = g−1.

PROOF. (a): For w =−1 the sequence (10.2.11.1) reads

0−→ H1(K)−→ H0(N)−→M −→ 0 .

If follows from 10.2.11 that K is isomorphic to ΣH1(K) in D(R), whence one hasGidR H1(K)− 1 = GidR K 6 w = −1. Consequently, the module H1(K) is Goren-stein injective. Similarly, one has N ' H0(N) and idR H0(N) = idR N = g.

(b): As g > 0 one can apply (10.2.11.1) with w = 0 to obtain the exact sequence

0−→M −→ H0(K)−→ H−1(N)−→ 0 .

If follows from 10.2.11 that K is isomorphic to H0(K) in D(R), whence one hasGidR H0(K) = GidR K 6 w = 0. That is, the module H0(K) is Gorenstein injective.Similarly, one has N ' Σ−1 H−1(N) and idR H−1(N)+1 = idR N = g.

10.2.13 Proposition. Let I and E be semi-injective R-complexes. If E has finiteprojective or finite injective dimension, then the morphism

HomR(E,τIĚ−n) : HomR(E, IĚ−n) −→ HomR(E, I)

is a quasi-isomorphism for every integer n> GidR I.

PROOF. Given a semi-injective R-complex J that is isomorphic to E in the derivedcategory, there is by 6.4.21 and 5.3.24 a homotopy equivalence ι : J→ E. From10.2.7 one gets −n 6 inf I, so the map τI

Ě−n : IĚ−n→ I is a quasi-isomorphism by4.2.4. Choose a semi-projective resolution π : P '−→ J. In the commutative diagram,

HomR(E, IĚ−n)Hom(ι,IĚ−n)

u//

Hom(E,τIĚ−n)

HomR(J, IĚ−n)

Hom(J,τIĚ−n)

Hom(π,IĚ−n)// HomR(P, IĚ−n)

Hom(P,τIĚ−n)'

HomR(E, I)Hom(ι,I)

u// HomR(J, I)

Hom(π,I)

'// HomR(P, I) ,

the left-hand horizontal maps are homotopy equivalences by 4.3.19, and HomR(π, I)and HomR(P,τI

Ě−n) are quasi-isomorphisms by semi-injectivity of I and semi-



projectivity of P. The diagram shows that it suffices to verify that HomR(π, IĚ−n)is a quasi-isomorphism. Set C = Coneπ and notice that it is an acyclic complex ofmodules that are direct sums of projective and injective modules. The goal is to showthat the complex HomR(C, IĚ−n)∼= Σ−1 ConeHomR(π, IĚ−n) is acyclic; cf. 4.1.15.

By (2.5.20.3) there is an exact sequence 0→ IĚ−n→ I→ IĎ−n−1→ 0. By 10.2.2and 10.2.8 these are complexes of Gorenstein injective modules, and it follows from7.3.16, in view of 10.2.5, that the sequence

0−→ HomR(C, IĚ−n)−→ HomR(C, I)−→ HomR(C, IĎ−n−1)−→ 0

is exact. The middle complex is acyclic as I is semi-injective and C is acyclic. Toshow acyclicity of HomR(C, IĚ−n) it suffices to see that HomR(C, IĎ−n−1) is acyclic;see 2.2.20. To that end, it suffices by A.12 to argue that HomR(C,G) is acyclic forevery Gorenstein injective module G. It is already know from 10.2.5 that one hasExtiR(Cv,G) = 0 for all v ∈ Z and all i > 0, so it suffices by A.11 to argue thatExtiR(Cv(C),G) = 0 holds for v 0 and all i > 0.

If E has finite injective dimension, then one can assume that the complexes J andP are bounded below; see 8.2.2 and 5.2.15. It follows that C is bounded below; inparticular, Cv(C) = 0 holds for v 0.

If E has finite projective dimension, then one can assume that the complexes Pand J are bounded above; see 8.1.2 and 5.3.26. It follows that C is bounded above; inparticular, one has Cv(C) = 0 for v 0. Let G be a Gorenstein injective R-module.For every v ∈ Z one has ExtiR(Cv,G) = 0 for all i > 0, so in view of 10.2.5 inductionon the exact sequences 0→Cv+1(C)→Cv→Cv(C)→ 0 yields ExtiR(Cv(C),G) = 0for all v ∈ Z and all i > 0.

10.2.14 Corollary. Let M be an R-complex of finite Gorenstein injective dimensionand N be an R-complex of finite projective or finite injective dimension. For everysemi-injective replacement I of M, every semi-injective replacement E of N, andevery integer n> GidR M there is an isomorhism in D(k),

RHomR(N,M) ' HomR(E, IĚ−n) .


10.2.15 Theorem. Let M be an R-complex of finite Gorenstein injective dimensionand let n be an integer. The following conditions are equivalent.

(i) GidR M 6 n.(ii) − infRHomR(N,M)6 n+ supN holds for every R-complex N with pdR N fi-

nite or idR N finite.(iii) − infRHomR(N,M)6 n holds for every injective R-module N.(iv) n > − infM and Extn+1

R (N,M) = 0 holds for every R-module N of finite in-jective dimension.

(v) n>− infM and for some, equivalently every, semi-injective replacement I ofM, the module Z−v(I) is Gorenstein injective for every v> n.

(vi) There is a semi-injective resolution M '−→ I with Iv = 0 for every v > supM,Hv(I) = 0 for every v <−n, and Z−v(I) Gorenstein injective for every v> n.




GidR M = sup− infRHomR(N,M) | N is an R-module with idR N finite= supm ∈ Z | ExtmR (N,M) 6= 0 for some injective R-module N .

PROOF. We start by establishing the equivalence of (i), (ii), and (iii).(i)=⇒ (ii): One can assume that N is in D@(R) and not acyclic; otherwise the

inequality is trivial. In this case, w = supN is an integer. By 5.3.26 there is a semi-injective resolution N '−→ J with Jv = 0 for v > w. If pdR N or idR N is finite, then10.2.14 implies that one has RHomR(N,M)' HomR(J, IĚ−n), where I is any semi-injective replacement of M. For every v <−n−w and p ∈ Z one of the inequalitiesp > w or p+ v6 w+ v <−n holds, so the module

HomR(J, IĚ−n)v =∏

p∈ZHomR(Jp,(IĚ−n)p+v)

is zero. In particular, one has Hv(RHomR(M,N)) = 0 for v <−n−w =−n− supN.(ii)=⇒ (iii): Trivial.(iii)=⇒ (i): By assumption, g = GidR M is finite. We must show that (iii) implies

g6 n. We can assume that M is not acyclic as otherwise the inequality is trivial. Bydefinition there exists a semi-injective replacement I of M with Z−g(I) Gorensteininjective. By 10.2.14 there is an isomorphism,

(?) RHomR(J,M) ' HomR(J, IĚ−g) ,

for every injective R-module J. Recall from 10.2.7 that g > − infM = − inf I. Weconsider two different cases:

First assume that g=− inf I holds; this implies H−g(I) 6= 0 so the homomorphismI−g+1→ Z−g(I) is not surjective. Since Z−g(I) is Gorenstein injective there exists,in particular, a surjection J Z−g(I) where J is an injective module. This map doesnot admit a factorization J→ I−g+1→ Z−g(I), as this would force I−g+1→ Z−g(I)to be surjective. Thus the map HomR(J, I−g+1)→HomR(J,Z−g(I)) is not surjective,so infHomR(J, IĚ−g) =−g. Now (?) and (iii) yield g =− infRHomR(J,M)6 n.

Next assume that g >− inf I. In this case there is an exact sequence of modules,0→ Z−g+1(I)→ I−g+1→ Z−g(I)→ 0. As g >− inf I and g = GidR M, the moduleZ−g+1(I) is not Gorenstein injective. Hence 10.2.4 yields Ext1R(J,Z−g+1(I)) 6= 0for some injective module J, and by 8.2.6 this means that H−g(RHomR(J,M)) 6= 0.Hence g6− infRHomR(J,M)6 n, where the last inequality holds by (iii).

To finish the proof we show (ii)=⇒(iv)=⇒ (v)=⇒(vi)=⇒(i).(ii)=⇒ (iv): The second assersion in (iv) is immediate from (ii). The inequality

n>− infM follows, in view of 10.2.7, from (i), which is equivalent to (ii).(iv)=⇒ (v): First note that by 8.2.13 and 10.2.4 the “some” version and the “ev-

ery” version of condition (v) are equivalent. By assumption, g = GidR M is finite,so in any semi-injective replacement I of M the module Z−v(I) Gorenstein injectivefor every integer v > g; see 10.2.8. Thus, to show (v) it is enough to prove n> g.Assume towards a contradiction that n < g holds. Notice that by the assumptionn > − infM, the module Z−n(I) can not be Gorenstein injective. There is an exactsequence 0→Z−n(I)→ I−n→···→ I−g+1→Z−g(I)→ 0, which shows that Z−n(I)



has finite Gorenstein injective dimension, as Iď−n is a semi-injective replacement ofΣ−n Z−n(I). Now 10.2.12 yields an exact sequence 0→Z−n(I)→G→N→ 0 whereG is Gorenstein injective and N has finite injective dimension. By 8.2.6 and (iv) onehas Ext1R(N,Z−n(I)) ∼= Extn+1

R (N,M) = 0, so the sequence splits by 7.3.15. Now10.2.4 implies that Z−n(I) is Gorenstein injective, which is a contradiction.

(v)=⇒ (vi): Immediate in view of 5.3.26.(vi)=⇒ (i): Immediate from the definition, 10.2.6, of GidR.

10.2.16 Theorem. Let M be an R-complex. If M has finite projective dimension,then one has GidR M = idR M.

PROOF. The inequality GidR M 6 idR M holds by 10.2.9. To show the opposite in-equality, we may assume that g = GidR M is an integer. Set w =−supM−1, whichis an integer by 8.1.3, and note that one has w < −supM 6 − infM 6 g by 10.2.7.Thus 10.2.11 yields a distinguished triangle in D(R),

(?) Σ−1 K −→ N −→M −→ K ,

with GidR K 6 w and idR N = g. As pdR M is finite, 10.2.15 shows that one has

− infRHomR(M,K) 6 GidR K + supM 6 w+ supM = −1 ,

and hence H0(RHomR(M,K)) = 0. By 7.3.10 this means that D(R)(M,K) = 0, inparticular, the morphism M→K in the distinguished triangle above is zero. By D.17this means the (?) splits, and hence N ' (Σ−1 K)⊕M holds in D(R). In particularone has idR M 6 idR N = g, as claimed.

10.2.17 Proposition. For every family of R-complexes Muu∈U one has,

GidR(∏

u∈U

Mu) = supu∈UGidR Mu .

PROOF. To prove the inequality “6”, one can assume that the right-hand side isfinite, say, s ∈ Z. By the definition, 10.2.6, of Gorenstein injective dimension andby 10.2.8 every Mu admits a semi-injective replacement Iu with Hv(Iu) = 0 for allv <−s and Zv(Iu) Gorenstein injective for all v 6 −s. Now I =

∏u∈U Iu is a semi-

injective replacement of∏

u∈U Mu with Hv(I) = 0 for all v < −s, see 5.3.21 and3.1.22; and Zv(I) =

∏u∈U Zv(Iu) Gorenstein injective for all v6−s, see 10.2.4.

To prove the opposite inequality “>” it suffices, as each Mu is a direct summandin∏

u∈U Mu, to argue that if M′ is a direct summand in an R-complex M, then onehas GidR M′ 6 GidR M. To this end, we may assume that M is not acyclic and thatg = GidR M is finite. Let M′′ be an R-complex with M = M′⊕M′′. Let I′ and I′′

be semi-injective replacements of M′ and M′′. Now I = I′⊕ I′′ is a semi-injectivereplacement of M, see 5.3.21. As Hv(I) = 0 holds for every v <−g, even for everyv < infM, one has Hv(I′) = 0 for every v <−g by 3.1.22. One gets from 10.2.8 thatZv(I) = Zv(I′)⊕Zv(I′′) is Gorenstein injective for all v 6 −g, and hence Zv(I′) isGorenstein injective for all v6−g by 10.2.4. Thus GidR M′ 6 g holds.



THE CASE OF MODULES

10.2.18. Notice from 10.2.15 that a non-zero R-module is Gorenstein injective ifand only if it has Gorenstein injective dimension 0 as an R-complex.

10.2.19 Theorem. Let M be an R-module of finite Gorenstein injective dimensionand let n> 0 be an integer. The following conditions are equivalent.

(i) GidR M 6 n.(ii) ExtmR (N,M) = 0 holds for every R-module N with pdR N finite or idR N finite

and all integers m > n.(iii) ExtmR (N,M) = 0 holds for every injective R-module N and all integers m > n.(iv) Extn+1

R (N,M) = 0 holds for every R-module N with idR N finite.(v) In some/every injective resolution 0→M→ I0→ ··· → I−(v−1)→ I−v→ ···

the module Ker(I−v→ I−(v+1)) is Gorenstein injective for every v> n.(vi) There is an exact sequence 0→ M → I0 → ·· · → I−(n−1) → G→ 0 of R-

modules with each Ii injective and G Gorenstein injective.In particular, there is an equality

GidR M = supm ∈ N0 | ExtmR (N,M) 6= 0 for some injective R-module N .PROOF. In an injective resolution 0→ M → I0 → ··· → I−(v−1) → I−v → ··· , themorphism M I0 yields a semi-injective resolution of M, considered as an R-complex; cf. 5.3.33. In particular, the complex 0→ I0→ ·· · → I−(v−1)→ I−v→ ·· ·is a semi-injective replacement of M. Conversely, a semi-injective resolution M '−→ Iwith Iv = 0 for all v > 0 yields an injective resolution of M. The equivalence of thesix conditions is now immediate from 10.2.15.

EXERCISES

E 10.2.1 Show that if the coproduct of every (countable) family of Gorenstein injective R-modules is Gorenstein injective, then R is left oetherian. Hint: 8.2.18.

E 10.2.2 Let n ∈ Z. Show that an R-complex M has GidR M 6 n if and only if there is a isa diagram M ι−→ I′ τ−→ I in C(R) where P is a totally acyclic complex of injectiveR-modules, τv is an isomorphism for v 0 and ι is a semi-injective resolution.

E 10.2.3 The invariant FGID(R)= supGidR M |M is an R-module with GidR M <∞ is calledthe finitistic Gorenstein injective dimension of R. Show that FGID(R) = FID(R).



10.3 Gorenstein Flat Dimension

SYNOPSIS. Totally acyclic complex of flat modules; Gorenstein flat module; Gorenstein flat dimen-sion; ∼ vs. Gorenstein injective dimension; ∼ is refinement of flat dimension; ∼ vs. Gorensteinprojective dimension; ∼ over Noetherian ring; ∼ of module.

GORENSTEIN FLAT MODULES

10.3.1 Definition. A complex F of flat R-modules is called totally acyclic if it isacyclic and E⊗R F is acyclic for every injective Ro-module E.

An R-module G is called Gorenstein flat if one has G ∼= C0(F) for some totallyacyclic complex F of flat R-modules.

Notice that if F is a totally acyclic complex of flat R-modules, then the moduleCv(F) is Gorenstein flat for every v ∈ Z.

10.3.2 Example. A flat module F is Gorenstein flat as the disc complex D0(F) =0→ F =−→ F → 0 is totally acyclic.

10.3.3. Since the Tor-functors 7.4.8 can be computed via semi-flat resolutions, see7.4.6, it follows directly from the definition, 10.3.1, that one has TorR

m(E,G) = 0 forevery injective Ro-module E, every Gorestein flat R-module G, and every m > 0.

10.3.4 Proposition. Let Guu∈U be a family of R-modules. If every Gu is Goren-stein flat, then the coproduct

∐u∈U Gu is Gorenstein flat.

PROOF. For each u ∈U there exists a totally acyclic complex Fu of flat R-moduleswith Gu ∼= C0(Fu). The complex F =

∐u∈U Fu consists of flat modules by 5.4.25

and it is acyclic by 3.1.10. For every Ro-module E there is by 3.1.12 an isomorphismE⊗R F ∼=

∐u∈U E⊗R Fu, and this complex is acyclic if E is injective. Thus, F is a

totally acyclic complex of flat R-modules. As C0(F) ∼=∐

u∈U C0(Fu) ∼=∐

u∈U Gu

holds by 3.1.8, it follows that∐

u∈U Gu is Gorenstein flat.

10.3.5 Proposition. Let G be an R-module. If G is Gorenstein flat, then the Ro-module Homk(G,E) is Gorenstein injective.

PROOF. There exists a totally acyclic complex F of flat R-modules with G∼=C0(F).The Ro-complex I = Homk(F,E) is acyclic, and it consists of injective modules; see1.3.43 and 2.5.6(b). For every Ro-module E adjunction ?? yields HomRo(E, I) ∼=Homk(E⊗R F ,E), so HomRo(E, I) is acyclic if E is injective. Therefore, I is a totallyacyclic complex of injective Ro-modules; see 10.2.1. By 2.2.18 one has Z0(I) ∼=Homk(C0(F),E)∼= Homk(G,E), so Homk(G,E) is Gorenstein injective.

10.3.6 Lemma. Assume that R is right Noetherian and let G be an R-module. If theRo-module HomR(G,E) is Gorenstein injective, then there exists an exact sequence,

η = 0−→ G−→ F0 −→ F−1 −→ ·· · ,

where each Fv is a flat R-module and HomR(η,P) is exact for every flat R-module P.Moreover, the sequence E⊗R η is exact for every injective Ro-module E.


10.3 Gorenstein Flat Dimension 379

PROOF. Let η be a sequence of R-modules and E an Ro-module. Adjunction 4.4.11yields an isomorphism Homk(E⊗R η,E)∼= HomR(η,Homk(E,E)). If E is injective,then Homk(E,E) is a flat R-module by 8.3.18. Thus if HomR(η,P) is exact for everyflat R-module P, then E⊗R η is exact for every injective Ro-module E; cf. 2.5.6(b).

To construct η it is now sufficient to construct a short exact sequence,

(?) η′ = 0−→ G−→ F −→ G′ −→ 0 ,

where F is flat, G′ has the same property as G—that is, Homk(G′,E) is Gorensteininjective—and HomR(η

′,P) is exact for every flat R-module P. To this end, chooseby E.20 a flat preenvelope ϕ : G→ F . As the Ro-module Homk(G,E) is Gorensteininjective there exists, in particular, a surjective homomorphism I Homk(G,E)where I is an injective Ro-module. There are thus injective homomorphisms,

G Homk(Homk(G,E),E) Homk(I,E) ,

where the first map is biduality 4.5.3. As above the R-module Homk(I,E) is flat, so Gmaps injectively into a flat R-module. This fact, combined with the defining propertyE.17 of preenvelopes, implies that ϕ is injective. Setting G′ = Cokerϕ one obtainsa short exact sequence (?). For every flat R-module P the sequence HomR(η

′,P)is exact as the functor HomR(–,P) is left exact and G→ F is a flat preenvelope. Itremains to see that Homk(G′,E) is Gorenstein injective. Consider the exact sequence

Homk(η′,E) = 0−→ Homk(G′,E)−→ Homk(F,E)−→ Homk(G,E)−→ 0 .

As the module Homk(G,E) is Gorenstein injective, by assumption, and Homk(F,E)is injective by 1.3.43, it suffices by 10.2.4 to show that Ext1Ro(E,Homk(G′,E)) = 0holds for all injective Ro-modules E. By 7.3.14 and injectivity of Homk(F,E), it isenough to show that the sequence HomRo(E,Homk(η′,E)) is exact for every injec-tive Ro-module E. By adjunction 1.2.5 and commutativity 1.2.2, this sequence isisomorphic to HomR(η

′,Homk(E,E)), which is exact as Homk(E,E) is flat.

10.3.7 Theorem. Assume that R is right Noetherian. An R-module G is Gorensteinflat if and only if the Ro-module Homk(G,E) is Gorenstein injective.

PROOF. The “only if” part is 10.3.5. To prove the “if” part assume that Homk(G,E)is Gorenstein injective. For every injective Ro-module E and all i > 0 one hasHomk(TorR

i (E,G),E) ∼= ExtiRo(E,Homk(G,E)) = 0 by 8.3.1 and 10.2.5. Thus alsoTorR

i (E,G) = 0 holds; cf. 2.5.6(b). It follows that every projective resolution,

(?) · · · −→ P1 −→ P0 −→ G−→ 0 ,

stays exact under application of the functor E⊗R – for every injective Ro-moduleE. Splicing together the sequences (?) and η from 10.3.6 one gets a totally acycliccomplex of flat R-modules in which G is a cokernel.

10.3.8. Let R be right Noetherian. It follows from 10.3.7 that the cokernels of allthe homomorphisms in the sequence η in 10.3.7 are Gorenstein flat.

REMARK. Theorem 10.3.7 above is crucial for our development of the theory of Gorenstein flatmodules/dimension. Indeed, several of the proofs in this section combine results from Sect. 10.2,



which hold over any ring, with 10.3.7, which require the ring to be right Noetherian. It is thereforeworth pointing out that 10.3.7 actually holds under the weaker assumption that R is right coherent.The reason is that the proof of 10.3.7 relies on the existence of flat preenvelopes E.20, whose proofin turn relies on 8.3.25; and as remarked there, that result holds if the ring is right coherent.

This means, for example, that 10.3.11 below holds with the same proof for a right coherent ring.Work of Štovícek and Šaroch [?] shows that most of these properties of Gorenstein flat moduleshold without assumptions on the ring.

One consequence of the result above is that, over a right Noetherian ring, the classof Gorenstein flat modules is closed under pure submodules and pure quotiens:

10.3.9 Corollary. Let R be right Noetherian and let 0→ G′→ G→ G′′→ 0 be apure exact sequence of R-modules. If G is Gorenstein flat then so are G′ and G′′.

PROOF. The sequence 0→Homk(G′′,E)→Homk(G,E)→Homk(G′,E)→ 0 splitsby E.2, so the assertion follows from 10.3.7 and 10.2.4.

10.3.10 Lemma. Let µvu : Mu→Mvu6v be a U-direct system of R-modules. If Uis filtered, then the canonical surjective homomorphism

∐u∈U Mu colimu∈U Mu

from 3.2.2 is a pure epimorphism.

PROOF. Write π for the canonical homomorphism. For every finitely presented R-module N the functor HomR(N,–) preserves coproducts and filtered colimits, see3.1.31 and 3.2.36, so HomR(N,π) is isomorphic to the canonical homomorphism∐

u∈U HomR(N,Mu)→ colimu∈U HomR(N,Mu) associated with the U-direct sys-tem HomR(N,µvu) : HomR(N,Mu)→ HomR(N,Mv)u6v. As this homomophismis surjective, so is HomR(N,π). Hence π is a pure epimorphism; see E.3.

10.3.11 Proposition. Assume that R is right Noetherian. The next assertions hold.(a) Let 0→ G′ → G→ G′′ → 0 be an exact sequence of R-modules. If G′′ is

Gorenstein flat, then G is Gorenstein flat if and only if G′ is so. If G′ and Gare Gorenstein flat, then G′′ is Gorenstein flat if and only if TorR

1 (E,G′′) = 0

holds for every injective Ro-module E.(b) Every direct summand of a Gorenstein flat R-module is Gorenstein flat.(c) Let µvu : Gu→ Gvu6v be a U-direct system of R-modules. If U is filtered

and every module Gu is Gorenstein flat, then colimu∈U Gu is Gorenstein flat.

PROOF. (a): The sequence 0→ Homk(G′′,E)→ Homk(G,E)→ Homk(G′,E)→ 0is exact. Furthermore, one has Homk(TorR

1 (E,G′′),E)∼= Ext1Ro(E,Homk(G′′,E)) for

every Ro-module E; see 8.3.1. The conclusions now follow from 10.3.7 and 10.2.4.(b): This is a special case of 10.3.9.(c): This follows from 10.3.4, 10.3.9, and 10.3.10.

10.3.12 Lemma. Let N be the class of Ro-modules N with TorRi (N,G) = 0 for every

Gorenstein flat R-module G and all i > 0. Let 0→ N′→ N → N′′→ 0 be an exactsequence of Ro-modules. If any two of the modules N′, N, and N′′ are in N, then sois the third. In particular, every Ro-module fdRo N or idRo N finite belongs to N.



PROOF. It follows from 7.4.13 that N′,N′′ ∈ N implies N ∈ N, and that N,N′′ ∈ N

implies N′ ∈ N. Now assume that N,N′ ∈ N. Another application of 7.4.13 yieldsTorR

i (N′′,G) = 0 for every Gorenstein flat R-module G and all i > 1. To show that

also TorR1 (N

′′,G) = 0 holds, note that every Gorenstein flat module G by definitionfits into an exact sequence 0→ G→ F → G′ → 0 with F flat and G′ Gorensteinflat. Application of 7.4.13 to this sequence shows that TorR

1 (N′′,G)∼= TorR

2 (N′′,G′),

and the right-hand side is zero, as just shown. The last assertion now follows as Ncontains every flat and every injective Ro-module, see 8.3.24 and 10.3.3.

GORENSTEIN FLAT DIMENSION

10.3.13 Definition. Let M be an R-complex. The Gorenstein flat dimension of M,written GfdR M, is defined as

GfdR M = inf

n ∈ Z∣∣∣∣ There exists a semi-flat replacement F of M withHv(F) = 0 for all v > n and Cn(F) Gorenstein flat

with the convention inf∅= ∞. One says that GfdR M is finite if GfdR M < ∞ holds.

10.3.14. Let M be an R-complex. For every semi-flat replacement F of M one hasH(F)∼= H(M); the next (in)equalities are hence immediate from the definition,

GfdR M > supM and GfdRΣs M = GfdR M+ s for every s ∈ Z .

Moreover, one has GfdR M =−∞ if and only if M is acyclic.

10.3.15 Proposition. Let M be an R-complex. There is an inequality,

GidRo Homk(M,E) 6 GfdR M ,

and equality holds if R is right Noetherian.

PROOF. To prove the inequality “6” we may assume that g = GfdR M is an integer.By definition there is a semi-flat replacement F of M with Hv(F) = 0 for all v > gand Cg(F) Gorenstein flat. The complex I = Homk(F,E) is a semi-injective replace-ment of Homk(M,E), see 5.4.9, and for v ∈ Z one has Hv(I) ∼= Homk(H−v(F),E)and Z−v(I) ∼= Homk(Cv(F),E) by 2.2.18. It follows that Hv(I) = 0 holds for allv <−g and, in view of 10.3.5, that Z−g(I) Gorenstein injective. Thus the inequalityGidRo Homk(M,E)6 g holds by the definition, 10.2.6.

Let R be right Noetherian. To prove the inequality “>” one can assume thatd = GidRo Homk(M,E) is an integer. Let F be a semi-flat replacement of M. As thecomplex I = Homk(F,E) is a semi-injective replacement of Homk(M,E), it followsfrom 10.2.8 that the module Z−d(I)∼= Homk(Cd(F),E) is Gorenstein injective, andhence Cd(F) is Gorenstein flat by 10.3.7. As d>− infHomk(M,E)= supM = supFholds by 10.2.7 and 2.5.6(b) one has Hv(F) = 0 for all v > d. Thus the desiredinequality GfdR M 6 d follows from the definition 10.3.13.



10.3.16 Lemma. Assume that R is right Noetherian and let M be an R-complex. Forevery semi-flat replacement F of M and every integer v>GfdR M the module Cv(F)is Gorenstein flat.

PROOF. Set g = GfdR M. By 10.3.15 one has GidRo Homk(M,E) = g, and by 5.4.9the complex I = Homk(F,E) is a semi-injective replacement of Homk(M,E). Thus10.2.8 and 2.2.18 imply that the module Z−v(I) ∼= Homk(Cv(F),E) is Gorensteininjective for every v> g. Thus Cv(F) is Gorenstein flat for v> g by 10.3.7.

Next we compare the Gorenstein flat dimension to the flat dimension and to theGorenstein projective dimension. The following result is sometimes expressed bysaying that GfdR is a refinement of fdR. It follows, in particular, that a Gorensteinflat module is either flat or has infinite flat dimension.


GfdR M 6 fdR M ,

and equality holds if M has finite flat dimension.

PROOF. The inequality is evident from the definitions of the dimensions, see 8.3.3and 10.3.13, and from the fact that every flat module is Gorenstein flat, see 10.3.2.If M has finite flat dimension, then 8.3.16, 10.2.9, and 10.3.15 yield,

fdR M = idRo Homk(M,E) = GidRo Homk(M,E) 6 GfdR M .

Under weak additional assumptions, equality also holds in 10.3.17 if M has finiteinjective dimension.

10.3.18 Theorem. Assume that R is left Noetherian and that every flat R-modulehas finite projective dimension. For every R-complex M of finite injective dimensionone has GfdR M = fdR M.

PROOF. Assume that M has finite injective dimension. Since R is left Noetherian,the Ro-complex Homk(M,E) has finite flat dimension by 8.3.18. The assumptionsand 8.3.8 now imply that Homk(M,E) has finite projective dimension. Thus thesecond equality in the computation below holds by 10.2.16; the first equality followsfrom 8.3.16 and the inequality from 10.3.15.

fdR M = idRo Homk(M,E) = GidRo Homk(M,E) 6 GfdR M .

The opposite inequality GfdR M 6 fdR M holds by 10.3.17.

The assumption in the next result about finiteness of projective dimnsion of flatR-modules holds if R has has finite finitistic projective dimension, see 8.5.18, or ifR has a dualizing complex, see 9.3.11.

10.3.19 Proposition. Assume that R is right Noetherian and that every flat R-modulehas finite projective dimension. For every R-complex M there is an inequality,

GfdR M 6 GpdR M .



PROOF. By the definitions, 10.1.8 and 10.3.13, of the dimensions, and by the factthat every semi-projective replacement of M is also a semi-flat replacement, see5.4.10, it suffices to show that every Gorenstein projective R-module is Goren-stein flat. To this end we argue that every totally acyclic complex P of projectiveR-modules is a totally acyclic per 10.3.1 when considered as a complex of flat R-modules. Certainly P is an acyclic complex of flat modules; see 1.3.38. For every in-jective Ro-module E the R-module Homk(E,E) is flat by 8.3.18. Hence, by assump-tion, Homk(E,E) has finite projective dimension. It now follows from 10.1.7 thatExtiR(G,Homk(E,E)) = 0 holds for every Gorenstein projective R-module G andall i > 0; consequently the complex HomR(P,Homk(E,E)) is acyclic; cf. A.11. Byadjunction 4.4.11 this complex is isomorphic to Homk(E⊗R P,E) which, therefore,is acyclic too. Finally, 2.5.6(b) yields acyclicity of E⊗R P, so P is totally acyclic asa complex of flat R-modules.

10.3.20 Example. The Z-module Q is flat, see 1.3.38, and hence Gorenstein flat by10.3.2, so GfdZQ= 0. However, one has GpdZQ= pdZQ= 1 by 8.1.6 and 10.1.11.Thus, the inequality in 10.3.19 can be strict.

10.3.21 Proposition. Let R be right Noetherian and M′→M→M′′→ ΣM′ be adistinguished triangle in D(R). With g′ = GfdR M′, g = GfdR M, and g′′ = GfdR M′′

there are inequalities,

g′ 6maxg,g′′−1 , g6maxg′,g′′ , and g′′ 6maxg′+1,g .

In particular, if two of the complexes M′, M, and M′′ have finite Gorenstein flatdimension, then so has the third.

PROOF. Application of the triangulated functor RHomk(–,E) ' Homk(–,E) to thegiven triangle yields a distinguished triangle in D(Ro),

Homk(M′′,E)→ Homk(M,E)→ Homk(M′,E)→ ΣHomk(M′′,E) .

The assertion now follows from 10.2.10 and 10.3.15.

10.3.22 Proposition. Let R be right Noetherian and let M be an R-complex of finiteGorenstein flat dimension g = GfdR M. For every semi-flat replacement F of M andevery integer u with g > u there is a distinguished triangle in D(R),

K −→M −→ N −→ ΣK ,

where the complexes K and N have the following properties:(a) There is a degreewise split exact sequence 0→ Fďu→K→ Σu G→ 0 in C(R)

where G is a Gorenstein flat R-module. Furthermore, one has

GfdR K 6 u and Hv(K)∼=

0 for v> u+1Hv(M) for v6 u−1 .


fdR N = g and Hv(N)∼=

Hv(M) for v> u+20 for v6 u .




(10.3.22.1) 0−→ Hu+1(M)−→ Hu+1(N)−→ Hu(K)−→ Hu(M)−→ 0 .

PROOF. If M is acyclic the statement is void as no integer u satisfies −∞ = g > u.Now assume that M is not acyclic, in which case g is an integer. By 10.3.16the module Cg(F) is Gorenstein flat and by 10.3.6 and 10.3.8 there exists anacyclic R-complex L = 0→ Cg(F)→ Lg−1 · · · → Lu→ G→ 0, concentrated in de-grees g, . . . ,u−1, where Lg−1, . . . ,Lu are flat, the cokernels are Gorenstein flat, andHomR(L,P′) is acyclic for every flat R-module P′. Set Lg = Cg(F) and Lu−1 = Gand notice that, in particular, the cokernel Cu(L)∼= G is Gorenstein flat.

As L is acyclic so is the truncated complex LĎg−1 and the sequence of R-modules0→Cg(F)→ Lg−1→Cg−1(L)→ 0 is exact. For every flat R-module P′ acyclicity ofHomR(L,P′) implies acyclicity of HomR(LĎg−1,P′) and exactness of the sequence

(?) 0−→ HomR(Cg−1(L),P′)−→ HomR(Lg−1,P′)−→ HomR(Cg(F),P′)−→ 0 .


(‡) 0−→ Σg−1 Cg(F)

α−−→ Lďg−1 −→ LĎg−1 −→ 0 ,

where α is a quasi-isomorphism as LĎg−1 is acyclic; see 4.2.6. Let P be a complex offlat R-modules. It follows from exatness of (?) that the functor HomR(–,Pn) leavesthe sequence (‡) exact for every n ∈ Z. It now follows from 2.3.18 that the functorHomR(–,P) also leaves the sequence (‡) exact. Furthermore, as HomR(LĎg−1,Pn)is acyclic for every n ∈ Z, it follows from A.12 that HomR(LĎg−1,P) is acyclic.Now 4.2.6 yields that HomR(α,P) is a surjective quasi-isomorphism, whence themorphism C(R)(α,P) is surjective as well by 4.2.7 and 2.3.10. Surjectivity ofC(R)(α,Fďg−1) yields a commutative diagram of R-complexes,

Σg−1 Cg(F)β//

'α

Fďg−1

Lďg−1γ// Fďg−1 ,

where β is induced by ∂Fg . This diagram—in conjunction with the definition of dis-

tinguished triangles in D(R), see 6.5.6, the axiom (TR3) in D.2, and 6.5.22—showsthat the complexes Coneβ and Coneγ are isomorphic in D(R). Evidently, one hasConeβ = FĎg, and this complex is isomorphic to F ' M in D(R). Consequently,Coneγ 'M holds in D(R).

Set C = Coneγ. By 2.5.17 and 6.5.19 there is a distinguished triangle in D(R),

Cďu −→C −→Cěu+1 −→ ΣCďu ,

which we argue is the desired one. As already noticed, C'M in D(R). The complex

K = Cďu = 0−→ Fu⊕G−→ Fu−1 −→ Fu−2 −→ ·· ·

fits into the degreewise split exact sequence in C(R),



() 0−→ Fďu −→ K −→ Σu G−→ 0 .

The complex F is semi-flat, and so is Fěu+1 by 5.4.8. Hence 5.4.12 applied to theexact sequence 0→ Fďu→ F → Fěu+1→ 0 shows that Fďu is semi-flat. It followsfrom 8.3.3 and 10.3.17 that GfdR(Fďu) = fdR(Fďu)6 u holds. As G is a Gorensteinflat module, one has GfdR(Σ

u G)6 u, with equality if G is non-zero, so application of6.5.19 and 10.3.21 to () shows that GfdR K 6 u. The assertion about the homologyof K =Cďu is immediate as C 'M in D(R). Notice that

N = Cěu+1 = 0−→ Lg−1 −→ Fg−1⊕Lg−2 −→ ·· · −→ Fu+1⊕Lu −→ 0

is a complex of flat R-modules concentrated in degrees g, . . . ,u+1. In the extremalcase u = g− 1 one has N = Σg Lg−1. Thus N is semi-flat by 5.4.8 with fdR N 6 g.Note that g 6 maxGfdR K,GfdR N holds by 10.3.21. As one has GfdR K 6 u < gwe get GfdR N > g, so fdR N = g holds by 10.3.17 that. The assertion about thehomology of N = Cěu+1 is immediate as C ' M in D(R). The exact sequence(10.3.22.1) follows by applying 6.5.22 to the constructed distinguished triangle.

10.3.23 Corollary. Let R be right Noetherian and let M be an R-module of finiteGorenstein flat dimension g = GfdR M. The following assertions hold.

(a) There is an exact sequence of R-modules 0→M→ X → G→ 0 where G isGorenstein flat and fdR X = g.

(b) If g > 0, then there is an exact sequence of R-modules 0→ X →G→M→ 0where G is Gorenstein flat and fdR X = g−1.

PROOF. (a): For u =−1 the sequence (10.3.22.1) reads

0−→M −→ H0(N)−→ H−1(K)−→ 0 .

If follows from 10.3.22 that K is isomorphic to Σ−1 H−1(K) in D(R), whence onehas GfdR H−1(K)− 1 = GfdR K 6 u = −1. Consequently, the module H−1(K) isGorenstein flat. Similarly, one has N ' H0(N) and fdR H0(N) = fdR N = g.

(b): As g > 0 one can apply (10.3.22.1) with u = 0 to obtain the exact sequence

0−→ H1(N)−→ H0(K)−→M −→ 0 .

If follows from 10.3.22 that K is isomorphic to H0(K) in D(R), whence one hasGfdR H0(K) = GfdR K 6 u = 0. That is, the module H0(K) is Gorenstein flat. Simi-larly, one has N ' ΣH1(N) and fdR H1(N)+1 = fdR N = g.

10.3.24 Proposition. Assume that R is right Noetherian. Let F be a semi-flat R-complex and E a semi-injective Ro-complex. If E has finite injective or finite flatdimension, then the morphism

E⊗R τFĎn : E⊗R F −→ E⊗R FĎn

is a quasi-isomorphism for every integer n> GfdR F .

PROOF. Given a semi-injective Ro-complex J that is isomorphic to E in the de-rived category there is by 6.4.21 and 5.3.24 a homotopy equivalence ι : J→ E. From10.3.14 one gets n > supF , so the map τF

Ďn : F → FĎn is a quasi-isomorphism by



4.2.4. Choose a semi-flat replacement P of E; by 6.4.21 there is a quasi-isomorphismπ : P '−→ J. In the commutative diagram,

P⊗R Fπ⊗F'

//

P⊗τFĎn '

J⊗R F

J⊗τFĎn

ι⊗F

u// E⊗R F

E⊗τFĎn

P⊗R FĎnπ⊗FĎn

// J⊗R FĎnι⊗FĎn

u// E⊗R FĎn ,

the right-hand horizontal maps are homotopy equivalences by 4.3.20, while P⊗R τFĎn

and π⊗R F are quasi-isomorphisms by semi-flatness of P and F . The diagramshows that it is sufficient to verify that π⊗R FĎn is a quasi-isomorphism. SetC = Coneπ and notice that it is an acyclic complex of modules that are directsums of flat and injective Ro-modules. The goal is to prove that the complexC⊗R FĎn ∼= Cone(π⊗R FĎn) is acyclic; see 4.1.18.

By (2.5.20.3) there is an exact sequence 0→ FĚn+1→ F → FĎn→ 0. By 10.3.2and 10.3.16 these are all complexes of Gorenstein flat modules, and it follows from7.4.14 and 10.3.12 that the sequence,

0−→C⊗R FĚn+1 −→C⊗R F −→C⊗R FĎn −→ 0 ,

is exact. The complex C⊗R F is acyclic as C is acyclic and F is semi-flat, and thusit is enough to show that the left-hand complex C⊗R FĚn+1 is acyclic; see 2.2.20.To that end, it suffices by A.14 to show that C⊗R G is acyclic for every Gorensteinflat module G. It is already known from 10.3.12 that one has TorR

i (Cv,G) = 0 for allv ∈ Z and all i > 0, so it suffices by A.13 to argue that TorR

i (Cv(C),G) = 0 holds forv 0 and all i > 0.

If E has finite injective dimension, then one can assume that J and P are boundedbelow; see 8.2.2, 5.2.15 and 5.4.10. It follows that C is bounded below; in particular,Cv(C) = 0 holds for v 0.

If E has finite flat dimension, then one can by 8.3.3 and 5.3.26 assume that thecomplexes P and J are bounded above. It follows that C is bounded above; in par-ticular, Cv(C) = 0 holds for v 0. Let G be a Gorenstein flat module. For everyv ∈ Z one has TorR

i (Cv,G) = 0 for all i > 0, so in view of 10.3.12, induction on theexact sequences 0→ Cv+1(C)→ Cv → Cv(C)→ 0 yields TorR

i (Cv(C),G) = 0 forall v ∈ Z and all i > 0.

10.3.25 Corollary. Assume that R is right Noetherian. Let M be an R-complexof finite Gorenstein flat dimension and N be an Ro-complex of finite injective orfinite flat dimension. For every semi-flat replacement F of M, every semi-injectivereplacement I of N, and every integer n> GfdR M there is in isomorphism D(k),

N⊗LR M ' I⊗R FĎn .


10.3.26 Theorem. Assume that R is right Noetherian. Let M be an R-complex offinite Gorenstein flat dimension and n an integer. The next conditions are equivalent.



(i) GfdR M 6 n.(ii) sup(N⊗L

R M)6 n+ supN holds for every Ro-complex N with fdRo N finite oridRo N finite.

(iii) sup(N⊗LR M)6 n holds for every injective Ro-module N.

(iv) n> supM and TorRn+1(N,M) = 0 for every Ro-module N with idRo N finite.

(v) n > supM and for some, equivalently every, semi-flat replacement F of M,the module Cv(F) is Gorenstein flat for every v> n.

(vi) There exists a semi-flat replacement F of M with Fv = 0 for every v < infM,Hv(F) = 0 for every v > n, and Cv(F) Gorenstein flat for every v> n.


GfdR M = supsup(N⊗LR M) | N is an Ro-module with idRo N finite

= supm ∈ Z | TorRm(N,M) 6= 0 for some injective Ro-module N .

PROOF. We start by establishing the equivalence of (i)–(iii).(i)=⇒ (ii): One can assume that N is in D@(Ro) and not acyclic; otherwise the

inequality is trivial. In this case, w = supN is an integer. By 5.3.26 there is a semi-injective resolution N '−→ I with Iv = 0 for v > w. If fdRo N or idRo N is finite, then10.3.25 yields N⊗L

R M ' I⊗R FĎn where F is any semi-flat replacement of M. Forv > n+w and p ∈ Z either p > w or v− p> v−w > n holds, so the module

(I⊗R FĎn)v =∐

p∈ZIp⊗R (FĎn)v−p

is zero. In particular, one has Hv(N⊗LR M) = 0 for v > n+w.

(ii)=⇒ (iii): Trivial.(iii)=⇒ (i): By 10.3.15 the Ro-complex Homk(M,E) has finite Gorenstein injec-

tive dimension. Furthermore, for every Ro-module N one has

− infRHomRo(N,Homk(M,E)) = − infHomk(N⊗LR M,E) = sup(N⊗L

R M) ,

where the first equality holds by adjunction 7.5.15 and the second one by 2.5.6(b).Thus condition (iii) implies, in view of 10.2.15, that one has GidR Homk(M,E)6 n,and hence GfdR M 6 n holds by another application of 10.3.15.

(ii)=⇒ (iv): The vanishing of Tor is immediate from (ii). In view of 10.3.14, theinequality n> supM follows from (i), which is known to be equivalent to (ii).

(iv)=⇒ (i): As already mentioned, the Ro-complex Homk(M,E) has finite Goren-stein injective dimension. Moreover, one has supM =− infHomk(M,E) by 2.5.6(b)and Homk(TorR

n+1(N,M),E)∼= Extn+1Ro (N,Homk(M,E)) by 8.3.1. Thus (iv) implies,

in view of 10.2.15, that GidR Homk(M,E)6 n. Hence GfdR M 6 n holds by 10.3.15.Before we connect the equivalent conditions (i)–(iv) to (v), note that by 8.3.17 and

10.3.11 the “some” version and the “every” version of condition (v) are equivalent.(i)=⇒ (v): Follows from 10.3.14 and 10.3.16.(v)=⇒ (vi): Immediate in view of 5.2.15 and 5.4.10.(vi)=⇒ (i): Immediate from the definition, 10.3.13, of GfdR.

10.3.27 Proposition. Assume that R is right Noetherian. For every family of R-complexes Muu∈U one has,



GfdR(∐

u∈U

Mu) = supu∈UGfdR Mu .

PROOF. By 10.3.15, 3.1.25, and 10.2.17 one has:

GfdR(∐

u∈U Mu) = GidRo Homk(∐

u∈U Mu,E)

= GidRo(∏

u∈U Homk(Mu,E))

= supu∈UGidRo Homk(Mu,E)= supu∈UGfdR Mu .

NOETHERIAN RINGS AND HOMOLOGICAL FINITENESS

10.3.28 Lemma. Let R be Noetherian and G be a finitely generated R-module. IfG is Gorenstein flat, then there exists an exact sequence 0→ G→ L→ G′→ 0 offinitely generated R-modules with L free and G′ Gorenstein flat.

PROOF. There exists, by definition, an exact sequence 0→ G→ F → G′′→ 0 withF flat and G′′ Gorenstein flat. As R is left Noetherian, it follows from C.16 thatthe map G→ F admits a factorization G

ϕ−→ L−→ F with L finitely generated free.Notice that ϕ is injective and that the module G′ = Cokerϕ is finitely generated.There is now a commutative diagram with exact rows,

0 // G // L //

G′ //

0

0 // G // F // G′′ // 0 .

For every injective Ro-module E one gets by 6.5.22, 7.4.8, and 7.4.13 a commutativediagram with exact rows,

0 // TorR1 (E,G

′)

// E⊗R G // E⊗R L

0 // TorR1 (E,G

′′) // E⊗R G // E⊗R F .

Per 10.3.3 one has TorR1 (E,G

′′) = 0, and it follows that also TorR1 (E,G

′) vanishes.As R is right Noetherian it now follows from 10.3.11 that G′ is Gorenstein flat.

10.3.29 Proposition. Assume that R is left Noetherian and let G be a finitely gener-ated R-module. If G is Gorenstein projective, then it is Gorenstein flat; the converseholds if R is Noetherian.

PROOF. If G is Gorenstein projective, then by 10.1.21 it is a cokernel in a totallyacyclic complex P of finitely generated free R-modules. It follows from 10.1.3 that Pis totally acyclic as a complex of flat modules; see 10.3.1. Thus G is Gorenstein flat.

Conversely, assume that R is Noetherian and that G is Gorenstein flat. It fol-lows from 10.3.28 that there is an injective quasi-isomorphism ι : G→ L, where L



is a complex of finitely generated free R-modules with Lv = 0 for v > 0 and Cv(L)Gorenstein flat for every v ∈ Z. By 5.1.13 there is a surjective semi-free resolutionπ : F '−→ G where F is degreewise finitely generated with Fv = 0 for v < 0. Thecomplex P = Σ−1 Cone ιπ is a complex of finitely generated free R-modules withC0(P) = C0(F) ∼= G. Moreover, for v 0 the module Cv(P) = Cv+1(L) is Goren-stein flat, so by 10.3.3 and A.13 the complex E⊗R P is acyclic for every injectiveRo-module E. That is, P is totally acyclic as a complex of flat modules.

10.3.30 Theorem. Let R be Noetherian. For every complex M in DfA(R) one has

GfdR M = GpdR M .

PROOF. Choose by 5.1.14 a degreewise finitely generated semi-free resolutionL '−→M. The complex L is, in particular, a semi-projective and a semi-flat replace-ment of M; see 5.2.11 and 5.4.10. From the definitions, 10.1.8 and 10.3.13, of thedimensions in question and from 10.1.17 and 10.3.26, it follows that one has

GpdR M = infn ∈ Z | n> supM and Cn(L) is Gorenstein projective andGfdR M = infn ∈ Z | n> supM and Cn(L) is Gorenstein flat

As every module Cn(L) is finitely generated, the assertion follows from 10.3.29.

THE CASE OF MODULES

10.3.31. Notice from 10.3.26 that a non-zero R-module is Gorenstein flat if andonly if it has Gorenstein flat dimension 0 as an R-complex.

10.3.32 Theorem. Let R be right Noetherian, let M be an R-module of finite Goren-stein flat dimension and n> 0 be an integer. The next conditions are equivalent.

(i) GfdR M 6 n.(ii) TorR

m(N,M) = 0 holds for every Ro-module N with fdRo N finite or idRo N finiteand all integers m > n.

(iii) TorRm(N,M) = 0 holds for every injective Ro-module N and all integers m > n.

(iv) TorRn+1(N,M) = 0 holds for every Ro-module N with idRo N finite.

(v) In some/every flat resolution · · · → Fv→ Fv−1→ ·· · → F0→M→ 0 the mo-dule Coker(Fv+1→ Fv) is Gorenstein flat for every v> n.

(vi) There is an exact sequence of R-modules 0→G→Fn−1→···→F0→M→ 0with each Fi flat and G Gorenstein flat.


GfdR M = supm ∈ N0 | TorRm(E,M) 6= 0 for some injective Ro-module E .

PROOF. In view of 8.3.23 the equivalence of the six conditions follows from10.3.26. The final equality follows from the equivalence of (i) and (iii).



EXERCISES

E 10.3.1 The invariant FGFD(R) = supGfdR M |M is an R-module with GfdR M < ∞ is calledthe finitistic Gorenstein flat dimension of R. Show that FGFD(R) > FFD(R) and thatequality holds if R is right Noetherian.

10.4 Gorenstein Dimensions and Foxby Equivalence

SYNOPSIS. Gorenstein projective/flat dimension and the Auslander category; Gorenstein injectivedimension and the Bass category; derived reflexive complex; Iwanaga-Gorenstein ring.

THE AUSLANDER CATEGORY

10.4.1 Proposition. Assume that R is flat as a k-module. Let M be an R-complex, Xa complex of R–So-bimodules, and N an S-complex. Assume that X ∈D@(R⊗k So)with idSo X and fdSo RHomR(M,X) finite. The tensor evaluation morphism 8.4.5,


S N) ,

is an isomorphism under each of the following conditions.(a) R is left Noetherian and M is in Df

A(R); S is right Noetherian and N has finiteGorenstein flat dimension.

(b) S is Noetherian and N is in DfA(S) and of finite Gorenstein flat dimension.

PROOF. Choose a semi-injective resolution X '−→ E over R⊗k So with Ev = 0 forv 0. As R is flat as a k-module, E is also semi-injective over So; see 7.3.18(c).

(a) Let π : P→M be a semi-free resolution with P bounded below and degree-wise finitely generated; see 5.1.14. Let F ' N be a semi-flat replacement over S andset n = GfdS N. By 8.4.6 it suffices to prove that the tensor evaluation morphismθPEF is a quasi-isomorphism in D(k). There is a commutative diagram,

HomR(P,E)⊗S F

Hom(P,E)⊗τFĎn '

θPEF// HomR(P,E⊗S F)

Hom(P,E⊗τFĎn)'

HomR(P,E)⊗S FĎnθPEFĎn

∼=// HomR(P,E⊗S FĎn) .

The left-hand morphism is a quasi-isomorphism by 10.3.24. Indeed, it follows from5.3.25 that HomR(P,E) is semi-injective So-complex, and fdSo HomR(P,E) is finiteby assumption as HomR(P,E) is RHomR(M,X). The morphism E⊗R τ

FĎn is a quasi-

isomorphism, also by 10.3.24, as idSo E = idSo X is finite by assumption, and sothe right hand morphism is a quasi-isomorphism by semi-projectivity of P. Finally,θPEFĎn is an isomorphism by 4.5.7(1,a).

(b) Let π : P→M be a semi-projective resolution and F '−→ N be a semi-projective resolution with F bounded below and degreewise finitely generated; see5.1.12 and 5.1.14. Set n = GfdS N. For these complexes there is a commutative dia-gram identical to the one above, and θPEFĎn is an isomorphism by 4.5.7(d).


10.4 Gorenstein Dimensions and Foxby Equivalence 391

10.4.2 Lemma. Assume that R and S are projective as k-modules. Let R be leftNoetherian, S be right Noetherian, and D be a dualizing complex for (R,So). Amodule G in A(S) is Gorenstein projective if ExtiS(G,L) = 0 holds for all projectiveS-modules L and all i > 0.

PROOF. Choose by 5.2.14 a surjective semi-projective resolution π : P '−→ G withPv = 0 for v < 0. It is sufficient to show that there is an injective quasi-isomorphismι : G→ F , where F is a complex of projective S-modules with Fv = 0 for v > 0and ExtiS(Cv(F),L) = 0 for all projective S-modules L and all i > 0. Indeed, P =Σ−1 Cone ιπ is then an acyclic complex of projective S-modules with C0(P) = G,and by A.11 the complex HomS(P,L) is acyclic for every projective module L.

To prove the existence of ι : G '−→ F , it suffices to show that there is an exactsequence of S-modules,

(?) 0−→ G−→ F ′ −→ G′ −→ 0 ,

where F ′ is projective and G′ has the same properties as G, that is, G′ belongs toA(S) and ExtiS(G

′,L) = 0 holds for all projective S-modules L and all i > 0. To thisend notice that by 7.3.14 the assumption of Ext vanishing yields

(‡) ExtiS(G,K) = 0 for every S-module K with pdS K < ∞ and all i > 0 .

First we argue that G maps injectively into a module of finite flat dimension. Chooseby 9.1.18 a bounded complex I of R⊗k So-bimodules that is semi-injective over Rand over So and isomorphic to D in D(R⊗k So). By assumption D⊗L

S G is homologi-cally bounded, so by 5.3.26 there is a semi-injective resolution D⊗L

S G '−→ J over Rwith Jv = 0 for v 0. In D(S) one now has G' RHomR(D,D⊗L

S G)'HomR(I,J),as G belongs to A(S). By 2.5.9 and 9.3.7(b) the complex X =HomR(I,J) is boundedabove and consists of flat S-modules. One has G∼= H0(X) and Hv(X) = 0 for v 6= 0,so G is per (2.2.11.4) isomorphic to a submodule of C = C0(X), and Xě0

'−→C is aflat resolution. As X is bounded above, C has finite flat dimension; see 8.3.24.

Next we argue that G maps into a flat module. There is an exact sequence,

0−→ K −→ F κ−−→C −→ 0 ,

of S-modules where F is free and fdS K is finite; see 1.3.11 and 8.3.13. In view of9.3.11 also pdS K is finite. By (‡) one has Ext1S(G,K) = 0, so the homomorphismHomS(G,κ) : HomS(G,F)→ HomS(G,C) is surjective; see 7.3.14. It follows thatthe injection GC factors through κ; i.e. there is an injection G F .

Finally we can construct (?). By E.20 there exists a flat preenvelope ϕ : G→ P′,and since G maps injectively into a flat module, ϕ is injective per E.18. As above,there is an exact sequence

0−→ K′ −→ F ′ψ−−→ P′ −→ 0

with F ′ free and pdS K′ finite, and it follows that ϕ factors through ψ. That is, there isan injective homomorphism ϕ′ : G→ F ′ with ψϕ′ = ϕ. Set G′ = Cokerϕ′; we arguethat this yields is the desired sequence (?). The free module F ′ is in A(S) by 9.3.8and G is in A(S) by assumption, so as A(S) is a triangulated subcategory of D(S),



see 9.3.4 it follows that G′ is in A(S), cf. 6.5.19. Let L be a projective S-module;one has ExtiS(G

′,L)∼= Exti−1S (G,L) = 0 for i > 2 by 7.3.14. To see that Ext1S(G

′,L)vanishes, consider the exact sequence of k-modules from 7.3.14,

() HomS(F ′,L)Hom(ϕ′,L)−−−−−−→ HomS(G,L)−→ Ext1S(G

′,L)−→ 0 .

As HomS(ϕ′,L) is surjective per E.18, exactness of () yields Ext1S(G

′,L) = 0.

10.4.3 Theorem. Assume that R and S are projective as k-modules. Let R be leftNoetherian, S be right Noetherian, and D be a dualizing complex for (R,So). Forevery S-complex N the following conditions are equivalent.

(i) GpdS N is finite.(ii) GfdS N is finite.

(iii) N and D⊗S N are homologically bounded above and N is in A(S).Moreover, if N is homologically bounded below, then (i)–(iii) are equivalent to(iii’) N is in A(S).

PROOF. An acyclic complex satisfies all four conditions, so one may assume that Nis not acyclic. The implication (i)=⇒ (ii) is immediate from 9.3.11 and 10.3.19.

(ii)=⇒ (iii): Let N be an S-complex of finite Gorenstein flat dimension. It fol-lows from 10.3.14 and 10.3.26 that both N and D⊗S N are homologiclly boundedabove. By 10.4.1(a) tensor evaluation θDDN is an isomorphism. Now it follows from(9.2.6.1) that the unit αN

D is an isomorphism, whence N belongs to A(S) per 9.3.3.(iii)=⇒ (i): Let N be in A(S) and choose a semi-projective resolution P '−→ N.

Set n=maxidR D+sup(D⊗LS N),supN; by assumption n is an integer. Per 10.1.8

it suffices to show that the module Cn(P) is Gorenstein projective. For every projec-tive S-module L one has − infRHomS(N,L) 6 n by 9.3.14(b), so by 8.1.7 one hasExtiS(Cn(P),L) = 0 for all i > 0. Moreover, in the exact sequence of S-complexes,

0−→ Pďn−1 −→ PĎn −→ Σn Cn(P)−→ 0 ,

the middle term PĎn 'N belongs to A(S) by assumption, and the left-hand complexis in A(S) by 9.3.5 as it has finite projective dimension. Thus, Cn(P) belongs toA(S), as it is a triangulated subcategory per 9.3.4, and it follows from 10.4.2 thatCn(P) is Gorenstein projective.

For a complex N in DA(S) condition (iii) is by 7.6.8 equivalent to N ∈A(S).

THE BASS CATEGORY

10.4.4 Proposition. Assume that S is flat as a k-module and right Noetherian. LetM be an R-complex, X a complex in D@(R⊗k So), and N a complex in Df

A(So). The

homomorphism evaluation morphism 8.4.10,


is an isomorphism if the dimensions GidR M, idR X and pdR RHomSo(N,X) are finite.



PROOF. Choose a semi-injective resolution X '−→ E over R⊗k So with Ev = 0 forv 0. As S is flat as a k-module, the complex E is also semi-injective over R;see 7.3.18(c). Let P '−→ N be a semi-free resolution with P bounded below anddegreewise finitely generated; see 5.1.14. Let M '−→ I be a semi-injective resolutionand set n = GidR M. By ?? it suffices to prove that the homomorphism evaluationmorphism ηPEI is a quasi-isomorphism in D(k). There is a commutative diagram,

P⊗S HomR(E, IĚ−n)ηPEIĚ−n

//

P⊗Hom(E,τIĚ−n) '

HomR(HomSo(P,E), IĚ−n)

Hom(Hom(P,E),τIĚ−n)'

P⊗S HomR(E, I)ηPEI

// HomR(HomSo(P,E), I) .

As idR E = idR X is finite by assumption, the left-hand vertical morphism is aquasi-isomorphism by 10.2.13 and semi-projectivity of P. The right-hand mor-phism is also a quasi-isomorphism by 10.2.13. Indeed, it follows from 5.3.25 thatHomSo(P,E) is a semi-injective R-complex, and by assumption pdR HomSo(P,E) =pdR RHomSo(N,X) is finite. Now ηPEIĚ−n is an isomorphism by 4.5.10(1,a).

10.4.5 Lemma. Assume that R and S are projective as k-modules. Let R be leftNoetherian, S be right Noetherian, and D be a dualizing complex for (R,So). Amodule G in B(R) is Gorenstein injective if ExtiR(E,G) = 0 holds for all injectiveR-modules E and all i > 0.

PROOF. Choose by 5.3.19 an injective semi-injective resolution ι : G '−→ I withIv = 0 for v > 0. It is sufficient to show that there is a surjective quasi-isomorphismπ : J→ G, where J is a complex of injective R-modules with Jv = 0 for v < 0 andExtiR(E,Zv(J)) = 0 for all injective R-modules E and all i > 0. Indeed, I = Cone ιπis then an acyclic complex of injective R-modules with Z0(I) = G, and by A.9 thecomplex HomR(E, I) is acyclic for every injective R-module E.

To prove the existence of π : J '−→ G, it suffices to show that there is an exactsequence of R-modules,

(?) 0−→ G′ −→ J′ −→ G−→ 0 ,

where J′ is injective and G′ has the same properties as G; that is, G′ belongs to B(R)and ExtiR(E,G

′) = 0 holds for all injective R-modules E and all i > 0. To this endnotice that by 7.3.14 the assumption of Ext vanishing yields

(‡) ExtiR(C,G) = 0 for every R-module C with idR C < ∞ and all i > 0 .

First we argue that G is a homomorphic image of a module of finite injective di-mension. Choose by 9.1.18 a bounded complex J of R⊗k So-bimodules that is semi-injective over R and over So and isomorphic to D in D(R⊗k So). Choose a semi-projective resolution P '−→ RHomR(D,G) over S with Pv = 0 for v 0. In D(R)one now has J⊗S P ' D⊗L

S RHomR(D,G) ' G, as G belongs to B(R). By 2.5.13and 9.3.7(a) the complex X = J⊗S P is bounded below and consists of injectiveR-modules. One has G ∼= H0(X) and Hv(X) = 0 for v 6= 0, so G is per (2.2.11.3) a



homomorphic image of Z = Z0(X), and Z '−→ Xď0 is an injective resolution. As Xis bounded below, Z has finite injective dimension; see 8.2.17.

Next we argue that G is a homomorphic image of an injective module. There isby 5.3.27 and 8.2.9 an exact sequence of R-modules,

0−→ Z ε−−→ I′ −→C −→ 0 ,

where I′ is injective and idR C is finite. By (‡) one has Ext1R(C,G) = 0, so the ho-momorphism HomR(ε,G) : HomR(I′,G)→ HomR(Z,G) is surjective; see 7.3.14. Itfollows that the surjection ZG factors through ε; i.e. there is a surjection I′G.

Finally we can construct (?). By F.9 there exists an injective precover ϕ : J′→ G,and since G is a homomorphic image of an injective module, ϕ is surjective perF.8. Set G′ = Kerϕ; we argue that this yields the desired sequence (?). The injectivemodule J′ is in B(R) by 9.3.5, and G is in B(R) by assumption, so as B(R) is atriangualted subcategory of D(R), see 9.3.4, it follows that G′ is in B(R); cf. 6.5.19.Let E be an injective R-module; one has ExtiR(E,G

′)∼=Exti−1R (E,G) = 0 for i> 2 by

7.3.14. To see that Ext1R(E,G′) vanishes, consider the exact sequence of k-modules,

() HomR(E,J′)Hom(E,ϕ)−−−−−−→ HomR(E,G)−→ Ext1R(E,G

′)−→ 0 .

As HomR(E,ϕ) is surjective, see F.8, exactness of () yields Ext1R(E,G′) = 0.

10.4.6 Theorem. Assume that R and S are projective as k-modules. Let R be leftNoetherian, S be right Noetherian, and D be a dualizing complex for (R,So). Forevery R-complex M the following conditions are equivalent.

(i) GidR M is finite.(ii) M and RHomR(D,M) are homologically bounded below and M is in B(R).

Moreover, if M is homologically bounded above, then (i) and (ii) are equivalent to(ii’) M is in B(R).

PROOF. An acyclic complex trivially satisfies all three conditions, so one may as-sume that M is not acyclic.

(i)=⇒ (ii): Let M be an R-complex of finite Gorenstein injective dimension. By10.2.7 and 10.2.15 both M and RHomR(D,M) are homologiclly bounded below.By 10.4.4 homomorphism evaluaion ηMDD is an isomorphism. Now (9.2.6.2) showsthat the counit βM

D is an isomorphism, whence M belongs to B(R) per 9.3.3.(ii)=⇒ (i): Let M be in B(R) and choose a semi-injective resolution M '−→ I. Set

n = maxidR D− infRHomR(D,M),− infM; by assumption n is an integer. Per10.2.6 it suffices to show that the module Z−n(I) is Gorenstein injective. For everyinjective R-module E one has − infRHomR(E,M)6 n by 9.3.14(a), so by 8.2.6 onehas ExtiR(E,Z−n(I)) = 0 for all i > 0. Moreover, in the exact sequence,

0−→ Σ−n Z−n(I)−→ IĚ−n −→ Iě−n+1 −→ 0 ,

of R-complexes the middle term IĚ−n 'M belongs to B(R) by assumption, and theright-hand complex is in B(R) by 9.3.5 as it has finite injective dimension. Thus,Z−n(I) belongs to B(R), as it is a triangulated subcategory per 9.3.4, and it followsfrom 10.4.5 that Z−n(I) is Gorenstein injective.



For a complex M in D@(R) condition (ii) is by 7.6.7 equivalent to M ∈B(R).

DERIVED REFLEXIVE COMPLEXES

10.4.7 Proposition. Assume that R is Noetherian. For every totally acyclic com-plex P of finitely generated projective R-modules the complex HomR(P,R) is a to-tally acyclic complex of finitely generated projective Ro-modules. In particular, fora finitely generated Gorenstein projective R-module G, the Ro-module HomR(G,R)is Gorenstein projective.

PROOF. Let P be a totally acyclic complex of finitely generated projective R-modules. By 4.5.4 and the assumption, HomR(P,R) is an acyclic complex of finitelygenerated projective Ro-modules. As one has P ∼= HomRo(HomR(P,R),R), still by4.5.4, it follows that HomR(P,R) is totally acyclic; see 10.1.3. The final assertionfollows from the isomorphism HomR(C0(P),R)∼= C1(HomR(P,R)).

A module that satisfies condition (2) in the next proposition is called reflex-ive , and a module that satisfies both conditions is often called totally reflex-ive.module!totally reflexive

10.4.8 Proposition. Assume that R is Noetherian. A finitely generated R-module Gis Gorenstein projective if and only if it satisfies the next conditions.

(1) ExtiR(G,R) = 0 = ExtiRo(HomR(G,R),R) holds for all i > 0, and(2) biduality δG

R : G→ HomRo(HomR(G,R),R) is an isomorphism.

PROOF. Assume that G is Gorenstein projective. From 10.1.7 and 10.4.7 one getsvanishing of ExtiR(G,R) and ExtiRo(HomR(G,R),R) for all i > 0. Choose by 10.1.21a totally acyclic complex P of finitely generated projective R-modules with C0(P)∼=G. Acyclicity of P∗ = HomR(P,R) and HomRo(P∗,R), see 10.4.7, yields

C0(HomRo(P∗,R)) ∼= HomRo(C1(P∗),R) ∼= HomRo(HomR(C0(P),R),R) ,

so δGR is a restriction of the biduality isomorphism δP

R; see 4.5.4.Conversely, assume that has ExtiR(G,R) = 0 = ExtiRo(HomR(G,R),R) holds for

all i > 0 and that δGR is an isomorphism. Let π : L '−→ G and π′ : L′ '−→ HomR(G,R)

be surjective semi-free resolutions over R and Ro with L and L′ degreewise finitelygenerated and Lv = 0 = L′v for v < 0; see 5.1.13. As ExtiRo(HomR(G,R),R) vanishesfor i> 0, the homology of HomRo(L′,R) is HomRo(HomR(G,R),R), so HomRo(π′,R)is an injective quasi-isomorphism. As HomRo(L′,R) by 4.5.4 is a complex of finitelygenerated projective R-modules, the complex P = Σ−1 Cone(HomRo(π′,R)δG

Rπ) isan acyclic complex of finitely generated free R-modules with C0(P) ∼= C0(L) ∼= G.For v 6 −1 one has Hv(HomR(P,R)) = Ext−v

R (G,R) = 0, and for v > 0 one hasHv(HomR(P,R)) = Hv(HomR(HomRo(L′,R),R)) ∼= Hv(L′) = 0 by biduality 4.5.4.Finally, one has Z0(P) = Ker(HomRo(π′0,R)δ

GRπ0)) = Kerπ0 as δG

R is an isomor-phism and HomRo(π′0,R) is injective. Further, one has Kerπ0 = B0(L) = B0(P), andit follows that also H0(HomR(P,R)) is zero.



10.4.9 Definition. Assume that R is flat as a k-module. An R-complex M is calledderived reflexive if both M and RHomR(M,R) have bounded homology and bidualityδM

R : M→ RHomRo(RHomR(M,R),R) is an isomorphism in D(R).

10.4.10 Example. Assume that R is flat as a k-module and left Noetherian. It followsfrom 9.2.11 and biduality 9.2.12 that every complex in Df

A(R) of finite projectivedimension is derived reflexive.

10.4.11 Lemma. Assume that R is flat as a k-module. Let G be a bounded complexof R-modules. If for every v ∈ Z and all i > 0 one has

ExtiR(Gv,R) = 0 = ExtiRo(HomR(Gv,R),R) ,

then G is derived reflexive if and only if biduality δGR in C(R) is a quasi-isomorphism.

PROOF. Let ι : R '−→ I be a semi-injective resolution over R⊗k Ro with Iv = 0 forv > 0. Since R is flat as a k-module, it follows from 7.3.18 that I is semi-injectiveboth as an R- and as an Ro-complex. Consider the commutative diagram,

GδG

I//

δGR

HomRo(HomR(G, I), I)

Hom(Hom(G,ι),I)'

HomRo(HomR(G,R),R)Hom(Hom(G,R),ι)

'// HomRo(HomR(G,R), I) .

To see that the right-hand map is a quasi-isomorphism, set C = Cone ι and no-tice that each module Cv is either R or injective. By the assumptions on G, it fol-lows from A.9 and A.10 that HomR(G,C) is acyclic. Thus, HomR(G, ι) is a quasi-isomorphism, see 4.1.15; in particular RHomR(G,R) ' HomR(G,R) has boundedhomology. By semi-injectivity of I as an Ro-complex, also HomRo(HomR(G, ι), I)is a quasi-isomorphism. Similarly it follows that HomRo(HomR(G,R), ι) is a quasi-isomorphism. By commutativity of the diagram, δG

R is a quasi-isomorphism if andonly if δG

I is a quasi-isomorphism. Per 8.4.1 biduality δGR is an isomorphism in D(R)

if an only if δGI is a quasi-isomorphism.

10.4.12 Theorem. Assume that R is flat as a k-module and Noetherian. For everycomplex M ∈Df

A(R) the following conditions are equivalent.(i) GpdR M is finite.

(ii) M is derived reflexive.

PROOF. One may assume that M is not acyclic. Let π : P '−→M be a semi-free re-solution with P bounded below and degreewise finitely generated; see 5.1.14.

(i)=⇒ (ii): If n = GpdR M is finite, then M has bounded homology; see 10.1.9.Now PĎn is by 10.1.17 a bounded complex of finitely generated Gorenstein projec-tive modules. It follows from 10.4.8 that PĎn satisfies the conditions in 10.4.11 andthat biduality δPĎn

R is an isomorphism, as each component (δPĎnR )v = δ

(PĎn)vR is so.

Since M ' PĎn in D(R) it now follows from 10.4.11 that M is derived reflexive.(ii)=⇒ (i): Set n =− infRHomR(M,R) and notice that n> supM holds by 7.6.7

and the isomorphism M ' RHomRo(RHomR(M,R),R) in D(R); the goal is to prove



that Cn(P) is Gorenstein projective. To this end, notice that biduality δPďnR is an iso-

morphim in D(R) by 9.2.12, as Pďn is a complex of finite projective dimension.Since one has M ' PĎn in D(R), biduality δPĎn

R is an isomorphism by assump-tion. The canonical exact sequence 0→ Pďn−1→ PĎn→ Σn Cn(P)→ 0 and bidual-ity 8.4.2 induce by 6.5.19 a commutative diagram in D(R), which by 6.5.22 showsthat also δ

Cn(P)R is an isomorphism. By assumption one has ExtiR(Cn(P),R) = 0

for all i > 0, see 8.1.7. It follows that there is an isomorphism HomR(Cn(P),R) 'RHomR(Cn(P),R) in D(R), and as δCn(P)

R is an isomorphism one has for every v 6= 0,

ExtvRo(HomR(Cn(P),R),R) ∼= H−v(RHomRo(RHomR(Cn(P),R),R))∼= H−v(Cn(P)) = 0 .

It now follows from 10.4.11 that biduality δCn(P)R is a quasi-isomorphism, and hence

an isomorphism as it is a map of modules. Finally it follows from 10.4.8 that Cn(P)is Gorenstein projective.

IWANAGA-GORENSTEIN RINGS

Recall from 9.1.19 that a Noetherian ring that satisfies condition (ii) in the nexttheorem is called Iwanaga-Gorenstein.

10.4.13 Theorem. Let R be Noetherian and n > 0 be an integer; the followingconditions are equivalent.

(i) For every finitely generated R-module M one has GpdR M 6 n, and for everyfinitely generated Ro-module N one has GpdRo N 6 n.

(ii) One has idR R6 n and idRo R6 n.

PROOF. (i)=⇒ (ii): For every left ideal a in R one has GpdR R/a 6 n, so 10.1.17yields ExtiR(R/a,R) = 0 for all i > n and then it follows from 8.2.8 that idR R 6 nholds. Similarly one gets idRo R6 n.

(ii)=⇒ (i): Let M be a finitely generated R-module and choose a semi-free reso-lution ϕ : L '−→M with L degreewise finitely generated and Lv = 0 for v < 0; see5.1.14. It suffices by 10.1.8 to show that the R-module Cn(L) is Gorenstein pro-jective, and that will be done by verifying that it satisfies the conditions in 10.4.8.It follows from 8.2.8 that for i > n one has 0 = ExtiR(M,R) = H−i(HomR(L,R)).In particular, ExtiR(Cn(L),R) = 0 holds for all i > 0, see 8.1.7, and the canonicalquasi-isomorphism Σ−n L Cn(L) induces a quasi-isomorphism,

ψ : HomR(Cn(L),R) −→ HomR(Σ−n Lěn,R) .

Choose an injective resolution ι : R '−→ I over Ro with Iv = 0 for v < −n and asemi-projective resolution π : P '−→ HomR(Cn(L),R) with Pv = 0 for v < 0. AsL′ = HomR(Σ

−n Lěn,R) by 4.5.4 is a complex of projective Ro-modules, it followsfrom A.10 that HomRo(L′,Cone ι) is acyclic. Thus by semi-injectivity of I and semi-projectivity of P there are quasi-isomorphisms of k-complexes,



HomRo(L′,R)Hom(L′,ι)−−−−−→HomRo(L′, I)

Hom(ψπ,I)−−−−−−→HomRo(P, I)Hom(P,ι)−−−−−→HomRo(P,R) .

For i > 0 one now has

ExtiRo(HomR(Cn(L),R),R) ∼= H−i(HomRo(P,R))∼= H−i(HomRo(L′,R))

= H−i(HomRo(HomR(Σ−n Lěn,R),R))

∼= H−i(Σ−n Lěn) = 0 ,

Notice that the argument above applies to the module Cm(L) for every m> n. In par-ticular, one has Ext1Ro(HomR(Cn+1(L),R),R) = 0 whence a straightforward compu-tation yields Cn(HomRo(HomR(L,R),R))∼= HomRo(HomR(Cn(L),R),R). It followsthat δCn(L)

R is a restriction of the biduality isomorphism δLR; see 4.5.4.

The same argument applies to show that every finitely generated Ro-module hasfinite Gorenstein projective dimension.

10.4.14 Corollary. Assume that R is projective as a k-module and Noetherian. Letn> 0 be an integer; the following conditions are equivalent.

(i) One has idR R6 n and idRo R6 n.(ii) Every finitely generated R-module and every finitely generated Ro-module has

Gorenstein projective dimension at most n.(iii) Every complex M in D@(R) and every complex M in D@(Ro) has Gorenstein

projective dimension at most n+ supM.(iv) Every complex M in D@(R) and every complex M in D@(Ro) has Gorenstein

flat dimension at most n+ supM.(v) Every complex M in DA(R) and every complex M in DA(Ro) has Gorenstein

injective dimension at most n− infM.

PROOF. If R is Iwanaga-Gorenstein, then R is a dualizing complex for R, and everyR-complex belongs to A(R) and to B(R); see 9.1.20 and 9.3.6. It thus follows from(7.5.1.1) and 10.4.3 that every complex M in D@(R) has finite Gorenstein projectiveand finite Gorenstein flat dimension. Moreover, it follows from 9.3.14, 10.1.17, and10.3.26 that n+ supM is an upper bound for GpdR M and GfdR M. Similarly it fol-lows from (7.5.1.2) and 10.4.6 that every complex M in DA(R) has finite Gorensteininjective dimension, and by 9.3.14 and 10.2.15 one has GidR M 6 n− infM. Thesame arguments apply to Ro, so (i) implies (iii)–(v); cf. 9.1.19. Further, both (iii) and(iv) imply (ii), see 10.3.30, and (ii) implies (i) by 10.4.13. Finally, idR R = GidR Rand idRo R = GidRo R by 10.2.16, so (v) implies (i).

EXERCISES

E 10.4.1 Show that there is a complex N in A(S) with GfdS N = ∞.E 10.4.2 Show that there is a complex M in B(R) with GidR M = ∞.E 10.4.3 Show that 10.4.1(b) is valid without boundedness assumptions on the complex X .



E 10.4.4 Assume that S is projective as a k-module and left Noetherian. Let M be an R-complex,X be a complex in DA(R⊗k So), and N be a complex in Df

A(S). Show that the tensorevaluation morphism θMXN from 8.4.5 is an isomorphism in D(k) if the dimensionsGpdR M, pdR X , idR(X⊗L

S N) are finite.E 10.4.5 Assume that R is left Noetherian and that S is flat as a k-module. Let M be a complex

in DfA(R), X be a complex of R–So-bimodules, and N be an S-complex. Show that

the tensor evaluation morphism θMXN from 8.4.5 is an isomorphism in D(k) if thedimensions GpdR M, fdR X , idR(X⊗L

S N) are finite.E 10.4.6 Assume that R is flat as a k-module and that S is right Noetherian. Let M be an R-

complex, X a complex of R–So-bimodules, and N a complex in DfA(S

o). Show thatthe homomorphism evaluation morphism ηMXN from 8.4.10, is an isomorphism if thedimensions fdSo X , fdS RHomR(X ,M), and GpdSo N are finite. Hint: E 10.1.4.

E 10.4.7 Let M be a complex of Gorenstein projective R-modules and let N be as in 10.1.7.(a) Show that HomR(M,–) preserves quasi-isomorphisms of bounded below com-plexes of modules from N. (b) Show that if M is bounded below, then HomR(M,–)preserves quasi-isomorphisms of bounded above complexes of modules from N.

E 10.4.8 Let M be a complex of Gorenstein injective R-modules and let N be as in 10.2.5.(a) Show that HomR(–,M) preserves quasi-isomorphisms of bounded above com-plexes of modules from N. (b) Show that if M is bounded above, then HomR(–,M)preserves quasi-isomorphisms of bounded below complexes of modules from N.

E 10.4.9 Let M be a complex of Gorenstein flat R-modules and let N be as in 10.3.12. (a) Showthat –⊗R M preserves quasi-isomorphisms of bounded above complexes of mod-ules from N. (b) Show that if M is bounded below, then –⊗R M preserves quasi-isomorphisms of bounded below complexes of modules from N.

E 10.4.10 Assume that R is flat as a k-module and left Noetherian. Show that the full subcategoryof D(R) whose objects are the derived reflexive complexes is triangulated.


Chapter 11Derived Torsion and Completion

In this chapter R is assumed to be commutative.

11.1 Torsion and Completion

SYNOPSIS. a-torsion functor; a-torsion subcomplex; a-adic completion functor; torsion(-free) mo-dule; flat module.

THE TORSION FUNCTOR

11.1.1 Definition. Let xxx = x1, . . . ,xn be a sequence in R. For u > 1 the symbol xxxu

denotes the sequence xu1, . . . ,x

un of uth powers.

11.1.2 Notation. For every ideal a in R there is a canonical tower of R-modules,

(11.1.2.1) ϑua : R/au R/au−1u>1 .

For a sequence xxx of elements in R there is another canonical tower of R-modules,

ϑuxxx : R/(xxxu) R/(xxxu−1)u>1 .

For a= (xxx) the inclusions (xxxu)⊆ (xxx)u = au yield a canonical comparison morphismof the towers above,

ξuxxx : R/(xxxu) R/(xxx)uu>1 .

11.1.3 Definition. Let a be an ideal in R. The tower (11.1.2.1) yields a functor

Γa(–) = colimu>1

HomR(R/au,–) : C(R)−→ C(R) ;

see 3.2.38. It is called the a-torsion functor.

11.1.4 Definition. Let a be an ideal in R and M be an R-complex. As the differential∂M is R-linear, it maps the graded submodule (0 :M\ a) of M\ to itself. Thus thesubmodule specifies a subcomplex of M, which we henceforth denote (0 :M a).

401

402 11 Derived Torsion and Completion

An element m of M is called a-torsion if aum = 0 holds for some u > 1. The a-torsion elements form a subcomplex

⋃u>1(0 :M au) of M, and M is called a-torsion

if this is all of M.

The next result justifies the name of the functor Γa, as it shows that Γa(M) canbe identified with the a-torsion subcomplex of M.

11.1.5 Proposition. Let a be an ideal in R. For every R-complex M one has

Γa(M) ∼=⋃

u>1

(0 :M au) .

PROOF. For every u > 1 there is an isomorphism HomR(R/au,M) ∼= (0 :M au) ofR-complexes; cf. 1.1.6 and 11.1.4. These subcomplexes form an increasing chain,so one identifies colimu>1 HomR(R/au,M) with

⋃u>1(0 :M au); cf. 3.2.40.

11.1.6. It follows from 11.1.5 that Γa(M) is isomorphic to a subcomplex of M andhence there is a natural transformation γa : Γa→ IdC(R). Evidently there is an equal-ity Γa γa = γaΓa of natural transformations ΓaΓa→ Γa, and this natural transforma-tion is a natural isomorphism, in particular, Γa is idempotent.

11.1.7. The functor Γa restricts to a functor M(R)→M(R), and it can be recoveredfrom this restriction by the procedure in 2.1.45. In this sense, Γa on C(R) is extendedfrom Γa on M(R). This follows from the definition, 11.1.3, in view of 2.3.1 and 3.2.6but, indeed, it is also evident from 11.1.5.

11.1.8 Example. For a module over an integral domain one has⋃

x 6=0Γ(x)(M) =MT.

11.1.9 Example. Let x be an element in R. An R-complex M is (x)-torsion, that is,Γ(x)(M) = M, if and only if Mx ∼= Rx⊗R M is the zero complex.

Recall that H(Rx⊗R M) ∼= Rx⊗R H(M) holds by flatness of Rx and 2.2.18. Con-sequently, if H(M) is (x)-torsion then, Rx⊗R M is acyclic.

11.1.10 Proposition. Let a be an ideal in R. The a-torsion functor Γa is left ex-act, k-linear, and way-out; it is also a \-functor and a Σ-functor, and it preservescoproducts. Moreover, if a is finitely generated, then Γa preserves filtered colimits.

PROOF. All but the last assertion are straightforward consequences of 11.1.5. Alter-natively, they follow from properties of Hom and extended functors: 2.1.45, A.16,2.1.48, 4.1.13, and 3.1.31. If a is finitely generated, then each module R/au is finitelypresented, so by 3.2.36 the functors HomR(R/au,–) preserve filtered colimits. Fi-nally, colimits commute.

11.1.11 Lemma. Let a be an ideal in R and Mu →Mvu6v a U-direct system inC(R). If U is filtered and H(Mu) is a-torsion for every u ∈U , then H(colimu∈U Mu)is a-torsion.

PROOF. The functors H and Γa preserve filtered colimits by 3.2.34 and 11.1.10.


11.1 Torsion and Completion 403

11.1.12 Proposition. Let xxx = x1, . . . ,xn be a sequence in R and M be an R-complex.With the identification from 11.1.5 there is an equality of subcomplexes

Γ(xxx)(M) =n⋂

i=1

Γ(xi)(M) .

Moreover, one has

Γ(xxx)(M) ∼= colimu>1

HomR(R/(xxxu),M) .

PROOF. For every u one has (xxx)nu ⊆ (xxxu) ⊆ (xxxu) and⋂n

i=1(0 :M xui ) = (0 :M (xxxu))

and, therefore,⋃u>1(0 :M (xxx)u) =

⋃u>1(0 :M (xxxu)) =

⋂ni=1⋃

u>1(0 :M xui ) .

In view of 11.1.5 these equalities yield the first assertion. The first equality aloneyields the second assertion, as one identifies HomR(R/(xxxu),M) with (0 :M (xxxu));cf. 1.1.6 and cf. 3.2.40.

THE COMPLETION FUNCTOR

11.1.13 Definition. Let a be an ideal in R. The tower (11.1.2.1) yields a functor

Λa(–) = lim

u>1(R/au⊗R –) : C(R)−→ C(R) ;

see 3.3.30. It is called the a-adic completion functor.

11.1.14. By 1.1.8 one can identify R/au⊗R M with M/auM and consider Λa(M) tobe the limit of the canonical tower M/auMM/au−1Mu>1. In particular, thereis a canonical natural transformation λa : IdC(R)→ Λa. By 3.3.32 one can thus iden-tify Λa(M) with the subcomplex of

∏u>1 M/auM whose elements are sequences

([mu]au)u>1 satisfying mu−mu+1 ∈ au for all u> 1.

11.1.15. The functor Λa restricts to a functor M(R)→M(R), and it can be recov-ered from this restriction by the procedure in 2.1.45. In this sense, Λa on C(R) isextended from Λa on M(R). This follows from the definition, 11.1.13, in view of2.4.1 and 3.3.6.

11.1.16. A priori, Λa(R) is an R-module, however, it is straightforward to see thatit is also a ring with multiplication given by ([ru]au)u>1([su]au)u>1 = ([rusu]au)u>1.That is, Λa(R) is a subring of the product ring

∏u>1 R/au. We write Ra for Λa(R).

11.1.17 Proposition. Let a be an ideal in R. The a-adic completion functor Λa isk-linear and way-out; it is also a \-functor and a Σ-functor. Moreover, if a is finitelygenerated, then Λa preserves products.

PROOF. All assertions but the last one follow from properties of tensor products andextended functors: 2.1.45, A.16, 2.1.48, and 4.1.13. If a is finitely generated, then



each module R/au is finitely presented, so by 3.1.28 the functors R/au⊗R – preserveproducts. Finally, limits and products commute.

11.1.18 Construction. Let a be an ideal in R and let 0−→ X α−→ Nβ−→M −→ 0 be

an exact sequence of R-complexes. Consider for every u> 1 the diagram

X/auX

πu

αu

&&

0 // X/α−1(auN)αu// N/auN

βu// M/auM // 0 ,

where αu, αu and βu are the morphisms induced by α and β, and πu is the morphisminduced by the inclusion auX ⊆ α−1(auN). It is straightforward to verify that thediagram is commutative and that the bottom row is an exact sequence.

Notice that there is a canonical tower X/α−1(auN) X/α−1(au−1N)u>1 andthat αuu>1, αuu>1, βuu>1, and πuu>1 are morphisms of towers.

11.1.19 Proposition. Let a be an ideal in R and let 0 −→ X α−→ Nβ−→M −→ 0 be

an exact sequence of R-complexes. With π = limu>1 πu and α = limu>1 α

u from11.1.18 there is a commutative diagram,

Λa(X)

π

Λa(α)

((

0 // limu>1(X/α−1(auN))α// Λa(N)

Λa(β)// Λa(M) // 0 ,

whose bottom row is an exact sequence. In particular, the sequence

Λa(X)Λa(α)

// Λa(N)Λa(β)

// Λa(M)

is exact if and only if π is surjective.

PROOF. The morphisms in the tower X/α−1(auN) X/α−1(au−1N)u>1 are sur-jective, so it satisfies the Mittag-Leffler Condition by 3.3.37. One now gets the de-sired diagram by taking limits of the morphism of towers in 11.1.18; see 3.3.42.

It is immediate from 11.1.19 that Λa preserves surjective morphisms. However,the next example shows that the Λa is, in general, not half-exact.

11.1.20 Example. Let p > 1 be an integer. For n> 1 consider the exact sequence

0−→ Z/pZpn−1−−−→ Z/pnZ−→ Z/pn−1Z−→ 0 .

of abelian groups. Forming coproducts yields the exact sequence,

0−→∐

n>1Z/pZ∐

n>1 pn−1

−−−−−−→∐

n>1Z/pnZ−→∐

n>1Z/pn−1Z−→ 0 ,


11.1 Torsion and Completion 405

Write α : X → N for the first map in this sequence. To show that the functor Λ(p) isnot half-exact we argue that the map π= limu>1 π

u from 11.1.19 is not surjective.Fix u > 1. As pX = 0 it follows that puX = 0 and hence one has X/puX ∼= X .

Next note that puN is the subgroup

0⊕·· ·⊕0⊕ puZ/pu+1Z⊕ puZ/pu+2Z⊕·· · ,

with u leading zeros, of∐

n>1Z/pnZ = N. Hence it follows that α−1(puN) is thesubgroup 0⊕·· ·⊕0⊕Z/pZ⊕Z/pZ⊕·· · , with u leading zeros, of X =

∐n>1Z/pZ

and, consequently, one has X/α−1(puN)∼= (Z/pZ)u. Via this isomorphism and thepreviously established X/puX ∼= X , the maps

X/puX X/pu−1X and X/α−1(puN) X/α−1(pu−1N)

in the given towers are identified with 1X : X → X and λu : (Z/pZ)u (Z/pZ)u−1

given by (z1, . . . ,zu−1,zu) 7→ (z1, . . . ,zu−1). Further, πu : X/puX X/α−1(puN) isidentified with the map πu : X (Z/pZ)u given by (zn)n>1 7→ (z1, . . . ,zu). In viewof 3.3.34 it is now elementary to verify that π= limu>1 π

u can be identified with thecanonical embedding X =

∐n>1Z/pZ

∏n>1Z/pZ, which is not surjective.

11.1.21 Proposition. Let xxx be a sequence of elements in R and M an R-complex.The map limu>1(ξ

uxxx ⊗M) : limu>1(R/(xxxu)⊗R M)→ limu>1(R/(xxx)u⊗R M) induced

by 11.1.2 and 3.3.30 is an isomorphism. In particular, one has

Λ(xxx)(M) ∼= lim

u>1(R/(xxxu)⊗R M) .

PROOF. For every u> 1 the morphism ξuxxx from 11.1.2 is surjective, and hence so is

ξuxxx ⊗R M by 2.4.9. It follows that there is a short exact sequence,

0−→ Ker(ξuxxx ⊗R M)

ιu−→ R/(xxxu)⊗R Mξuxxx⊗M−−−−→ R/(xxx)u⊗R M −→ 0 ,

where ιu is the embedding. Recall from 11.1.2 that the families R/(xxxu)⊗R Mu>1and R/(xxx)u⊗R Mu>1 constitute towers, whose morphisms are ϑu

xxx⊗R Mu>1 andϑu

(xxx)⊗R Mu>1, and that ξuxxx ⊗R Mu>1 is a morphism of these towers. Write χu

for the morphisms on kernels induced by ϑuxxx⊗R M; it is elementary to check that

they form a tower and that ιuu>1 is a morphism of towers.Let u > 1 be given, let n be the number of elements in xxx, and set v = nu.

One has (xxx)v ⊆ (xxxu), so the canonical map R/(xxxv) R/(xxxu) admits a factoriza-tion R/(xxxv) R/(xxx)v π R/(xxxu). In the notation above this means ϑu+1

xxx · · ·ϑvxxx = πξv

xxx .Consequently, one has

ιuχu+1 · · ·χv = (ϑu+1xxx ⊗M) · · ·(ϑv

xxx⊗M)ιv = (π⊗M)(ξvxxx⊗M)ιv = 0 ,

and hence χu+1 · · ·χv = 0 as ιu is injective. This shows that the tower of kernelssatisfies the trivial Mittag-Leffler Condition; see 3.3.36. Now it follows from 3.3.39and 3.3.42 that the map limu>1(ξ

uxxx ⊗R M) is an isomorphism.



TORSION AND FLATNESS

First a result that compares to 1.3.27.

11.1.22 Proposition. Let R be an integral domain with field of fractions Q. Everyflat R-module is torsion-free, and for every R-module M there is an exact sequence

0−→MT −→M −→ Q⊗R M .

PROOF. Application of –⊗R M to the embedding ι : R Q yields a homomorphismι⊗R M from R⊗R M ∼= M to Q⊗R M. As this can be identified with the canonicalmap M → M(0), it follows that Ker(ι⊗R M) is the torsion submodule MT. If M isflat, then ι⊗R M is injective, so one has MT = 0.

Torsion-freeness does not imply flatness.

11.1.23 Example. Let k be a field and consider the integral domain R = k[x,y]. Asan R-module, the ideal M = (x,y) is evidently torsion-free. However, it is not flatsince the homomorphism obtained by applying –⊗R M to the embedding ι : M Ris not injective. Indeed, in R⊗R M one has

(ι⊗R M)(x⊗ y− y⊗ x) = x⊗ y− y⊗ x = 1⊗ xy−1⊗ yx = 0 ,

but in M⊗R M the element t = x⊗ y− y⊗ x is non-zero. To see why, note that thehomomorphism M⊗R M→ k given by f ⊗g 7→ f ′x(0,0)g

′y(0,0) maps t to 1 6= 0.

11.1.24 Proposition. Let R be an integral domain. Every finitely generated torsion-free R-module can be embedded into a finitely generated free R-module.

PROOF. Let M be a torsion-free R-module generated by a finite set m1, . . . ,mpof elements. Consider the injective homomorphism α : M→ Q⊗R M from 11.1.22.The elements 1⊗m1, . . . ,1⊗mp generate the vector space Q⊗R M, so some subset,say, 1⊗m1, . . . ,1⊗mt is a basis. Write 1⊗mv = ∑

ti=1 qvi(1⊗mi) with qvi ∈ Q,

and chose x 6= 0 in R such that xqvi is in R for all 1 6 v 6 p and 1 6 i 6 t. The R-submodule R〈1/x⊗m1, . . . ,1/x⊗mt〉 of Q⊗R M is free and contains Imα∼= M.

The next theorem compares to 1.3.17 and 1.3.28.

11.1.25 Theorem. Let R be a principal ideal domain. An R-module is flat if andonly if it is torsion-free. In particular, every submodule of a flat R-module is flat.

PROOF. Every flat R-module is torsion-free by 11.1.22. Let M be a torsion-free R-module. It follows from 3.2.25 that M is the filtered colimit of its finitely generatedsubmodules; so by 5.4.24 it suffices to argue that every finitely generated submoduleof M is flat. By 11.1.24 every such module F is a submodule of a free R-module. AsR is a principal ideal domain, F is free by 1.3.10, in particular F is flat.

It transpires from the proof above that every flat module over a principal idealdomain is a colimit of finitely generated free modules. This is true in general andknown as Govorov and Lazard’s Theorem; see Appn. C.


11.2 Right Derived Torsion Functor 407

EXERCISES

E 11.1.1 Show that for every R-module M one has MT =⋃Γ(x)(M) where the union is over all

non-zerodivisors x ∈ R.E 11.1.2 Let a be an ideal in R and let M and N be R-complexes. (a) Show that if M is bounded

and degreewise finitely presented, then one has Γa HomR(M,N) ∼= HomR(M,Γa(N)).(b) Show that if a is finitely generated, N is a complex of flat R-modules, and M or N isbounded, then one has Γa(M⊗R N)∼= Γa(M)⊗R N.

E 11.1.3 Let a be a finitely generated ideal in R, and let M and N be R-complexes. (a) Showthat if M is a complex of projective modules and M or N is bounded, then one hasΛa HomR(M,N) ∼= HomR(M,Λa N). (b) Show that if N is a complex of injective mod-ules and M or N is bounded, then one has Λa HomR(M,N)∼= HomR(Γa M,N).

E 11.1.4 Let a be a finitely generated ideal in R. Show that in an exact sequence of R-modules0→M′→M→M′′→ 0 the module M is a-torsion if and only if M′ and M′′ are so.

E 11.1.5 Let k be an integral domain. Show that a torsion-free and divisible k-module is injective.Conclude that the field of fractions of k is an injective k-module.

11.2 Right Derived Torsion Functor

SYNOPSIS. RΓa; local cohomology; derived a-torsion complex; Koszul complex; Cech complex;weakly proregular sequence.

The a-torsion functor has by the theory developed in Chap. 7 a right derived functorRΓa that can be computed by way of semi-injective resolutions.

LOCAL COHOMOLOGY

11.2.1 Proposition. Let a be an ideal in R. The right derived a-torsion functorRΓa : D(R)→D(R) is k-linear and triangulated.

PROOF. From 11.1.5 and 4.3.18 it follows that the functor Γa preserves homotopy.It induces by 11.1.7 and 6.1.20 a k-linear functor on K(R), which is triangulated by6.2.17. Finally, invoke 7.2.12.

11.2.2 Definition. Let a be an ideal in R. For m ∈ Z denote by Hma (–) the functor

H−m(RΓa(–)) : D(R) −→ M(R) ;

it is called the mth local cohomology functor with support in a.

It follows from 11.2.1 and 6.5.20 that the local cohomology functors are k-linear.

11.2.3 Theorem. Let a be an ideal in R and let M′ α′−→ M α−→ M′′ −→ ΣM′ be adistinguished triangle in D(R). There is an exact sequence of R-modules,

· · · → Hm−1a (M′′)→ Hm

a (M′)

Hma (α

′)−−−−→ Hma (M)

Hma (α)−−−−→ Hm

a (M′′)→ Hm+1

a (M′)→ ··· .



PROOF. The functor RΓa is triangulated, see 11.2.1, so applied to the distinguishedtriangle M′→M→M′′→ ΣM′ it yields another distinguished triangle. Now 6.5.22yields the exact sequence of local cohomology modules.

11.2.4 Definition. Let a be an ideal in R. The transformation γa : Γa→ IdK(R) from11.1.6 induces by 7.2.6 a natural transformation γa = Rγa : RΓa→ IdD(R), cf. 7.2.9.An R-complex M is called derived a-torsion if γM

a : RΓa(M)→M is an isomorphismin D(R). The full subcategory of D(R) whose objects are derived a-torsion is de-noted Da-tor(R). It is evidently a triangulated subcategory.

THE KOSZUL COMPLEX

11.2.5. For every element x in R the complex

(11.2.5.1) 0−→ R x−−→ R−→ 0 ,

concentrated in degrees 1 and 0, is isomorphic to the Koszul complex KR(x); see2.2.8. We henceforth identify KR(x) with the complex (11.2.5.1), and for a sequencex1, . . . ,xn we make per 4.4.7 the identification

(11.2.5.2) KR(x1, . . . ,xn) = KR(x1)⊗R · · ·⊗R KR(xn) .

By commutativity 4.4.3 of the tensor product, the complex KR(x1, . . . ,xn) is invari-ant, up to isomorphism, under permutation of the elements x1, . . . ,xn. Notice thatKR(x1, . . . ,xn) is a complex of finitely generated free R-modules, concentrated indegrees n, . . . ,0; in particular it is semi-free: see 5.1.3.

The gist of 11.2.5 is that we ignore the product on the Koszul complex to considerit, simply, as a complex of free R-modules.

11.2.6 Example. Representing the maps between free R-modules by matrices thatact by left multiplication on column vectors, the Koszul complexes on sequences oftwo and three elements look as follows with natural orderings of the bases:

KR(x1,x2) = 0−→ R

(−x2x1

)−−−−→ R2 (x1 x2 )−−−−→ R−→ 0

and

KR(x1,x2,x3) = 0−→ R

(x3−x2

x1

)−−−−→ R3

(−x2 −x3 0x1 0 −x30 x1 x2

)−−−−−−−−−−→ R3 (x1 x2 x3 )−−−−−−→ R−→ 0 .

11.2.7 Lemma. Let x ∈ R and M be an R-complex. There is an isomorphism ofR-complexes KR(x)⊗R M ∼= ConexM .

PROOF. There is an isomorphism of complexes KR(x) ∼= ConexR, and the unitorR⊗R M

∼=−→M from (4.4.1.1) identifies xR⊗R M with xM . Now apply 4.1.18.

11.2.8 Proposition. Let xxx be a sequence of elements in R and M be an R-complex.One has (xxx)H(KR(xxx)⊗R M) = 0 and H(KR(xxxu)⊗R M) is (xxx)-torsion for every u> 1.



PROOF. Set xxx= x1, . . . ,xn. For every element xi one has KR(xxx)⊗R M∼=KR(xi)⊗R M′

for some complex M′, see 11.2.5. Thus it follows from 11.2.7 and 4.1.3 that thecomplex H(KR(xxx)⊗R M) is annihilated by each xi, and hence by the ideal (xxx).

The homology complex H(KR(xxxu)⊗R M) is, as shown above, annihilated by (xxxu),and hence by (xxx)nu ⊆ (xxxu), so the complex is (xxx)-torsion.

The isomorphism in the lemma below is often referred to as “self-duality” of theKoszul complex, where duality is understood to be with respect to the ring.

11.2.9 Lemma. For every sequence xxx = x1, . . . ,xn in R there are isomorphisms,

HomR(KR(xxx),R) ∼= HomR(KR(x1),R)⊗R · · ·⊗R HomR(KR(xn),R) ∼= Σ−n KR(xxx) .

PROOF. For an element x ∈ R, the complex HomR(KR(x),R) is concentrated in de-grees 0 and −1, and after identification of HomR(R,R) with R the non-trivial differ-ential is multiplication by−x; cf. 2.3.1. Thus one has HomR(KR(x),R)∼= Σ−1 KR(x).Now invoke (11.2.5.2), tensor evaluation 4.5.7, adjunction 4.4.11, and 2.4.12.

11.2.10 Construction. For every x∈R and every u> 1 there is a morphism in C(R),

KR(xu)

κux

0 // R xu//

x

R //

=

0

KR(xu−1) 0 // R xu−1// R // 0 .

The dual morphism of complexes, concentrated indegrees 0 and −1, takes the form

HomR(KR(xu−1),R)

Hom(κux ,R)

0 // R−xu−1

//

=

R //

x

0

HomR(KR(xu),R) 0 // R−xu// R // 0 ;

see 11.2.9. For a sequence xxx = x1, . . . ,xn in R set

κuxxx = κu

x1⊗R · · ·⊗R κ

uxn and κu

xxx = HomR(κu+1xxx ,R) .

The family

(11.2.10.1) κuxxx : KR(xxxu) −→ KR(xxxu−1)u>1

is a tower, and the dual family

(11.2.10.2) κuxxx : HomR(KR(xxxu),R) −→ HomR(KR(xxxu+1),R)u>1 .

is a telescope. Notice that in view of 11.2.9 one has

(11.2.10.3) κuxxx∼= κu

x1⊗R · · ·⊗R κ

uxn .



THE CECH COMPLEX

The powers of an element x ∈ R form a multiplicative subset, and Mx denotes thelocalization of an R-module M at this set.

11.2.11 Definition. For an element x in R, the complex

CR(x) = 0−→ Rρx−−→ Rx −→ 0 ,

concentrated in degrees 0 and −1 and with ρx given by r 7→ r/1, is called the Cechcomplex on x. The Cech complex on a sequence x1, . . . ,xn in R is

(11.2.11.1) CR(x1, . . . ,xn) = CR(x1)⊗R · · ·⊗R CR(xn).

Notice that by commutativity 4.4.3 the complex CR(x1, . . . ,xn) is invariant, up toisomorphism, under permutation of the elements x1, . . . ,xn.

11.2.12. The Cech complex CR(x1, . . . ,xn) is a complex of flat R-modules concen-trated in degrees 0, . . . ,−n, in particular, it is semi-flat; see 1.3.37 and 5.4.8.

We procede to show that the Cech complex can be obtained as a colimit of dualKoszul complexes.

11.2.13 Construction. For every x∈R and every u> 1 there is a morphism in C(R),

HomR(KR(xu),R)

%ux

0 // R−xu//

=

R //

0

CR(x) 0 // Rρx// Rx // 0 ,

where the map in degree −1 maps r to −r/xu. Evidently, one has %ux = %u+1

x κux .

Let xxx = x1, . . . ,xn be a sequence in R. It follows in view of (11.2.10.3) that themorphisms %u

x1⊗·· ·⊗%u

xn induce a family

(11.2.13.1) %uxxx : HomR(KR(xxxu),R) −→ CR(xxx)u>1

of morphisms that are compatible with the morphisms in the telescope (11.2.10.2),that is, %u

xxx = %u+1xxx κu

xxx for all u> 1.

11.2.14 Proposition. Let xxx be a sequence of elements in R. The unique morphism

colimu>1

HomR(KR(xxxu),R)−→ CR(xxx)

determined by the family (11.2.13.1) and 3.2.4 is an isomorphism.

PROOF. By 3.2.16 it suffices to consider the case where xxx = x is a single element.Let % : colimu>1 HomR(KR(xu),R)→ CR(x) be the unique morphism that every %u

xfactors through. The component %−1 is an isomorphism by 3.2.6 and 3.2.41, and %0is an isomorphism as well as one has (κu

x)0 = 1R = (%ux)0 for all u> 1.



11.2.15 Lemma. Let xxx be a sequence of elements in R. For every R-complex M thehomology complex H(CR(xxx)⊗R M) is (xxx)-torsion.

PROOF. By 11.2.9, 11.2.14, and 3.2.16 the complex CR(xxx)⊗R M is isomorphic, upto a shift, to a filtered colimit of the complexes KR(xxxu)⊗R M. The homology of eachof these complexes is (xxx)-torsion by 11.2.8, so the claim follows from 11.1.11.

11.2.16 Notation. Let x be an element in R. Denote by εx the canonical mapCR(x)→ R. For a sequence xxx = x1, . . . ,xn set εxxx = εx1 ⊗R · · ·⊗R εxn : CR(xxx)→ R.

11.2.17 Lemma. Let xxx be a sequence of elements in R and M be an R-complex. IfH(M) is (xxx)-torsion, then εxxx⊗R M : CR(xxx)⊗R M→ R⊗R M is a quasi-isomorphism.

PROOF. By the definitions, (11.2.11.1) and 11.2.16, and semi-flatness of Cech com-plexes, see 11.2.12, it is enough to consider the case where xxx = x is a single element.The sequence 0−→ Σ−1 Rx −→ CR(x) εx−→ R−→ 0 is degreewise split exact, so it re-mains exact after application of –⊗R M. Now apply 4.2.6 and 11.1.9.

It follows from 2.2.8 that H1(KZ(2)) is zero while H1(KZ(2,4)) is non-zero as[4]2Z = 0. That is, even though the sequences 2 and 2,4 generate the same ideal inZ, the Koszul complexes KZ(2) and KZ(2,4) are not isomorphic in D(Z). The Cechcomplex does not exhibit this phenomenon.

11.2.18 Proposition. Let xxx and yyy be sequences of elements in R. If they generatethe same ideal, (xxx) = (yyy), then there is an isomophism CR(xxx)' CR(yyy) in D(R).

PROOF. The homology complex H(CR(yyy)) is (xxx)-torsion by 11.2.15, so 11.2.17yields a quasi-isomorphism CR(xxx)⊗R CR(yyy) −→ CR(yyy). The isomorphism in D(R)now follows by symmetry in xxx and yyy.

WEAKLY PROREGULAR SEQUENCES

11.2.19 Definition. A sequence xxx of elements in R is called weakly proregular if forevery v > 0 the tower

Hv(κuxxx) : Hv(KR(xxxu))→ Hv(KR(xxxu−1))u>1 ,

induced by (11.2.10.1), satisfies the trivial Mittag-Leffler Condition; see 3.3.36.

11.2.20 Example. Let x ∈ R; one has H1(KR(xu)) = (0 :R xu) for every u > 1, andthese ideals form an ascending chain in R. If R is Noetherian, then this chain stabi-lizes; that is, there is a v > 1 such that (0 :R xw) = (0 :R xw+1) holds for all w > v.Each morphism in the tower H1(κ

ux) : H1(KR(xu))→ H1(KR(xu−1))u>1 is multi-

plication by x, so for every u > 1 the composite H1(κux) · · ·H1(κ

u+vx ) is zero. Thus,

in a Noetherian ring every element forms a proregular sequence.

In fact, every finite sequence of elements in a commutative Noetherian ring isproregular; that is proved in ?? via the next lemma.



11.2.21 Proposition. A sequence xxx of elements in R is weakly proregular if andonly if Hv(CR(xxx)⊗R I) = 0 holds for every injective R-module I and every v < 0.

PROOF. For every R-module I, the telescope κuxxx ⊗R Iu>1, cf. (11.2.10.2), is by

tensor evaluation 4.5.7 isomorphic to the telescope HomR(κuxxx , I)u>1.

Fix v < 0. For every injective R-module I one has

Hv(CR(xxx)⊗R I) ∼= Hv((colimu>1 HomR(KR(xxxu),R))⊗R I)∼= Hv(colimu>1(HomR(KR(xxxu),R)⊗R I))∼= Hv(colimu>1 HomR(KR(xxxu), I))∼= colimu>1 HomR(H−v(KR(xxxu)), I) .

Indeed, the first two isomorphisms follow from 11.2.14 and 3.2.16. The third followsfrom the isomorphism of telescopes induces by tensor evaluation 4.5.7, and the finalisomorphism holds as both colim and the exact functor HomR(–, I) commute withhomology; see 3.2.34 and 2.2.18. The final colimit is thus of the telescope inducedby the tower H−v(κ

uxxx) : H−v(KR(xxxu))→ H−v(KR(xxxu−1))u>1.

If xxx is proregular, then this tower satisfies the trivial Mittag-Leffler Condi-tion, and it follows that the colimit is zero; see 3.2.26(a). Conversely, for everyu there is by 5.3.27 an injective R-module I and an injective homomorphism ι inHomR(H−v(KR(xxxu)), I). If the colimit above is zero, then 3.2.26(b) yields a w > usuch that one has

0 = (HomR(H−v(κwxxx ), I) · · ·HomR(H−v(κ

u+1xxx ), I))(ι) = ιH−v(κ

u+1xxx ) · · ·H−v(κ

wxxx ) .

As ι is injective, this implies that the composite H−v(κu+1xxx ) · · ·H−v(κ

wxxx ) is zero, so xxx

is weaky proregular.

DERIVED TORSION VIA THE CECH COMPLEX

11.2.22 Construction. Let xxx = x1, . . . ,xn be a sequence in R. From 2.2.8 one getsH0(KR(xxx))∼= R/(xxx), so 2.5.7 yields a canonical morphism πxxx : KR(xxx) R/(xxx).

For every R-complex M and every u> 1 there is a commutative diagram,

HomR(R/(xxxu),M)(θK(xxxu)RM)−1 Hom(πxxxu ,M)

//

Hom(ϑu+1xxx ,M)

HomR(KR(xxxu),R)⊗R M

κuxxx⊗M

HomR(R/(xxxu+1),M)(θK(xxxu+1)RM)−1 Hom(πxxxu+1 ,M)

// HomR(KR(xxxu+1),R)⊗R M .

Thus, the morphisms $Mxxxu = (θKR(xxxu)RM)−1 HomR(πxxxu ,M) form a morphism of tele-

scopes. Let $Mxxx be the morphism defined by commutativity of the diagram,



colimu>1 HomR(R/(xxxu),M)colimu>1$

Mxxxu//

∼=

colimu>1(HomR(KR(xxxu),R)⊗R M)

∼=

Γ(xxx)(M)$M

xxx// CR(xxx)⊗R M ,

where the vertical isomorphisms come from 11.1.12, 3.2.16, and 11.2.14.

11.2.23 Lemma. Let xxx be a sequence of elements in R. For every R-complex M themorphism

$Mxxx : Γ(xxx)(M) −→ CR(xxx)⊗R M

constructed in 11.2.22 is natural in M, and as a natural transformation of functors,$xxx is a Σ-transformation. Moreover, for every R-module M, the induced map

H0($Mxxx ) : Γ(xxx)(M) −→ H0(CR(xxx)⊗R M)

is an isomorphism.

PROOF. It is evident from the construction that $xxx is a Σ-transformation.Let M be an R-module. It suffices by 3.2.34 to show that each H0($

Mxxxu) is an

isomorphism. To this end it is by the definition of $Mxxxu enough to prove that each

H0(HomR(πxxxu ,M)) : H0(HomR(R/(xxxu),M)) −→ H0(HomR(KR(xxxu),M))

is an isomorphism, and that follows from 2.5.7.

11.2.24 Lemma. Let xxx be a weakly proregular sequence R. For every complex I ofinjective R-modules the morphism

$Ixxx : Γ(xxx)(I) −→ CR(xxx)⊗R I

from 11.2.23 is a quasi-isomorphism.

PROOF. It follows from 2.4.11, 4.1.17, and A.18 that CR(xxx)⊗R – is a \-functor, a Σ-functor, and way-out; by 11.1.5 the functor Γ(xxx) has the same properties. Moreover,$xxx is a Σ-transformation by 11.2.23. Thus, by A.19 we may assume that I is aninjective module, and the assertion now follows from 11.2.21 and 11.2.23.

An immediate consequence of the next theorem and 11.2.18 is that the deriveda-torsion functor is idempotent. A more precise statement can be found in 11.4.1.

11.2.25 Theorem. Let a be an ideal in R generated by a weakly proregular sequencexxx and let M be an R-complex. For every complex J of injective R-modules withM ' J in D(R) there are isomorphisms in D(R),

Γa(J) ' RΓa(M) ' CR(xxx)⊗R M ,

and if M is a module, then one has H0a(M)∼= Γa(M).

Moreover, RΓa is way-out and preserves coproducts, and one has γa ∼= εxxx⊗R –.



PROOF. Let M '−→ I be a semi-injective resolution. By 11.2.24 and semi-flatness ofthe Cech complex, see 11.2.12, there are quasi-isomorphisms

Γa(I)$I

xxx−−→ CR(xxx)⊗R I←− CR(xxx)⊗R M .

By definition, Γa(I) is RΓa(M), so the diagram above accounts for the isomorphismRΓa(M) ' CR(xxx)⊗R M, and the assertion about modules is now immediate from11.2.23. Furthermore, it follows from 11.2.12 and A.26 that RΓa is way-out andfrom 7.1.8 and 6.4.31 that RΓa preserves coproducts. By 6.4.21 there is a quasi-isomorphism J '−→ I in C(R), and the next quasi-isomorphisms account for the otherisomorphism in D(R),

Γa(J)$J

xxx−−→ CR(xxx)⊗R J −→ CR(xxx)⊗R I$I

xxx←−− Γa(I) .

It remains to show that there is an isomorphism between the natural transfor-mations γa and εxxx⊗R –. In D(R) every R-complex is naturally isomorphic to itssemi-injective resolution by 6.3.14, so it suffices to argue that the morphisms γI

a

and εxxx⊗R I are isomorphic in D(R) for every semi-injective R-complex I, and thatfollows from commutativity of the diagram:

Γa(I)

$Ixxx '

γIa

// I

CR(xxx)⊗R Iεxxx⊗I// R⊗R I .

∼= µIR

OO

11.2.26 Proposition. Let a be an ideal in R generated by a weakly proregular se-quence. An R-complex M is derived a-torsion if and only if the homology complexH(M) is a-torsion. That is M ∈Da-tor(R) if and only if H(M)∼= Γa(H(M)).

PROOF. The “only if” is immediate from 11.2.25 and 11.2.15. Conversely, if H(M)is a-torsion, then the morphism εxxx⊗R M : CR(xxx)⊗R M→ R⊗R M is an isomorphismin D(R) by 11.2.17, and hence so is γM

a by 11.2.25.

EXERCISES

E 11.2.1 Write the dual of the complex K =KR(x1,x2) from 11.2.6 explicitly with the differentialrepresented by matrices and determine an isomorphism HomR(K,R)→ Σ−2 K.

E 11.2.2 Compute the ranks of the free modules in the Koszul complex KR(x1, . . . ,xn).E 11.2.3 Let xxx be a sequence in R. Prove without recourse to 11.2.23 that Z0(CR(xxx)) is Γ(xxx)(R).


11.3 Left Derived Completion Functor 415

11.3 Left Derived Completion Functor

SYNOPSIS. LΛa; local homology; derived a-adically complete complex; semi-free resolution ofCech complex.

The a-adic completion functor has by the theory developed in Chap. 7 a right derivedfunctor RΓa that can be computed by way of semi-projective resolutions.

LOCAL HOMOLOGY

11.3.1 Proposition. Let a be an ideal in R. The left derived a-adic completion func-tor LΛa : D(R)→D(R) is k-linear and triangulated.

PROOF. From 11.1.17 and 4.3.18 it follows that the functor Λa preserves homotopy.It induces by 11.1.15 and 6.1.20 a k-linear functor on K(R), which is triangulatedby 6.2.17. Finally, invoke 7.2.12.

11.3.2 Definition. Let a be an ideal in R. For m ∈ Z denote by Ham(–) the functor

Hm(LΛa(–)) : D(R) −→ M(R) ;

it is called the mth local homology functor with support in a.

It follows from 11.3.1 and 6.5.20 that the local homology functors are k-linear.

11.3.3 Theorem. Let a be an ideal in R and let M′ α′−→ M α−→ M′′ −→ ΣM′ be adistinguished triangle in D(R). There is an exact sequence of R-modules,

· · · → Ham+1(M

′′)→ Ham(M

′)Ha

m(α′)−−−−→ Ha

m(M)Ha

m(α)−−−−→ Ham(M

′′)→ Ham−1(M

′)→ ··· .PROOF. The functor LΛa is triangulated, see 11.3.1, so applied to the distinguishedtriangle M′→M→M′′→ ΣM′ it yields another distinguished triangle. Now 6.5.22yields the exact sequence of local homology modules.

11.3.4 Definition. Let a be an ideal in R. The transformation λa : IdK(R)→ Λa from11.1.14 induces by 7.2.6 a natural transformation λa = Lλa : IdD(R)→ LΛa, cf.7.2.9. An R-complex M is called derived a-adically complete if λaM : M→ LΛa(M)is an isomorphism in D(R). The full subcategory of D(R) whose objects are deriveda-adically complete is denoted Da-com(R). It is evidently a triangulated subcategory.

A SEMI-FREE RESOLUTION OF THE CECH COMPLEX

11.3.5 Construction. Let x ∈ R. Consider the complex of free R-modules,

L(x) = 0−→ R〈E〉 ðx−−→ R〈E〉 −→ 0 ,

where E = ei | i ∈ N0 and the differential ðx is given by



ðx(ei) =

e0 for i = 0ei−1− xei for i> 1 .

The complex L(x) is concentrated in degrees 0 and −1.For every u > 1 the differential ðx maps the submodule R〈e0, . . . ,eu〉 of R〈E〉 to

itself, and hence one has the following subcomplex of L(x),

Lu(x) = 0−→ R〈e0, . . . ,eu〉ðx−−→ R〈e0, . . . ,eu〉 −→ 0 .

Write ιux : Lu(x) Lu+1(x) and ιux : Lu(x) L(x) for the canonical embeddings.For a sequence xxx = x1, . . . ,xn in R and u> 1 set

L(xxx) = L(x1)⊗R · · ·⊗R L(xn) ,

Lu(xxx) = Lu(x1)⊗R · · ·⊗R Lu(xn) ,

ιuxxx = ιux1⊗R · · · ⊗R ι

uxn , and

ιuxxx = ιux1⊗R · · · ⊗R ι

uxn .

Note that L(xxx) and Lu(xxx) are bounded complexes of free modules concentrated indegrees 0, . . . ,−n while ιuxxx and ιuxxx are morphisms Lu(xxx)→ Lu+1(xxx) and Lu(xxx)→ L(xxx).

11.3.6 Lemma. Let xxx be a sequence of elements in R. For every u> 1 the morphismsιuxxx and ιuxxx are injective and the exact sequences

0−→ Lu(xxx)ιuxxx−→ Lu+1(xxx)−→ Lu+1(xxx)/Lu(xxx)−→ 0 and

0−→ Lu(xxx)ιuxxx−→ L(xxx)−→ L(xxx)/Lu(xxx)−→ 0

are degreewise split.

PROOF. For a sequence of one element, this is clear from the definitions in 11.3.5.The general case follows by induction.

11.3.7 Lemma. Let xxx be a sequence of elements in R. The colimit of the telescopeιuxxx : Lu(xxx) Lu+1(xxx)u>1 is L(xxx) with canonical morphisms ιuxxx : Lu(xxx) L(xxx).

PROOF. The claims are evident for a sequence of one element. As tensor productspreserve colimits, see 3.2.16 and 3.2.17, the general case follows by induction.

11.3.8 Construction. Let x ∈ R and u> 1. Consider the morphism of complexes

Lu(x)

ϕux

0 // R〈e0, . . . ,eu〉

(ϕux)0

ðx// R〈e0, . . . ,eu〉

(ϕux)−1

// 0

HomR(KR(xu),R) 0 // R−xu

// R // 0

given by

(ϕux)0(ei) =

1 for i = 00 for 16 i6 u

and (ϕux)−1(ei) = −xu−i for 06 i6 u .



Furthermore, let

L(x)

ϕx

0 // R〈E〉

(ϕx)0

ðx// R〈E〉

(ϕx)−1

// 0

CR(x) 0 // Rρx

// Rx // 0

be the morphism given by

(ϕx)0(ei) =

1 for i = 00 for i> 1

and (ϕx)−1(ei) = 1/xi for i> 0 .

Finally, for a sequence xxx = x1, . . . ,xn in R set

ϕuxxx = ϕu

x1⊗R · · ·⊗R ϕ

uxn and ϕxxx = ϕx1

⊗R · · ·⊗R ϕxn .

Note that ϕxxx is a morphism L(xxx)→ CR(xxx). In view of 11.2.9 consider ϕuxxx as a

morphism Lu(xxx)→ HomR(KR(xxxu),R).

11.3.9 Lemma. Let xxx be a sequence of elements in R. The morphisms from 11.3.8,11.3.5, and (11.2.10.2) fit into a commutative diagram,

Lu(xxx)

ϕuxxx

ιuxxx// Lu+1(xxx)

ϕu+1xxx

HomR(KR(xxxu),R)κuxxx// HomR(KR(xxxu+1),R) .

Thus, ϕuxxxu>1 is a morphism between the telescopes from 11.3.7 and (11.2.10.2).

PROOF. By the definitions of the maps it is sufficient to consider the case wherexxx = x is a single element, and in that case commutativity is straightforward to verifyfrom the definitions.

11.3.10 Proposition. Let xxx be a sequence of elements in R.(a) For every u> 1 the morphism ϕu

xxx is a homotopy equivalence.(b) There is an equality ϕxxx = colimu>1ϕ

uxxx of morphisms from L(xxx) to CR(xxx).

(c) The morphism ϕxxx : L(xxx) '−→ CR(xxx) is a semi-free resolution; in particular,CR(xxx) has finite projective dimension.

PROOF. (a): By 4.3.20 it suffices to consider the case where xxx= x is a single element.Since ϕu

x is morphism between bounded complexes of free modules, it suffices by5.2.8 and 5.2.20 and to argue that ϕu

x is a quasi-isomorphism.The homomorphism H0(ϕ

ux) is the (co)restriction of (ϕu

x)0 to kernels, i.e. the map

R〈e0, . . . ,eu〉∩Kerðx −→ (0 :R xu)

that maps z = r0e0 + · · ·+ rueu to (ϕux)0(z) = r0; see 11.2.13 and 11.3.8. From

0 = ðx(z) = (r0 + r1)e0 +(r2− xr1)e1 + · · ·+(ru− xru−1)eu−1− xrueu



one gets r1 =−r0, r2 =−xr0, . . . ,ru =−xu−1r0, and−xru = 0. Hence xur0 = 0 andz has the form z = r0(e0− e1− xe2− ·· ·− xu−1eu). Thus H0(ϕ

ux) is bijective with

inverse given by r 7→ r(e0− e1− xe2−·· ·− xu−1eu).The homomorphism H−1(ϕ

ux) is the by (ϕu

x)−1 induced map on cokernels

Cokerðx −→ R/(xu) ,

where ðx is considered as an endomorphism on R〈e0, . . . ,eu〉. Since (ϕux)−1 is sur-

jective, so is H−1(ϕux). To prove injectivity, let z = r0e0 + · · ·+ rueu be an element in

R〈e0, . . . ,eu〉 with (ϕux)−1(z) ∈ (xu). As one has

(ϕux)−1(z) = −xur0− xu−1r1−·· ·− xru−1− ru

the equality −xur0− xu−1r1−·· ·− xru−1− ru = xus holds for some s ∈ R. The goalis to construct an element w = s0e0+ · · ·+ sueu in R〈e0, . . . ,eu〉 with ðx(w) = z. Thisamounts to finding s0, . . . ,su ∈ R with

s0 + s1 = r0 , si+1− xsi = ri for i = 1, . . . ,u−1 , and − xsu = ru .

Set s0 =−s. Solving the equations we get s1 = r0 + s, s2 = r1 +xs1 = r1 +xr0 +xs,and generally si+1 = ri + · · ·+ xi−1r1 + xir0 + xis for i = 1, . . . ,u− 1. It remains tonote that the last of these elements, that is, su = ru−1+ · · ·+xu−2r1+xu−1r0+xu−1s,satisfies −xsu = ru, but this follows from the defining property of s.

(b): By 3.2.8, 11.2.14, and 11.3.7 it is sufficient to argue that the diagram

Lu(xxx)

ϕuxxx

ιuxxx// L(xxx)

ϕxxx

HomR(KR(xxxu),R)%u

xxx// CR(xxx)

is commutative for every u > 1. By the definitions of the morphisms involved, itsuffices consider the case where xxx = x is a single element. That is, it must be verifiedthat (ϕx)v(ι

ux )v = (%u

x)v(ϕux)v holds for v = 0,−1. For v = 0 one has

(ϕx)0(ιux )0(e0) = (ϕx)0(e0) = 1 = (%u

x)0(1) = (%ux)0(ϕ

ux)0(e0) and

(ϕx)0(ιux )0(ei) = (ϕx)0(ei) = 0 = (%u

x)0(0) = (%ux)0(ϕ

ux)0(ei) for 16 i6 u .

For v =−1 the equality also holds since for 06 i6 u one has:

(ϕx)−1(ιux )−1(ei) = (ϕx)−1(ei) = 1/xi = (%u

x)−1(−xu−i) = (%ux)−1(ϕ

ux)−1(ei) .

(c): The complex L(xxx) is bounded and semi-free by 5.1.3 and 11.3.5, and themorphism ϕxxx is a quasi-isomorphism by parts (a,b) and 4.2.11.

DERIVED COMPLETION VIA THE CECH COMPLEX

11.3.11 Construction. Let xxx be a sequence of elements in R. For u> 1 consider thecomposite of morphisms



KR(xxxu)δ

K(xxxu)R−−−−→∼= HomR(HomR(KR(xxxu),R),R)

HomR(ϕuxxx ,R)−−−−−−−→

uHomR(Lu(xxx),R) ,

where the first is an isomorphism by 4.5.4 and the second is a homotopy equivalenceby 11.3.10 and 4.3.19. The resulting map is a homotopy equivalence denoted ϕu

xxx .Naturality of biduality combined with 11.3.9 yields a commutative diagram,

(11.3.11.1)

KR(xxxu)ϕu

xxx

u//

κuxxx

HomR(Lu(xxx),R)

HomR(ιu−1xxx ,R)

KR(xxxu−1)ϕu−1

xxx

u// HomR(Lu−1(xxx),R) .

The complex HomR(Lu(xxx),R) is concentrated in degrees v> 0; this explains the firstmorphism below,

HomR(Lu(xxx),R)−→ H0(HomR(Lu(xxx),R))H0(ϕ

uxxx)−1

−−−−−−→∼=H0(KR(xxxu)) = R/(xxxu) ;

see 2.5.7 and 2.2.8. Denote by τuxxx the composite HomR(Lu(xxx),R)→ R/(xxxu) and note

that H0(τuxxx) is an isomorphism.

11.3.12 Construction. Let xxx be a sequence of elements in R and M an R-complex.Denote by φM

xxx the unique morphism that makes the following diagram commutative,

HomR(L(xxx),M)φM

xxx//

∼=

Λ(xxx)(M)

HomR(colimLu(xxx),M)

∼=

lim(R/(xxxu)⊗R M)

lim(ξuxxx⊗M)∼=

OO

limHomR(Lu(xxx),M) lim(HomR(Lu(xxx),R)⊗R M) .limθLu(xxx)RM

∼=oo

lim(τuxxx⊗M)

OO

All (co)limits are over u> 1. The left-hand vertical isomorphisms come from 11.3.7and 3.3.21, and the morphisms on the right come from 11.1.21 and 11.3.11. Thehorizontal isomorphism is tensor evaluation 4.5.7(d).

11.3.13 Lemma. Let xxx be a sequence of elements in R. For every R-complex M themorphism

φMxxx : HomR(L(xxx),M) −→ Λ

(xxx)(M)

constructed in 11.3.12 is natural in M, and as a natural transformation of functors,φxxx is a Σ-transformation.

PROOF. The claims are evident from the construction 11.3.12 of φMxxx .

11.3.14 Lemma. Let xxx be a sequence of elements in R. In the tower,

HomR(ιu−1xxx ,R) : HomR(Lu(xxx),R)→ HomR(Lu−1(xxx),R)u>1 ,



induced by the telescope in 11.3.7, every morphism is surjective. If xxx is weaklyproregular, then for every v > 0 the induced tower in homology

Hv(HomR(ιu−1xxx ,R)) : Hv(HomR(Lu(xxx),R))→ Hv(HomR(Lu−1(xxx),R))u>1

satisfies the trivial Mittag-Leffler Condition.

PROOF. The first assertion follows from 11.3.6. Fix v > 0; the tower induced byHv(–) is by (11.3.11.1) isomorphic to Hv(κ

uxxx) : Hv(KR(xxxu))→ Hv(KR(xxxu−1))u>1,

so the last assertion follows by the definition of weak proregularity; see 11.2.19.

11.3.15 Lemma. Let xxx be a weakly proregular sequence in R. For every complex Fof flat R-modules the morphism

φFxxx : HomR(L(xxx),F) −→ Λ

(xxx)(F)

from 11.3.12 is a quasi-isomorphism.

PROOF. It follows from 2.3.12, 4.1.15, and A.17 that HomR(L(xxx),–) is a \-functor,a Σ-functor, and way-out; by 11.1.17 the functor Λ(xxx)(–) has the same properties.Furthermore, φxxx is a Σ-transformation by 11.3.13. Thus, by A.19 we may assumethat F is a flat module.

Proving that φFxxx is a quasi-isomorphism is by 11.3.12 equivalent to showing that

limu>1

(τuxxx⊗R F) : lim

u>1(HomR(Lu(xxx),R)⊗R F) −→ lim

u>1(R/(xxxu)⊗R F)

is a quasi-isomorphism. The codomain of this map is a module, so one must prove:(†) Hv(limu>1(HomR(Lu(xxx),R)⊗R F)) = 0 for every v 6= 0, and(‡) H0(limu>1(τ

uxxx⊗R F)) is an isomorphism.

The homology modules in (†) are zero for v< 0; cf. 11.3.5. The next step is to verifythe assumptions in 3.3.45 in order to conclude that the canonical morphism,

ψv : Hv(limu>1

(HomR(Lu(xxx),R)⊗R F)) −→ limu>1

Hv(HomR(Lu(xxx),R)⊗R F) ,

is an isomorphism for v> 0. The maps in the tower HomR(Lu(xxx),R)⊗R Fu>1 areHomR(ι

uxxx ,R)⊗R Fu>1, and they are surjective by 11.3.14. In particular, this tower

satisfies the Mittag-Leffler Condition; see 3.3.37. As F is a flat module, 2.2.18 yields

Hv(HomR(Lu(xxx),R)⊗R F) ∼= Hv(HomR(Lu(xxx),R))⊗R F ,

and, therefore, the tower Hv(HomR(Lu(xxx),R)⊗R F)u>1 is isomorphic to the towerHv(HomR(Lu(xxx),R))⊗R Fu>1, whose maps are Hv(HomR(ι

uxxx ,R))⊗R Fu>1. For

v > 0 it follows from 11.3.14 that this tower satisfies the trivial Mittag-Leffler Con-dition. This establishes that ψv is an isomorphism for v > 0, and by 3.3.39 thecodomain of ψv is zero for v > 0, so (†) follows.

The isomorphism ψ0 fits by 3.3.16 into the commutative diagram,



H0(limu>1(HomR(Lu(xxx),R)⊗R F))H0(limu>1(τ

uxxx⊗F))

//

ψ0 ∼=

H0(limu>1(R/(xxxu)⊗R F))

limu>1 H0(HomR(Lu(xxx),R)⊗R F)limu>1 H0(τ

uxxx⊗F)

// limu>1 H0((R/(xxxu)⊗R F)) ,

where the equality on the right holds as R/(xxxu) and F are modules. Thus, to establish(‡) it suffices to verify that limu>1 H0(τ

uxxx⊗R F) is an isomorphism. To that end, note

that each map H0(τuxxx) and hence also H0(τ

uxxx)⊗R F ∼=H0(τ

uxxx⊗R F) is an isomorphism

by 11.3.11.

An immediate consequence of the next theorem and 11.2.18 is that the deriveda-adic completion functor is idempotent. A precise statement is made in 11.4.1.

11.3.16 Theorem. Let a be an ideal in R generated by a weakly proregular sequencexxx and let M be an R-complex. For every complex F of flat R-modules with F 'Min D(R) there are isomorphisms in D(R),

Λa(F) ' LΛa(M) ' RHomR(CR(xxx),M) .

Moreover, LΛa is way-out and preserves products, and one has λa∼=HomR(εxxxϕxxx,–).

PROOF. Let P '−→M be a semi-projective resolution. By 11.3.15 and semi-freenessof L(xxx), see 11.3.10, there are quasi-isomorphisms

Λa(P)

φPxxx←−− HomR(L(xxx),P)−→ HomR(L(xxx),M) .

By definition,Λa(P) is LΛa(M). Furthermore, HomR(L(xxx),M) is RHomR(CR(xxx),M)as L(xxx) '−→ CR(xxx) is a semi-free resolution; see 11.3.10. Thus the diagram aboveaccounts for the second isomorphism in the statement. It follows that LΛa is way-out, see 11.3.10 and A.25, and preserves products, see. 7.3.1. By 6.4.20 there is aquasi-isomorphism P '−→ F in C(R), and the next quasi-isomorphisms account forthe first isomorphism in the statement,

Λa(F)

φFxxx←−− HomR(L(xxx),F)←− HomR(L(xxx),P)

φPxxx−−→ Λ

a(P) .

It remains to show that there is an isomorphism between the natural transforma-tions λa and HomR(εxxxϕxxx,–). In D(R) every R-complex is naturally isomorphic to itssemi-projective resolution by 6.3.10, so it suffices to argue that the morphisms λaPand HomR(εxxxϕxxx,P) are isomorphic in D(R) for every semi-projective R-complex P,and that follows from commutativity of the diagram:

P

εPR

∼=

λaP// Λa(P)

HomR(R,P)Hom(εxxxϕxxx,P)

// HomR(L(xxx),P) .

' φPxxx

OO

11.3.17. Let a be an ideal in R generated by a weakly proregular sequence. Forevery flat R-module F one has Ha

0 (F)∼=Λa(F) and Hai (F) = 0 for i > 0 by 11.3.16.



EXERCISES

E 11.3.1 Let x ∈ R be an element. Show directly from the defintion that ϕx : L(x)→ CR(x) is aquasi-isomorphism.

E 11.3.2 Let p > 0 be a natural number. Calculate H(p)i (Q/Z) for i = 0,1.

11.4 Greenlees–May Equivalence

SYNOPSIS. Idempotence of RΓa and LΛa; Greenlees–May Equivalence.

11.4.1 Proposition. Let a be an ideal in R generated by a weakly proregular se-quence and let M be an R-complex. The following morphisms in D(R) are isomor-phisms:

RΓa(γMa ) : RΓa(RΓa(M)) −→ RΓa(M)(11.4.1.1)

LΛa(λaM) : LΛa(M) −→ LΛa(LΛa(M))(11.4.1.2)RΓa(λ

aM) : RΓa(M) −→ RΓa(LΛ

a(M))(11.4.1.3)

LΛa(γMa ) : LΛa(RΓa(M)) −→ LΛa(M)(11.4.1.4)

Furthermore, there are isomorphisms of morphisms in D(R),

(11.4.1.5) γRΓa(M)a

∼= RΓa(γMa ) and λaLΛa(M)

∼= LΛa(λaM) .

In particular, the complex RΓa(M) is derived a-torsion and the LΛa(M) is deriveda-adically complete.

PROOF. Let xxx be a weakly proregular sequence that generates a, and let ωxxx denotethe composite εxxxϕxxx : L(xxx)→ R; see 11.2.16 and 11.3.8. The functors and naturaltransformations from 11.2.4 and 11.3.4 can be realized as follows:

RΓa(–) = L(xxx)⊗R – , γa = ωxxx⊗R – ,

LΛa(–) = HomR(L(xxx),–) , and λa = HomR(ωxxx,–) ;

see 11.2.25, 11.3.10, and 11.3.16.By associativity 7.5.7 and commutativity 7.5.5 one has

RΓa(γMa ) = L(xxx)⊗R (ωxxx⊗R M) ∼= ωxxx⊗R (L(xxx)⊗R M) = γ

RΓa(M)a ;

this proves the first isomorphism in (11.4.1.5). The second follows from swap 7.5.11:

LΛa(λaM) = HomR(L(xxx),HomR(ωxxx,M)) ∼= HomR(ωxxx,HomR(L(xxx),M)) = λaLΛa(M) .

The functor Hom preserves homotopy, see 4.3.19, so to prove that RΓa(γMa ) is an

isomorphism in D(R), it suffices, by the expression above, to show that L(xxx)⊗Rωxxxis a homotopy equivalence in C(R). In the commutative diagram,


11.4 Greenlees–May Equivalence 423

L(xxx)⊗R L(xxx)L(xxx)⊗ωxxx

//

'ϕxxx⊗ϕxxx

L(xxx)⊗R R

' ϕxxx⊗R

CR(xxx)⊗R CR(xxx) 'CR(xxx)⊗εxxx

// CR(xxx)⊗R R ,

the vertical maps are quasi-isomorphisms by 11.3.10 and semi-flatness of L(xxx) andCR(xxx). The lower horizontal morphism is a quasi-isomorphism by 11.2.17. The dia-gram shows that L(xxx)⊗Rωxxx is a quasi-isomorphism and hence a homotopy equiva-lence as L(xxx) is a bounded complex of free modules; see 5.2.11 and 5.2.20.

The expression above, combined with adjunction 4.4.11 and commutativity 4.4.3,yields LΛa(λaM)∼= HomR(L(xxx)⊗Rωxxx,M). Thus, the fact that L(xxx)⊗Rωxxx is a homo-topy equivalence in C(R) also implies that LΛa(λaM) is an isomorphism in D(R).

We now turn to (11.4.1.3). Recall from 11.3.7 that L(xxx) is colimu>l Lu(xxx). Hence,to prove that

RΓa(λaM) = L(xxx)⊗R HomR(ωxxx,M)

is an isomorphsim in D(R), it suffices by 3.2.16 and 4.2.11 to show that each mor-phism Lu(xxx)⊗R HomR(ωxxx,M) is a quasi-isomorphism in C(R). To this end considerthe commutative diagram,

Lu(xxx)⊗R HomR(R,M)Lu(xxx)⊗Hom(ωxxx,M)

//

ηMRLu(xxx) ∼=

Lu(xxx)⊗R HomR(L(xxx),M)

∼=

HomR(HomR(Lu(xxx),R),M)

Hom(ϕuxxx ,M) u

HomR(HomR(Lu(xxx),R),HomR(L(xxx),M))

Hom(ϕuxxx ,Hom(L(xxx),M))u

HomR(KR(xxxu),M)

Hom(µMR ,M) ∼=

HomR(KR(xxxu),HomR(L(xxx),M))

HomR(R⊗R KR(xxxu),M)Hom(ωxxx⊗KR(xxxu),M)

// HomR(L(xxx)⊗R KR(xxxu),M) ,

ρML(xxx)KR(xxxu)∼=

OO

where ϕuxxx is the homotopy equivalence from 11.3.11 and the unlabled isomorphism

is induced by the counitor (4.4.1.2) and homomorphism evaluation 4.5.10(3,a). Tosee that HomR(ωxxx⊗R KR(xxxu),M) is a quasi-isomorphism, notice that ωxxx⊗R KR(xxxu)is a homotopy equivalence. Indeed, it is by 11.3.10, 11.2.8, and 11.2.17 a compositeof quasi-isomorphisms, and it is a morphism of bounded complexes of free modules.

Finally, we deal with (11.4.1.4). By 11.3.7 and 3.3.21 one has

LΛa(γMa ) = HomR(L(xxx),ωxxx⊗R M) ∼= lim

u>1HomR(Lu(xxx),ωxxx⊗R M) .

To see that the morphisms ωuxxx = HomR(Lu(xxx),ωxxx⊗R M) are quasi-isomorphisms,

consider the commutative diagram,



KR(xxxu)⊗R L(xxx)⊗R M

ϕuxxx⊗L(xxx)⊗M u

KR(xxxu)⊗ωxxx⊗M

u// KR(xxxu)⊗R R⊗R M

ϕuxxx⊗µM

Ru

HomR(Lu(xxx),R)⊗R L(xxx)⊗R M

∼=

HomR(Lu(xxx),R)⊗R M

θLu(xxx)RM∼=

HomR(Lu(xxx),L(xxx)⊗R M)ωu

xxx// HomR(Lu(xxx),R⊗R M) ,

where ϕuxxx is the homotopy equivalence from 11.3.11 and the unlabled isomorphism

is induced by the unitor (4.4.1.1) and tensor evaluation 4.5.7(3,a). The top horizontalmorphism is a homotopy equivalence as shown above. To see that this implies thatLΛa(γM

a ) = limu>1ωuxxx is a quasi-isomorphism, notice that the degreewise split exact

sequence 0−→ Kxxxαxxx−→ L(xxx) ωxxx−→ R−→ 0, with Kxxx = Kerωxxx, induces exact sequences,

0→Hom(Lu(xxx),Kxxx⊗M)αu

xxx−→Hom(Lu(xxx),L(xxx)⊗M)ωu

xxx−→Hom(Lu(xxx),R⊗M)→ 0 .

The families αuxxxu>1 and ωu

xxxu>1 are morphisms of towers. The morphisms inthe towers are surjective as they are induced by the degreewise split embeddingsLu(xxx) Lu+1(xxx) from 11.3.6; in particular, all three towers satisfy the Mittag-Leffler Condition by 3.3.37. Since each ωu

xxx is a quasi-isomorphism, the complexesHom(Lu(xxx),Kxxx⊗M) are acyclic by 4.2.6. Now it follows from 3.3.42 and 3.3.47that limu>1ω

uxxx is a quasi-isomorphism.

11.4.2 Theorem. Let a be an ideal in R generated by a weakly proregular sequence.For all R-complexes M and N there is an isomorphism,

RHomR(RΓa(M),N) ' RHomR(M,LΛa(N)) ,

which is natural in M and N. In particular, RΓa is left adjoint for LΛa. The unit andcounit of this adjunction are given by the composites

MλaM−−→ LΛa(M)

(LΛa(γMa ))−1

−−−−−−−−→ LΛa(RΓa(M)) and

RΓa(LΛa(M))

(RΓa(λaM))−1

−−−−−−−→ RΓa(M)γMa−−→M .

PROOF. The isomorphism follows from 11.2.25, 11.3.16, and adjunction 7.5.15.The last assertion is straightforward to check.

11.4.3 Corollary. Let a be an ideal in R generated by a weakly proregular sequence.There is an adjoint equivalence of k-linear triangulated categories,

Da-com(R)RΓa

//Da-tor(R) .

LΛaoo

PROOF. It is part of 11.4.1 that the image of RΓa is contained in Da-tor(R) and theimage of LΛa is contained in Da-com(R). It is immediate from 11.4.2 that the unit ofthe adjuction is an isomorphism for every complex in Da-com(R) and that the counitis an isomorphism for every complex in Da-tor(R).


11.4 Greenlees–May Equivalence 425

EXERCISES

E 11.4.1 Let a be an ideal in R generated by a weakly proregular sequence. Show that for R-complexes M and N, the following four complexes are isomorphic in D(R):• RHomR(RΓa(M),RΓa(N))

• RHomR(RΓa(M),N)

• RHomR(RΓa(M),LΛa(N))

• RHomR(M,LΛa(N))

• RHomR(LΛa(M),LΛa(N))


Appendices


Appendix AVanishing and Boundedness of Functors

SYNOPSIS. Vanishing of functor on module category; acyclicity of Hom complex; acyclicity oftensor product complex; way-out functor.

We collect here some technical results about vanishing of functors on module cat-egories (A.1–A.8), about acyclicity of Hom and tensor product complexes (A.10–A.14), and about way-out functors on categories of complexes (A.15–A.23) and onderived categories (A.24–A.32).

VANISHING OF FUNCTORS

The proof of 8.1.14 demonstrates a typical application of the next lemma.

A.1 Lemma. Let U be an Abelian category and let F: M(R)→ U be a half exactfunctor. One has F(M) = 0 for every finitely generated R-module M if and only ifone has F(R/a) = 0 for every left ideal a in R.

PROOF. The “only if” part is trivial. Assume that F(R/a) = 0 holds for every leftideal a in R. Let M be an R-module generated by elements x1, . . . ,xn and proceed byinduction on the number, n, of generators. For n = 1 one has M ∼= R/(0 :R x1),whence F(M) = 0 holds by assumption. For n > 1 set N = R〈x1, . . . ,xn−1〉; thequotient module M/N is then generated by [xn]N . Apply F to the exact sequence0→ N → M → M/N → 0 to get the exact sequence F(N) −→ F(M) −→ F(M/N)where F(M/N) = 0 holds by assumption and F(N) = 0 holds by the induction hy-pothesis. It follows that also F(M) is zero.

A.2. One obtains the same statement for a half exact functor M(R)op → U by ap-plying A.1 to the opposite functor M(R)→ Uop.

Over a commutative Noetherian ring R the notion of support ensures that van-ishing of functors on finitely generated R-modules can be tested on cyclic modulesR/p, where p is a prime ideal.

429

430 Appendix A

A.3 Lemma. Let U be an Abelian category, let R be commutative and Noetherian,and let F: M(R)→ U be a half exact functor. For every finitely generated R-moduleM with F(M) 6= 0 there exists a prime ideal p in SuppR M with F(R/p) 6= 0.

In particular, one has F(M) = 0 for every finitely generated R-module M if andonly if one has F(R/p) = 0 for every prime ideal p in R.

PROOF. Choose a filtration 0 = M0 ⊂ M1 ⊂ ·· · ⊂ Mn−1 ⊂ Mn = M such that foreach index u one has Mu/Mu−1 ∼= R/pu for some pu ∈ SuppR M. The canonicalexact sequences 0→Mu−1→Mu→ R/pu→ 0 yield exact sequences

F(Mu−1)−→ F(Mu)−→ F(R/pu) .

Recall that F is additive so that one has F(0) = 0. As F(M) is non-zero, it followsfrom these sequences that one has F(R/pu) 6= 0 for at least one u ∈ 1, . . . ,n.

A.4. One obtains the same statement for a half exact functor M(R)op → U by ap-plying A.3 to the opposite functor M(R)→ Uop.

For linear endofunctors on the category of modules over a commutative Noether-ian ring, non-vanishing on a finitely generated module persists after localization.

A.5 Proposition. Let R be commutative and Noetherian and let F: M(R)→M(R)be a left exact and R-linear functor. For every finitely generated R-module M withF(M) 6= 0 there exists a prime ideal p in SuppR M with F(R/p)p 6= 0.

PROOF. By A.3 there exists a prime ideal p in SuppR M with F(R/p) 6= 0. For everyx in R\p it follows from the assumptions on F that the next sequence is exact,

0−→ F(R/p) x−−→ F(R/p) .

Thus the canonical homomorphism F(R/p)→ F(R/p)p is injective; in particular,one has F(R/p)p 6= 0.

A.6 Proposition. Let R be commutative and Noetherian and let F: M(R)op→M(R)be a half exact and R-linear functor. For every finitely generated R-module M withF(M) 6= 0 there exists a prime ideal p in SuppR M with F(R/p)p 6= 0.

PROOF. By A.4 the set p ∈ SuppR M | F(R/p) 6= 0 is non-empty; as R is Noe-therian, the set has a maximal element q. Set N = R/q; for every element x ∈ R\ qit follows from A.4 that F is zero on N/xN = R/(q+(x)). Application of F to theexact sequence 0−→ N x−→N −→ N/xN −→ 0 now yields an exact sequence,

0−→ F(N)x−−→ F(N) .

It follows that the canonical homomorphism F(N)→ F(N)q is injective; in particu-lar, one has F(R/q)q = F(N)q 6= 0.

A.7 Proposition. Let R be commutative and Noetherian with Jacobson radical J,and let F: M(R)→M(R) be a half exact and R-linear functor such that F(R/p) isfinitely generated for every prime ideal p in R. For every finitely generated R-moduleM with F(M) 6= 0 there exists a prime ideal p in SuppR M with p⊇ J and F(R/p) 6= 0.


Vanishing and Boundedness of Functors 431

PROOF. By A.3 the set p ∈ SuppR M | F(R/p) 6= 0 is non-empty; as R is Noether-ian, the set has a maximal element q. Set N = R/q and let x be any element in R\q.It follows from A.3 that F is zero on N/xN = R/(q+(x)). Application of F to theexact sequence 0−→ N x−→N −→ N/xN −→ 0 yields an exact sequence

F(N)x−−→ F(N)−→ 0 .

Nakayama’s lemma B.30 now implies that x is not in J, whence q contains J.

A.8 Corollary. Let R be commutative Noetherian and local with unique maximalideal m, and let F: M(R)→M(R) be a half exact and R-linear functor such thatF(R/p) is finitely generated for every prime ideal p in R. If there exists a finitelygenerated R-module M with F(M) 6= 0, then one has F(R/m) 6= 0.

ACYCLICITY OF HOM AND TENSOR PRODUCT COMPLEXES

Note that condition (b) in the next lemma is trivially satisfied if N is bounded above.

A.9 Lemma. Let M be an R-module and let N be an acyclic R-complex. The com-plex HomR(M,N) is acyclic if the following conditions are satisfied.

(a) ExtiR(M,Nv) = 0 for every v ∈ Z and all i > 0; and(b) ExtiR(M,Zv(N)) = 0 for every v 0 and all i > 0.

PROOF. As N is acyclic, the sequences 0→ Zv(N)→ Nv→ Zv−1(N)→ 0 are exact,and by 7.3.14 it is enough to show that ExtiR(M,Zv(N))= 0 holds for every v ∈ Z andall i > 0. In view of (b) one can procede by descending induction on v. Fix v ∈ Z andassume that ExtiR(M,Zv(N)) = 0 holds for all i > 0. As one has ExtiR(M,Nv) = 0 forall i> 0, another application of 7.3.14 yields ExtiR(M,Zv−1(N))= 0 for all i> 0.

A standard application of the next lemma is to an acyclic complex N and a com-plex M of projective R-modules; see for example 5.2.8.

A.10 Lemma. Let M and N be R-complexes such that M or N is bounded below. Ifthe complex HomR(Mv,N) is acyclic for every v ∈ Z, then HomR(M,N) is acyclic.

PROOF. For every n ∈ Z we must verify the equality Hn(HomR(M,N)) = 0. Fix n;by 2.2.15 and 2.3.15 there is an isomorphism

Hn(HomR(M,N)) ∼= H0(HomR(M,Σ−n N)) .

Since the assumptions are invariant under a shift of N, it suffices to argue that onehas H0(HomR(M,N)) = 0. Thus, we need to show that every morphism α : M→ Nis null-homotopic; see 2.3.10. Given a morphism α, we must construct a degree 1homomorphism σ : M→ N, such that

(?) αv = ∂Nv+1σv +σv−1∂

Mv

holds for every v ∈ Z. As M or N is bounded below, one has αv = 0 for v 0. Thus,one can choose σv = 0 for v 0. Now, proceed by induction. Given that (?) holds


432 Appendix A

for v, it follows that αv+1−σv∂Mv+1 is a cycle in HomR(Mv+1,N) of degree v+ 1;

indeed, one has

∂Nv+1(αv+1−σv∂

Mv+1) = (αv−∂N

v+1σv)∂Mv+1 = (σv−1∂

Mv )∂M

v+1 = 0 .

As the complex HomR(Mv+1,N) is acyclic, αv+1−σv∂Mv+1 is a boundary. Thus, there

exists an element σv+1 in HomR(Mv+1,Nv+2) with ∂Nv+2σv+1 = αv+1−σv∂

Mv+1.

Condition (b) in the next lemma is trivially satisfied if M is bounded below.

A.11 Lemma. Let M be an acyclic R-complex and N be an R-module. The complexHomR(M,N) is acyclic if the following conditions are satisfied.

(a) ExtiR(Mv,N) = 0 for every v ∈ Z and all i > 0; and(b) ExtiR(Cv(M),N) = 0 for every v 0 and all i > 0.

PROOF. As M is acyclic, the sequences 0→ Cv+1(M)→ Mv → Cv(M)→ 0 areexact, and by 7.3.14 it is enough to argue that ExtiR(Cv(M),N) = 0 holds for everyv ∈ Z and all i > 0. Now procede by induction in parallel with the proof of A.9.

The prototypical application of the next lemma is to an acyclic complex M and acomplex N of injective modules; see for example 5.3.12.

A.12 Lemma. Let M and N be R-complexes such that M or N is bounded above. Ifthe complex HomR(M,Nv) is acyclic for every v ∈ Z, then HomR(M,N) is acyclic.

PROOF. Parallel to the proof of A.10.

A.13 Lemma. Let M be an acyclic Ro-complex and N be an R-module. The complexM⊗R N is acyclic if the following conditions are satisfied.

(a) TorRi (Mv,N) = 0 for every v ∈ Z and all i > 0; and

(b) TorRi (Cv(M),N) = 0 for every v 0 and all i > 0.

PROOF. As M is acyclic, the sequences 0→ Cv+1(M)→ Mv → Cv(M)→ 0 areexact, and by 7.4.13 it is enough to show that TorR

1 (Cv(M),N) = 0 holds for everyv ∈ Z. Part (b) yields the base for a proof by induction on v that TorR

i (Cv(M),N) = 0holds for all v∈Z and all i> 0. Fix v ∈ Z and assume that TorR

i (Cv(M),N) = 0 holdsfor all i > 0. As by (a) one has TorR

i (Mv,N) = 0 for all i > 0, another application of7.4.13 yields TorR

i (Cv+1(M),N) = 0 for all i > 0.

A standard application of the next lemma is to an acyclic complex M and a com-plex N of flat R-modules; see for example 5.4.8.

A.14 Lemma. Let M be an Ro-complex and let N be an R-complex, such that M isbounded above or N is bounded below. If the complex M⊗R Nv is acyclic for everyv ∈ Z, then M⊗R N is acyclic.

PROOF. By adjunction 4.4.11 there is an isomorphism of k-complexes

Homk(M⊗R N,E) ∼= HomR(N,Homk(M,E)) .

Recall from 2.5.6(b) that Homk(M,E) is bounded below if M is bounded above. AsE is a faithfully injective k-module, the claim follows immediately from A.10.



WAY-OUT FUNCTORS

A way-out functor F: C(R)→ C(S) is one where the shape of F(M) depends on theshape of M in a controlled manner.

A.15 Definition. Let F: C(R)→ C(S) be a functor.The functor F is called way-out above if for every n ∈ Z there is a w ∈ Z such

that inf(F(M))\ > n holds for every R-complex M with infM\ > w.The functor F is called way-out below if for every n ∈ Z there is a u ∈ Z such that

sup(F(M))\ 6 n holds for every R-complex M with supM\ 6 u.The functor F is called way-out if it is way-out above and way-out below.

REMARK. Other words for way-out above, way-out below, and way-out are way-out left, way-outright, and way-out in both directions.

A.16 Example. A functor C(R)→ C(S) that is extended from an additive functoron modules, as described in 2.1.45, is per construction way-out.

A.17 Proposition. Let M be an R-complex.(a) If M is bounded below, then the functor HomR(M,–) is way-out below.(b) If M is bounded above, then the functor HomR(M,–) is way-out above.(c) If M is bounded, then HomR(M,–) is way-out.

PROOF. Part (a) is immediate from 2.5.9(a), and part (b) follows from an argumentsimilar to the proof of loc. cit. Part (c) follows from (a) and (b).

A.18 Proposition. Let M be an R-complex.(a) If M is bounded below, then the functor –⊗R M is way-out above.(b) If M is bounded above, then the functor –⊗R M is way-out below.(c) If M is bounded, then –⊗R M is way-out.


A class U of R-complexes that satisfies the assumptions in the next lemma is, forexample, the class of all complexes of projective R-modules.

A.19 Lemma. Let F,G: C(R)→ C(S) be \- and Σ-functors and let τ : F→ G be aΣ-transformation. Let U be a class of R-complexes that is closed under shift andhard truncations. If τM is a quasi-isomorphism for every module M in U, then thefollowing assertions hold.

(a) The morphism τM is a quasi-isomorphism for every bounded complex M in U.(b) If F and G are way-out above, then τM is a quasi-isomorphism for every

bounded below complex M in U.(c) If F and G are way-out below, then τM is a quasi-isomorphism for every

bounded above complex M in U.(d) If F and G are way-out, then τM is a quasi-isomorphism for every M in U.


434 Appendix A

PROOF. (a): After a shift, M is concentrated in degrees u,u−1, . . . ,0. If M is a mo-dule, i.e. u = 0, then τM is a quasi-isomorphism. Let u > 0 and assume that τ is aquasi-isomorphism when evaluated at a complex in U concentrated u degrees. Con-sider the degreewise split exact sequence 0→Mďu−1→M→ Σu Mu→ 0; cf. 2.5.17.All three complexes in the sequence belong to U, and together with τ it inducesby 2.1.49 an obvious commutative diagram in which the morphisms τMďu−1 andτΣ

u Mu ∼= Σu τM are quasi-isomorphisms by assumption. It follows from 4.2.5 thatthe third map, τM , is a quasi-isomorphism. Thus, the assertion follows by induction.

(b): To prove that τM is a quasi-isomorphism we fix an integer n and show thatHn(τ

M) is an isomorphism. Choose w ∈ Z such that one has inf(F(X))\ > n+2 andinf(G(X))\ > n+2 for all complexes X with infX \ > w. All three complexes in thedegreewise split exact sequence 0→Mďw−1→M→Měw→ 0 from 2.5.17 belongto U, and it induces by 2.1.49 and 2.2.21 a commutative diagram,

Hn+1(F(Měw)) // Hn(F(Mďw−1)) ∼=//

Hn(τMďw−1 )∼=

Hn(F(M)) //

Hn(τM)

Hn(F(Měw))

Hn+1(G(Měw)) // Hn(G(Mďw−1)) ∼=// Hn(G(M)) // Hn(G(Měw))

where the horizontal isomorphisms are forced by the vanishing of F(Měw)v andG(Měw)v for v6 n+1. As Mďw−1 is a bounded complex in U, it follows from part(a) that τMďw−1 is a quasi-isomorphism, which explains the vertical isomorphism inthe diagram. By commutativity of the diagram, Hn(τ

M) is an isomorphism.(c): Similar to the proof of part (b).(d): The degreewise split exact sequence 0→ Mď0 → M→ Mě1 → 0 in U and

the natural transformation τ induce by 2.1.49 an obvious commutative diagram.It follows from parts (b) and (c) that the morphisms τMě1 and τMď0 are quasi-isomorphisms, and 4.2.5 then implies that τM is a quasi-isomorphism.

A.20 Definition. Let G: C(R)op→ C(S) be a functor.The functor G is called way-out above if for every n ∈ Z there is a u ∈ Z such

that inf(G(M))\ > n holds for every R-complex M with supM\ 6 u.The functor G is called way-out below if for every n ∈ Z there is a w ∈ Z such

that sup(G(M))\ 6 n holds for every R-complex M with infM\ > w.The functor G is called way-out if it is way-out above and way-out below.

REMARK. A functor C(R)→ C(S)op is called way-out above/below if and only if the oppositefunctor C(R)op → C(S) is way-out below/above according to A.20. For such functors the resultbelow holds with “way-out above” and “way-out below” interchanged; see E A.5.

A.21 Example. A functor C(R)op→ C(S) that is extended from an additive functoron modules, as described in 2.1.45, is per construction way-out.

A.22 Proposition. Let N be an R-complex.(a) If N is bounded above, then HomR(–,N) is way-out below.(b) If N is bounded below, then HomR(–,N) is way-out above.



(c) If N is bounded, then HomR(–,N) is way-out.


A.23 Lemma. Let F,G: C(R)op→ C(S) be \- and Σ-functors and let τ : F→ G bea Σ-transformation. Let U be a class of R-complexes that is closed under shift andhard truncations. If τM is a quasi-isomorphism for every module M in U, then thefollowing assertions hold.

(a) The morphism τM is a quasi-isomorphism for every bounded complex M in U.(b) If F and G are way-out above, then τM is a quasi-isomorphism for every

bounded above complex M in U.(c) If F and G are way-out below, then τM is a quasi-isomorphism for every

bounded below complex M in U.(d) If F and G are way-out, then τM is a quasi-isomorphism for every M in U.

PROOF. Parallel to the proof of A.19

WAY-OUT FUNCTORS ON THE DERIVED CATEGORY

A way-out functor F: D(R)→D(S) is one where the shape of the homology ofF(M) depends on the shape of the homology of M in a controlled manner.

A.24 Definition. Let F: D(R)→D(S) be a functor.The functor F is called way-out above if for every n ∈ Z there is a w ∈ Z such

that infF(M)> n holds for every R-complex M with infM > w.The functor F is called way-out below if for every n ∈ Z there is a u ∈ Z such that

supF(M)6 n holds for every R-complex M with supM 6 u.The functor F is called way-out if it is way-out above and way-out below.

A.25 Proposition. Let M be an R-complex.(a) If M is in DA(R), then the functor RHomR(M,–) is way-out below.(b) If M has finite projective dimension, then RHomR(M,–) is way-out above.(c) If M in D@A(R) has finite projective dimension, then RHomR(M,–) is way-out.

PROOF. Parts (a) and (b) follow from 7.6.7 and 8.1.9, and they imply (c).

A.26 Proposition. Let M be an R-complex.(a) If M is in DA(R), then the functor –⊗L

R M is way-out above.(b) If M has finite flat dimension, then –⊗L

R M is way-out below.(c) If M is in D@A(R) and has has finite flat dimension, then –⊗L

R M is way-out.


If R is Noetherian, then the class U=Df(R) satisfies the assumptions in the nextlemmas; cf. 7.6.4.


436 Appendix A

A.27 Lemma. Let F,G: D(R)→D(S) be triangulated functors and let τ : F→ Gbe a triangulated natural transformation. Let U be a class of R-complexes that isclosed under shift and soft truncations. If τM is an isomorphism for every moduleM in U, then the following assertions hold.

(a) The morphism τM is an isomorphism for every complex M in U∩D@A(R).(b) If F and G are way-out above, then τM is an isomorphism for every complex

M in U∩DA(R).(c) If F and G are way-out below, then τM is an isomorphism for every complex

M in U∩D@(R).(d) If F and G are way-out, then τM is an isomorphism for every M in U.

PROOF. (a): Because of the isomorphism M'MĎsupM , see 4.2.4, it suffices to showthat τMĎn is an isomorphism for every n ∈ Z. For n 0 one has MĎn ' 0 in D(R),and hence τMĎn is an isomorphism. Now proceed by induction on n. The complexMĎn−1 belongs to U; assume that τMĎn−1 is an isomorphism and consider the distin-guished triangle from (7.6.6.1),

(?) Σn Hn(M)−→MĎn −→MĎn−1 −→ Σ(Σn Hn(M)) .

One has Σn Hn(M)' (MĚn)Ďn, in particular the module Hn(M) belongs to U, so themorphism τΣ

n Hn(M) ∼= Σn τHn(M) is an isomorphism by assumption. Now it followsfrom 6.5.22, applied to the commutative diagram induced by the natural transfor-mation τ and (?), that τMĎn is an isomorphism.

(b): The morphism τM is an isomorphism in D(S) if and only if Hn(τM) is an

isomorphism for all n ∈ Z; cf. 6.5.20. Fix an integer n; choose w ∈ Z such thatone has infF(X) > n+1 and infG(X) > n+1 for all complexes X with infX > w.Together with τ, the distinguished triangle from (7.6.6.3),

MĚw −→M −→MĎw−1 −→ Σ(MĚw) ,

induces by 6.5.22 the following commutative diagram in D(S),

Hn(F(MĚw)) // Hn(F(M)) ∼=//

Hn(τM)

Hn(F(MĎw−1)) //

Hn(τ(MĎw−1))∼=

Hn−1(F(MĚw))

Hn(G(MĚw)) // Hn(G(M)) ∼=// Hn(G(MĎw−1)) // Hn−1(G(MĚw)) .

The horizontal isomorphisms are forced by the vanishing of Hv(F(MĚw)) andHv(G(MĚw)) for all v 6 n. The complex MĎw−1 is in U∩D@A(R), so τ(MĎw−1) isan isomorphism by part (a), and it follows that Hn(τ

M) is an isomorphism.(c): Similar to the proof of part (b).(d): By parts (b) and (c) the morphisms τMĚ1 and τMĎ0 are isomorphisms. Now

it follows from 6.5.22, applied to the commutative diagram induced by τ and thetriangle MĚ1→M→MĎ0→ Σ(MĚ1), that also τM is an isomorphism.

A.28 Lemma. Let F: D(R)→D(S) be a triangulated functor and let U be a classof R-complexes that is closed under shift and soft truncations. If S is left Noetherianand F(M) belongs to Df(S) for every module M in U, then the next assertions hold.



(a) The complex F(M) belongs to Df(S) for every complex M in U∩D@A(R).(b) If F is way-out above, then F(M) belongs to Df(S) for every M in U∩DA(R).(c) If F is way-out below, then F(M) belongs to Df(S) for every M in U∩D@(R).(d) If F is way-out, then F(M) belongs to Df(S) for every M in U.

PROOF. Similar to the proof of A.27; see also the proof of A.32.

A.29 Definition. Let G: D(R)op→D(S) be a functor.The functor G is called way-out above if for every n ∈ Z there is a u ∈ Z such

that infG(M)> n holds for every R-complex M with supM 6 u.The functor G is called way-out below if for every n ∈ Z there is a w ∈ Z such

that supG(M)6 n holds for every R-complex M with infM > w.The functor G is called way-out if it is way-out above and way-out below.

REMARK. A functor D(R)→D(S)op is called way-out above/below if and only if the oppositefunctor D(R)op→D(S) is way-out below/above according to A.29. One gets results about suchfunctors by interchanging “way-out above” and “way-out below” in the lemmas below; see E A.6.

A.30 Proposition. Let N be an R-complex.(a) If N is in D@(R), then the functor RHomR(–,N) is way-out below.(b) If N has finite injective dimension, then RHomR(–,N) is way-out above.(c) If N in D@A(R) has finite injective dimension, then RHomR(–,N) is way-out.


A.31 Lemma. Let F,G: D(R)op→D(S) be triangulated functors, and let τ : F→ Gbe a triangulated natural transformation. Let U be a class of R-complexes that isclosed under shift and soft truncations. If τM is an isomorphism for every module Min U, then the following assertions hold.

(a) The morphism τM is an isomorphism for every complex M in U∩D@A(R)op.(b) If F and G are way-out above, then τM is an isomorphism for every complex

M in U∩D@(R)op.(c) If F and G are way-out below, then τM is an isomorphism for every complex

M in U∩DA(R)op.(d) If F and G are way-out, then τM is an isomorphism for every M in U.

PROOF. Parallel to the proof of A.27.

A.32 Lemma. Let G: D(R)op→D(S) be a triangulated functor and let U be a classof R-complexes that is closed under shift and soft truncations. If S is left Noetherianand G(M) belongs to Df(S) for every module M in U, then the next assertions hold.

(a) The complex G(M) is in Df(S) for every complex M in U∩D@A(R)op.(b) If G is way-out above, then G(M) is in Df(S) for every M in U∩D@(R)op.(c) If G is way-out below, then G(M) is in Df(S) for every M in U∩DA(R)op.(d) If G is way-out, then G(M) is in Df(S) for every M in U.


438 Appendix A

PROOF. (a): Because of the isomorphism M ' MĎsupM in D(R), see 4.2.4, it suf-fices to show that G(MĎn) is in Df(S) for every n ∈ Z. For n 0 one has MĎn ' 0in D(R) and, therefore, G(MĎn) ' 0 ∈ Df(S). Now proceed by induction on n.The complex MĎn−1 is in U; assume that G(MĎn−1) belongs to Df(S). One hasΣn Hn(M)' (MĚn)Ďn, so the module Hn(M) belongs to U and thus G(Σn Hn(M))∼=Σ−n G(Hn(M)) is in Df(S) by assumption. Apply the triangulated functor G to thedistinguished triangle (7.6.6.1),

Σn Hn(M)−→MĎn −→MĎn−1 −→ Σ(Σn Hn(M)) .

As the category Df(S) is triangulated, see 7.6.4, it follows that G(MĎn) is in Df(S).(b): Fix n ∈ Z and choose u ∈ Z such that infG(X) > n+1 holds for all X with

supX 6 u. Consider the distinguished triangle MĚu+1 → M→ MĎu → Σ(MĚu+1);cf. (7.6.6.3). It induces by 6.5.22 an exact sequence of S-modules

Hn(G(MĎu))−→ Hn(G(M))∼=−−→ Hn(G(MĚu+1))−→ Hn−1(G(MĎu)) ,

where the isomorphism is forced by the vanishing of Hv(G(MĎu)) for v 6 n. Asthe complex MĚu+1 is in U∩D@A(R)op, it follows from part (a) that G(MĚu+1) is inDf(S). In particular, Hn(G(MĚu+1)) is finitely generated, and hence so is Hn(G(M)).

(c): Similar to the proof of part (b).(d): Consider the distinguished triangle MĚ1→M→MĎ0→ Σ(MĚ1) in U from

(7.6.6.3). It follows from parts (b) and (c) that G(MĎ0) and G(MĚ1) belong to Df(S).Since G is a triangulated functor, and Df(S) is a triangulated category, it follows thatG(M) belongs to Df(S).

EXERCISES

E A.1 Give a proof of A.12.E A.2 Give a direct proof of A.14, i.e. one that does not invoke A.10 or A.12.E A.3 Let τ : F→ G be a natural transformation of functors from M(R) to M(S). Show that if F

and G are right exact and τR is an isomorphism, then τM is an isomorphism for all finitelypresented R-modules M.

E A.4 Let τ : F→ G be a natural transformation of functors from M(R)op to M(S). Show that ifF and G are left exact and τR is an isomorphism, then τM is an isomorphism for all finitelypresented R-modules M.

E A.5 Let F: C(R)→C(S)op be a functor, and let U be a class of R-complexes. Define F beway-out above/below on U if and only if the opposite functor C(R)op→ C(S) is way-outbelow/above on U. State and prove the result that parallels A.23.

E A.6 Let F: D(R)→D(S)op be a functor. Define F to be way-out above/below if and only ifthe opposite functor D(R)op→D(S) is way-out below/above. State and prove the resultsthat correspond to A.31 and A.32.


Appendix BMinimality

SYNOPSIS. Minimal complex; injective envelope; minimal semi-injective resolution; Nakayama’slemma; projective cover; minimal semi-projective resolution; semi-perfect ring; perfect ring.

If I is a semi-injective complex, then every quasi-isomorphism I → I is a homo-topy equivalence by 5.3.24; minimality of I would imply that every such quasi-isomorphism would even be an isomorphism. After giving a few characterizationsof minimal complexes, we prove existence of minimal semi-injective resolutions forall complexes over any ring. In contrast, existence of minimal semi-projective reso-lutions for all complexes holds only over perfect rings. For complexes with boundedabove and degreewise finitely generated homology, minimal semi-projective reso-lutions are more accessible: They can be constructed for every such complex over aNoetherian semi-perfect ring, in particular over any Noetherian local ring.

B.1 Lemma. Let 0 −→ K α−→Mβ−→ N −→ 0 be a degreewise split exact sequence

of R-complexes.(a) The complex K is contractible if and only if β is a homotopy equivalence, and

in that case the sequence is split in C(R).(b) The complex N is contractible if and only if α is a homotopy equivalence, and

in that case the sequence is split in C(R).

PROOF. It follows from 2.3.12 that the following sequences are exact.

0−→ HomR(K,K)−→ HomR(K,M)Hom(K,β)−−−−−−→ HomR(K,N)−→ 0 ,(?)

0−→ HomR(N,M)Hom(β,M)−−−−−−→ HomR(M,M)−→ HomR(K,M)−→ 0 , and(‡)

0−→ HomR(N,K)−→ HomR(N,M)Hom(N,β)−−−−−−→ HomR(N,N)−→ 0 .()

If β is a homotopy equivalence, then so is HomR(K,β) by 4.3.19; in particular, itis a quasi-isomorphism. It follows from (?) and 4.2.6 that HomR(K,K) is acyclic,whence K is contractible by 4.3.29.

Conversely, if K is contractible, then HomR(K,M) and HomR(N,K) are acyclicby 4.3.29. It follows from () that HomR(N,β) is a quasi-isomorphism. It is sur-

439

440 Appendix B

jective on cycles by 4.2.7, so there exists a morphism γ : N→M with βγ = 1N . Inparticular, the original sequence is split in C(R). Moreover, it follows from (‡) thatHomR(β,M) is a quasi-isomorphism, whence there exists a morphism γ′ : N→Msuch that γ′β ∼ 1M . Thus β is a homotopy equivalence; see 4.3.3. This proves part(a), and a similar argument proves (b).

B.2 Definition. An R-complex M is called minimal if every homotopy equivalenceε : M→M is an isomorphism.

Minimality for complexes extends no interesting notion for modules; indeed, ev-ery module is minimal when viewed as a complex. Here are a couple less trivialexamples of minimal complexes.

B.3 Example. The Z-complex I = 0→ Q→ Q/Z→ 0 is minimal, as there are nonon-zero homomorphisms Q/Z→Q and hence no non-zero homomorphisms I→ Iof degree 1. It follows that every homotopy equivalence I→ I is an isomorphism.

B.4 Example. Let k be a field and consider the local ring R = k[x]/(x2) of dualnumbers. The R-complex

X = · · · x−−→ R x−−→ R x−−→ R x−−→ ·· ·

is minimal. Indeed, every homomorphism R→ R is given by multiplication by anelement in R. A homotopy equivalence α : X → X is, therefore, given by a sequenceof ring elements (av)v∈Z; so is its homotopy inverse β= (bv)v∈Z and the homotopy% = (rv)v∈Z from 1X to αβ. For every v ∈ Z one has 1− avbv = xrv + rv−1x. Theelement avbv = 1− x(rv + rv−1) is not in the maximal ideal (x), so it is invertible,whence αv is invertible and α is an isomorphism.

B.5 Lemma. Let M be an R-complex. If M is minimal, then the zero complex is theonly contractible direct summand of M.

PROOF. Let M be minimal and assume that it is a direct sum M = M′⊕C where C iscontractible. The morphism 1M′⊕0C is a homotopy equivalence M→M; see 4.3.6.As M is minimal, 1M′⊕0C is an isomorphism and hence C is the zero complex.

Minimality can be characterized in several different ways.

B.6 Proposition. For an R-complex M, the following conditions are equivalent.(i) M is minimal.

(ii) Every morphism ε : M→M with ε∼ 1M is an isomorphism.(iii) Every homotopy equivalence α : K→M has a right inverse.

(iii’) Every homotopy equivalence ε : M→M has a right inverse.(iv) Every homotopy equivalence β : M→ N has a left inverse.

(iv’) Every homotopy equivalence ε : M→M has a left inverse.When these conditions hold, the complexes Kerα and Cokerβ are contractible forevery homotopy equivalence α : K→M and every homotopy equivalence β : M→ N.


Minimality 441

PROOF. First notice that if α : K→M is a homotopy equivalence and (iii) holds,then there is a split exact sequence 0−→ Kerα−→ K α−→M −→ 0, and Kerα is con-tractible by B.1. Similarly, if β : M→ N is a homotopy equivalence and (iv) holds,then Cokerβ is contractible.

(i)=⇒ (ii): If the endomorphisms ε : M→M and 1M are homotopic, then one hasε2 ∼ ε ∼ 1M; see 4.3.3. It follows that ε is its own homotopy inverse; in particular,ε is a homotopy equivalence, so by (i) it is an isomorphism.

(ii)=⇒ (iii): Let α : K→M be a homotopy equivalence with homotopy inverseγ : M→ K. By (ii) the morphism ε= αγ is an isomorphism, and it follows that γε−1

is a right inverse of α.(iii’)=⇒ (i): Let ε : M→M be a homotopy equivalence. It has a right inverse,

so the sequence 0 −→ Kerε ι−→ M ε−→ M −→ 0 is split exact. Consider the splitexact sequence 0 −→M σ−→M τ−→ Kerε −→ 0, where τι = 1Kerε and εσ = 1M . Asalready noted, Kerε is contractible, whence σ is a homotopy equivalence by B.1 andsurjective by (iii’). Thus σ is an isomorphism and, therefore, ε is an isomorphism.

(ii)=⇒ (iv): Let β : M→ N be a homotopy equivalence with homotopy inverseγ : N→M. By (ii) the morphism ε= γβ is an isomorphism, and it follows that ε−1γis a left inverse of β.

(iv’)=⇒ (i): Let ε : M→M be a homotopy equivalence. It has a left inverse, sothe sequence 0 −→M ε−→M π−→ Cokerε −→ 0 is split exact. Consider the split ex-act sequence 0 −→ Cokerε σ−→ M τ−→ M −→ 0, where πσ = 1Cokerε and τε = 1M .As noted, Cokerε is contractible, whence τ is a homotopy equivalence by B.1 andinjective by (iv’). Thus τ is an isomorphism and, therefore, ε is an isomorphism.

B.7 Corollary. Let M and M′ be minimal R-complexes.(a) Every homotopy equivalence α : M→M′ is an isomorphism.(b) If there exist contractible R-complexes C and C′ such that M⊕C and M′⊕C′

are homotopy equivalent, then the complexes M and M′ are isomorphic.

PROOF. (a): It follows from B.6 that α has a right inverse as well as a left inverse,whence it is an isomorphism.

(b): It follows from B.1 that the injection M M ⊕C and the projectionM′⊕C′M′ are homotopy equivalences. Therefore, the composite map

MM⊕C u−−→M′⊕C′M′

is a homotopy equivalence, and hence an isomorphism by (a).

INJECTIVE ENVELOPES

Every graded module can by 5.3.4 be embedded into a graded-injective one; theimmediate goal is to show that there is a unique way to do this minimally.

B.8 Definition. Let M be a graded R-module. A graded submodule N of M is calledessential if M′∩N 6= 0 holds for every graded submodule M′ 6= 0 of M.


442 Appendix B

REMARK. Another word for essential submodule is large submodule.

B.9 Example. Every non-zero ideal in Z is essential.

B.10. A graded direct summand N of a graded R-module M is essential if and onlyif N = M.

B.11 Lemma. Let M be a graded R-module with graded submodules N and N′.(a) Assume that there is an inclusion N ⊆ N′. If N is essential in N′, and N′ is

essential in M, then N is essential in M.(b) If N and N′ are essential in M, then N∩N′ is essential in M.(c) Let α : M→ X be a homomorphism of graded R-modules. If α is injective and

N is essential in M, then α(N) is essential in α(M).

PROOF. (a): Let M′ 6= 0 be a graded submodule of M. The submodule N ∩M′ =N∩ (N′∩M′) is non-zero, as N′ is essential in M and N is essential in N′.

(b): Let M′ 6= 0 be a graded submodule of M. The submodule (N ∩N′)∩M′ =N∩ (N′∩M′) is non-zero, as N′ and N are both essential in M.

(c): Immediate as α : M→ α(M) is an isomorphism.

B.12 Definition. An injective envelope of a graded R-module M is an injectivemorphism ι : M→ E of graded R-modules where E is graded-injective and Im ι isessential in E.

REMARK. Another word for injective envelope is injective hull.

B.13 Example. The Z-module Q is divisible and hence injective by 1.3.28. As Z isan essential submodule of Q, the embedding Z Q is an injective envelope.

For every prime number p, the Prüfer p-group Z(p∞) = (1, p, p2, . . .−1Z)/Zis divisible, and hence injective. To see this, it suffices to show that multiplicationby q is surjective on Z(p∞) for every prime number q. For q = p this is clear as[ x

pn ]Z = p [ xpn+1 ]Z. Given q 6= p and [ x

pn ]Z in Z(p∞), choose integers a and b withaq+bpn = 1; one then has [ x

pn ]Z = q [ axpn ]Z. The homomorphism ι : Z/pZ→ Z(p∞)

given by [x]pZ 7→ [ xp ]Z is evidently injective. To see that its image is essential in

Z(p∞), let y = [ xpn ]Z be a non-zero element in Z(p∞). We may assume that x is not

divisible by p, and thus [ xp ]Z = pn−1y is a non-zero element in Im ι ∩ Z〈y〉. It follows

that ι is an injective envelope.

We shall see that injective envelopes always exist and that they are unique up toisomorphism. We begin by proving uniqueness, which is the easiest.

B.14 Proposition. Let M be a graded R-module. If ι : M E and ι′ : M E ′ areinjective envelopes, then there exists an isomorphism γ : E→ E ′ with γι= ι′.

PROOF. By graded-injectivity of E ′ there is a morphism γ : E→ E ′ with γι = ι′;see 5.3.6. Since one has Kerγ∩ Im ι = 0 and Im ι is essential in E, the map γ isinjective. In particular, Imγ ∼= E is graded-injective, whence there is an equality


Minimality 443

E ′ = Imγ⊕C by 5.3.6. As Im ι′ is contained in Imγ one has Im ι′ ∩C = 0; conse-quently, C is zero as Im ι′ is essential in E ′. It follows that γ is surjective and hencean isomorphism.

B.15 Theorem. Every graded R-module has an injective envelope.

PROOF. Let M be a graded R-module. By 5.3.4 there exists an injective morphismof graded R-modules ι : M→ I, where I is graded-injective. The set of graded sub-modules of I that contains ι(M) as an essential submodule is inductively ordered byinclusion. Hence, by Zorn’s lemma, it has a maximal element E. It remains to provethat E is graded-injective. By another application of Zorn’s lemma, choose a gradedsubmodule Z of I maximal with the property Z∩E = 0. Denote by ζ the morphismE I I/Z. It is sufficient to prove that ζ is an isomorphism. Indeed, the com-posite of I I/Z and ζ−1 : I/Z→ E will then be a left inverse of the embeddingE I. As I is graded-injective, so is the graded direct summand E.

To prove that ζ is an isomorphism, notice first that it is injective as one hasKerζ = Z∩E = 0. It follows from maximality of Z that Imζ is essential in I/Z. Bythe lifting property of the graded-injective module I there is a morphism α : I/Z→ Isuch that αζ : E→ I is the inclusion; see 5.3.6. As Kerα∩ Imζ = 0 and Imζ isessential in I/Z, it follows that α is injective. Thus B.11 yields that αζ(E) = E is es-sential in α(I/Z), whence ι(M) is essential in α(I/Z). By maximality of E one getsα(I/Z) = E, and thus α : I/Z→ E satisfies αζ = 1E . It follows that the essentialsubmodule Imζ is a direct summand of I/Z, so one has Imζ = I/Z.

B.16 Definition. Let M be a graded R-module. The injective module in an injectiveenvelope of M, which by B.14 is unique up to isomorphism, is denoted ER(M).

REMARK. In spite of what the notation might suggest, ER(M) is not natural in M; that is, theassignment of injective envelopes is not a functor. The obstruction is the non-uniqueness of lifts ofmorphisms M→ N to ER(M)→ ER(N); Goodearl makes it explicit in [30, 1.12].

B.17 Corollary. Let I be a graded-injective R-module. For every graded submoduleZ of I there exist graded-injective submodules E and V of I, such that Z is essentialin E and there is an equality I = E⊕V of graded R-modules. In particular, E is aninjective envelope of Z.

PROOF. By B.15 the graded module Z has an injective envelope ι : Z E ′. By thelifting property of I there is a morphism α : E ′→ I such that αι : Z→ I is the inclu-sion; see 5.3.6. As one has Kerα∩ Im ι= 0 and Im ι is essential in E ′, it follows thatα is injective. Set E = Imα; note that E ∼= E ′ is graded-injective and hence a directsummand of I by 5.3.6. It follows from B.11 that αι(Z) = Z is essential in E, so theassertion holds with V = I/E.

MINIMAL COMPLEXES OF INJECTIVE MODULES

For a complex of injective modules, minimality means that the subcomplex of cyclesis essential.


444 Appendix B

B.18 Lemma. Let I be a complex of injective R-modules. If the graded submoduleZ(I)\ of I\ is essential, then I is minimal.

PROOF. Assume that Z(I)\ is essential in I\. Let ε : I→ I be an endomorphism ho-motopic to 1I ; to prove that I is minimal, it suffices by B.6 to show that ε is anisomorphism. By assumption there exists a homomorphism σ : I→ I of degree 1with 1I−ε= ∂Iσ+σ∂I . Set X = Z(I)∩Kerε; for every x in X one has x = ∂Iσ(x)and, therefore, σ(X)∩Z(I) = 0. As Z(I)\ is essential in I\, it follows that σ(X) is 0,and then X is zero. From the definition of X it follows that ε is injective.

The exact sequence 0 −→ I ε−→ I −→ Cokerε −→ 0 is degreewise split by 5.3.6.Since ε is homotopic to 1I , it is a homotopy equivalence, and it follows from B.1that the sequence is split in C(R). Let % : I→ I be a morphism such that %ε = 1I .It follows that % is homotopic to 1I , so by the argument above % is injective and,therefore, it is an isomorphism. Hence, also ε is an isomorphism.

B.19 Theorem. Let I be a complex of injective R-modules. There is an equalityI = I′⊕ I′′, where I′ and I′′ are complexes of injective R-modules, I′ is minimal, andI′′ is contractible. Moreover, the following assertions hold.

(a) The complex I′ is unique in the following sense: if one has I = J′⊕J′′, whereJ′ is minimal and J′′ is contractible, then J′ is isomorphic to I′.

(b) I is minimal if and only if Z(I)\ is essential in I\.(c) If I is semi-injective, then I′ and I′′ are semi-injective.

PROOF. By B.17 there is an equality I\ = E⊕V of graded R-modules, where Z(I)\

is essential in E. As one has V ∩ Z(I)\ = 0, the differential induces an isomor-phism V ∼= Σ1 ∂I(V ), in particular ∂I(V ) is a graded-injective R-module. As ∂I(V )is contained in Z(I)\ and hence in E, there is a graded R-module U such thatE = U ⊕ ∂I(V ); see 5.3.6. The differential ∂I restricts to a homomorphism fromthe graded-injective module V ⊕ ∂I(V ) to itself. Denote by I′′ the subcomplex of Igiven by V ⊕ ∂I(V ) and notice that it is contractible with contraction given by theinverse of the isomorphism ∂ : V → Σ∂I(V ). Now it follows from B.1 that there is anequality I = I′⊕ I′′in C(R), where I′ is isomorphic to the quotient complex I = I/I′′.Since I is a complex of injective modules, so is I′. To prove that the complex I′ ∼= Iis minimal, it suffices by B.18 to prove that Z(I)\ is essential in I\. Denote by π thecanonical map I I and note that its restriction to U is injective. Let x 6= 0 be anelement in I and choose an element u ∈U with π(u) = x. As Z(I)\ is essential inE =U⊕∂I(V ), there exists an element r in R such that ru 6= 0 is in Z(I) and, there-fore rx = π(ru) 6= 0 is in Z(I). Thus Z(I) is essential in I. This proves the existenceof complexes I′ and I′′ with the desired properties. The assertions (a) and (c) followfrom B.7 and 5.3.21, respectively.

(b): The “if” part is B.18. For the converse, assume that I is minimal. By the argu-ments above, there is an equality of R-complexes I = I′⊕ I′′, where I′′ is contractibleand Z(I′)\ is essential in I′\; in particular I′ is minimal. The surjection I I′ is a ho-motopy equivalence by B.1 and hence an isomorphism by B.7. Thus, one has I = I′

so Z(I)\ is essential in I\.


Minimality 445

B.20 Corollary. A complex I of injective R-modules is minimal if and only if thezero complex is the only contractible direct summand of I.

PROOF. The “if” part is immediate from the decomposition I = I′⊕ I′′ from B.19.and the “only if” follows from B.5.

MINIMAL SEMI-INJECTIVE RESOLUTIONS

Every quasi-isomorphism with a semi-injective domain has by 5.3.23 a homotopyleft inverse. Under the additional assumption that the domain complex is minimal,such a quasi-isomorphism has a genuine left inverse.

B.21 Proposition. For an R-complex I, the following conditions are equivalent.(i) I is semi-injective and minimal.

(ii) Every quasi-isomorphism I→M has a left inverse.

PROOF. Assume that I is semi-injective and minimal. If β : I→M is a quasi-isomorphism, then there exists by 5.3.23 a morphism γ : M→ I with γβ ∼ 1I . Setε= γβ; by assumption ε has an inverse, so ε−1γ is a left inverse for β.

Assume that every quasi-isomorphism I → M has a left inverse. In particular,every homotopy equivalence I→ I has a left inverse and hence I is minimal by B.6.It follows from 5.3.16 that I is semi-injective.

B.22 Corollary. A semi-injective R-complex I is minimal if and only if the zero-complex is the only acyclic subcomplex of I.

PROOF. A semi-injective complex I with no non-zero acyclic subcomplex is mini-mal per B.20. On the other hand, if A is an acyclic subcomplex of I and I is minimal,then the quasi-isomorphism π : I I/A has a left inverse by B.21, so π is an isomor-phism and A = 0.

B.23 Definition. Let M be an R-complex. A semi-injective resolution M '−→ I iscalled minimal if the semi-injective complex I is minimal.

B.24 Theorem. Every R-complex M has a minimal semi-injective resolution M '−→E; the complex E is unique up to isomorphism and has the next properties.

(a) Ev = 0 holds for all v > supM.(b) For every semi-injective resolution M '−→ I the complex E is isomorphic to a

direct summand of I.

PROOF. By 5.3.26 there is a semi-injective resolution ι : M '−→ I with Iv = 0 forv > supM. By B.19 one has I = E⊕ I′′, where E is semi-injective and minimal, andI′′ is contractible. Let π be the projection I E; by 4.2.6 it is a quasi-isomorphism,as I′′ is acyclic. Now the composite πι : M '−→ E is a minimal semi-injective resolu-tion with Ev = 0 for all v > supM. This proves existence and part (a).


446 Appendix B

Let M '−→ I be any semi-injective resolution. It follows from 5.3.22 that thereis a quasi-isomorphism α : E→ I, and by B.21 it has left inverse β. Thus E is iso-morphic to a direct summand of I; this proves part (b). The morphism β is a quasi-isomorphism, so if I is minimal, then β has a left inverse by the same argument, andthen it is an isomorphism with β−1 = α. This proves the uniqueness statement.

B.25 Example. If M is an acyclic R-complex, then M '−→ 0 is the minimal semi-injective resolution; the morphism is only injective if M is the zero complex.

NAKAYAMA’S LEMMA

Intersection and sum of submodules are categorically dual notions. Indeed, if M andN are submodules of same module, then the commutative diagram

M∩N

// // N

M // // N +M .

is both a pushout and a pullback square. Thus, the following notion of a superfluoussubmodule is dual to that of an essential submodule from B.8.

B.26 Definition. Let M be a graded R-module. A graded submodule N of M iscalled superfluous if N +M′ 6= M holds for every graded submodule M′ 6= M of M.

REMARK. Another word for superfluous submodule is small submodule.

B.27 Example. The only superfluous ideal in Z is 0. The maximal ideal 2Z/4Z issuperfluous in Z/4Z.

B.28. A graded direct summand N of a graded R-module M is superfluous if andonly if N = 0.

B.29 Lemma. Let M be a graded R-module with graded submodules N and N′.(a) Assume that there is an inclusion N ⊆ N′. If N is superfluous in N′, then N is

superfluous in M.(b) If N and N′ are superfluous in M, then N +N′ is superfluous in M.(c) If α : M→ X is a homomorphism of graded R-modules and N is superfluous

in M, then α(N) is superfluous in X .

PROOF. (a): If M′ is a graded submodule of M such that N +M′ = M holds, thenone has N′ = (N +M′)∩N′ = N +(M′ ∩N′). Thus, if N is superfluous in N′, thenone has M′∩N′ = N′ and, therefore, N′ ⊆M′. In particular, N is then a submoduleof M′, whence one has M′ = N +M′ = M.

(b): If M′ is a graded submodule of M such that (N +N′)+M′ = N +(N′+M′)is M, then one has N′+M′ = M because N is superfluous in M, and then M = M′

because N′ is superfluous in M as well.


Minimality 447

(c): By part (a) it suffices to show that α(N) is superfluous in α(M), so assumewithout loss of generality that α is surjective. If X ′ is a submodule of X such thatα(N)+X ′ = X holds, then one has N +α−1(X ′) = M and, therefore, α−1(X ′) = Mas N is superfluous in M. Thus, one has X ′ = α(α−1(X ′)) = α(M) = X .

The next result is known as Nakayama’s lemma.

B.30 Lemma. Let J denote the Jacobson radical of R. For a left ideal a in R thefollowing conditions are equivalent.

(i) The left ideal a is a superfluous submodule of R.(ii) There is an inclusion a⊆ J.

(iii) For every finitely generated R-module M 6= 0 one has aM 6= M.(iv) For every graded R-module M and every graded submodule N ⊆M such that

the quotient (M/N)v is non-zero and finitely generated for some v ∈ Z, onehas N +aM 6= M.

(v) For every degreewise finitely generated graded R-module M, the submoduleaM is superfluous.

PROOF. Condition (i) is a special case of (v).(i)=⇒ (ii): If a is not contained in J, then one has a*M for at least one maximal

left ideal M. Thus one has a+M= R, and hence a can not be superfluous in R.(ii)=⇒ (iii): Let M 6= 0 be a finitely generated R-module and choose a set of

generators m1, . . . ,mt for M with t least possible. Assume towards a contradictionthat one has aM = M. Then there exist elements a1, . . . ,at in a such that m1 =

∑ti=1 aimi holds. Since a1 is in J, the element 1− a1 is invertible, whence m1 is a

linear combination of m2, . . . ,mt , which contradicts the minimality of t.(iii)=⇒ (iv): Assume that (M/N)v is non-zero and finitely generated. By (iii) one

has a(M/N)v 6= (M/N)v and, therefore (N +aM)v 6= Mv.(iv)=⇒ (v): For every proper graded submodule N ⊂M it follows from (iv) that

N +aM 6= M holds. Thus, aM is superfluous in M.

PROJECTIVE COVERS

B.31 Definition. A projective cover of a graded R-module M is a surjective mor-phism π : P→M of graded R-modules where P is graded-projective and Kerπ issuperfluous in P.

B.32 Example. Let P be a graded-projective R-module. The identity morphism1P is a projective cover of P. Moreover, if P is degreewise finitely generated, thenit follows from Nakayama’s lemma B.30 that the canonical map P P/aP is aprojective cover for every left ideal a contained in the Jacobson radical of R.

The notion of a projective cover is dual to that of an injective envelope. As estab-lished in B.15, injective envelopes exist for all graded modules, however, projectivecovers do not always exist. In fact, a ring over which every graded module has aprojective cover is perfect and vice versa; see B.51.


448 Appendix B

B.33 Lemma. Let M be a graded R-module with a projective cover π : PM. Letπ′ : P′→M be a surjective morphism with P′ graded-projective. There is a mor-phism γ : P→ P′ with π= π′γ, and for every such morphism the following hold.

(a) The morphism γ has a left inverse.(b) If π′ is a projective cover, then γ is an isomorphism.

For the direct summand P′′ = γ(P) of P′, the restriction π′|P′′ : P′′M is a pro-jective cover, and there is an equality of graded modules, P′ = P′′⊕K, where K iscontained in Kerπ′.

PROOF. By 5.2.2 there exist morphisms γ : P→ P′ and γ′ : P′→ P with π= π′γ andπ′ = πγ′. Set ε= γ′γ. Then one has πε= π, so there is an equality P = ε(P)+Kerπ.As Kerπ is superfluous in P, it follows that ε is surjective. Since P is graded-projective, Kerε is a direct summand in P; again by 5.2.2. By B.29, the moduleKerε is superfluous as it is contained in Kerπ, whence it is zero. Thus, ε is an iso-morphism, and ε−1γ′ is a left inverse of γ. This proves part (a), and with P′′ = γ(P)and K = Kerγ′ it follows that there is an equality P′ = P′′ ⊕K. The submoduleP′′ ∩Kerπ′ is superfluous in P′′, because the isomorphism γ′|P′′ : P′′→ P maps itto Kerπ, which is superfluous in P. Thus, π′|P′′ is a projective cover. Moreover, theequality π′ = πγ′ implies that K is contained in Kerπ′.

To prove part (b), note that one has P′′+Kerπ′ = P′. Thus, if π′ is a projectivecover, then P′ = P′′ holds, whence γ is an isomorphism.

As already mentioned, a module might not have a projective cover, however, anytwo covers of a given module are necessarily isomorphic.

B.34 Proposition. Let M be a graded R-module. If π : PM and π′ : P′M areprojective covers, then there exists an isomorphism γ : P→ P′ with π′γ = π. More-over, if M is degreewise finitely generated, then so are P and P′.

PROOF. The existence of an isomorphism γ with π′γ = π is part of B.33. It alsofollows from B.33 that if PM is a projective cover and P′→M is any surjectivemorphism with P′ graded-projective, then P is isomorphic to a graded direct sum-mand in P′. If M is degreewise finitely generated, then P′ can be chosen degreewisefinitely generated by 2.5.23, and hence the direct summand P is degreewise finitelygenerated as well.

B.35 Lemma. Let J denote the Jacobson radical of R and let P 6= 0 be a graded-projective R-module. One has JP 6= P, and every superfluous submodule of P iscontained in JP. In particular, if M is a graded R-module and π : PM is a projec-tive cover, then the induced morphism π : P/JP→M/JM is an isomorphism.

PROOF. By 5.2.2 the module P is a graded direct summand of a graded free R-module L. Let E be a graded basis for L. Fix a homogeneous element p 6= 0 in P;it is a unique linear combination of some basis elements, p = ∑

mi=1 riei. Let ε be

the composition of canonical morphisms L P L. For each i ∈ 1, . . . ,m writeε(ei) = ∑

nj=1 ai je j, where also em+1, . . . ,en are elements in E. Suppose the equality

P = JP holds, then all the elements ai j belong to J. The equality p = ε(p) yields


Minimality 449

(?)m∑

i=1riei =

m∑

i=1ri

n∑j=1

ai je j .

Let A be the m×m matrix with entries ai j for 1 6 i, j 6 m, and let Im denote them×m identity matrix. From (?) one gets the equality (r1, . . . ,rm)(Im−A) = 0. It iselementary to verify that the Jacobson radical of the matrix ring Mm×m(R) contains(in fact, equality holds) the ideal Mm×m(J). Since A has entries in J, it follows thatIm−A is invertible in Mm×m(R). Hence (r1, . . . ,rm) is the zero row and one hasp = 0, a contradiction.

Let N be a superfluous graded submodule of P and thereby of L; cf. B.29. Itfollows that N is contained in every maximal submodule of L. In particular, N iscontained in the module Je′+R〈E \e′〉 for every e′ ∈ E. Indeed, this is the inter-section of the maximal submodules Me′+R〈E \e′〉, where M is a maximal leftideal in R. Thus, for an element x = ∑e∈E ree in N one has re ∈ J for all e ∈ E. Itfollows that N is contained in JL∩P = JP, where the equality holds because P is adirect summand of L.

The last assertion is now immediate as the kernel of a cover PM is superfluousin P and hence contained in JP.

REMARK. For a graded-projective R-module P, the submodule JP itself may not be superfluous;see E B.15.

SEMI-PERFECT MODULES

B.36 Definition. A graded R-module M is called semi-perfect if every homomorphicimage of M has a projective cover.

The notion of semi-perfect modules is an auxiliary; its utility comes to fore inTheorems B.44 and B.51 and, implicitly, in Theorems B.58 and B.59.

B.37 Example. If R is semi-simple, then every R-module is projective by 1.3.24,and it follows that every R-module is semi-perfect.

The Z-module Z has a projective cover, namely 1Z, but it is not semi-perfect,since the quotient Z/nZ has no projective cover for n > 1. Indeed, suppose thatπ : P Z/nZ is a projective cover and let π′ : Z Z/nZ be the canonical map. ByB.33 there is an isomorphism of Z-modules, Z ∼= P⊕K, and since Z is indecom-posable, it follows that K = 0, P = Z, and π = π′. However, π′ : Z Z/nZ is not aprojective cover since its kernel is not superfluous; see B.27.

B.38 Lemma. Let J denote the Jacobson radical of R and let M be a graded R-module. If M is semi-perfect, then the following assertions hold.

(a) If M is non-zero, then JM 6= M holds.(b) The submodule JM is superfluous in M.(c) If PM is a projective cover, then P is semi-perfect.(d) The graded R/J-module M/JM is semi-simple.


450 Appendix B

Let α : L→M be a morphism of graded R-modules and denote by α the inducedmorphism L/JL→M/JM of R/J-modules.

(e) If α is surjective, then α is surjective.(f) If α is bijective and L and M are graded-projective, then α is bijective.

PROOF. Let π : PM be a projective cover.(a): If JM = M holds, then one has π(JP) = M, whence there is an equality

Kerπ+JP = P. As P is graded-projective and Kerπ is superfluous in P, it followsfrom B.35 that P is zero, whence M = 0.

(b): Let M′ be a graded submodule of M such that the equality JM +M′ = Mholds. The quotient module N = M/M′ then satisfies JN = N. By assumption themodule N is semi-perfect, so part (a) yields N = 0, whence one has M′ = M.

(c): To prove that P is semi-perfect, it must be show that P/N has a projec-tive cover for every graded submodule N of P. Let N be such a submodule andset K = Kerπ, then P/(N + K) is a homomorphic image of P/K ∼= M, so thereis a projective cover π′ : P′ P/(N +K). As P′ is graded-projective, there existsby 5.2.2 a morphism γ : P′→ P/N with π′ = βγ, where β is the canonical mor-phism P/N P/(N + K). To see that γ is a projective cover, notice first thatKerβ = (N +K)/N is a superfluous submodule of P/N, because it is the imageof the superfluous submodule K of P; see B.29. On the other hand, since π′ is sur-jective, one has γ(P′)+Kerβ = P/N, so γ is surjective. Finally, Kerγ is containedin Kerπ′, which is superfluous in P′.

(d): The morphism of graded R/J-modules P/JP→ M/JM, induced by π, issurjective; it is, therefore, sufficient to prove that P/JP is semi-simple. To that end,let X/JP be a proper graded submodule of P/JP; the goal is to show that thissubmodule is a graded direct summand. The module P/X has a projective coverκ : F P/X by part (c). Let β be the canonical morphism P P/X . By B.33 thereis a morphism γ : F → P with κ = βγ and P = γ(F)⊕K, where K is contained inKerβ = X and the restriction of β to γ(F) is a projective cover. Replacing κ withβ|γ(F) one has P = F⊕K and, therefore,

(?)PJP

=F +JPJP

+XJP

.

The submodule F ∩ X is contained in Kerκ, so it is superfluous in F and hencecontained in JF ⊆ JP by B.35. Thus, the sum in (?) is direct; indeed, one has

F +JPJP

∩ XJP

=(F +JP)∩X

JP=

(F ∩X)+JPJP

= 0 .

(e): Set C = Cokerα and consider the exact sequence L α−→ M −→ C −→ 0 ofgraded R-modules. It induces an exact sequence L/JL α−→ M/JM −→ C/JC −→ 0with C/JC = 0 by assumption. The graded module C is semi-perfect, because it isa homomorphic image of M. Thus, part (a) yields C = 0, whence α is surjective.

(f): It follows from part (e) that α is surjective. Set K = Kerα and consider theexact sequence 0 −→ K −→ L α−→M −→ 0 of graded R-modules. By assumption, Land M are graded-projective, so the sequence is split and K is graded-projective by


Minimality 451

5.2.2 and 5.2.3. Now the induced sequence 0−→ K/JK −→ L/JL α−→M/JM −→ 0is exact. By assumption α is injective, so one has K/JK = 0. Lemma B.35 nowyields K = 0, so α is injective.

B.39 Lemma. Let J denote the Jacobson radical of R and set k = R/J. Considerevery k-module as an R-module via the canonical homomorphism κ : R k.

(a) Let e ∈ R be an idempotent and set u = κ(e). The map κu : Re ku given byre 7→ κ(r)u is a projective cover of the R-module ku, and one has Kerκu = Je.

(b) If a graded k-module generated by a homogeneous element u has a projectivecover as a graded R-module, then it has one of the form κu : Σ|u|Reu ku,where eu is an idempotent in R.

(c) Let M be a graded R-module such that the graded k-module M/JM is semi-simple. Let U be a set of homogeneous elements with M/JM =

∐u∈U ku.

Assume that each R-module ku has a projective cover κu : Σ|u|Reu ku witheu an idempotent in R. The R-module F =

∐u∈U Σ

|u|Reu is graded-projective,and the morphism κ=

∐u∈U κ

u : F →M/JM is surjective with Kerκ= JF .

PROOF. (a): Let e be an idempotent in R. The ideal Re is a projective R-module asone has R = Re⊕R(1− e). The inclusion Je ⊆ Kerκu holds by the definition ofκu. To prove the reverse inclusion, let re be an element in Kerκu. The equalities0 = κ(r)u = κ(re) in ku show that re is in J, and since e is an idempotent one hasre = (re)e ∈ Je. By Nakayama’s lemma B.30, the Jacobson radical J is superfluousin R, so Je = Kerκu is superfluous in Re by B.29. Thus, κu is a projective cover.

(b): Let π : P ku be a projective cover of ku as a graded R-module. SetP′ = Σ|u|R and let p denote the generator 1 in P′. Let π′ : P′→ ku be the surjec-tive morphism of graded R-modules that maps p to u. By B.33 there is a morphismγ : P→ P′ with π= π′γ, P′′ = γ(P), and P′ = P′′⊕K, such that κu = π′|P′′ : P′′ kuis a projective cover. Restrict the codomain of γ so that it becomes an isomorphismγ : P→ P′′ and let υ : P′→ P denote the canonical morphism with υγ= 1P. One hasγυ(p) = ep for some e ∈ R. The identity γυ= γ1Pυ= (γυ)(γυ) yields ep = e2 p, soe is an idempotent. The equality R = Re⊕R(1−e) now yields P′′ = Σ|u|Re, so κu isthe desired projective cover.

(c): Each graded R-module Σ|u|Reu is graded-projective as eu is an idempotent.It follows from 5.2.12 and 5.2.17 that F is graded-projective, and κ is surjective byconstruction. By 3.1.5 and (a) one has Kerκ=

∐u∈U Kerκu =

∐u∈U Jeu = JF .

B.40 Theorem. Let M be a semi-perfect graded R-module. If PM is a projectivecover, then P is isomorphic to a module of the form

∐u∈U Σ

nu Reu, where each eu isan idempotent in R.

PROOF. Let J denote the Jacobson radical of R and set k = R/J. By parts (c) and(d) in B.38 the graded R-module P is semi-perfect, and the k-module P/JP is semi-simple. By a standard application of Zorn’s lemma, choose a set U of homogeneouselements with P/JP =

∐u∈U ku. Each graded module ku is a homomorphic image

of P and, therefore, it has a projective cover. By B.39 it then has a projective coverof the form Σ|u|Reu ku, where eu is an idempotent in R. Now it follows, still from


452 Appendix B

B.39, that there is a surjective morphism κ : F → P/JP with Kerκ = JF , where Fis the graded-projective R-module

∐u∈U Σ

|u|Reu. By graded-projectivity of F , thereis a morphism γ : F → P such that the composite F

γ−→ P P/JP equals κ; see5.2.2. The induced morphism κ : F/JF → P/JP is an isomorphism; it follows thatalso γ : F/JF → P/JP is an isomorphism, so γ is an isomorphism by B.38(f).

SEMI-PERFECT RINGS

B.41 Definition. Let J denote the Jacobson radical of R. The ring R is called semi-perfect if R/J is semi-simple and idempotents lift from R/J to R.

REMARK. Because semi-simplicity is a left/right symmetric property so is semi-perfection. A ringR with R/J semi-simple is sometimes called semi-local. A commutative ring is semi-local if andonly it it has finitely many maximal ideals.

B.42 Example. If R is local with unique maximal ideal m, then k=R/m is a divisionring, so 1 and 0 are the only idempotents in k. Moreover, k is simple, so R is semi-perfect.

If R is left (or right) Artinian with Jacobson radical J, then R/J is semi-simple.Moreover, if r is in R and [r]J is an idempotent, then r− r2 is in J. Since J isnilpotent, there is an integer n > 1 with (r− r2)n = 0. Powers of r commute, soone has 0 = (r(1− r))n = rn(1− r)n = rn− rn+1x for an element x with rx = xr. Itfollows that one has (rx)n = rnxn = rn+1xn+1 = (rx)n+1, so the element e = (rx)n isan idempotent in R. One has [r]J = [rn]J = [rn+1x]J = [rn+1]J[x]J = [rx]J and hence[e]J = [rx]nJ = [r]nJ = [r]J. Thus R is semi-perfect.

REMARK. A ring R is semi-perfect with R/J simple if and only if it is isomorphic to a matrix ringMn×n(S) where S is local; see [51, thm. (23.10)]. A commutative ring is semi-perfect if an only ifit is a finite product of commutative local rings; see [51, thm. (23.11)].

B.43 Lemma. Let J denote the Jacobson radical of R and let M be a graded R-module. If R is semi-perfect, then there exists a graded-projective R-module F and asurjective morphism κ : F →M/JM with Kerκ= JF . Moreover, if M is degreewisefinitely generated, then one can choose F degreewise finitely generated.

PROOF. Set k = R/J and consider the graded k-module M/JM. By assumption, kis semi-simple, so by a standard application of Zorn’s lemma one can choose a setU of homogeneous elements with M/JM =

∐u∈U Σ

|u| ku. As every cyclic k-moduleis a direct sum of simple modules generated by idempotents, one can assume thateach u is an idempotent in k. By assumption, each u now lifts to an idempotent eu inR, so by B.39 the canonical surjection κu : Reu→ ku is a projective cover of ku, andthe desired morphism is κ=

∐u∈U κ

u. Finally, if M is degreewise finitely generated,then one can choose U with only finitely many elements of each degree.

We can now link semi-perfectness of rings to existence of projective covers.

B.44 Theorem. The following conditions are equivalent.


Minimality 453

(i) The ring R is semi-perfect.(ii) Every degreewise finitely generated graded R-module has a projective cover.

(iii) Every degreewise finitely generated graded R-module is semi-perfect.(iv) The R-module R is semi-perfect.

Moreover, if R is semi-perfect and P M is a projective cover of a degreewisefinitely generated graded R-module, then P is degreewise finitely generated.

PROOF. Let J denote the Jacobson radical of R and set k = R/J. The implication(ii)=⇒ (iii) follows from the definition of semi-perfect modules B.36, and (iv) is aspecial case of (iii).

(i)=⇒ (ii): Let M be a degreewise finitely generated graded R-module. By B.43there exists a graded-projective and degreewise finitely generated R-module F andsurjective morphism κ : F →M/JM with Kerκ = JF . Let β be the canonical mor-phism MM/JM; by 5.2.2 there exists a morphism γ : F →M with βγ= κ. Thus,one has M = γ(F)+JM. It follows from Nakayama’s lemma B.30 that JM is super-fluous in M, so γ is surjective. To see that γ is a projective cover, it remains to verifythat Kerγ is superfluous in F . This follows as Kerγ is contained in Kerκ = JF ,which is a superfluous submodule by Nakayama’s lemma. As a projective cover isunique up to isomorphism, see B.34, the final assertion in the theorem also follows.

(iv)=⇒ (i): By B.38(d) the k-module k is semi-simple, so k is a semi-simple ring.From B.38(b) it follows that J is superfluous in R, whence the canonical map R kis a projective cover. Let u be an idempotent in k; the goal is to lift u to R. There isan equality k = ku⊕ k(1− u). As R is a semi-perfect R-module, its homomorphicimages ku and k(1− u) have projective covers π : P ku and π′ : P′ k(1−u).The morphism π⊕ π′ is a projective cover of k. By B.33 there is an isomorphismγ : P⊕P′→ R such that γ followed by the canonical map R k is π⊕π′. Thus, themodules P and P′ are isomorphic to ideals e and e′ in R, and one has R = e⊕ e′.Choose elements e ∈ e and e′ ∈ e′ with 1 = e+e′ in R. It is elementary to verify thate and e′ are orthogonal idempotents and that e maps to u in k.

REMARK. Every finitely generated flat module over a semi-perfect ring is projective; see E B.19.It is a result of Kaplansky [45] that the finiteness hypothesis in the next corollary can be omitted.

B.45 Corollary. Let R be local. Every degreewise finitely generated graded-projectiveR-module is graded-free.

PROOF. Let P be a degreewise finitely generated and graded-projective R-module.The identity 1P is a projective cover of P. By B.44 the graded module P is semi-perfect, so by B.40 it has the form

∐u∈U Σ

nu Reu, where each eu is an idempotent inR. As R is local, 1 and 0 are the only idempotents in R, so P is graded-free.

PERFECT RINGS

B.46 Definition. A left ideal a in R is called left T-nilpotent if for every sequenceof elements (ai)i∈N in a there is an integer n> 1 such that a1a2 · · ·an = 0 holds.


454 Appendix B

Every nilpotent left ideal is left T-nilpotent; in particular the Jacobson radical ofa left (or right) Artinian ring is left T-nilpotent. A T-nilpotent ideal has propertiessimilar to the Jacobson radical as captured by Nakayama’s lemma.

B.47 Lemma. For a left ideal a in R, the following conditions are equivalent.(i) The left ideal a is left T-nilpotent.

(ii) For every Ro-module N 6= 0 one has (0 :N a) 6= 0.(iii) For every R-module M 6= 0 one has aM 6= M.(iv) For every graded R-module M and every proper graded submodule M′ ⊂ M

one has M′+aM 6= M.(v) For every graded R-module M the submodule aM is superfluous.

PROOF. (i)=⇒ (ii): Assume that there exists an Ro-module N 6= 0 with (0 :N a) = 0.For every element x 6= 0 in N there exists then an element a ∈ a with xa 6= 0. Itfollows by induction that there exists a sequence (ai)i∈N in a with a1a2 · · ·an 6= 0 forevery n ∈ N. Thus, a is not left T-nilpotent.

(ii)=⇒ (iii): Let M 6= 0 be an R-module; consider the ideal b= (0 :R M) and theright ideal B= (b :R a) = r ∈ R | ra⊆ b. Now (ii) yields B/b= (0 :R/b a) 6= 0.It follows that b is strictly contained in B, whence BM is non-zero. However, onehas Ba⊆ b and, therefore, B(aM) = 0, so aM 6= M holds.

(iii)=⇒ (iv): Apply (iii) to the non-zero R-module M/M′.(iv)=⇒ (v): Immediate from the definition B.26 of superfluous submodules.(v)=⇒ (i): Let a sequence (ai)i∈N of elements in a be given. For j > i > 1 let

α ji be the homothety given by right multiplication on R with ai · · ·a j−1 and setαii = 1R. These maps form a direct system of R-modules; let A denote its colimit.For every element a in A there is a ring element r and an integer i> 1 with a= αi(r),where αi is the canonical morphism R→ A; see 3.2.26. From the equalities αi(r) =αi+1(rai) = raiα

i+1(1) one gets aA = A. By (v) the submodule aA is superfluousin A, so the colimit A is zero. In particular, α1(1) is zero, so by 3.2.26 one hasα j1(1) = a1 · · ·a j−1 = 0 some j > 1. Thus, a is left T-nilpotent.

B.48 Definition. Let J be the Jacobson radical of R. The ring R is called left perfectif R/J is semi-simple and J is left T -nilpotent.

B.49 Example. As a nilpotent ideal is left T-nilpotent, every left (or right) Artinianring is left perfect.

REMARK. Perfectness is not a left/right symmetric property; Bass gives an example in [10]. SeeE 5.4.19 for an example of a perfect ring that is not Artinian.

B.50. If the Jacobson radical J of R is left T-nilpotent, then every element in J isnilpotent, whence idempotents lift from R/J to R; see B.42. Thus, every left perfectring is semi-perfect.

B.51 Theorem. The following conditions are equivalent.(i) The ring R is left perfect.

(ii) Every graded R-module has a projective cover.


Minimality 455

(iii) Every graded R-module is semi-perfect.

PROOF. Let J denote the Jacobson radical of R and set k = R/J. The implication(ii)=⇒ (iii) follows from the definition B.36 of semi-perfect modules.

(i)=⇒ (ii): Let M be a graded R-module. Since R is semi-perfect, it follows fromB.43 that there exists a graded-projective R-module F and a surjective morphismκ : F →M/JM with Kerκ= JF . Let β be the canonical morphism MM/JM; by5.2.2 there is a morphism γ : F →M with βγ= κ. Thus, one has M = γ(F)+JM. Itfollows from B.47 that JM is superfluous in M, so γ is surjective. To see that γ is aprojective cover, it remains to verify that Kerγ is superfluous in F . This follows asKerγ is contained in Kerκ= JF , which is a superfluous submodule by B.47.

(iii)=⇒ (i): It follows from B.38(d) that k is semi-simple as a k-module; i.e. it isa semi-simple ring. For every graded R-module M the submodule JM is superfluousin M by B.38(b), so J is left T-nilpotent by B.47. Thus, R is left perfect.

Another homological characterization of perfect rings is given in 5.4.28.

MINIMAL COMPLEXES OF PROJECTIVE MODULES

For a complex of projective modules, minimality means that the subcomplex ofboundaries is superfluous.

B.52 Lemma. Let J be the Jacobson radical of R and let P be a complex of projec-tive R-modules. If P\ is semi-perfect and one has ∂P(P)⊆ JP, then P is minimal.

PROOF. Let ε : P→ P be a morphism with ε∼ 1P; it suffices by B.6 to prove that ε isan isomorphism. By assumption there exists a homomorphism σ : P→ P of degree1 such that 1P−ε= ∂Pσ+σ∂P holds. For every p∈P, the element p−ε(p) belongsto JP+σ(JP) = JP. It follows that the induced morphism ε : P/JP→ P/JP is theidentity 1P/JP, so ε is an isomorphism by B.38(f).

B.53 Theorem. Let J denote the Jacobson radical of R and let P be a complex ofprojective R-modules such that P\ is semi-perfect. There is an equality P = P′⊕P′′,where P′ and P′′ are complexes of projective R-modules, P′ is minimal, and P′′ iscontractible. Moreover, the following assertions hold.

(a) The complex P′ is unique in the following sense: if one has P = F ′⊕F ′′,where F ′ is minimal and F ′′ is contractible, then F ′ is isomorphic to P′.

(b) P is minimal if and only if B(P)\ is superfluous in P\ if and only if theinclusion ∂P(P)⊆ JP holds.

(c) If P is semi-projective, then P′ and P′′ are semi-projective.

PROOF. Set k = R/J; the graded k-module (P/JP)\ is semi-simple by B.38(d). SetB = B(P/JP) and H = H(P/JP); by 4.2.16 there is a split exact sequence of k-complexes,

(?) 0−→ H −→ P/JP τ−−→ Cone1B −→ 0 .


456 Appendix B

Because it is a homomorphic image of the semi-perfect graded R-module P\, thegraded module B\ is semi-perfect. In particular, B\ has a projective cover κ : F B\.

Set P′′ = Cone1F and C = Cone1B; both complexes are contractible by 4.3.31.The projective cover κ induces a surjective morphism χ= κ⊕Σκ : P′′→C. It iselementary to verify that χ : P′′\C\ is a projective cover. By (?) the graded moduleC\ is a homomorphic image of P\, so it is semi-perfect and, therefore, P′′\ is semi-perfect by B.38(c).

Let π denote the canonical map P P/JP. The complexes HomR(P,P′′) andHomR(P,C) are contractible by 4.3.29 so HomR(P,χ) is a quasi-isomorphism. More-over, HomR(P,χ) is surjective by 5.2.2 and hence surjective on cycles; see 4.2.7.Thus, there exists a morphism γ : P→ P′′ with χγ = τπ. Let π′ be the restriction ofπ to the subcomplex P′ = Kerγ and consider the commutative diagram

0 // P′

π′

// P

π

γ// P′′

χ

// 0

0 // H // P/JP τ// C // 0 .

(‡)

The bottom row is the split exact sequence (?), and by construction the top row isexact at P′ and at P. To see that it is exact at P′′, notice that χ : P′′/JP′′→C/JC isbijective by B.35 and π is the identity morphism 1P/JP, while τ = τ is surjective.It follows that γ : P/JP→ P′′/JP′′ is surjective, and then γ is surjective by B.38(f).As P′′\ is a graded-projective R-module, the top row in (‡) is degreewise split by5.2.2 and, therefore, split by B.1 as P′′ is contractible. Thus, one has P = P′⊕P′′.

To see that P′ is minimal, note that (‡) now yields a commutative diagram,

0 // P′/JP′

π′

// P/JP

1P/JP

γ// P′′/JP′′

χ∼=

// 0

0 // H // P/JP τ// C // 0 ,

with exact rows. It follows from the Five Lemma that π′ is an isomorphism. Thedifferential on H is zero, so ∂P′(P′) is contained in JP′. As the top row in (‡) issplit, the graded-projective R-module P′\ is a homomorphic image of P\ and hencesemi-perfect. It follows from B.52 that P′ is a minimal complex, which finishes theproof of the first assertion. Parts (a) and (c) follow from B.7 and 5.2.17, respectively.

(b): Since P\ is graded-projective and semi-perfect, it follows from B.35 andB.38(b) that B(P)\ = (∂P(P))\ is superfluous in P\ if and only if the inclusion∂P(P)⊆ JP holds. In view of B.52 it remains to prove that ∂P(P)⊆ JP holds if P isminimal. Assume that P is minimal; by the arguments above one has P = P′⊕P′′,where P′′ is contractible and ∂P′(P′) is contained in JP′. The surjection P→ P′ isa homotopy equivalence by B.1 and hence an isomorphism by B.7. Thus ∂P(P) iscontained in JP.


Minimality 457

MINIMAL SEMI-PROJECTIVE RESOLUTIONS

Every quasi-isomorphism with a semi-projective codomain has by 5.2.19 a homo-topy right inverse. Under the additional assumption that the codomain complex isminimal, such a quasi-isomorphism has a genuine right inverse.

B.54 Proposition. For an R-complex P, the following conditions are equivalent.(i) P is semi-projective and minimal.

(ii) Every quasi-isomorphism M→ P has a right inverse.

PROOF. Assume that P is semi-projective and minimal. If α : M→ P is a quasi-isomorphism, then there exists by 5.2.19 a morphism γ : P→M with αγ ∼ 1P. Setε= αγ; by assumption, ε has an inverse, so γε−1 is a right inverse for α.

Assume that every quasi-isomorphism M→ P has a right inverse. In particular,every homotopy equivalence P→ P has a right inverse and hence P is minimal byB.6. It follows from 5.2.10 that P is semi-projective.

B.55 Definition. Let M be an R-complex. A semi-projective resolution P '−→M iscalled minimal if the semi-projective complex P is minimal.

B.56 Theorem. Let M be an R-complex. If L '−→ M is a minimal semi-projectiveresolution, then L is unique up to isomorphism, and it has the following properties.

(a) Lv = 0 holds for all v < infM.(b) For every semi-projective resolution P '−→M the complex L is isomorphic to

a direct summand of P.

PROOF. Part (a) follows from part (b), as one by 5.2.15 can choose a semi-projectiveresolution P '−→M with Pv = 0 for all v< infM. To prove part (b), let P '−→M be anysemi-projective resolution. It follows from 5.2.18 that there is a quasi-isomorphismα : P→ L, and by B.54 it has right inverse β. Thus L is isomorphic to a direct sum-mand of P. The morphism β is a quasi-isomorphism, so if P is minimal, then β hasa right inverse by the same argument, and then it is an isomorphism with β−1 = α.This proves the uniqueness statement.

B.57 Example. If M is an acyclic R-complex, then 0 '−→ M is the minimal semi-projective resolution; this morphism is only surjective if M is the zero complex.

B.58 Theorem. If R is left perfect, then every R-complex has a minimal semi-projective resolution.

PROOF. Let M be an R-complex, by 5.2.14 there is a semi-projective resolutionπ : P '−→M. By B.51 the graded R-module P\ is semi-perfect, so by B.53 one hasP = L⊕P′′, where L is minimal and semi-projective, and P′′ is contractible. Let ιbe the embedding L P; by 4.2.6 it is a quasi-isomorphism as P′′ is acyclic. Thusπι : P '−→M is the desired resolution.

B.59 Theorem. Let R be left Noetherian and semi-perfect. Every R-complex Mwith H(M) bounded below and degreewise finitely generated has a minimal semi-projective resolution L '−→M with L degreewise finitely generated.


458 Appendix B

PROOF. By 5.1.14 there is a semi-projective resolution π : P '−→M with P degree-wise finitely generated. By B.44 the graded R-module P\ is semi-perfect, so by B.53one has P = L⊕P′′, where L is minimal and semi-projective, and P′′ is contractible.Let ι be the embedding L P; by 4.2.6 it is a quasi-isomorphism as P′′ is acyclic.Thus πι : L '−→M is the desired resolution.

EXERCISES

E B.1 Show that the conclusions in B.1 may fail if the exact sequence is not degreewise split.E B.2 Let R be left Noetherian. Show that every injective R-module has an indecomposable

direct summand.E B.3 Let k be an integral domain with field of fractions Q. Show that the embedding k Q is

an injective envelope of the k-module k.E B.4 Let ι : M I be an injective preenvelope of an R-module. Show that ι is an injective

envelope if and only if every endomorphism γ : I→ I with γι= ι is an automorphism.E B.5 Let ι : M I and ι′ : M′ I′ be injective envelopes of graded R-modules. Show that the

direct sum ι⊕ ι′ : M⊕M′→ I⊕ I′ is an injective envelope.E B.6 Assume that R is left Noetherian and let ιn : Mn Inn∈N be a family of injective en-

velopes. Show that the coproduct∐

n∈N ιn :∐

n∈NMn→∐

n∈N In is an injective envelope.Hint: See [88, thm. 1.4.6].

E B.7 Let p be a prime and consider the injective envelope Z/pZ Z(p∞) from B.13. Showthat the product (Z/pZ)N→ Z(p∞)N is not an injective envelope.

E B.8 Consider a complex M = 0→M′→M′′→ 0. Show that if one has HomR(M′′,M′) = 0,then M is minimal.

E B.9 Let I be a minimal semi-injective R-complex and let M be a simple R-module. Show thatthe complex HomR(M, I) has zero-differential, and conclude that HomR(N, I) has zero-differential for every semi-simple R-module N.

E B.10 Show how to construct a minimal injective resolution of an R-module by taking successiveinjective envelopes.

E B.11 Show that in a minimal injective resolution ι : M→ I the map ι need not be injective.E B.12 Let p be a prime number. Show that the canonical ring homomorphism Z(p)→ Z/pZ is a

projective cover of the Z(p)-module Z/pZ.

E B.13 Let M be a graded R-module with graded submodules N ⊆ M and N′ ⊆ M′ ⊆ M. Showthat if N′ is superfluous in M′ and one has M′ ⊂ N +N′, then M′ is contained in N.

E B.14 Show that the unique maximal ideal in a local ring is both essential and superfluous.E B.15 Set R = s/t ∈ Q | (s, t) = 1 and t odd. Show that the Jacobson radical of R is J= R(2).

Set F = R(N) and let µ : F → Q be the map given by (an)n∈N 7→ ∑n∈N an/2n. Show thatµ(JF) is not superfluous in Q and conclude that JF is not superfluous in F .

E B.16 Let a be an ideal in R such that R/a is local. Show that idempotents lift from R/a to R.E B.17 Let π : PM be a projective precover of an R-module. Show that π is a projective cover

if and only if every endomorphism γ : P→ P with πγ = π is an automorphism.E B.18 Show that every flat module over a perfect ring is projective. Hint: 1.3.39.E B.19 Show that every finitely generated flat module over a semi-perfect ring is projective. Give

an example of a semi-perfect ring that is not left Noetherian.E B.20 Let π : PM and π′ : P′M′ be projective covers of graded R-modules. Show that the

direct sum π⊕π′ : P⊕P′→M⊕M′ is a projective cover.


Minimality 459

E B.21 Assume that R is left perfect and let πn : PnMnn∈N be a family of projective covers.Show that the product

∏n∈N π

n :∏

n∈NPn→∏

n∈NMn is a projective cover. Hint: See[88, thm. 1.4.7 and cor. 1.4.8].

E B.22 Let p be a prime number and consider the projective cover Z(p) → Z/pZ from E B.12.Show that the coproduct (Z(p))

(N)→ (Z/pZ)(N) is not a projective cover.E B.23 Let P be a minimal semi-projective R-complex and let M be a simple R-module. Show

that the complex HomR(P,M) has zero-differential.E B.24 Assume that R is left perfect. Show how to construct a minimal projective resolution of

an R-module by taking successive projective covers.E B.25 Show that in a minimal projective resolution π : P→M the map π need not be surjective.


Appendix CStructure of Semi-flat Complexes

SYNOPSIS. Degreewise finitely presented complex; Govorov and Lazard’s theorem; categoricallyflat complex.

The purpose of this appendix is to prove Theorem C.1 below. In the case of modules,see C.16, it was obtained by Govorov [31] and Lazard [53].

C.1 Theorem. For an R-complex F the following conditions are equivalent.(i) F is semi-flat.

(ii) Every morphism of R-complexes, ϕ : N→ F , with N bounded and degreewisefinitely presented admits a factorization in C(R),

N

κ

ϕ// F

L ,λ

??

where L is a bounded complex of finitely generated free modules.(iii) F is isomorphic to a filtered colimit of bounded complexes of finitely gener-

ated free R-modules.

The proof comes after Lemma C.12; we start by collecting a couple of corollaries.

C.2 Corollary. Let F be an R-complex. If every morphism N→ F with N boundedand degreewise finitely presented factors through a semi-flat R-complex, then F issemi-flat.

The next result characterizes semi-flat complexes by a lifting property akin to5.2.10(v) for semi-projective complexes.

C.3 Corollary. For an R-complex F the following conditions are equivalent.(i) F is semi-flat.

(ii) For every morphism ϕ : N→ F with N bounded and degreewise finitely pre-sented and for every surjective quasi-isomorphism α : M→ F there is a mor-phism β : N→M with ϕ= αβ.

461

462 Appendix C

PROOF. (i)=⇒ (ii): It follows from C.1 that there is a bounded complex L of finitelygenerated free R-modules and morphisms κ : N→ L and λ : L→ F with ϕ= λκ. AsL is semi-projective, see 5.2.8, there exists by 5.2.10 a morphism γ : L→M withλ= αγ, so with β= γκ one has ϕ= αβ.

(ii)=⇒ (i): Let α : P '−→ F be a surjective semi-projective resolution; see 5.2.14.For every morphism ϕ : N→ F with N bounded and degreewise finitely presented,there exists by (ii) a morphism β : N→ P with ϕ= αβ. As P is semi-flat, see 5.4.10,it follows from C.2 that F is semi-flat.

BOUNDED AND DEGREEWISE FINITELY PRESENTED COMPLEXES

For the proof of C.1 we need a result of independent interest, Theorem C.7, whichsays that every complex is a filtered colimit of bounded and degreewise finitelypresented complexes. The next construction is the first step towards this result.

C.4 Construction. Let M be an R-complex and let ∆ be a set of R-complexes.Denote by Ξ the set of all morphisms ξ : Dξ→M whose domain Dξ belongs to ∆.Set X =

∐ξ∈ΞDξ and let π : X →M be the unique morphism whose composite with

the injection Dξ X equals ξ for every ξ ∈ Ξ; see 3.1.2. Set X = X (N), and letπ : X →M be the morphism whose composite with every injection X X is π.

A family K = Knn∈N of finite subsets Kn ⊆ Ξ, such that Kn is non-empty foronly finitely many n∈N, is called a string. To a string K one associates the followingsubcomplex of X ,

XK =∐n∈N

(⊕ξ∈Kn

Dξ) .

Let U be the set of pairs (K,Y ) where K is a string and Y is a degreewise finitelygenerated subcomplex of XK ∩Ker π. For elements (K,Y ) and (K′,Y ′) in U declare(K,Y ) 6 (K′,Y ′) if one has Kn ⊆ K′n for all n ∈ N and Y ⊆ Y ′. Evidently, the set(U,6) is partially ordered. It is also filtered as (K∪K′,Y +Y ′), where K∪K′ is thestring Kn∪K′nn∈N, dominates both (K,Y ) and (K′,Y ′). For every element (K,Y )in U set M(K,Y ) = XK/Y . For (K,Y )6 (K′,Y ′) there is a morphism,

µ(K′,Y ′)(K,Y ) : M(K,Y ) −→ M(K′,Y ′) given by [x]Y 7−→ [x]Y ′ .

It is straightforward to verify that these morphisms form a U-direct system in C(R).For every (K,Y ) in U the restriction of π : X →M to XK is zero on Y , and thus thereis a morphism,

α(K,Y ) : M(K,Y ) −→ M given by [x]Y 7−→ π(x) .

For elements (K,Y ) 6 (K′,Y ′) in U one evidently has α(K,Y ) = α(K′,Y ′)µ(K

′,Y ′)(K,Y ),so by the universal property of colimits 3.2.4 there is a morphism,

α : colim(K,Y )∈U

M(K,Y ) −→ M ,

that satisfies αµ(K,Y ) = α(K,Y ) for all (K,Y ) ∈U .


Structure of Semi-flat Complexes 463

C.5 Proposition. Let M be an R-complex and let ∆ be a set of R-complexes. Thecomplexes and morphisms constructed in C.4 have the following properties.

(a) The morphism α is injective.(b) If every homogeneous element m ∈ M belongs to the image of a morphism

D→M with D ∈ ∆, then the morphism α is surjective.(c) If every complex D in ∆ is bounded (above/below), then the complex M(K,Y )

is bounded (above/below) for every (K,Y ) ∈U .(d) If every complex D in ∆ is degreewise finitely presented, then the complex

M(K,Y ) is degreewise finitely presented for every (K,Y ) ∈U .

PROOF. (a): Since α is, in particular, a morphism of graded modules, it sufficesto show that z = 0 is the only homogeneous element in colim(K,Y )∈U M(K,Y ) withα(z) = 0. Set w= |z| and assume that one has α(z) = 0. By 3.2.26 there is an element(K,Y ) in U and a z′ in M(K,Y ) with |z′|= w and µ(K,Y )(z′) = z. It follows that one has

(?) α(K,Y )(z′) = αµ(K,Y )(z′) = α(z) = 0 .

Write z′ = [x]Y for some homogeneous element x ∈ XK with |x|= w. Now

Y ′ = 0−→ Rx∂X

w−−→ R∂Xw(x)−→ 0

is a subcomplex of XK , concentrated in degrees w and w−1; evidently it is degree-wise finitely generated. By definition, one has α(K,Y )(z′) = π(x), so it follows from(?) that x belongs to Ker π. As π : X →M is a morphism of complexes, it followsthat Y ′ is a subcomplex of Ker π. Consequently, (K,Y +Y ′) is an element in U . Byconstruction one has

µ(K,Y+Y ′)(K,Y )(z′) = [x]Y+Y ′ = 0 ,

and therefore also

z = µ(K,Y )(z′) = µ(K,Y+Y ′)µ(K,Y+Y ′)(K,Y )(z′) = µ(K,Y+Y ′)(0) = 0 .

(b): Let m be in M. We may assume that m is homogeneous, and it is enough toargue that m is in the image of one of the morphisms α(K,Y ); see 3.2.5. By definition,α(K,0) is the restricted morphism π : XK →M, so it suffices to show that m is inthe image of this map for some string K. By assumption there exists a morphismξ : Dξ→M with Dξ ∈∆ such that m is in Imξ. Define a string K by setting K1 = ξand Kn = ∅ for n > 1. For this string one has XK = Dξ, and the restriction of π tothis subcomplex is the morphism ξ, which has m in its image.

(c): For every (K,Y ) in U the complex M(K,Y ) is a quotient of a direct sum ofcomplexes from ∆, so if they are all bounded (above/below) then so is M(K,Y ).

(d): Assume that every complex D in ∆ is degreewise finitely presented. For astring K, the complex XK is a direct sum of complexes from ∆ and hence degreewisefinitely presented. For every degreewise finitely generated subcomplex of Y of X itfollows from 1.3.35 that M(K,Y ) = XK/Y is degreewise finitely presented.


464 Appendix C

C.6. Let M be an R-complex and let m be a homogeneous element in M. Denoteby M′ the subcomplex of M whose underlying graded module is R〈m,∂M(m)〉. Itfollows from 2.5.25 that m is in the image of a morphism D|m|(R)−→M′M.

C.7 Theorem. Every R-complex is isomorphic to a filtered colimit of bounded anddegreewise finitely presented R-complexes.

PROOF. Let ∆ be the set Du(R) | u ∈ Z of bounded degreewise finitely presentedR-complexes. Let M be an R-complex; by C.6 every homogeneous element m in Mis in the image of a morphism D|m|(R)→M. The claim now follows from C.5.

C.8 Corollary. Every R-module is isomorphic to a filtered colimit of finitely pre-sented R-modules.

PROOF. Let M be an R-module. By C.7 there is an isomorphism of R-complexesM ∼= colimu∈U Mu where µvu : Mu→Mvu6v is a filtered U-direct system of de-greewise finitely presented complexes. The degree 0 component yields an isomor-phism between M and the colimit of the U-direct system µvu

0 : Mu0 →Mv

0u6v.

REMARK. The corollary also follows directly from Proposition C.5 with ∆= R.

We make a minor digression to supplement 3.2.36.

C.9 Theorem. For an R-complex M, the following conditions are equivalent.(i) M is bounded and degreewise finitely presented.

(ii) The functor HomR(M,–) preserves filtered colimits.(iii) The functor C(R)(M,–) preserves filtered colimits.

PROOF. The implication (i)=⇒ (ii) is proved in 3.2.36.(ii)=⇒ (iii): By 3.2.33 the functor Z0(–) preserves filtered colimits. From this

fact and (ii), it follows that the composite functor Z0(HomR(M,–)) preserves filteredcolimits, and by 2.3.10 this functor is nothing but C(R)(M,–).

(iii)=⇒ (i): By C.7 there is a filtered U-direct system µvu : Mu→Mvu6v ofbounded degreewise finitely presented R-complexes with M ∼= colimu∈U Mu. By as-sumption, the canonical morphism

ϕ : colimu∈U

C(R)(M,Mu) −→ C(R)(M,colimu∈U

Mu) ∼= C(R)(M,M)

is an isomorphism. Write

µu : Mu −→ colimu∈U

Mu ∼= M and λu : C(R)(M,Mu) −→ colimu∈U

C(R)(M,Mu)

for the canonical morphisms. By 3.2.12 one has ϕλu = C(R)(M,µu) for all u ∈U .Surjectivity of ϕ yields an element κ ∈ colimu∈U C(R)(M,Mu) with ϕ(κ) = 1M . By3.2.26 one has κ = λu(ψu) for some u ∈U and ψu ∈ C(R)(M,Mu). Consequently,there are equalities µuψu = C(R)(M,µu)(ψu) = ϕλu(ψu) = ϕ(κ) = 1M . It followsthat M is a direct summand of Mu, and since Mu is bounded and degreewise finitelypresented, so is M.



REMARK. Following the standard definition—see for example Crawley-Boevey [21] or Krause[47]—the finitely presented objects in C(R) are exactly the complexes M that satisfy condition(iii) in C.9. By C.7 the category C(R) is hence locally finitely presented.

PROOF OF GOVOROV AND LAZARD’S THEOREM

C.10 Proposition. Let M be an R-complex and let Λ be a class of (bounded(above/below) and) degreewise finitely presented R-complexes. If every morphismof R-complexes ϕ : N→M where N is (bounded (above/below) and) degreewisefinitely presented admits a factorization in C(R),

N

κ

ϕ// M

Lλ

??

with L in Λ, then M is isomorphic to a filtered colimit of complexes from Λ.

PROOF. Every homogeneous element m ∈M is, as noted in C.6, in the image of amorphism D|m|(R)→M. It follows from the assumptions that m is in the image of amorphism L→M with L ∈Λ. Let ∆ be a set of representatives for the isomorphismclasses in Λ. It follows from C.5 that M is isomorphic to the colimit of the filteredU-direct system of degreewise finitely presented R-complexes

µ(K′,Y ′)(K,Y ) : M(K,Y ) −→ M(K′,Y ′)(K,Y )6(K′,Y ′)

constructed in C.4. To prove the assertion, it suffices by 3.2.29 to show that thesubset of U of elements (K,Y ) such that M(K,Y ) belongs to ∆ is cofinal in U . To thisend, let (K,Y ) be in U ; by assumption there exists a factorization,

M(K,Y )

κ

α(K,Y )// M

Lλ

??

with L in Λ. We may assume that L is in ∆. Pick any p ∈N such that Kp =∅ holds.Define a string K′ by setting K′p = λ and K′n =Kn for n 6= p. With the notation fromC.4 one has XK′ = XK⊕L, where L is considered as a subcomplex of the pth copyof X in X . Let % : XK XK/Y = M(K,Y ) be the canonical morphism. The morphism

(?) XK′ = XK⊕L−→ L given by (x, l) 7−→ κ%(x)+ l

has as kernel the subcomplex

Y ′ = (x,−κ%(x)) ∈ XK⊕L | x ∈ XK .

Since XK is degreewise finitely generated, so is Y ′. By definition, π= α(K,Y )% on XK ,and the restriction of π to L is λ. For every (x,−κ%(x)) in Y ′ there are now equalities,

π(x,−κ%(x)) = π(x)− πκ%(x) = π(x)−λκ%(x) = π(x)−α(K,Y )%(x) = 0 .


466 Appendix C

Thus Y ′ is contained in Ker π and, therefore, (K′,Y ′) is an element in U . For x inY one has κ%(x) = κ(0) = 0, so Y is contained in Y ′ and (K,Y ) 6 (K′,Y ′) holds.It remains to see that M(K′,Y ′) belongs to ∆. As the morphism (?) is surjective withkernel Y ′, it follows that there is an isomorphism M(K′,Y ′) = XK′/Y ′ ∼= L.

C.11. Let L be a complex of finitely generated free R-modules. The dual complexHomR(L,R) is one of finitely generated free Ro-modules, so the biduality morphism

δHomR(L,R)R : HomR(L,R) −→ HomR(HomRo(HomR(L,R),R),R)

is an isomorphism of Ro-modules by 4.5.4. Its inverse is HomR(δLR,R); to prove this

it is sufficient to verify that HomR(δLR,R)δ

HomR(L,R)R is the identity on HomR(L,R),

and that follows from the next computation. For ϕ in HomR(L,R) and l in L one has

(HomR(δLR,R)δ

HomR(L,R)R )(ϕ)(l) = (δ

HomR(L,R)R (ϕ)δL

R)(l)

= δHomR(L,R)R (ϕ)(δL

R(l))

= δLR(l)(ϕ)

= ϕ(l) .

C.12 Lemma. Let F be a semi-flat R-complex. Every morphism of R-complexes,ϕ : N→ F , with N bounded and degreewise finitely presented admits a factorization,

N

κ

ϕ// F

L ,λ

??

in C(R) where L is a bounded complex of finitely generated free modules.

PROOF. By 2.5.26 there is an exact sequence L′′ψ′′−→ L′

ψ′−→ N −→ 0 of R-complexeswhere L′ and L′′ are bounded complexes of finitely generated free modules. Considerthe exact sequence of Ro-complexes,

(?) 0−→ K ι−−→ HomR(L′,R)Hom(ψ′′,R)−−−−−−−→ HomR(L′′,R) ,

where K is the kernel of HomR(ψ′′,R) and ι is the embedding. The functor Z0(–)

is left exact, so since F is a complex of flat R-modules, it follows from 5.4.1 thatthe functor Z0(–⊗R F) leaves the sequence (?) exact. As L′ is bounded, so is K; setu = infK\. By 5.1.7 there is an exact sequence,

(‡) P π−−→ K −→ 0 ,

where π is a quasi-isomorphism and P is a semi-free Ro-complex with Pv = 0 forall v < u. As F is semi-flat, π⊗R F is a surjective quasi-isomorphism by 5.4.9, andit follows from 4.2.7 that the functor Z0(–⊗R F) leaves the sequence (‡) exact. Intotal, there is an exact sequence,

Z0(P⊗R F)(ιπ)⊗F−−−−→ Z0(HomR(L′,R)⊗R F)

Hom(ψ′′,R)⊗F−−−−−−−−−→ Z0(HomR(L′′,R)⊗R F) .



For every R-complex M, denote by ξM the composite morphism

HomR(M,R)⊗R F θMRF−−−→ HomR(M,R⊗R F)

Hom(M,µFR )−−−−−−−→∼=

HomR(M,F) ,

where θMRF is tensor evaluation 4.5.6 and µFR is the unitor (4.4.1.1). The morphism

ξM is natural in M; recall from 4.5.7(3,c) that θMRF , and hence ξM , is an isomor-phism if M is a bounded complex of finitely presented modules. From the exactsequence above, one now gets another exact sequence,

() Z0(P⊗R F)ξL′((ιπ)⊗F)−−−−−−−−→ Z0(HomR(L′,F))

Hom(ψ′′,F)−−−−−−−→ Z0(HomR(L′′,F)) .

As ϕψ′ : L′→ F is a morphism, it is an element in Z0(HomR(L′,F)); see 2.3.10.Since one has HomR(ψ

′′,F)(ϕψ′) = ϕψ′ψ′′ = 0, exactness of () yields an elementx in Z0(P⊗R F) with

(§) (ξL′ ((ιπ)⊗R F))(x) = ϕψ′ .

The graded module P\ has a graded basis E, and x has the form x = ∑ni=1 ei⊗ fi with

ei ∈E and fi ∈F . Set w=max|e1|, . . . , |en|; since one has Pv = 0 for all v< u, eachelement ei satisfies u6 |ei|6w. For v ∈ Z set Ev = e∈ E | |e|= v. Next we definea bounded subcomplex P′ of P such that each module P′v is finitely generated andfree: For v /∈ u, . . . ,w set P′v = 0; for v∈ u, . . . ,w the modules P′v are constructedrecursively. Let P′w be the finitely generated free submodule of Pw generated by theset E ′w = e1, . . . ,en∩Ew. For v 6 w assume that a free submodule P′v of Pv withfinite basis E ′v has been constructed. As the subset B′v−1 = ∂P(e) | e ∈ E ′v of Pv−1is finite, there is a finite subset G′v−1 of Ev−1 with B′v−1 ⊆ Ro〈G′v−1〉. Now let P′v−1be the submodule of Pv−1 generated by the finite set

E ′v−1 = G′v−1∪ (e1, . . . ,en∩Ev−1) .

By construction, one has ∂P(P′v) ⊆ P′v−1 for all v ∈ Z, so P′ is a subcomplex of P.Note that x = ∑

ni=1 ei⊗ fi belongs to P′⊗R F . As F consists of flat modules, P′⊗R F

is a subcomplex of P⊗R F ; since x is in Z0(P⊗R F), it is also in Z0(P′⊗R F).Set L = HomRo(P′,R). As P′ is a bounded complex of finitely generated free

Ro-modules, L is a bounded complex of finitely generated free R-modules. Letε : P′ P be the embedding and let κ′ : L′→ L be the composite morphism

L′δL′

R−−→ HomRo(HomR(L′,R),R)Hom(ιπε,R)−−−−−−−→ HomRo(P′,R) = L .

In the commutative diagram

P′

δP′R

∼=

ιπε// HomR(L′,R)

δHom(L′,R)R

∼=

HomR(HomRo(P′,R),R)Hom(Hom(ιπε,R),R)

// HomR(HomRo(HomR(L′,R),R),R)

the vertical morphisms are isomorphisms by 4.5.4, and δHom(L′,R)R is by C.11 the

inverse of HomR(δL′R ,R). One now gets


468 Appendix C

(#) HomR(κ′,R)δP′

R = ιπε .

It follows that there are equalities,

HomR(κ′ψ′′,R)δP′

R = HomR(ψ′′,R)ιπε = 0πε = 0 ,

and since δP′R is an isomorphism, the morphism HomR(κ

′ψ′′,R) is zero. In particular,HomRo(HomR(κ

′ψ′′,R),R) is zero, and hence the commutative diagram

L′′

δL′′R

∼=

κ′ψ′′// L

δLR

∼=

HomRo(HomR(L′′,R),R)Hom(Hom(κ′ψ′′,R),R)

// HomRo(HomR(L,R),R)

shows that one has κ′ψ′′ = 0. Again the vertical morphisms are isomorphisms by4.5.4. Since κ′ is zero on Imψ′′ = Kerψ′ there is a unique morphism κ : N→ L withκψ′ = κ′. Finally, consider the diagram,

(??)

P′⊗R F

ε⊗F

δP′R ⊗F

// HomR(L,R)⊗R F

Hom(κ′,R)⊗F

ξL// HomR(L,F)

HomR(κ′,F)

P⊗R F(ιπ)⊗F

// HomR(L′,R)⊗R FξL′

// HomR(L′,F) ,

where the left-hand square is commutative by (#) and the right-hand square is com-mutative by naturality of ξ. Set

λ = (ξL (δP′R ⊗ F))(x) ;

it is an element in HomR(L,F), and since x belongs to Z0(P′⊗R F), also λ is a cycle;that is, λ is a morphism. From (??), from the definition of λ, and from (§) one getsλκ′ = ϕψ′. The identity κ′ = κψ′ and surjectivity of ψ′ now yield λκ = ϕ.

We can now prove the main result in this appendix.

PROOF OF C.1. The implication (i)=⇒ (ii) is proved in C.12. For the implication(ii)=⇒ (iii), let Λ be the class of bounded complexes of finitely generated free R-modules and apply C.10. Finally, (i) follows from (iii) by 5.4.13.

THE CASE OF GRADED MODULES

C.13 Theorem. For a graded R-module F the following conditions are equivalent.(i) F is graded-flat.

(ii) Every morphism of graded R-modules, ϕ : N→ F , with N bounded and de-greewise finitely presented admits a factorization in Mgr(R),



N

κ

ϕ// F

L ,λ

??

where L is a bounded and degreewise finitely generated graded-free module.(iii) F is isomorphic to a filtered colimit of bounded and degreewise finitely gen-

erated graded-free R-modules.

PROOF. (i)=⇒ (ii): By 5.4.11 a graded-flat R-module is semi-flat as an R-complex.Thus, it follows from C.1 that ϕ factors in C(R) as N→ L→F , where L is a boundedcomplex of finitely generated free R-modules. Now N → L\ → F is the desiredfactorization in Mgr(R).

(ii)=⇒ (iii): Let Λ be the class of bounded and degreewise finitely generatedgraded-free R-modules. By 3.2.6 it suffices to show that F is isomorphic to a fil-tered colimit in C(R) of objects from Λ. To apply C.10, it must be argued thatevery morphism N → F , where N is a bounded and degreewise finitely presentedR-complex, admits a factorization in C(R) through an object from Λ. Since F is agraded R-module, every morphism N → F in C(R) factors through the morphismN → C(N). The graded module C(N) is bounded, and by 1.3.35 it is degreewisefinitely presented, so every morphism C(N)→ F factors by assumption through anobject in Λ.

(iii)=⇒ (i): A graded-free R-module is graded-flat by 2.5.22 and 1.3.38. A fil-tered colimit of graded-flat modules is graded-flat by 5.4.11, 5.4.13, and 3.2.6.

C.14 Corollary. Let F be a graded R-module. If every morphism of graded R-modules N→ F , with N bounded and degreewise finitely presented, factors througha graded-flat R-module, then F is graded-flat.

C.15 Corollary. For a graded R-module F the following conditions are equivalent.(i) F is graded-flat.

(ii) For every morphism ϕ : N→ F of graded R-modules with N bounded anddegreewise finitely presented and for every surjective morphism α : M→ Fof graded R-modules there is a morphism β : N→M with ϕ= αβ.

PROOF. (i)=⇒ (ii): Immediate from C.3.(ii)=⇒ (i): Let α : L→ F be a surjective morphism with L graded-free; see

2.5.23. For every morphism ϕ : N→ F with N bounded and degreewise finitely pre-sented, there is by (ii) a morphism β : N→ L with ϕ = αβ. As L is graded-flat, see2.5.22 and 1.3.38, it follows from C.14 that F is graded-flat.

THE CASE OF MODULES

C.16 Theorem. For an R-module F the following conditions are equivalent.(i) F is flat.


470 Appendix C

(ii) Every homomorphism of R-modules ϕ : N→ F with N finitely presented ad-mits a factorization in M(R),

N

κ

ϕ// F

L ,λ

??

where L is a finitely generated free module.(iii) F is isomorphic to a filtered colimit of finitely generated free R-modules.

PROOF. Immediate from C.13 in view of 3.2.6.

C.17 Corollary. Let F be an R-module. If every homomorphism N → F with Nfinitely presented factors through a flat R-module, then F is flat.

C.18 Corollary. For an R-module F the following conditions are equivalent.(i) F is flat.

(ii) For every homomorphism ϕ : N→ F of R-modules with N finitely presentedand for every surjective homomorphism α : M→ F there is a homomorphismβ : N→M with ϕ= αβ.

C.19. Govorov and Lazard’s theorem C.16 yields a short proof of 1.3.42. Indeed,if F is a finitely presented and flat R-module, then the identity 1F factors through afinitely generated free R-module L, whence F is a direct summand of L.

CATEGORICALLY FLAT COMPLEXES

C.20 Theorem. For an R-complex F the following conditions are equivalent.(i) F is semi-flat and acyclic.

(ii) Every morphism of R-complexes ϕ : N→ F , with N bounded and degreewisefinitely presented, admits a factorization N → L→ F in C(R), where L is abounded and contractible complex of finitely generated free modules.

(iii) F is isomorphic to a filtered colimit of bounded and contractible complexesof finitely generated free R-modules.

PROOF. (i)=⇒ (ii): Let π : P′ '−→ F be a surjective semi-projective resolution; see5.2.14. It follows from C.3 that ϕ factors through π; that is, there is a morphismκ : N→ P′ with ϕ = πκ. As P′ is acyclic, the complex HomR(P′,P′) is acyclic by5.2.10, whence P′ is contractible by 4.3.29. By 4.3.32 there is a graded R-moduleP with P′ ∼= Cone1P =

∐v∈ZDv+1(Pv). In particular, P is graded-projective, so for

every v there is a module Qv and a set Ev, such that one has Pv⊕Qv ∼= R(Ev). Set

L′ = P′⊕∐

v∈ZDv+1(Qv)∼=

∐v∈Z

Dv+1(R)(Ev) .

The morphism κ factors trivially, κ=$P′(εP′κ), through the complex L′ of free R-modules. Since N is bounded and degreewise finitely generated, the morphism εP′κ,



and hence ϕ, factors through a direct sum L =⊕n

i=1 Dvi+1(R)(E′vi) where each E ′vi

isa finite subset of Evi ; see 3.1.31. Evidently, L is a bounded and contractible complexof finitely generated free R-modules.

(ii)=⇒ (iii): Follows from C.10.(iii)=⇒ (i): By assumption there is an isomorphism F ∼= colimu∈U Lu where each

Lu is a bounded contractible complex of free R-modules. In particular, each complexLu is semi-flat and acyclic; see 5.4.7. It follows from 5.4.13 that F is semi-flat andfrom 3.2.35 that F is acyclic.

C.21 Corollary. Let F be an R-complex. If every morphism N→F with N boundedand degreewise finitely presented factors through a semi-flat acyclic R-complex,then F is semi-flat and acyclic.

C.22 Corollary. For an R-complex F the following conditions are equivalent.(i) F is semi-flat and acyclic.

(ii) For every morphism ϕ : N→ F with N bounded and degreewise finitely pre-sented and for every surjective morphism α : M→ F there is a morphismβ : N→M with ϕ= αβ.

PROOF. (i)=⇒ (ii): It follows from C.20 that there is a contractible complex L offree R-modules and morphisms κ : N→ L and λ : L→ F with ϕ= λκ. The map

HomR(L,α) : HomR(L,M) −→ HomR(L,F)

is by 5.2.2 and 4.3.29 a surjective morphism of acyclic complexes. In particular, itis a surjective quasi-isomorphism, so it is surjective on cycles by 4.2.7. In view of2.3.10 there is a morphism β′ : L→M with λ= αβ′; now set β= β′κ.

(ii)=⇒ (i): Let α : P→ F be a surjective morphism where P is a contractible com-plex of free R-modules; see 4.3.24. For every morphism ϕ : N→ F with N boundedand degreewise finitely presented, there exists by (ii) a morphism β : N→ P withϕ = αβ. As P is semi-flat and acyclic, see 5.4.7, it follows from C.21 that F issemi-flat and acyclic.

The complexes described in C.20–C.22 are referred to as categorically flat; thenext theorem gives further characterizations of these complexes.

C.23 Theorem. For an R-complex F the following conditions are equivalent.(i) F is semi-flat and acyclic.

(ii) F is a complex of flat R-modules, and for every Ro-complex M the complexM⊗R F is acyclic.

(iii) F is a complex of flat R-modules, and for every Ro-module M the complexM⊗R F is acyclic.

(iv) F is a complex of flat R-modules, and for every finitely presented R-moduleN the complex HomR(N,F) is acyclic.

(v) F is a complex of flat R-modules, and for every bounded below and degree-wise finitely presented R-complex N the complex HomR(N,F) is acyclic.

(vi) F is an acyclic complex, and B(F) = Z(F) is a complex of flat R-modules.


472 Appendix C

(vii) The Ro-complex Homk(F,E) is a contractible complex of injective modules.

PROOF. (i)=⇒ (ii): By C.20 an acyclic semi-flat complex F is isomorphic to a fil-tered colimit colimu∈U Pu where each Pu is a bounded contractible complex of pro-jective R-modules. It follows from 5.4.13 that F is a complex of flat R-modules.For every Ro-complex M the complexes M⊗R Pu are acyclic by 4.3.20 and 4.3.27,whence and M⊗R F is acyclic by 3.2.17 and 3.2.35.

(ii)=⇒ (iv): Tensor evaluation 4.5.7(3,c) yields an isomorphism of k-complexesHomR(N,R)⊗R F ∼= HomR(N,F), and the left-hand complex is acyclic by (ii).

(iv)=⇒ (v): Immediate in view of A.10.(v)=⇒ (vi): By assumption HomR(R,F) and hence F is acyclic. Fix v in Z. For

every finitely presented R-module N, a homomorphism ϕ : N→ Zv(F) yields a chainmap of R-complexes N→ F . As the complex HomR(N,F) is acyclic it follows from2.3.3 that the homomorphism ϕ factors as

N −→ Fv−1∂F

v−1−−→ Zv(F) .

As Fv−1 is flat by assumption, it follows from C.17 that Zv(F) is flat.(vi)=⇒ (iii): Apply 5.4.11 and 5.4.12 to 0→ Z(F)→ F → Σ−1 B(F)→ 0 to

conclude that F is a complex of flat R-modules. For every v ∈ Z the sequence

(?) 0−→M⊗R Zv(F)−→M⊗R Fv −→M⊗R Bv−1(F)−→ 0

is exact by flatness of Bv−1(F), see 5.4.3, whence M⊗R F is acyclic.(iii)=⇒ (i): By assumption F is a complex of flat R-modules; moreover, R⊗R F

and hence F is acyclic. Furthermore it follows from A.14 that M⊗R F is acyclic forevery bounded Ro-complex M. To prove that F is semi-flat, one has to verify thatA⊗R F is acyclic for every acyclic Ro-complex A; see 5.4.9. Given such a complex,C.7 yields A∼= colimu∈U Mu for a U-direct system of bounded Ro-complex Mu. Nowit follows from 3.2.16 and 3.2.35 that A⊗R F is acyclic.

(i)=⇒ (vii): The character complex Homk(F,E) is acyclic, and by 5.4.9 it is semi-injective. Hence, then complex HomRo(Homk(F,E),Homk(F,E)) is acyclic, and itfollows from 4.3.29 that Homk(F,E) is contractible.

(vii)=⇒ (i): Immediate from 5.4.9 and 2.5.6(b).

REMARK. The contractible complexes of injective Ro-modules are precisely the injective objectsin C(Ro), see E 5.3.3, so while flatness is not a categorical notion, condition (vii) in Theorem C.23explains why the complexes that satisfy the conditions in that theorem are called categorically flat.

The next corollary supplements 5.4.16.

C.24 Corollary. Let α : F → F ′ be a quasi-isomorphism of semi-flat R-complexes.For every bounded below and degreewise finitely presented R-complex N, the mor-phism HomR(N,α) is a quasi-isomorphism.

PROOF. The complex C =Coneα is acyclic by 4.2.15, and it follows from 5.4.12 ap-plied to the mapping cone sequence (4.1.5.1) that it is semi-flat. For every boundedbelow and degreewise finitely presented R-complex N the complex HomR(N,C) isacyclic by C.23, so HomR(N,α) is a quasi-isomorphism by 4.1.15.



REMARK. Emmanouil [26] proves that no boundedness condition is needed on the complex N inC.23(v) and C.24.

EXERCISES

E C.1 Without recourse to 5.2.11 and 5.4.10, show that every semi-free R-complex is semiflat.Hint: C.1 and E 5.1.6.

E C.2 Show that every acyclic complex can be written as a filtered colimit of bounded belowacyclic complexes, and show that not every acyclic complex can be written as a filteredcolimit of bounded acyclic complexes.

E C.3 Let η= 0→ F ′→ F→ F ′′→ 0 be an exact sequence of R-complexes and assume that F issemi-flat. Show that if HomR(N,η) is exact for every bounded complex of finitely presentedR-modules, then F ′ and F ′′ are semi-flat as well. Hint: C.3.

E C.4 Show that a complex F of flat R-modules is semi-flat if and only if HomR(F,A) is acyclicfor every acyclic complex A with Z(A) is degreewise cotorsion.

E C.5 Let 0→ F → P→ F → 0 be an exact sequence with F flat and P projective. Show that ifFPD(R) is finite, then F is projective.

The assumption on FPD(R) is superfluous; see Benson and Goodearl [12].E C.6 Let P be an acyclic complex of projective R-modules such that B(P) is a complex of flat

R-modules. Show that if FPD(R) is finite, then P is contractible.The assumption on FPD(R) in is superfluous; see remark after 5.4.10.

E C.7 Show that if FPD(R) is finite, then every semi-flat R-complex of projective R-modules issemi-projective. Hint: Consider a semi-projective resolution and use C.20 and E C.6 toshow it is a homotopy equivalence.

The assumption on FPD(R) in is superfluous; see remark after 5.4.10.


Appendix DTriangulated Categories

SYNOPSIS. Axioms for triangulated category; triangulated functor; triangulated subcategory; FiveLemma.

In this appendix T is an additive category equipped with an additive autoequivalenceΣ. We describe the conditions for (T,Σ) to form a triangulated category; the first stepis to settle on a collection of triangles.

D.1 Definition. A candidate triangle in T is a diagram

M α−−→ Nβ−−→ X

γ−−→ ΣM ,

such that the composites βα, γβ, and (Σα)γ are all zero. An morphism (ϕ,ψ,χ) ofcandidate triangles is a commutative diagram in T,

M

ϕ

α// N

ψ

β// X

χ

γ// ΣM

Σϕ

M′ α′// N′

β′// X ′

γ′// ΣM′ ;

it is called an isomorphism if ϕ, ψ, and χ are isomorphisms in T.

D.2. For a collection4 of candidate triangles in T, consider the next conditions.(TR0) For every M in T, the candidate triangle

M 1M−−→M −→ 0−→ ΣM

is in4. Every candidate triangle that is isomorphic to one from4 is in4.(TR1) Every morphism α : M→ N in T fits in a candidate triangle from4,

M α−−→ N −→ X −→ ΣM .

(TR2) For every candidate triangle M α−→ Nβ−→ X

γ−→ ΣM in 4, the followingtwo candidate triangles belong to4 as well,

Nβ−−→ X

γ−−→ ΣM −Σα−−−→ ΣN and Σ−1 X

−Σ−1 γ−−−−→M α−−→Nβ−−→ X .

475

476 Appendix D

(TR2’) Consider two candidate triangles,

M α−−→ Nβ−−→ X

γ−−→ ΣM and N−β−−→ X

−γ−−→ ΣM −Σα−−−→ ΣN .

If one belongs to4 then so does the other.(TR3) For every commutative diagram

(D.2.1)

M

ϕ

α// N

ψ

β// X

γ// ΣM

Σϕ

M′ α′// N′

β′// X ′

γ′// ΣM′ ,

where the rows are candidate triangles in4, there exists a (not necessarilyunique) morphism χ : X → X ′, such that (ϕ,ψ,χ) is a morphism of candi-date triangles.

(TR4’) For every commutative diagram (D.2.1), where the rows are candidate tri-angles in 4, there exists a (not necessarily unique) morphism χ : X → X ′

such that (ϕ,ψ,χ) is a morphism of candidate triangles, and such that thefollowing candidate triangle belongs to4,

(D.2.2)M′

⊕N

(α′ ψ0 −β

)//

N′

⊕X

(β′ χ0 −γ

)//

X ′

⊕ΣM

(γ′ Σϕ0 −Σα

)//

ΣM′

⊕ΣN

.

The candidate triangle (D.2.2) is called the mapping cone of (ϕ,ψ,χ).

Condition (TR4’) is evidently stronger than (TR3), and it is proved in D.5 belowthat (TR2) and (TR2’) are equivalent under assumption of (TR0). The conditions inD.2 supply the axioms for a triangulated category.

D.3 Definition. A triangulated category is an additive category T equipped with anadditive autoequivalence Σ and a collection 4 of candidate triangles, called distin-guished triangles, such that (TR0), (TR1), (TR2’), and (TR4’) are satisfied.

In a triangulated category (T,Σ), the functor Σ is usually called the translation.Axiom (TR4’) is perhaps the less intuitive of the lot; here is a simple application.

D.4 Example. Let (T,Σ) be a triangulated category. For M and X in T it followsfrom (TR0) and (TR2’) that there are distinguished triangles

∆X = Σ

−1 X −→ 0−→ X −1X−−→ X and ∆

M = M 1M−−→M −→ 0−→ ΣM .

The only morphism from ∆X to ∆M is the zero morphism, and its mapping cone,

M

(1M

0

)−−−−→

M⊕X

(0 1X )−−−−→ X 0−−→ ΣM ,

is a distinguished triangle by (TR4’).


Triangulated Categories 477

REMARK. If the collection4 in (T,Σ) satisfies only (TR0), (TR1), (TR2’), and (TR3), then T iscalled pretriangulated. It can be proved that for a pretriangulated category the so-called octahedralaxiom, which is usually denoted (TR4), is equivalent to (TR4’); see Neeman [62]. That is, a trian-gulated category is a pretriangulated category that satisfies the octahedral axiom. This perspectivegoes back to Verdier’s thesis on derived categories [84] from the mid 1960s—it was published 30years late and only after Verdier’s passing. Indeed, triangulated categories in algebra were orig-inally defined through axiomatization of the properties of derived categories; the axioms being(TR0), (TR1), (TR2), (TR3), and (TR4); usually with (TR0) included in (TR1). The contemporaryformulation of the definition in D.3 follows Neeman’s monograph [63].

D.5 Lemma. Let4 be a collection of candidate triangles in (T,Σ) such that (TR0)is satisfied. Condition (TR2) is then satisfied if and only if (TR2’) is satisfied.

PROOF. Assume that (TR2’) is satisfied. Let M α−→ Nβ−→ X

γ−→ ΣM be a candidatetriangle in4. Consider the following isomorphism of candidate triangles,

(?)N

β// X

−1X∼=

γ// ΣM

−Σα// ΣN

N−β// X

−γ// ΣM

−Σα// ΣN .

By (TR2’) the lower row in (?) belongs to 4, and hence so does the upper row by(TR0). To show that the candidate triangle

Σ−1 X

−Σ−1 γ−−−−→M α−−→ Nβ−−→ X

is in 4, is by (TR2’) equivalent to showing that M −α−→ N−β−→ X

γ−→ ΣM is in 4;and that follows from (TR0) and the next isomorphism of candidate triangles,

M−α// N

−1N∼=

−β// X

γ// ΣM

M α// N

β// X

γ// ΣM .

Similar arguments show that (TR2) implies (TR2’).

Although triangulated categories are seldom Abelian, they are close enough toAbelian categories that one can do homological algebra in much the same fash-ion. The object X in a distinguished triangle M α−→ N −→ X −→ ΣM is a pseudo-cokernel—and Σ−1 X is a pseudo-kernel—of α. In this perspective, the distinguishedtriangles correspond to short exact sequences.

The autoequivalence Σ on T yields an autoequivalence Σop on the opposite cate-gory Top; for notational bliss its inverse (Σop)−1 is denoted Σ−1.

D.6 Proposition. Let (T,Σ) be a triangulated category. The opposite category(Top,Σ−1) is triangulated in the following canonical way: A candidate triangleM→ N→ X → Σ−1 M in Top is distinguished if and only if the corresponding dia-gram Σ−1 M→ X → N→M is a distinguished triangle in T.

PROOF. It is evident that a diagram in Top is a candidate triangle if and only ifthe corresponding diagram in T is a candidate triangle. Let 4 be the collection of


478 Appendix D

distinguished triangles in T. It is elementary to verify that the collection of diagramsM→ N→ X → Σ−1 M in Top such that the corresponding diagram in T belongs to4 satisfies the axioms in D.3. As an example, we provide the details for (TR0).

Let M be an object in Top, and hence in T. The candidate triangle

(?) M 1M−−→M −→ 0−→ Σ

−1 M

in Top is distinguished if and only if the corresponding candidate triangle in T,

(‡) Σ−1 M −→ 0−→M 1M

−−→M ,

belongs to 4. By (TR2’), applied twice, (‡) is in 4 if and only if the followingcandidate triangle is in4,

() M 1M−−→M −→ Σ0−→ ΣM .

There is an isomorphism Σ0∼= 0 in T, so by (TR0) the triangle () is in4, whence(?) is distinguished in Top. Next, let

(§)

M′

ϕ∼=

α′// N′

ψ∼=

β′// X ′

χ∼=

γ′// Σ−1 M′

Σ−1 ϕ∼=

M α// N

β// X

γ// Σ−1 M

be an isomorphism of candidate triangles in Top, and assume that the bottom row isdistinguished. In the corresponding diagram in T,

Σ−1 M

Σ−1 ϕ∼=

γ// X

χ∼=

β// N

ψ∼=

α// M

ϕ∼=

Σ−1 M′γ′// X ′

β′// N′ α′

// M′ ,

the top row belongs to4, and by (TR0) so does the bottom row. Hence, the top rowin (§) is a distinguished triangle in Top.

TRIANGULATED FUNCTORS AND TRIANGULATED SUBCATEGORIES

D.7 Definition. Let (T,ΣT) and (U,ΣU) be triangulated categories. A functorF: T→ U is called triangulated if it is additive and there is a natural isomorphismφ : FΣT → ΣU F such that for every distinguished triangle in T,

M α−−→ Nβ−−→ X

γ−−→ ΣT M ,

the induced candidate triangle in U,

F(M)F(α)−−−→ F(N)

F(β)−−→ F(X)φM F(γ)−−−−→ ΣU F(M) ,

is distinguished.



In the suggested analogy with Abelian categories, triangulated functors corre-spond to exact functors. The translation functor, for example, is triangulated.

D.8 Example. Let (T,Σ) be a triangulated category. Given a distinguished triangle

M α−−→ Nβ−−→ X

γ−−→ ΣM

in T, successive applications of (TR2’) yield distinguished triangles

N−β// X

−γ// ΣM

−Σα// ΣN ,

Xγ// ΣM Σα

// ΣNΣβ// ΣX , and

ΣM−Σα

// ΣN−Σβ

// ΣX−Σγ// Σ2 M .

Now it follows from the isomorphism of candidate triangles,

ΣM−Σα// ΣN

∼= −1N

−Σβ// ΣX

−Σγ// Σ2 M

ΣM Σα// ΣN

Σβ// ΣX

−Σγ// Σ2 M ,

that ΣM Σα−−→ ΣNΣβ−−→ ΣX

−Σγ−−−→ Σ2 M is distinguished. Thus, the functor Σ : T→ T

is triangulated with −1Σ2 M for the natural isomorphism Σ(ΣM)→ Σ(ΣM).

D.9. Let T, U, and V be triangulated categories; let F : T→ U and G: U→ V be tri-angulated functors with natural isomorphisms φ and ψ. It is straightforward to verifythat the composite GF: T→ V is triangulated with natural isomorphism ψFGφ.

D.10. Let (T,ΣT) and (U,ΣU) be triangulated categories. It is straightforwardto verify that if F : T→ U is a triangulated functor with natural isomorphismφ : FΣT → ΣU F, then the opposite functor Fop : Top→ Uop is triangulated with nat-ural isomorphism (Σ−1

U φΣ−1T )op : Σ−1

U Fop→ FopΣ−1T .

D.11 Definition. For triangulated categories T and U, an equivalence T U iscalled an equivalence of triangulated categories if the functors are triangulated.

It is not automatic that a natural transformations between triangulated functorscommutes with translation; hence one makes the following definition.

D.12 Definition. Let (T,ΣT) and (U,ΣU) be triangulated categories, and let F andG be triangulated functors T→ U with natural isomorphisms φ : FΣT → ΣU F andψ : GΣT → ΣU G. A natural transformation τ : F→ G is called triangulated if thefollowing diagram is commutative for every X in T,

F(ΣT X)

τΣX

∼=φX// ΣU F(X)

ΣτX

G(ΣT X) ∼=ψX// ΣU G(X) .


480 Appendix D

That is, τΣT X and ΣU τX are isomorphic, and the isomorphism is natural in X .

REMARK. Another name for triangulated natural transformation is graded natural transformation;see Bondal and Orlov [15]. It is proved ibid. that the unit and counit in an adjunction of triangulatedfunctors are automatically triangulated.

D.13 Example. Let (T,ΣT) and (U,ΣU) be triangulated categories and let F : T→ U

be a triangulated functor with natural isomorphism φ : FΣT → ΣU F. It follows fromD.8 and D.9 that the composite functors FΣT and ΣU F are triangulated with natu-ral isomorphisms−φΣT : FΣ2

T → ΣU FΣT and−ΣUφ : ΣU FΣT → Σ2U F. It follows

that the natural isomorphism φ is triangulated.

D.14 Definition. Let (T,Σ) be a triangulated category. A triangulated subcategoryof T is a full additive subcategory S that satisfies the following conditions.

(1) If N and N′ are isomorphic objects in T, then N is in S if and only if N′ is in S.(2) An object N is in S if an only if ΣN is in S.(3) For every distinguished triangle M→N→X→ΣM in T, such that the objects

M and N are in S, also X is in S.

Colloquially, the conditions (1)–(3) above are expressed as: S is closed underisomorphisms, Σ, and distinguished triangles in T.

Notice that if S is a triangulated subcategory of (T,Σ), then (S,Σ) is on its own atriangulated category.

THE FIVE LEMMA

The next result is the triangulated analogue of the Five Lemma.

D.15 Lemma. Let (T,Σ) be a triangulated category and consider a morphism,

M

ϕ

α// N

ψ

β// X

χ

γ// ΣM

Σϕ

M′ α′// N′

β′// X ′

γ′// ΣM′ ,

of distinguished triangles. If two of the morphisms ϕ, ψ, and χ are isomorphisms,then so is the third.

PROOF. By axiom (TR2’) it suffices to argue that if ϕ and ψ are isomorphisms, thenso is χ. To this end, it is enough to show that for every object Y in T, the inducedhomomorphism T(Y,χ) : T(Y,X)→ T(Y,X ′) of Abelian groups is an isomorphism.Indeed, by surjectivity of T(X ′,χ) there is a morphism ξ : X ′→ X with χξ = 1X ′ .Now the equalities χ(ξχ) = 1X ′χ= χ1X and injectivity of T(X ,χ) yield ξχ= 1X ,whence ξ is an inverse of χ.

Let Y be an object in T. To prove that T(Y,χ) is an isomorphism, apply the functorT(Y,–) : T→M(Z) to the given diagram. This yields a commutative diagram,



T(Y,M)

T(Y,ϕ)∼=

T(Y,α)// T(Y,N)

T(Y,ψ)∼=

T(Y,β)// T(Y,X)

T(Y,χ)

T(Y,γ)// T(Y,ΣM)

T(Y,Σϕ)∼=

T(Y,Σα)// T(Y,ΣN)

T(Y,Σψ)∼=

T(Y,M′)T(Y,α′)

// T(Y,N′)T(Y,β′)

// T(Y,X ′)T(Y,γ′)

// T(Y,ΣM′)T(Y,Σα′)

// T(Y,ΣN′) .

By the Five Lemma 1.1.2, it suffices to show that the rows in this diagram are exact.To this end it is, by another application of (TR2’), enough to see that the sequence

T(Y,M)T(Y,α)−−−−→ T(Y,N)

T(Y,β)−−−−→ T(Y,X)

is exact. As one has βα = 0, it follows that the inclusion ImT(Y,α) ⊆ KerT(Y,β)holds. Conversely, assume that ϑ : Y → N is in KerT(Y,β), that is, one has βϑ = 0.It follows that the next diagram is commutative,

Y 1Y// Y

ϑ

// 0

// ΣY

M α// N

β// X

γ// ΣM .

The rows are distinguished triangles by (TR1) and by assumption. By (TR2’) and(TR4’) there is a morphism υ : Y →M with αυ= ϑ, whence ϑ is in ImT(Y,α).

In the suggested analogy with Abelian categories, distinguished triangles of theform M −→ N −→ X 0−→ ΣM correspond to split exact sequences.

D.16 Definition. Let (T,Σ) be a triangulated category. A distinguished triangleM α−→ N

β−→ Xγ−→ ΣM in T is called split if there exist morphisms % : N→M and

σ : X → N such that the following hold,

%α = 1M, α%+σβ = 1N , and βσ = 1X .

D.17 Proposition. Let (T,Σ) be a triangulated category. For a distinguished triangle∆ = M α−→ N

β−→ Xγ−→ ΣM in T, the following conditions are equivalent.

(i) The distinguished triangle ∆ is split.(ii) There exists a morphism % : N→M such that %α= 1M .

(iii) There exists a morphism σ : X → N such that βσ= 1X .(iv) ∆ is isomorphic to the distinguished triangle M ε−→M⊕X $−→ X 0−→ ΣM,

where ε and $ are the injection and the projection, respectively.(v) The morphism γ is zero.

Moreover, if ∆ is split, then the diagram X σ−→ N%−→M 0−→ ΣX , where % and σ are

as in D.16, is a split distinguished triangle in T.

PROOF. Conditions (ii) and (iii) follow from (i). Condition (v) follows from (iii) asone has γβ= 0, and similarly (v) follows from (ii), as one has (Σα)γ = 0.

(v)=⇒ (iv): Consider the commutative diagram


482 Appendix D

(?)

M α// N

β// X

γ// ΣM

M εM// M⊕X $X

// X 0// ΣM .

The lower row in (?) is a distinguished triangle by D.4. By (TR4’) there exists amorphism ψ : N→M⊕X such that the resulting diagram is commutative, and itfollows from the Five Lemma D.15 that ψ is an isomorphism.

(iv)=⇒ (i): By assumption there is an isomorphism of distinguished triangles,

M

ϕ'

α// N

ψ'

β// X

χ'

γ// ΣM

Σϕ'

M εM// M⊕X $X

// X 0// ΣM .

With %= ϕ−1$Mψ and σ= ψ−1εXχ, one gets %α= ϕ−1$Mψα= 1M and, similarly,βσ= βψ−1εXχ= 1X . Finally, one has

α%+σβ= αϕ−1$Mψ+ψ−1εXχβ= ψ−1(εM$M +εX$X )ψ= 1N .

Now assume that ∆ is split. The diagram ∆′ = X σ−→ N%−→M 0−→ ΣX is a can-

didate triangle; indeed, pre and post composing the identity α%+σβ= 1N by σ and% one gets %σ= 0. If ∆′ is distinguished, then it is split because M→ ΣX is the zeromorphism. In the commutative diagram

(‡)

X σ// N

%// M 0

// ΣX

X εX// X⊕M $M

// M 0// ΣM ,

the lower row is a distinguished triangle by D.4. Since ∆ is split, the morphism ψ=(β %)T from N to X ⊕M is an isomorphism with inverse ψ−1 = (σ α); and it yieldsan isomorphism between ∆′ and the lower row in (‡). Thus ∆′ is distinguished.

The next result facilitates verification of axiom (TR4’).

D.18 Proposition. Let T be an additive category equipped with an additive auto-equivalence Σ. Consider a commutative diagram in T,

(D.18.1)

M1

ϕ

α1// N1

ψ

β1// X1

χ

γ1// ΣM1

Σ ϕ

M1

µ1

∼=

??

ϕ

α1// N1

ν1∼=

??

ψ

β1// X1

κ1∼=

??

χ

γ1// ΣM1

Σµ1

∼=

??

Σϕ

M2 α2// N2 β2

// X2 γ2// ΣM2 ,

M2

µ2∼=??

α2// N2

ν2∼=??

β2// X2

κ2∼=??

γ2// ΣM2

Σµ2∼=??



where (ϕ,ψ,χ) and (ϕ, ψ, χ) are morphisms of candidate triangles, while (µ1,ν1,κ1)and (µ2,ν2,κ2) are isomorphisms of candidate triangles. The mapping cone candi-date triangles of (ϕ,ψ,χ) and (ϕ, ψ, χ) are isomorphic.

PROOF. The commutative diagram

M2

⊕N1

(α2 ψ

0 −β1

)//

∼=(µ2 00 ν1

)

N2

⊕X1

(β2 χ

0 −γ1

)//

∼=(ν2 00 κ1

)

X2

⊕ΣM1

(γ2 Σϕ

0 −Σα1

)//

∼=(κ2 00 Σµ1

)

ΣM2

⊕ΣN1

∼=(Σµ2 0

0 Σν1

)

M2

⊕N1

(α2 ψ

0 −β1

)//

N2

⊕X1

(β2 χ

0 −γ1

)//

X2

⊕ΣM1

(γ2 Σ ϕ

0 −Σ α1

)//

ΣM2

⊕Σ N1

is an isomorphism from the mapping cone candidate triangle of (ϕ,ψ,χ) to the map-ping cone candidate triangle of (ϕ, ψ, χ).

EXERCISES

E D.1 Show that a morphism in a triangulated category is a monomorphism (epimorphism) if andonly if it has a left (right) inverse. Conclude that every object in a triangulated category isboth injective and projective.

E D.2 Let α : M→ N be a morphism in a triangulated category (T,Σ). Show that the complex Xin a distinguished triangle M α−→ N −→ X −→ ΣM in T is unique up to isomorphism.

E D.3 (Cf. D.6) Let (T,Σ) be a triangulated category. Show that (Top,Σ−1) is triangulated in thecanonical way: A candidate triangle M→ N→ X → Σ−1 M in Top is distinguished if andonly if the candidate triangle Σ−1 M→ X → N→M is distinguished in T.

E D.4 Two homomorphisms of R-modules α,β : M→ N are called stably equivalent if α−β fac-tors through a projective R-module. The stable module category M(R) has as objectsall R-modules. The hom-set M(R)(M,N), often written as HomR(M,N), is the set ofclasses of stably equivalent homomorphisms M → N. (a) Show that M(R) is a k-linearcategory with coproducts. (b) For an R-module M, let Ω(M) be the kernel of any pro-jective precover P M. Show that Ω is a well-defined k-linear endofunctor on M(R).(c) Show that the category M(R) is triangulated if R is quasi-Frobenius.

E D.5 Show that the stable module category M(Q[x]/(x2)) is not Abelian.


Appendix EProjective Dimension of Flat Modules

SYNOPSIS. Jensen’s theorem; Homotopy Lemma; pure exact sequence; ℵ-generated module;countably related flat module; continuous chain of modules; transfinite extension of projectivemodules; flat preenvelopes.

The goal is to prove the following result of Jensen [41]; stated in 8.5.16.

Theorem. For every flat R-module F one has pdR F 6 FPD(R).

The proof is based on an in-depth study of kernels of free precovers of flat modules.As part of it emerges a result of independent interest, E.8, which asserts that a flatmodule has projective dimension at most 1 if it can be generated by a set of elementswith only countably many relations among them.

PURITY AND FLATNESS

The first result is known as the Homotopy Lemma.

E.1 Lemma. Consider a commutative diagram in M(R) with exact rows,

M′

ϕ′

α′// M

ϕ

α// M′′

ϕ′′

// 0

0 // N′β′// N

β// N′′ .

The following conditions are equivalent.(i) There exists a homomorphism σ : M′′→ N with βσ= ϕ′′.

(ii) There exists a homomorphism % : M→ N′ with %α′ = ϕ′.

PROOF. (i)=⇒ (ii): Apply the functor HomR(M,–) to the lower row in the diagramto get an exact sequence

0−→ HomR(M,N′)Hom(M,β′)−−−−−−→ HomR(M,N)

Hom(M,β)−−−−−−→ HomR(M,N′′) .

485

486 Appendix E

If βσ= ϕ′′ holds, then one has β(ϕ−σα) = βϕ−ϕ′′α= 0, so there is a homomor-phism % : M→ N′ with β′%= ϕ−σα. Consequently, one has β′%α′ = (ϕ−σα)α′ =ϕα′ = β′ϕ′, and hence %α′ = ϕ′, as β′ is injective.

(ii)=⇒ (i): Apply HomR(–,N) to the upper row and proceed as above.

E.2 Theorem. For a short exact sequence of R-modules,

η = 0−→M′ α′−−→M α−−→M′′ −→ 0 ,

the following conditions are equivalent.(i) The sequence HomR(N,η) is exact for every finitely presented R-module N.

(ii) The sequence K⊗R η is exact for every Ro-module K.(iii) The sequence Homk(η,E) of Ro-modules is split exact every injective k-

module E, equivalently, for some faithfully injective k-module E.(iv) For every commutative diagram in M(R),

L′

ϕ′

κ// L

ϕ

M′ α′// M ,

where L and L′ are finitely generated free R-modules, there is a homomor-phism % : L→M′ with %κ= ϕ′.

PROOF. (i)⇐⇒ (ii): It follows from C.8, 3.2.16, and 3.2.32 that K⊗R η is exact forevery Ro-module K if and only if it is exact for every finitely presented Ro-module K.For every m×n matrix α with entries in R, a finitely presented R-module Nα and afinitely presented Ro-module Kα are defined by the exact sequences

να = Rm ·α−−→ Rn −→ Nα −→ 0 and κα = Rn α·−−→ Rm −→ Kα −→ 0 .

Every finitely presented R-module has the form Nα, and every finitely presentedRo-module has the form Kα. Hence it suffices to prove, for every matrix α, thatHomR(Nα,η) is exact if and only if Kα⊗R η is exact. Let X be an R-module; one has

HomR(να,X) = 0−→ HomR(Nα,X)−→ HomR(Rn,X)Hom(·α,X)−−−−−−→ HomR(Rm,X) ,

and under the counitor (1.2.1.2) the homomorphism of k-modules HomR(·α,X) cor-responds to Xn α·−→ Xm as HomR(Rn,X) and HomR(Rm,X) get their R-module struc-tures from RR. Splicing together the exact sequences HomR(να,X) and κα⊗R X , onegets an exact sequence of k-modules,

0−→ HomR(Nα,X)−→ Xn α·−−→ Xm −→ Kα⊗R X −→ 0 .

This sequence depends naturally on X , so η induces an exact sequence,

0−→ HomR(Nα,η)−→ ηn −→ η

m −→ Kα⊗R η−→ 0 ,

of k-complexes. As η is exact, so are ηn and ηm, and it follows that HomR(Nα,η) isexact if and only if Kα⊗R η is exact; cf. 2.2.20 .


Projective Dimension of Flat Modules 487

(ii)⇐⇒ (iii): Let K be an Ro-module and E be a k-module. By adjunction 1.2.5and commutativity 1.2.2 one has

HomRo(K,Homk(η,E)) ∼= Homk(K⊗R η,E) .

Thus, if E is injective and the sequence K⊗R η is exact for every Ro-module K,then HomRo(K,Homk(η,E)) is exact for every Ro-module K, whence the exact se-quence HomR(η,E) of Ro-modules is split. Conversely, if E is faithfully injectiveand Homk(η,E) is split exact, then HomRo(K,Homk(η,E)), and hence K⊗R η, isexact for every Ro-module K.

(iv)⇐⇒ (i): Let N be a finitely presented R-module and consider a presentation

(?) L′ κ−−→ L π−−→ N −→ 0

where L and L′ are finitely generated free R-modules. To show that the sequenceHomR(N,η) is exact it suffices to prove that for every homomorphism ϕ′′ : N→M′′

there exists a homomorphism σ : N→M with ασ = ϕ′′. Given a homomorphismϕ′′ : N→M′′, the extension property 1.3.5 yields a commutative diagram in M(R),

(‡)

L′

ϕ′

κ// L

ϕ

π// N

ϕ′′

// 0

0 // M′ α′// M α

// M′′ // 0 .

By (iv) there is a homomorphism % : L→M′ with %κ= ϕ′, and the existence of thedesired σ now follows from E.1. Conversely, consider the commutative square in(iv), define a finitely presented R-module N by exactness of (?), and consider theinduced commutative diagram (‡). By (i) there is a homomorphism σ : N→M withασ= ϕ′′, so E.1 yields a homomorphism % : L→M′ with %κ= ϕ′, as desired.

REMARK. The conditions in E.2 are also equivalent to the exact sequence η being isomorphic toa filtered colimit of split exact sequences.

E.3 Definition. An exact sequence of R-modules, 0→M′→M→M′′→ 0, is calledpure if it satisfies the conditions in E.2. A surjective homomorphism α : M→M′′ iscalled a pure epimorphism if the exact sequence 0→Kerα→M→M′′→ 0 is pure.An injective homomorphism α : M′→M is called a pure monomorphism if the exactsequence 0→M′→M→Cokerα→ 0 is pure. A submodule M′ of M is called pureif the embedding M′M is a pure monomorphism.

REMARK. A module N with the property that Hom(N,η) is exact for every pure exact sequenceη = 0→ M′ → M → M′′ → 0 is called pure-projective. Dually one defines pure-injective, alsocalled algebraically compact, modules. These notions form the basis for pure homological algebra.

An exact sequence 0→ M → N → F → 0 with F flat is pure; that is immedi-ate from 5.4.3. Soon we establish a partial converse to this statement, but first weinterject a definition and some observations on submodules of free modules.

E.4 Definition. Let ℵ be an infinite cardinal. An R-module is called ℵ-generated ifit has a set E of generators with cardE 6 ℵ, and it is called ℵ<-generated if it has


488 Appendix E

a set E of generators with cardE < ℵ. For an ordinal α the notation α < ℵ meansα < ωℵ, where ωℵ is the first ordinal of cardinality ℵ. For an ℵ-generated modulethis convention allows us to write a set of generators on the form E = eαα<ℵ.

An ℵ0-generated module is called countably generated; notice that a module isℵ<0 -generated if and only if it is finitely generated.

E.5. Let L be a free R-module with basis euu∈U . Every element l in L has arepresentation l = ∑u∈Ul

rueu for a uniquely determined finite subset Ul of U anduniquely determined elements ru 6= 0 in R. Note that U0 is the empty set.

For a submodule K of L, set UK =⋃

k∈K Uk, then K is contained in the free sub-module L′ = R〈UK〉 of L, and the quotient L/L′ is free with basis U \UK . Note thatif K is generated by k1, . . . ,kn, then one has UK =Uk1 ∪ . . .∪Ukn ; that is, UK is finite.Similarly, if ℵ is an infinite cardinal and K is ℵ-generated, then one has cardUK 6ℵ.

E.6 Proposition. For an exact sequence η= 0−→K κ−→ L π−→F −→ 0 of R-moduleswhere L is free, the following conditions are equivalent.

(i) F is flat.(ii) The sequence η is pure.

(iii) For every finitely generated submodule H of K there is a homomorphism% : L→ K that satisfies %κ|H = 1H .

PROOF. (i)⇐⇒ (ii): As L is free, one has TorR1 (−,L) = 0 by 8.3.2, and it follows

from E.2, 7.4.13, and 7.4.11 that η is pure exact if and only if TorR1 (−,F) = 0 holds,

i.e. if and only if F is flat.(ii)=⇒ (iii): Let H be a finitely generated submodule of K. Choose by 1.3.11

a surjective homomorphism π′ : L′→ H with L′ finitely generated and free. AsIm(κπ′) = κ(H) is a finitely generated submodule of L, it is contained in a finitelygenerated direct summand L′′ of L; cf. E.5. Now consider the commutative diagram,

L′

π′

κπ′// L′′

ε

K κ// L .

By (ii) the sequence η is pure exact, so E.2 yields a homomorphism %′ : L′′→ K with%′(κπ′) = π′. As L′′ is a direct summand of L, there is a projection $ : L L′′ with$ε= 1L′′ . Set %= %′$ and note that one has %κπ′ = %′$εκπ′ = %′κπ′ = π′. As Imπ′

is H it follows that %κ|H = 1H holds, as desired.(iii)=⇒ (ii): Consider a commutative diagram,

L′

ϕ′

κ′// L′′

ϕ′′

K κ// L ,

where also L′ and L′′ are finitely generated free R-modules. To show that η is pure,it is by E.2 sufficient to show the existence of a homomorphism %′ : L′′→ K with



%′κ′ = ϕ′. As Imϕ′ is a finitely generated submodule of K, it follows from (iii) thatthere is a homomorphism % : L→ K with %κϕ′ = ϕ′. Now set %′ = %ϕ′′, and note thatone has %′κ′ = %ϕ′′κ′ = %κϕ′ = ϕ′, as desired.

E.7. Consider a short exact sequence of R-modules, 0−→K κ−→ L π−→F −→ 0 whereL is free with basis euu∈U and F is flat. Let K′ be a submodule of K and let L′ bea free submodule of L with κ(K′)⊆ L′; cf. E.5. Let κ′ : K′→ L′ be the restriction ofκ, set F ′ = Cokerκ′, and let π′ : L′→ F ′ be the canonical homomorphism. If K′ is apure submodule of K, then F ′ is flat. Indeed, there is a commutative diagram

(E.7.1)

0 // K′

ι

κ′// L′

π′// F ′

// 0

0 // K κ// L π

// F // 0 .

It is elementary to verify that a composite αβ of monomorphisms is pure if both αand β are pure and only if β is pure. By E.6 the monomorphism κ is pure. Thus, if ιis pure, then κι is pure, whence κ′ is pure by commutativity of (E.7.1), and F ′ is flat.

A module M is called countably related if there exists a short exact sequence0 → K → L → M → 0 where L is free and K is countably generated. The nexttheorem asserts: A countably related flat module has projective dimension at most 1.

E.8 Theorem. Let 0→ K→ L→ F→ 0 be an exact sequence of R-modules whereL is free and F is flat. If K is countably generated, then K is projective.

PROOF. By 8.1.19 the module K is projective if and only if one has pdR F 6 1.As in E.7 denote the homomorphisms in the exact sequence by κ and π. Since Kis countably generated, one has a diagram as (E.7.1) with L′ countably generatedand L′′ = L/L′ free; see E.5. As one has F = F ′⊕ L′′, we may assume that L iscountably generated in the first place. To prove the inequality pdR F 6 1, we write Kas the union of a chain K1⊆K2⊆ ·· · of finitely generated submodules and constructendomorphisms ϕu : L→ L with the following properties.

(1) πϕu = π for all u in N.(2) ϕuκ|Ku = 0 for all u in N.(3) ϕwϕv = ϕw for all w > v in N.

From these data we construct an exact sequence,

(?) 0−→ L(N) Φ−−→ L(N) Π−−→ F −→ 0 ,

showing that pdR F 6 1 holds as desired.Assume for the moment that the submodules Ku and the endomorphisms ϕu have

been constructed. The sequence (?) is then obtained by setting

Φ((lu)u∈N) = (lu−ϕu−1(lu−1))u∈N and Π((lu)u∈N) = π(∑u∈N lu) ,

with the ingenuous convention ϕ0(l0) = 0. It is evident that Φ is injective and thatΠ is surjective; furthermore, (1) yields


490 Appendix E

ΠΦ((lu)u∈N) = π( ∑u∈N

(lu−ϕu−1(lu−1)) = π( ∑u∈N

lu)−π( ∑u∈N

ϕu(lu)) = 0 .

To show that (?) is exact, it remains to see that KerΠ is contained in ImΦ. Letl = (lu)u∈N be an element in KerΠ , then ∑u∈N lu is in Kerπ = Imκ, so there is anelement k in K =

⋃u∈NKu with κ(k) = ∑u∈N lu. As the submodules Ku form an

ascending chain, one has k ∈ Ku for all u 0. Choose v such that k is in Kv andsuch that lu = 0 holds for all u > v. Define x = (xu)u∈N in L(N) by setting x1 = l1,

xu = lu +ϕu−1(u−1∑

i=1li) for 1 < u6 v and xu = 0 for u > v .

One has Φ(x) = l; that is, xu−ϕu−1(xu−1) = lu holds for all u ∈ N. It is evident foru= 1, and for u= 2 it is a short computation, x2−ϕ1(x1) = l2+ϕ1(l1)−ϕ1(l1) = l2.For 2 < u6 v one has ϕu−1ϕu−2 = ϕu−1 by (3) and, consequently,

xu−ϕu−1(xu−1) = lu +ϕu−1(u−1∑

i=1li)−ϕu−1(lu−1 +ϕu−2(

u−2∑

i=1li)) = lu .

For u > v+1 one has xu−ϕu−1(xu−1) = 0 = lu and, finally, (3) and (2) yield

xv+1−ϕv(xv) = 0−ϕv(lv +ϕv−1(v−1∑

i=1li)) = −ϕv(

v∑

i=1li) = ϕvκ(k) = 0 = lv+1 .

It remains to construct the submodules Ku and the endomorphisms ϕu. Recallthat L is countably generated and let euu∈N be a basis; we proceed with a con-struction that applies to every finitely generated submodule H of K. Assume that His generated by elements h1, . . . ,hn. By E.6 there is a homomorphism % : L→ K with%κ|H = 1H . Choose v 0 such that κ(H) is contained in R〈e1, . . . ,ev〉, and denoteby H the submodule of K generated by %(e1), . . . ,%(ev). The assignments

eu 7→ eu−κ%(eu) for 16 u6 v and eu 7→ eu for u > v

define an endomorphism ϕH : L→ L with the following properties,(I) For every u ∈ N one has ϕH(eu) = eu + xu for some xu ∈ κ(H).

(II) πϕH = π.(III) ϕHκ|H = 0.(IV) H ⊆ H.

Property (I) is immediate from the definitions of ϕH and H, and then (II) holds asthe composite πκ is the zero morphism. To establish (III) and (IV) write

κ(h ) =v∑

u=1r ueu for ∈ 1, . . . ,n .

For every ∈ 1, . . . ,n one has h = %κ(h ) = ∑vu=1 r u%(eu), which proves (IV).

The next computation establishes (III); the penultimate equality uses the identity%κ|H = 1H .



ϕHκ(h ) = ϕH(v∑

u=1r ueu)

=v∑

u=1r u(eu−κ%(eu))

= κ(h )−κ%κ(h )= κ(h −%κ(h ))= 0

Being countably generated, K is the union of an ascending chain H1 ⊆ H2 ⊆ ·· ·of finitely generated submodules. Set K1 = H1 and Ku = Ku−1 +Hu for u > 1. Bythese definitions and (IV) one has K =

⋃u∈NKu and

K1 ⊆ K1 ⊆ K2 ⊆ K2 ⊆ K3 ⊆ ·· · .

For every u ∈ N set ϕu = ϕKu; the properties (1) and (2) follow immediately from

(II) and (III). To prove (3), let w > v and apply (I) to write ϕKv(eu) = eu + xu for

some xu ∈ κ(Kv)⊆ κ(Kw). As ϕKwκ|Kw = 0 holds by (2) one has ϕKw

(xu) = 0 and,therefore, ϕKw

ϕKv(eu) = ϕKw

(eu + xu) = ϕKw(eu).

E.9 Proposition. Let 0−→ K κ−→ L π−→ F → 0 be an exact sequence of R-moduleswhere L is free and F is flat. For every countably generated submodule H of K thereis a countably generated submodule H ′ of K such that H is contained in H ′ and κ|H ′is a pure monomorphism.

PROOF. Let euu∈U be a basis for L and let hnn∈N be a set of generators for H.Set H0 = 0; we construct a family of modules Hnn∈N and a family of homomor-phisms %n : L→ Hnn∈N with the following properties.

(1) Hn is a finitely generated submodule of K with hn ∈ Hn and Hn−1 ⊆ Hn.(2) %nκ|Hn−1 = 1Hn−1

.Assuming that such families have been constructed, set H ′ =

⋃n∈NHn. By construc-

tion, H ′ is a submodule of K and countably generated. Moreover, H ′ contains H aseach generator hn is in H ′ by (1). To see that the monomorphism κ|H ′ is pure, itsuffices by E.6 to argue that for every finitely generated submodule H ′′ of H ′ thereis a homomorphism % : L→ H ′ with %κ|H ′′ = 1H ′′ . Thus, let H ′′ be a finitely gener-ated submodule of H ′ and choose by (1) an m in N such that H ′′ is contained in Hm.Since %m+1 : L→ Hm+1 ⊆ H ′ satisfies %m+1κ|Hm = 1Hm

by (2), the homomorphism%= %m+1 has the sought after property.

To construct the submodules Hn and the homomorphisms %n, set %0 = 0 andproceed recursively. Assuming that Hn−1 and %n−1 have been constructed, the sub-module Kn = Hn−1 +Rhn of K is finitely generated, so there exists by E.6 a homo-morphism %′ : L→ K with %′κ|Kn = 1Kn

. Let Uκ(Kn) be the set constructed in E.5.The assignments

eu 7→ %′(eu) for u ∈Uκ(Kn) and eu 7→ 0 for u /∈Uκ(Kn)

define a homomorphism %n : L→ K, and with Hn = Im%n both (1) and (2) hold.Indeed, the set Uκ(Kn) is finite, and thus Hn is finitely generated. For h ∈ Kn one has


492 Appendix E

κ(h) = ∑u∈Uκ(h) rueu, and since Uκ(h) is contained in Uκ(Kn), one obtains

(?) h = %′κ(h) = ∑u∈Uκ(h)

ri%′(eu) = ∑

u∈Uκ(h)ri%

n(eu) = %nκ(h) ;

in particular, h belongs to Hn. With Kn contained in Hn, one has both hn ∈ Hn andHn−1 ⊆Hn, which establishes (1). Furthermore, as Hn−1 is contained in Kn one has%nκ|Hn−1 = 1Hn−1

, also by (?), whence (2) holds as well.

JENSEN’S THEOREM

The proof of Jensen’s theorem uses transfinite induction, and the preceding results,E.8 and E.9, furnish the base cases for the final induction arguments.

E.10 Definition. Let M be an R-module and let λ be an ordinal. A family Mαα<λof submodules of M is called a continuous chain if the next conditions are satisfied.

(1) For α6 β< λ one has Mα ⊆Mβ.(2) If β< λ is a limit ordinal, then one has Mβ =

⋃α<βMα.

The next result is due to Auslander [1]; the case n = 0 says that the class ofprojective modules is closed under transfinite extensions.

E.11 Proposition. Let M be an R-module and let n > 0 be an integer. If M isthe union of a continuous chain Mαα<λ of submodules with pdR M0 6 n andpdR(M

α+1/Mα)6 n for every ordinal α with α+1 < λ, then one has pdR M 6 n.

PROOF. We proceed by induction on n; first assume n = 0. For each ordinal α withα+ 1 < λ there is a short exact sequence 0 → Mα → Mα+1 → Mα+1/Mα → 0,which splits as Mα+1/Mα is projective. Thus, there is a projective submodule Nα+1

of Mα+1 with Nα+1 ∼= Mα+1/Mα and Mα+1 = Mα⊕Nα+1. By transfinite inductionon 16 α< λ we now establish an equality

(?) Mα = M0⊕∐γ<α

Nγ+1 .

For α = 1 the equality holds as one has M1 = M0⊕N1. Now let 1 < α < λ andassume that (?) holds for all β< α. If α= β+1 is a successor ordinal, then one has

Mα = Mβ+1 = Mβ⊕Nβ+1 = (M0⊕∐γ<β

Nγ+1)⊕Nβ+1 = M0⊕∐γ<α

Nγ+1 ;

here the second equality holds by the definition of Nβ+1, and the third one uses theinduction hypothesis. If α is a limit ordinal, then one has

Mα =⋃β<α

Mβ =⋃β<α

(M0⊕∐γ<β

Nγ+1) = M0⊕∐γ<α

Nγ+1.

As M is the union⋃α<λMα, one gets as above M = M0⊕

∐γ<λ′ Nγ+1, where λ′ = λ

if λ is a limit ordinal and λ′ = λ−1 if λ is a successor ordinal. As M0 is projectiveand each Nγ+1 is projective, it follows from 1.3.20 that M is projective.



Now let n > 0 and assume that the statement holds for n− 1. Consider thecanonical surjection π : L = R(M)→M and set K = Kerπ. It suffices to establishthe inequality pdR K 6 n− 1, as pdR M 6 n then holds by 8.1.10. To this end, setLα = R(Mα) ⊆ L and Kα = Ker(π|Lα) = Lα ∩K for α < λ. As M is the union of thecontinuous chain Mαα<λ, it follows that K is the union of the continuous chainKαα<λ. Thus, by the induction hypothesis, it is sufficient to show that one haspdR K0 6 n− 1, and pdR(K

α+1/Kα) 6 n− 1 for all ordinals α with α+ 1 < λ. AspdR M06 n holds by assumption, the first inequality follows by application of 8.1.10to the exact sequence 0→ K0→ L0→M0→ 0. To prove the second inequality, ap-ply the Snake Lemma 1.1.4 to the commutative diagram

0 // Kα

// Lα

π|Lα//// Mα

// 0

0 // Kα+1 // Lα+1π|Lα+1

// // Mα+1 // 0

to obtain the exact sequence,

0−→ Kα+1/Kα −→ Lα+1/Lα −→Mα+1/Mα −→ 0 .

The quotient module Lα+1/Lα is free with basis Mα+1 \Mα, and by assumption onehas pdR(M

α+1/Mα)6 n, so 8.1.10 yields pdR(Kα+1/Kα)6 n−1.

E.12 Lemma. Let κ : M→ N be a homomorphism of R-modules, and assume thatM is the union of a continuous chain Mαα<λ of submodules. If κ|Mα

: Mα→ N isa pure monomorphism for every α< λ, then κ is a pure monomorphism.

PROOF. For α< β< λ there is a commutative diagram with exact rows,

ηα = 0 // Mα

κ|Mα// N // Coker(κ|Mα

)

// 0

ηβ = 0 // Mβ

κ|Mβ// N // Coker(κ|Mβ

) // 0 .

These diagrams yield a filtered direct systems of R-complexes with

η = colimα<λ

ηα = 0−→M κ−−→ N −→ Cokerκ−→ 0 ,

cf. 3.2.40, and η is exact by 3.2.35. Let K be an Ro-module; by 3.2.17 one has

K⊗R η = K⊗R (colimα<λ

ηα) ∼= colimα<λ

(K⊗R ηα) .

Each sequence K⊗R ηα is exact as κ|Mαis a pure monomorphism, see E.2, so it

follows, again by 3.2.35, that K⊗R η is exact. Thus, K⊗R η is exact for every Ro-module K, whence η is pure exact; that is, κ is a pure monomorphism.

E.13 Proposition. Let 0−→K κ−→ L π−→ F→ 0 be an exact sequence of R-moduleswhere L is free and F is flat. Let ℵ be an infinite cardinal, and let H be an ℵ-generated submodule of K. The following assertions hold.


494 Appendix E

(a) There is an ℵ-generated submodule H ′ of K such that H is contained in H ′

and κ|H ′ is a pure monomorphism.(b) If ℵ is uncountable, then there is a continuous chain Hαα<ℵ of submod-

ules in K, such that each module Hα is maxcardα,ℵ0-generated, eachmonomorphism κ|Hα is pure, and one has H ⊆

⋃α<ℵHα.

PROOF. We proceed by induction on ℵ. For ℵ = ℵ0 part (a) holds by E.9. Assumenow that ℵ is uncountable and that (a) holds for all ℵ<-generated submodules of K.The goal is to prove (a) and (b) for all ℵ-generated submodules H of K; we start bynoting that (a) follows from (b). Indeed, assume (b) holds and set H ′ =

⋃α<ℵHα.

This module contains H, and it follows from E.12 that κ|H ′ is a pure monomor-phism. Since each module Hα has a set Eα of generators with cardEα 6 ℵ (actually,cardEα < ℵ) the set E ′ =

⋃α<ℵEα of generators for H ′ has cardE ′ 6 ℵ2 = ℵ.

To prove (b), let H be an ℵ-generated submodule of K, and let hαα<ℵ be a setof generators for H. We construct a continuous chain Hαα<ℵ of maxcardα,ℵ0-generated submodules of K such that each monomorphism κ|Hα is pure, and one hashα ∈Hβ for all α< β. The module H ′ =

⋃α<ℵHα then contains the family hαα<ℵ

and hence it contains H.Set H0 = 0; it is generated by ∅, it is a submodule of every other submodule

of K, and the zero morphism is pure. Now, let γ < ℵ and assume that there is acontinuous chain Hαα<γ of maxcardα,ℵ0-generated submodules of K suchthat each monomorphism κ|Hα is pure, and one has hα ∈ Hβ for all α< β< γ.

If γ is a limit ordinal, then Hγ =⋃α<γHα contains hα for all α < γ, the family

Hαα<γ+1 is a continuous chain of submodules, and κ|Hγ is pure by E.12. To seethat Hγ is maxcardγ,ℵ0-generated, note that each module Hα for α < γ has aset of generators Eα with cardEα 6 maxcardα,ℵ0 6 cardγ. Thus, the set Eγ =⋃α<γEα of generators for Hγ has cardEγ 6 (cardγ)2 = cardγ= maxcardγ,ℵ0.If γ = α+ 1 is a successor ordinal, consider the submodule K′ = Hα+Rhα of

K. By assumption, Hα and, therefore, K′ has a generating set of cardinality at mostmaxcardα,ℵ0 = maxcardγ,ℵ0 < ℵ. By the assumption that (a) holds for allℵ<-generated submodules of K, there is a maxcardγ,ℵ0-generated submodule

Hγ of K such that κ|Hγ is pure and K′ is contained in Hγ. This ensures that hα is inHγ for all α< γ and that Hαα<γ+1 is a continuous chain of submodules.

The next result is essentially a restatement of Jensen’s theorem.

E.14 Theorem. If FPD(R) is finite, then every pure submodule of a free R-modulehas finite projective dimension.

PROOF. Let L be a free R-module, let ℵ be an infinite cardinal, and let H be an ℵ-generated pure submodule of L. By transfinite induction we prove that H has finiteprojective dimension. Note that L/H is a flat module by E.6.

If H is ℵ0-generated, then it is projective by E.8. Now let ℵ> ℵ0, and assumethat all ℵ<-generated pure submodules of L have finite projective dimension. Anapplication of E.13 to the exact sequence 0 → H → L → L/H → 0 shows thatH is the union of a continuous chain Hαα<ℵ of ℵ<-generated pure submodulesof L. Set p = FPD(R); by the induction hypothesis the modules in the chain have



pdR Hα 6 p. For every α with α+ 1 < ℵ it follows from 8.1.10 applied to the se-quence 0→Hα→Hα+1→Hα+1/Hα→ 0 that also pdR(H

α+1/Hα)6 p holds. Thedesired conclusion, pdR H 6 p, is now immediate from E.11.

PROOF OF 8.5.16. We may assume that FPD(R) is finite. Let F be a flat R-moduleand choose by 1.3.11 a surjective homomorphism π : L→ F with L free. The exactsequence 0→ Kerπ→ L→ F → 0 is pure by E.6, so Kerπ has finite projectivedimension by E.14. It follows from 8.1.10 that F has finite projective dimension aswell, whence pdR F 6 FPD(R) holds.

FLAT PREENVELOPES

E.15. Let α be an infinite cardinal and let Muu∈U be a family of R-modules withcardMu 6 α for every u ∈U . For the coproduct one has

card(∐

u∈U Mu) 6 maxα,cardU .

If each Mu is a submodule of a common R-module M, then one can consider the sum∑u∈U Mu. There is a canonical surjective homomorphism

∐u∈U Mu→∑u∈U Mu, and

hence one also has card(∑u∈U Mu)6maxα,cardU.

E.16 Lemma. Given a cardinal α there is a cardinal α, depending on α and cardR,with the following property: For every R-module M and every submodule N ⊆ Mwith cardN 6 α, there exists a pure submodule N ⊆M with card N 6 α and N ⊆ N.

PROOF. Let ∆ be the set of all triples (m,n,κ) where m,n ∈ N and κ : Rm→ Rn is ahomomorphism. Let κ : F → F ′ be the coproduct over ∆ of the homomorphisms κ.That is, one has F =

∐(m,n,κ)∈∆Rm and F ′ =

∐(m,n,κ)∈∆Rn, and κ is uniquely deter-

mined by commutativity of the diagrams,

Rm

ε(m,n,κ)

κ// Rn

ε′(m,n,κ)

F κ// F ′ ,

where ε and ε′ denote the canonical embeddings.We now construct a sequence N = N0 ⊆ N1 ⊆ N2 ⊆ ·· · of submodules of M. Set

N0 = N. Given Nu we construct Nu+1 as follows. Let Ψu be the set of all homomor-phisms ψ : F → Nu for which there exists some commutative diagram,

(?)

F

ψ

κ// F ′

Nu //νu// M ,

where νu is the embedding. Choose for every ψ∈Ψu a homomorphism ψ′ : F ′→M,that makes the diagram (?) commutative and set Nu+1 = Nu +∑ψ∈Ψu Imψ′. We


496 Appendix E

show that N =⋃

∞u=0 Nu has the desired properties. Notice that each νu factors as a

composite of embeddings νu : Nu N and ν : NM.Evidently one has N ⊆ N ⊆M. To see that N is a pure submodule of M it suffices,

by E.2, to show that for every commutative diagram,

(‡)

Rm

ϕ

κ// Rn

ϕ′

N // ν// M ,

there exists a homomorphism % : Rn→ N with %κ = ϕ. Let (‡) be given and noticethat (m,n,κ) is an element in ∆. As Rm is finitely generated, ϕ factors through someNu, that is, ϕ can be written as a composite

Rm ϕ// Nu //

νu// N .

Let ψ : F → Nu and ξ : F ′→M be the homomorphisms given by

ψε(m,n,κ) =

ϕ (m, n, κ) = (m,n,κ)0 (m, n, κ) 6= (m,n,κ)

and ξε′(m,n,κ) =

ϕ′ (m, n, κ) = (m,n,κ)0 (m, n, κ) 6= (m,n,κ)

for (m, n, κ) ∈ ∆. Now one has νuψ= ξκ, that is, the diagram (?) with ξ as the dottedarrow is commutative. Indeed, one has

νuψε(m,n,κ) = νuϕ = ν νuϕ = νϕ = ϕ′κ = ξε′(m,n,κ)κ = ξκε(m,n,κ)

and νuψε(m,n,κ) = 0 = ξε′(m,n,κ)κ= ξκε(m,n,κ) for all (m, n, κ) 6= (m,n,κ) in ∆. Thismeans that the homomorphism ψ : F → Nu belongs to the set Ψu. By construction ofNu+1 there exists a homomorphism ψ′ that makes the diagram below commutative,

F

ψ

κ// F ′

ψ′

Nu //νu,u+1

// Nu+1 ;

here νu,u+1 denotes the embedding. We argue that %= νu+1ψ′ε′(m,n,κ) : Rn→ N sat-

isfies %κ = ϕ as required. As ν is injective, it suffices to show that ν%κ = νϕ holds.To verify this is a computation based on commutativity of the diagrams above:

ν%κ = ν νu+1ψ′ε′(m,n,κ)κ = νu+1ψ

′ε′(m,n,κ)κ = νu+1ψ′κε(m,n,κ)

= νu+1νu,u+1ψε(m,n,κ) = νuψε(m,n,κ) = νuϕ = ν νuϕ = νϕ .

It remains to prove the assertion about cardinality. Without loss of generality onecan assume that α is infinite. Set λ= maxℵ0,cardR and α= αλ. We show that ifcardN 6 α holds, then card N 6 α. As N is the countable union

⋃∞u=0 Nu, it suffices

to argue that cardNu 6 α holds for every u. We prove this by induction. For u = 0we have N0 = N and hence cardN0 6 α 6 α. Now assume that cardNu 6 α holds



for some u. Note that card∆ = λ whence the free modules F and F ′ have bases ofcardinality λ. It follows that

HomR(F,Nu) ∼= HomR(R(λ),Nu) ∼= N λu ,

and consequently, card(HomR(F,Nu))6 αλ = (αλ)λ = αλ·λ = αλ = α, where the in-equality comes from the assumption cardNu 6 α. As Ψu is a subset of HomR(F,Nu)one also has cardΨu 6 α. Recall that Nu+1 = Nu +∑ψ∈Ψu Imψ′ where each ψ′ is ahomomorphism F ′→ X . In particular, one has

card(Imψ′) 6 cardF ′ = cardR(λ) 6 maxλ,cardR = λ 6 α .

Thus Nu+1 is a sum, over a set of cardinality at most α, of modules of cardinality atmost α. It follows that cardNu+1 6 α; see E.15.

E.17 Definition. Let X be a class of R-modules and let M be an R-module. An X-preenvelope of M is a homomorphism ϕ : M→ X with X ∈ X such that for everyhomomorphism ϕ′ : M→ X ′ with X ′ ∈ X there is a homomorphism χ : X → X ′ thatmakes the following diagram commutative,

M

ϕ′

ϕ// X

χ~~

X ′ .

E.18. Let X be a class of R-modules and let M be an R-module. A homomorphismϕ : M→ X with X ∈ X is an X-preenvelope if and only if HomR(ϕ,X ′) is surjectivefor every X ′ ∈ X. Notice that if M can be embedded in some module from X, thenevery X-preenvelope M→ X is injective.

E.19 Theorem. Let X be a class of R-modules. If the next two conditions are satis-fied, then every R-module has an X-preenvelope.

(1) Every pure submodule of a module in X belongs to X.(2) For every family Xuu∈U of modules in X the product

∏u∈U Xu is in X.

PROOF. Let M be an R-module. Set α = cardM and let α be as in E.16. The col-lection of isomorphism classes of modules in X of cardinality 6 α constitutes a set;let Xuu∈U be a set of representatives for these classes. That is, every X ∈ X withcardX 6 α is isomorphic to some Xu. By assumption the product

X =∏u∈U

XHomR(M,Xu)u

belongs to X. For every u ∈U and β ∈ HomR(M,Xu) let πu,β : X → Xu be the pro-jection. Now, let ϕ : M→ X be the unique homomorphism that satisfies πu,βϕ = βfor every u and β; see 1.1.15. To see that ϕ is an X-preenvelope, let ϕ′ : M→ X ′ be ahomomorphism with X ′ ∈X. Note that the submodule Imϕ′ ⊆ X ′ has card(Imϕ′)6cardM = α, so E.16 yields a pure submodule Y ⊆ X ′ with Imϕ′ ⊆Y and cardY 6 α.By assumption, Y belongs to X. Hence there is an isomorphism Y ∼= Xu for some u.


498 Appendix E

As ϕ′ factors through Imϕ′, it also factors through Y ∼= Xu. Thus there exist homo-morphisms β : M→ Xu and γ : Xu→ X ′ with γβ= ϕ′. Therefore the homomorphismχ= γπu,β : X → X ′ satisfies χϕ= γπu,βϕ= γβ= ϕ′.

E.20 Corollary. If R is right Noetherian, then every R-module has a flat preenve-lope, that is, an X-preenvelope as in E.17 with X the class of flat R-modules.

PROOF. A pure submodule of a flat R-module is flat; this follows from e.g. E.2(iii)and 1.3.43. Moreover, if R is right Noetherian, then a product of flat modules is flatby 8.3.25. The assertion now follows from E.19.

EXERCISES

E E.1 Let ϕ : R→ S be a ring homomorphism. (a) Show that ϕ is a pure monomorphism of Ro-modules if and only if S is faithfully flat over Ro. (b) Show that if S is faithfully flat overRo and I is injective over R, then I is a direct summand of the R-module HomR(S, I).

E E.2 Show that the sequence 0→ M(U) → MU → MU/M(U) → 0 is pure exact for everyR-module M and every set U .

E E.3 Show that biduality δME : M→ Homk(Homk(M,E),E) is a pure monomorphism in M(R)

for every R-module M.E E.4 Let Λ be a complete set of representatives of isomorphism classes of finitely presented R-

modules. Show that the canonical map∐

L∈Λ L(HomR(L,M))→M is a pure epimorphismfor every R-module M.

E E.5 Let F be a flat R-module. Show that an exact sequence 0→M→ F → N→ 0 is pure ifand only if M and N are flat.

E E.6 Show that an R-module E is fp-injective if and only if every short exact sequence0→ E→M→ N→ 0 is pure.

E E.7 Let E be an fp-injective R-module. Show that an exact sequence 0→M→ E → N→ 0is pure if and only if M is fp-injective.

E E.8 Show that a pure exact sequence 0→M′→M→M′′→ 0 of modules with M′′ finitelypresented is split.

E E.9 Let R be Noetherian. Show that if I is an injective R-module, then every pure exact se-quence 0→M→ I→ N→ 0 splits.

E E.10 Show that an R-module is pure-injective if and only if it is a direct summand of a charactermodule Homk(M,E). Conclude that a pure-injective module is cotorsion. Hint: E E.3.

E E.11 Show that an R-module is pure-projective if and only if it is a summand of a coproduct offinitely presented R-modules. Hint: E E.4.

E E.12 Show that an R-module is injective if and only if it is pure injective and fp-injective.E E.13 Show that an R-module is projective if and only if it is pure projective and flat.E E.14 An R-complex M is called pure-acyclic if 0→ Zv(M)→Mv → Zv−1(M)→ 0 is a pure

exact sequence for every v ∈ Z. Show that the following conditions are quivalent.(i) M is pure-acyclic.

(ii) The complex HomR(N,M) is acyclic for every finitely presented R-module N.(iii) The complex N⊗R M is acyclic for every Ro-module N.(iv) M is a filtered colimit of bounded and degreewise finitely presented contractible

complexes.(v) The complex N⊗R M is acyclic for every Ro-complex N.

Hint: Proof of C.20 and/or Emmanouil [26].



E E.15 Show that a pure-acyclic degreewise pure subcomplex of a graded-flat complex is semi-flat.

E E.16 Show that the following conditions are equivalent.(i) R is von Neumann regular.

(ii) Every acyclic R-complex is pure-acyclic.(iii) Every R-complex is semi-flat.(iv) Every complex of pure-projective R-modules is semi-projective.(v) Every complex of pure-injective R-modules is semi-injective.


Appendix FStructure of Injective Modules

SYNOPSIS. Hom-vanishing Lemma; faithfully injective module; structure of injective module overNoetherian ring; injective precover; indecomposable injective module; endomorphism of ∼.

In his thesis [56] Matlis developed a structure theory for injective modules overNoetherian rings. Our goal is to expose its main points, which are that every injectivemodule is a coproduct of indecomposable modules (F.4 below), and that these havesimple descriptions over Artinian rings (F.6) and over commutative rings (F.10).

We open with a general result that is known as the Hom-vanishing Lemma.

F.1 Lemma. Let M and N be R-modules. A necessary condition for HomR(M,N) tobe non-zero is existence of elements m in M and n 6= 0 in N with (0 :R m)⊆ (0 :R n).If N is injective, then this condition is also sufficient.

PROOF. If ϕ : M→ N is a non-zero homomorphism, then there exists an elementm ∈ M with ϕ(m) 6= 0, and by R-linearity of ϕ one has (0 :R m) ⊆ (0 :R ϕ(m)).Conversely, if m and n 6= 0 are elements with (0 :R m)⊆ (0 :R n), then the assignmentrm 7→ rn defines a non-zero homomorphism from the submodule R〈m〉 ⊆M to N,and if N is injective, then it extends to a homomorphism M→ N.

F.2 Proposition. Every indecomposable injective R-module is isomorphic to theinjective envelope ER(R/a) for some left ideal a in R. In particular, the collection ofisomorphism classes of indecomposable injective R-modules constitutes a set.

PROOF. Let E be an indecomposable injective R-module. Every non-zero elementin E generates a submodule isomorphic to R/a for some left ideal a; see 1.3.1. As Eis injective it follows from B.17 that the injective envelope ER(R/a) is isomorphicto a direct summand of E, so by indecomposability of E one has ER(R/a)∼= E.

F.3 Proposition. Let Max R denote the set of maximal left ideals of R. The R-module

∏m∈Max R ER(R/m) is faithfully injective.

PROOF. The module E =∏

m∈Max R ER(R/m) is injective by 1.3.23. To see that thefunctor HomR(–,E) is faithful, notice that for every R-module M and for everym 6= 0 in M one has (0 :R m) ⊆ m for some m in Max R. Now it follows from F.1that there is a non-zero homomorphism M→ ER(R/m) E.

501

502 Appendix F

If R is left Noetherian, then also the coproduct∐

m∈Max R ER(R/m) of the mod-ules from F.3 is a faithfully injective R-module; cf. 8.2.18.

LEFT NOETHERIAN RINGS

F.4 Theorem. Let R be left Noetherian. Every injective R-module is a coproduct ofindecomposable injective modules.

PROOF. We first argue that every injective R-module I 6= 0 contains an indecompos-able injective submodule. Let i 6= 0 be an element in I; by B.17 there is an injectivesubmodule E of I such that Ri is essential in E. As R is left Noetherian, the moduleRi cannot accommodate an infinite independent family of non-zero submodules, see1.1.20, and since Ri is essential in E, neither can E. It follows that E contains anindecomposable direct summand. Indeed, if E is indecomposable, then that is thedesired module. Otherwise, one has E = M1⊕E1 for non-zero modules M1 and E1.If E1 is indecomposable, then that is the desired module, and if not then one hasE1 = M2⊕E2, etc. As the family M1,M2, . . . is independent, the process terminatesafter finitely many iterations with an indecomposable direct summand En of E. Inparticular, En is an injective submodule of I.

Given an injective R-module I, consider the set of all independent families ofindecomposable injective submodules of I. By the argument above, this set is non-empty, and furthermore it is inductively ordered under inclusion. By Zorn’s lemmathere exists a maximal such family Iuu∈U . The submodule ∑u∈U Iu ∼=

∐u∈U Iu is

injective by 8.2.18, so one has I = E⊕∑u∈U Iu for some R-module E, which is alsoinjective. If E were non-zero, then it would contain an indecomposable submoduleE ′, and the family Iuu∈U ∪E ′ would be independent, which would contradictthe maximality of Iuu∈U . Thus, one has E = 0 and I ∼=

∐u∈U Iu.

REMARK. The decomposition of injective modules described in F.4 is unique to Noetherian rings;this is also a result of Matlis [57].

For the next result, recall that a direct sum is finite by convention.

F.5 Corollary. Let R be left Noetherian. For every finitely generated R-module Mthe injective envelope ER(M) is a direct sum of indecomposable injective modules.

PROOF. By F.4 the module ER(M) is the sum ∑u∈U Eu of an independent family ofindecomposable injective submodules. As M is finitely generated, it is contained inthe sum E ′ = Eu1 + · · ·+Eun ∼= Eu1 ⊕·· ·⊕Eun of finitely many of these modules.Now, since M is an essential submodule of ER(M), one has E ′ = ER(M).

It is evident that the injective envelope of a simple module is indecomposable.To fully appreciate the next result, recall that—up to isomorphism—a left Artinianring has only finitely many simple modules.

F.6 Theorem. Let R be left Artinian. The assignment N 7→ ER(N) yields a one-to-one correspondence between isomorphism classes of simple R-modules and isomor-phism classes of indecomposable injective R-modules.


Structure of Injective Modules 503

PROOF. Let E be an indecomposable injective R-module. For every element e 6= 0in E the finitely generated submodule Re has a composition series, so it contains asimple R-module N. Because E is indecomposable, is is an injective envelope of Nby B.17. To see that non-isomorphic simple modules have non-isomorphic injectiveenvelopes, let N and N′ be simple R-modules with E = ER(N)∼= ER(N′). There arethen injective homomorphisms ι : N→ E and ι′ : N′→ E such that Im ι and Im ι′ areessential in E. By B.11 the submodule Im ι∩ Im ι′ is essential in E; in particular, it isnon-zero. It is also isomorphic to a submodule of N and to a submodule of N′, andhence isomorphic to both N and N′. Thus, one has N ∼= N′.

REMARK. For every maximal left ideal m in R the quotient R/m is a simple R-module, and everysimple R-module is isomorphic to one of those. However, if R is not commutative, then distinctmaximal left ideals may yield isomorphic simple modules; see also E 8.2.13 and ??.

F.7 Definition. Let X be a class of R-modules and let M be an R-module. An X-precover of M is a homomorphism ϕ : X →M with X ∈X such that for every homo-morphism ϕ′ : X ′→M with X ′ ∈X there is a homomorphism χ : X ′→ X that makesthe following diagram commutative,

X ′

ϕ′

χ

~~

Xϕ// M .

F.8. Let X be a class of R-modules and let M be an R-module. A homomorphismϕ : X →M with X ∈ X is an X-precover if and only if HomR(X ′,ϕ) is surjective forevery X ′ ∈ X. Notice that if M is a homomorphic image of some module from X,then every X-precover X →M is is surjective.

F.9 Proposition. If R is left Noetherian, then every R-module has an injective pre-cover, i.e. an X-precover as in F.7 with X the class of injective R-modules.PROOF. Let Euu∈U be a set of representatives for the isomorphism classes of in-decomposable injective R-modules, see F.2, and let M be an R-module. Considerthe module E =

∐u∈U E(HomR(Eu,M))

u , which is injective by 8.2.18. For u ∈ U andα ∈ HomR(Eu,M) write εu,α : Eu→ E for the embedding. Let ϕ : E→M be theunique homomorphism that satisfies ϕεu,α = α for every u and α; cf. 1.1.16. For ev-ery u ∈U the homomorphism HomR(Eu,ϕ) : HomR(Eu,E)→ HomR(Eu,M) mapsεu,α to α, and thus HomR(Eu,ϕ) is surjective. Every injective R-module E ′ is byF.4 a coproduct of modules Eu, so HomR(E ′,ϕ) is a product of homomorphismsof the form HomR(Eu,ϕ); see 3.1.25. Since each map HomR(Eu,ϕ) is surjective,so is HomR(E ′,ϕ). This means that every homomorphism ϕ′ : E ′→M has the formϕ′ = ϕχ for some χ : E ′→ E.

COMMUTATIVE NOETHERIAN RINGS

If R is commutative and Artinian with maximal ideals m1, . . . ,mn, then by F.6 andF.3 the indecomposable injective R-modules are precisely the injective envelopes


504 Appendix F

ER(R/m1), . . . ,ER(R/mn), and the direct sum D =⊕n

u=1 ER(R/mu) is a faithfullyinjective R-module. The endomorphism ring of D is by 9.1.10 isomorphic to R, thatis, every R-linear map D→ D is a homothety. This is just one striking consequenceof the structure of injective R-modules.

F.10 Theorem. Let R be commutative and Noetherian. An indecomposable injectiveR-module has exactly one associated prime ideal, and the assignment p 7→ ER(R/p)yields a one-to-one correspondence between prime ideals of R and isomorphismclasses of indecomposable injective R-modules.

PROOF. If E is an indecomposable injective R-module, then E is by B.17 an injec-tive envelope of every submodule N 6= 0 of E. Since R is Noetherian, the moduleE has an associated prime p, that is, it contains a submodule N isomorphic to R/p.Suppose p′ is also an associated prime of E, then there is a submodule N′ ∼= R/p′ ofE, and the intersection N ∩N′ is non-zero. It follows that there is a non-zero sub-module of R/p that is annihilated by p′, whence one has p′ ⊆ p and, by symmetry,p= p′. Thus, p is the only associated prime of E, and one has E ∼= ER(R/p).

Conversely, let p be a prime ideal of R. To see that E = ER(R/p) is indecompos-able, assume that one has E = E ′⊕E ′′. There are ideals a′ ⊇ p and a′′ ⊇ p in R withE ′∩R/p= a′/p and E ′′∩R/p= a′′/p. The intersection a′/p∩a′′/p in R/p is trivial,so one has a′a′′ ⊆ a′∩a′′ ⊆ p. As p is prime this implies that, say, a′ is contained inp, whence E ′∩R/p is trivial. As R/p is essential in E, this forces E ′ = 0.

F.11 Example. It follows from F.10 and B.13 that the indecomposable injective Z-modules are precisely EZ(Z)∼= Q and EZ(Z/pZ)∼= Z(p∞) where p is a prime. It isroutine to verify that the Z-modules Q/Z and

∐p primeZ(p∞) are isomorphic.

For a prime ideal p, every element in the module ER(R/p) is p-torsion.

F.12 Proposition. Let R be commutative and Noetherian and let p be a prime idealin R. Every element in ER(R/p) is annihilated by a power of p.

PROOF. Let e be a non-zero element of E = ER(R/p) and set a= (0 :R e); the goal isto prove that some power pn is contained in a. As p is finitely generated, it suffices toshow that every element x∈ p has a power in a. Fix an x in p; the ideals bn = (a :R xn)form an ascending chain, so one has bn = bn+1 for some n.

As p is the only associated prime of E, see F.10, the submodule Re∼=R/a of E hasan element re with (0 :R re) = p, whence there is an element y in R with (a :R y) = p.One has a = (a+Ry)∩ (a+Rxn). Indeed, an element z in the intersection has theform a+ry = z = a′+r′xn, so xz = xa+rxy is in a and hence so is xz−xa′ = r′xn+1.Thus, the element r′ is in bn+1 = bn, so r′xn and hence z is in a. The ideal a+Rystrictly contains a, so it corresponds to a non-zero submodule of R/a ∼= Re ⊆ E.Every non-zero submodule of E is essential, cf. B.17, so it follows that a+ Rxn

corresponds to the zero submodule, i.e. xn is in a.

F.13 Corollary. Let R be commutative and Noetherian. For prime ideals p and q inR one has HomR(ER(R/p),ER(R/q)) 6= 0 if and only if p⊆ q.



PROOF. If q contains p, then the canonical map R/p R/q ER(R/q) extends byinjectivity of ER(R/p) to a non-zero homomorphism ER(R/p)→ ER(R/q). On theother hand, existence such a homomorphism implies by F.1 and F.12 that a powerof p is contained in the annihilator of an element in the module ER(R/q) and hencecontained in its unique associated prime q; cf. F.10. Thus, q contains p.

F.14 Lemma. Let R be commutative and Noetherian and let p be a prime ideal inR. For every x ∈ R\p the homothety xER(R/p) is an automorphism.

PROOF. Multiplication by x ∈ R \ p is injective on the essential submodule R/p ofE = ER(R/p) and hence injective on E. The submodule xE ∼= E of E is injective,whence it is a direct summand of E, and since E is indecomposable it is all of E.

Via the next result, questions about indecomposable injective modules can oftenbe dealt with in a local setting.

F.15 Proposition. Let R be commutative and Noetherian and let p be a prime idealin R. The R-module ER(R/p) has a canonical structure of an Rp-module, and as suchit is isomorphic to ERp(Rp/pRp).

PROOF. It follows from F.14 that E = ER(R/p) is an Rp-module when one setsrx e = r(xE)−1(e) for r ∈ R, x ∈ R\p, and e ∈ E. With this structure, E is an injectiveRp-module; indeed, by 1.2.5 there are natural isomorphisms of functors on M(Rp),

HomRp(–,E)∼= HomRp(Rp⊗R –,E)∼= HomR(–,HomRp(Rp,E))∼= HomR(–,E) .

Moreover, since E is indecomposable as an R-module it is indecomposable as anRp-module. Denote by e the element [1]p ∈ R/p ⊆ E. The assignment r

x 7→rx e de-

fines a non-zero Rp-linear map Rp→ E with kernel pRp. Thus, there is an injectivehomomorphism of Rp-modules Rp/pRp→ E, and it follows from B.17 that E is aninjective envelope of Rp/pRp.

F.16 Construction. Let R be commutative and Noetherian and let p be a prime idealin R. Set E = ER(R/p) and for n> 0 set

En = (0 :E pn);

by F.12 these submodules yield a filtration 0 = E0 ⊆ E1 ⊆ ·· · with E =⋃

n>0 En.

F.17 Proposition. Let R be a commutative Noetherian and local with unique maxi-mal ideal m; set k = R/m and E = ER(k).

(a) For a k-vector space V of finite rank, one has HomR(V,E) = HomR(V,k)∼=V .(b) With the notation from F.16 there are isomorphisms of R-modules,

En/En−1 ∼= mn−1/mn and HomR(En,E) ∼= R/mn .

PROOF. (a): The image of a homomorphism V → E is contained in E1, cf. F.1, andE1 is a k-vector space of rank 1, as k is essential in E. Thus one has E1 = k andHomR(V,E) = HomR(V,E1)∼= Homk(V,k)∼=V .


506 Appendix F

(b): For every n > 0 one has HomR(R/mn,E) ∼= En; see 1.1.6. Apply the exactfunctor HomR(–,E) to the sequence 0→mn−1/mn→ R/mn→ R/mn−1→ 0 to getan exact sequence

0−→ En−1 −→ En −→ HomR(mn−1/mn,E)−→ 0 .

By (a) one now has En/En−1 ∼= HomR(mn−1/mn,E)∼=mn−1/mn.

The module HomR(En,E) is isomorphic to HomR(HomR(R/mn,E),E), so toprove the second isomorphism claimed in (b), it suffices to show that the bidual-ity homomorphism from 1.4.2,

δR/mn: R/mn −→ HomR(HomR(R/mn,E),E) ,

is an isomorphism. Set (–)∨ = HomR(–,E). Note that δmn−1/mn

is injective by 4.5.3and hence an isomorphism, as the k-vector spaces mn−1/mn and (mn−1/mn)∨∨ havethe same rank by (a). In particular, δR/m is an isomorphism. Let n > 1 and assumethat δR/mn−1

is an isomorphism. An application of the Five Lemma 1.1.2 to thecommutative diagram below shows that δR/mn

is an isomorphism.

0 // mn−1/mn //

δmn−1/mn∼=

R/mn //

δR/mn

R/mn−1 //

δR/mn−1∼=

0

0 // (mn−1/mn)∨∨ // (R/mn)∨∨ // (R/mn−1)∨∨ // 0

F.18. Let R be commutative Artinian and local with unique maximal ideal m. Forsome integer n one has mn = 0, so F.17 yields HomR(ER(R/m),ER(R/m))∼= R.

Here is the corresponding result for a general local ring.

F.19 Theorem. Let R be a commutative Noetherian and local with unique maximalideal m and set k = R/m. The endomorphism ring HomR(ER(k),ER(k)) is isomor-phic to R, the m-adic completion of R.

PROOF. Set E = ER(k); with the notation from F.16 one has by 3.2.40, 3.3.21, F.17,and ?? the following isomorphisms of R-modules,

HomR(E,E) = HomR(∪n>0En,E)∼= HomR(colimn>0 En,E)∼= limn>0(HomR(En,E))∼= limn>0(R/mn)

= Λm(R)

= R .

Recall from 11.1.14 that an element in R is a family ([ru]mu)u>1 with ru−ru+1 ∈mu

for all u ∈ N. Tracing the isomorphisms above bottom up one sees that an element([ru]mu)u>1 is mapped to the homomorphism E → E given by e 7→ rue for e ∈ Eu,and it is straightforward to verify that it is a homomorphism of rings.



COMMUTATIVE ARTINIAN RINGS

F.20 Proposition. Let R be commutative and Noetherian, and let m be a maximalideal in R. The following hold for E = ER(R/m).

(a) Each submodule En = (0 :E mn) has finite length.(b) Every finitely generated submodule of E has finite length.

PROOF. Part (b) follows from part (a) as one has E =⋃

n>0 En. To prove part (a)one may by F.15 assume that R is local with maximal ideal m, and then it followsfrom F.17 that each subquotient Eu/Eu−1 is an R/m-vector space of finite rank.

F.21 Lemma. Let R be a integral domain with field of fractions Q. If Q is finitelygenerated as an R-module, then R is a field, i.e. R = Q.

PROOF. Let x1, . . . ,xn be elements in R such that the fractions 1x1, . . . , 1

xngenerate Q

as an R-module. In particular, one has x21x2

2 · · ·x2n(∑

ni=1

rixi) = 1 for elements r1, . . . ,rn

in R. It follows that each fraction 1xi

is an element of R. Thus, one has R = Q.

F.22 Theorem. Let R be commutative and Noetherian. The following conditionsare equivalent.

(i) R is Artinian.(ii) Every indecomposable injective R-module has finite length.

(iii) Every indecomposable injective R-module is finitely generated.

PROOF. It is evident that (ii) implies (iii).(i)=⇒ (ii): Let E be an indecomposable injective R-module; by F.10 it has the

form E = ER(R/p) for some prime ideal p in R. As R is Artinian, p is maximal andthere is an n > 1 such that pn = pn+1. With the notation from F.16 one now hasE = En, and this module has finite length by F.20.

(iii)=⇒ (i): To prove that R is Artinian, it is sufficient to prove that every primeideal p in R is maximal. Set E = ER(R/p), it is an indecomposable injective R-module by F.10, and hence it is finitely generated by assumption. By F.15 thereis an isomorphism of R-modules E ∼= ERp(Rp/pRp), so in particular the submoduleRp/pRp of E is finitely generated over R. That is, the field of fractions of the integraldomain R/p is finitely generated as an R-module and hence as an R/p-module. Nowit follows from F.21 that R/p is a field, whence the ideal p is maximal.

F.23 Corollary. Let R be commutative and Artinian. For every finitely generatedR-module M the injective envelope ER(M) is finitely generated.

PROOF. By F.5 the module ER(M) is a direct sum of indecomposable injective R-modules, and each of these is finitely generated by F.22.

EXERCISES

E F.1 (Cf. F.11) Show that the Z-modules Q/Z and∐

p primeZ(p∞) are isomorphic.

E F.2 An R-module E is called Σ-injective if E(U) is injective for every set U . Show that the fieldof fractions of an integral domain is Σ-injective. Hint: Baer’s criterion 1.3.26.


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31. Govorov, V.E.: On flat modules. Sibirsk. Mat. Ž. 6, 300–304 (1965)32. Green, E.L., Kirkman, E., Kuzmanovich, J.: Finitistic dimensions of finite-dimensional mono-

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33. Grothendieck, A.: Sur quelques points d’algèbre homologique. Tôhoku Math. J. (2) 9, 119–221 (1957)

34. Grothendieck, A.: Éléments de géométrie algébrique. IV. Étude locale des schémas et desmorphismes de schémas I–IV. Inst. Hautes Études Sci. Publ. Math. (20, 24, 28, 32) (1967)

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36. Head, T.: Preservation of coproducts by HomR(M,−). Rocky Mountain J. Math. 2(2), 235–237 (1972)

37. Hovey, M.: Model categories, Mathematical Surveys and Monographs, vol. 63. AmericanMathematical Society, Providence, RI (1999)

38. Iacob, A., Iyengar, S.B.: Homological dimensions and regular rings. J. Algebra 322(10),3451–3458 (2009). DOI 10.1016/j.jalgebra.2009.08.006

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Glossary

Here we recapitulate, briefly, the definitions of several key notions. For other stan-dard concepts that we use but do not define, we refer to the following textbooks.• “Categories for the Working Mathematician” [54] by MacLane for notions in

category theory,• “Lectures on Modules and Rings” [50] and “A First Course in Noncommutative

Rings” [51] by Lam for notions in general ring theory, and• “Commutative Ring Theory” [58] by Matsumura for notions in commutative

algebra.Abelian category. An additive category in which every morphism has a kernel and a cokernel,every monomorphism is the kernel of a morphism, and every epimorphism is the cokernel of amorphism. See [54, VIII.3].

In the Abelian categories M(R), Mgr(R), and C(R), a kernel is identified with its domain and acokernel is identified with its codomain. The opposite category of an Abelian category is Abelian.

Adjoint functors. Functors F: U→V and G: V→U are adjoint if for every pair of objects M,Nthere is an isomorphism of hom-sets ΦM,N : V(F(M),N)→U(M,G(N)) which is natural in Mand N. The functor F is then said to be a left adjoint for G, and G is said to be a right adjoint for F.

For objects M in U the images of 1F(M) under ΦM,F(M) are natural morphisms M→ GF(M),whence there is a natural transformation IdU → GF, called the unit of adjunction. Similarly, forobjects N in V the images of 1G(N) under (ΦG(N),N)−1 yield a natural transformation FG→ IdV,called the counit of adjunction. See [54, IV.1].

k-algebra. A unital associative ring A and a homomorphism of unital rings k→ A that maps k tothe center of A.

Artinian ring. A ring that is both left and right Artinian. A ring is left Artinian if it satisfies thedescending chain condition on left ideals. A ring R is right Artinian if Ro is left Artinian. A leftArtinian ring is left Noetherian, and the quotient of a left Artinian ring by its Jacobson radical is asemi-simple ring. See [51, §1, (4.14), (4.15)].

A commutative ring R is Artinian if and only if it is Noetherian and every prime ideal in R ismaximal. Such a ring has only finitely many maximal ideals, m1, . . . ,mn and Jacobson radical J=⋂n

u=1mu =∏n

u=1mu; it is isomorphic to the direct product of Artinian local rings×nu=1

(R/mpu )∼=

×nu=1

Rmu , where p is minimal with Jp = 0. See [51, (23.12)] and [58, proof of 8.15].

Center of a ring. The subring Rc of all the elements in a ring R that commute with every elementof R. Elements in the center are called central.

513

514 Glossary

Coherent ring. A ring that is both left and right coherent. A ring R is left coherent if every finitelygenerated left ideal is finitely presented, equivalently, every finitely generated submodule of afinitely presented R-module is finitely presented. A ring R is right coherent if Ro is left coherent.See [50, §4G].

Colon submodule. Let M be an R-module and M′ ⊆M be a submodule. For a subset a⊆ R the set(M′ :M a) = m ∈M | am⊆M′

is a subgroup of the Abelian group M; if a is a right ideal, then (M′ :M a) is an R-submodule of M.For a subset X ⊆M, the set

(M′ :R X) = r ∈ R | rX ⊆M′is a left ideal, and if X is a submodule, then (M′ :R X) is an ideal.

Coproduct. Let V be a category. The coproduct of a set-indexed family Muu∈U of objects inV is an object M ∈ V together with morphisms Mu → M, called injections, with the followinguniversal property: For every family of morphisms αu : Mu→ Nu∈U there is a unique morphismfrom M to N that for every u ∈U makes the next diagram commutative

Mu

αu

// M

~~

N .

Such a coproduct M is unique up to isomorphism in V and usually denoted∐

u∈U Mu.Let αu : Mu→ Nuu∈U be a family of morphisms in V. If Mu→Mu∈U is a coproduct of

Muu∈U and Nu → Nu∈U is a coproduct of Nuu∈U , then the unique morphism from M toN that makes the diagram

Mu

αu

// M

Nu // N

commutative for every u∈U is called the coproduct of αu : Mu→ Nuu∈U and denoted∐

u∈U αu.One says that V has coproducts if every set-indexed family of objects in V has a coproduct.A functor F: V→W between categories that have coproducts is said to preserve coproducts

if given any family Muu∈U of objects in V the unique morphism that makes the diagram

F(Mu)

//∐

u∈U F(Mu)

ww

F(∐

u∈U Mu)

in W commutative for every u ∈U , is an isomorphism. Here the horizontal morphism is the injec-tion while the vertical morphism is the image of the injection Mu→

∐u∈U Mu under F.

Let F: V→W be a functor between categories that have coproducts. For every family ofmorphisms αu : Mu→ Nuu∈U in V there is a commutative diagram in W,

∐u∈U

F(Mu)

∐u∈U F(αu)

//∐

u∈UF(Nu)

F(∐

u∈UMu)

F(∐

u∈U αu)// F(

∐u∈U

Nu) ,

so if F preserves coproducts, then the morphisms∐

u∈U F(αu) and F(∐

u∈U αu) are isomorphic.

Division ring. A ring in which every non-zero element has a multiplicative inverse. See [51, §13].


Glossary 515

Domain. A R that has the zero-product property, i.e. for elements r,r′ ∈ R the product rr′ is zero(if and) only if r = 0 or r′ = 0. A commutative domain is called an integral domain.

Equivalence of categories. A pair of functors F: U→V and G: V→U such that there are naturalisomorphisms GF→ IdU and FG→ IdV. See [54, I.4].

Faithful functor. A functor that is injective on hom-sets. See [54, I.3].

Filtered set. A preordered set (U,6) with the property that for any two elements u and v in U thereis a w ∈U with u6 w and v6 w. See [54, IX.1]. A filtered set is also called a directed set.

Full functor. A functor that is surjective on hom-sets. See [54, I.3].

Hereditary ring. A ring that is both left and right hereditary. A ring R is left hereditary if every leftideal is projective as an R-module; R is right hereditary if Ro is left hereditary. See also [50, §2E].

Idempotent element of a ring. An element x with x2 = x.

Inductively ordered set. A (partially) ordered set such that every totally ordered subset has an upperbound. By Zorn’s lemma an inductively ordered set has a maximal element.

Invariant basis number (IBN). A left/right symmetric property: R has (left) IBN if for any twonatural numbers m and n one has Rn ∼= Rm as R-modules if and only if m = n. See [50, §1].

Indecomposable module. A module M 6= 0 with no other direct summands than 0 and M. See [51,§7].

Isomorphism of categories. A functor F: U→V that has an inverse, that is, there exists a functorG: V→U such that GF = IdU and FG = IdV hold. See [54, I.3].

Integral domain. A commutative domain.

Jacobson radical of a ring. The intersection of all maximal left ideals or, equivalently, all maximalright ideals. In particular, the Jacobson radical is an ideal. See [51, §4].

Local ring. A ring with a unique maximal left ideal or, equivalently, a unique (the same) maximalright ideal. See [51, §19].

Localization of a module/ring. The localization U−1M of a k-module M at a multiplicative subsetU of k is the k-module of fractions m/u with m ∈M and u ∈U . The k-module U−1k is a ring andU−1M is a module over U−1k. Localization U−1(–) is an exact functor. See [58, §4].

Middle R-linear map. A map ϕ : M×N→ X , where M is an Ro-module, N is an R-module, and Xis a k-module, such that ϕ(mr,n) = ϕ(m,rn) holds for all m ∈M, n ∈ N, and r ∈ R.

Natural transformation. For functors F,G: U→V a natural transformation τ : F→ G assigns toeach object M in U a morphism τM : F(M)→ G(M) in V which is natural in M; that is, one hasτN F(α) = G(α)τM for every morphism α : M→ N in U. If each morphism τM is an isomorphism,then τ is called a natural isomorphism. See [54, I.4].

Nilpotent element/ideal of a ring. An element x (an ideal a) with xn = 0 (an = 0) for some n ∈ N.The same terminology is used for left and right ideals. Every element of a nilpotent (left/right)ideal is nilpotent. A (left/right) ideal is called nil if all its elements are nilpotent. See [51, §4].

Noetherian ring. A ring that is both left and right Noetherian. A ring R is left Noetherian if itsatisfies the ascending chain condition on left ideals, equivalently, every submodule of a finitelygenerated R-module is finitely generated. A ring R is right Noetherian if Ro is left Noetherian.See [51, §1].

Opposite category. For a category U, the opposite category Uop has the same objects as U. Ahom-set Uop(M,N) is identified with U(N,M) under interchange of domains and codomains.Composition Uop(M,N)×Uop(L,M)→Uop(L,N) is given by U(M,L)×U(N,M)→U(N,L);that is, for α ∈Uop(M,N) and β ∈Uop(L,M) the composite α β in Uop(L,N) is βα ∈U(N,L)with domain and codomain interchanged. See [54, II.2].


516 Glossary

Opposite functor. For a functor F: U→V, the opposite functor Fop : Uop→Vop is definedon objects by Fop(M) = F(M). For every morphism α in Uop(M,N) the morphism Fop(α)in Vop(Fop(M),Fop(N)) is F(α) in V(F(N),F(M)) with domain and codomain interchanged.For a natural transformation τ : F→ G of functors U→ V, the opposite natural transformationτop : Gop→ Fop of the opposite functors Uop → Vop assigns to each object M the morphism inVop(Gop(M),Fop(M)) obtained from τM in V(F(M),G(M)) by interchanging the domain andcodomain. See [54, II.2].

Opposite ring. The opposite ring Ro of a ring R has the same underlying additive group whilemultiplication is given by ab = b ·a, where ‘·’ denotes multiplication in R. See [51, §1].

Partially ordered set. A set endowed with a binary relation that is reflexive, transitive, and anti-symmetric. See [54, I.2].

Preordered set. A set endowed with a reflexive and transitive binary relation. See [54, I.2].

Principal (left/right) ideal domain. A domain R is a principal left ideal domian if every left idealin R is principal, i.e. of the form Rx for some x ∈ R. A domian R is a principal right ideal domainif Ro is principal left ideal domain. A commutative principal left (=right) ideal domain is called aprincipal ideal domain.

Product. Let V be a category. The product of a set-indexed family Nuu∈U of objects in V is anobject N ∈ V together with morphisms N → Nu, called projections, with the following universalproperty: For every family of morphisms αu : M→ Nuu∈U there is a unique morphism from Mto N that for every u ∈U makes the next diagram commutative

N

M

>>

αu// Nu .

Such a product N is unique up to isomorphism in V and usually denoted∏

u∈U Nu.Let αu : Mu→ Nuu∈U be a family of morphisms in V. If M → Muu∈U is a product of

Muu∈U and N → Nuu∈U is a product of Nuu∈U , then the unique morphism from M to Nthat makes the diagram

M

// N

Mu αu// Nu

commutative for every u ∈U is called the product of αu : Mu→ Nuu∈U and denoted∏

u∈U αu.One says that V has products if every set-indexed family of objects in V has a product.A functor F: V→W between categories that have products is said to preserve products if

given any family Nuu∈U of objects in V the unique morphism that makes the diagram∏u∈U F(N)

F(∏

u∈U Nu)

88

// F(Nu)

in W commutative for every u∈U , is an isomorphism. Here the vertical morphism is the projectionwhile the horizontal morphism is the image of the projection

∏u∈U Nu→ Nu under F.

Let F: V→W be a functor between categories that have products. For every family of mor-phisms αu : Mu→ Nuu∈U in V there is a commutative diagram in W,


Glossary 517

F(∏

u∈UMu)

F(∏

u∈U αu)// F(

∏u∈U

Nu)

∏u∈U

F(Mu)

∏u∈U F(αu)

//∏

u∈UF(Nu) ,

so if F preserves products, then the morphisms F(∏

u∈U αu) and∏

u∈U F(αu) are isomorphic.

Quasi-Frobenius ring. A ring R such that an R-module (equivalently, an Ro-module) is projectiveif and only if it is injective. Such a ring is Artinian, in particular Noetherian. See [51, §15].

Self-injective ring. A ring that is both left and right self-injective. A ring R is left self-injective ifit is sinjective as a module over itself. A ring R is right self-injective if Ro is left self-injective

Semi-simple module. A module whose every submodule is a direct summand. A module is semi-simple if and only if it is a coproduct of simple modules. See [51, §2].

Semi-simple ring. A ring R such that every R-module (equivalently: every Ro-module, the R-module R, or the Ro-module R) is semi-simple. A cyclic module over a semi-simple ring is isomor-phic to a direct sum of simple ideals generated by idempotents. A semi-simple ring is Artinian, inparticular Noetherian. See [51, §§2–3].

Simple module. A module M 6= 0 with no other submodules than 0 and M. See [51, §2].

Simple ring. A non-zero ring R with no other ideals than (0) and R. See [51, §1].

Torsion. An element m of a k-module M is torsion if one has xm = 0 for some non-zerodivisor xin k. The torsion elements in M form a submodule MT of M called the torsion submodule. If onehas MT = M, then M is torsion, and M is torsion-free if MT = 0 holds. See [50, §4B].

von Neumann regular ring. A ring R such that for every x ∈ R there is an r ∈ R with x = xrx;equivalently, every finitely generated left ideal (equivalently, every finitely generated right ideal)in R is generated by an idempotent. See [51, thm. (4.23)].

Zerodivisor. An element z ∈ k is a zerodivisor if one has zr = 0 for some r 6= 0 in k. An elementthat is not a zerodivisor is called a non-zerodivisor. See [51, §1].


List of Symbols

Page numbers refer to definitions.

0 sufficiently small xxiii 0 sufficiently large xxiii injective map xxiii surjective map xxiii⊂ proper subset xxiii⊆ subset xxiii\ difference of sets xxiii⊎

disjoint union of sets xxiii

∼ homotopy 60' quasi-isomorphism

in category of complexes 133in homotopy category 220

isomorphism in derived category 230u homotopy equivalence 138

isomorphism in homotopy cat. 200∼= isomorphism xxiii

∏product 516

in category of modules 6in category of complexes 86∐

coproduct 514in category of modules 7in category of complexes 84∧

exterior algebra 42

⊕ biproduct 5direct sum

in category of modules 7in category of graded modules 41in category of complexes 89

u pullback 112t pushout 97

⊗ tensor productin category of modules 5in category of complexes 68

⊗L derived tensor product 275∧ wedge product 42

| · | degree of homogeneous element 39[ · ] homology class 57

[ · ]N coset w.r.t. submodule N 4〈·〉 generators of submodule 18

basis for free module 19, 80( : ) colon submodule 514(·)\ underlying graded module 43(·)c center of ring 513(·)e enveloping algebra 9(·)o opposite ring xxiii(·)op opposite category/functor xxiii(·)U U-indexed product 7

(·)(U) U-indexed coproduct 7(·)⊗p pth tensor power 42(·)T torsion submodule 517

(·)ď/ě hard truncation above/below 79(·)Ď/Ě soft truncation above/below 79

∇ superdiagonal 93∂M differential 43

Γa a-torsion 401Λa a-adic completion 403Σ shift 52

αMX unit of (a certain) adjunction 338βM

X counit of (a certain) adjunction 338δM

X biduality

519

520 List of Symbols

in category of modules 32in category of complexes 155in homotopy category 259

δMX in derived category 318εM counitor


εM in derived category 281ζMNX swap isomorphism


ζMNX in derived category 286ηXNM homomorphism evaluation


ηXNM in derived category 321θMXN tensor evaluation


θMXN in derived category 319ιM injective resolution 223µM unitor


µM in derived category 281πM projective resolution 220

ρXMN adjunction isomorphismin category of modules 17in category of complexes 153in homotopy category 259

ρXMN in derived category 288ςM degree shift chain map 53

τMĎ/Ě

truncation morphism 80

υMN commutativity isomorphismin category of modules 15in category of complexes 148in homotopy category 258

υMN in derived category 282χM homothety formation

in category of complexes 329in homotopy category 330

χM in derived category 332ωMXN associativity isomorphism


ωMXN in derived category 284

C complex numbers xxiiiE faithfully injective module 26N natural numbers xxiiiN0 non-negative integers xxiiiQ rational numbers xxiiiR real numbers xxiiiZ integers xxiii

k commutative ground ring xxiii

A,A (gross) Auslander Category 343B,B (gross) Bass Category 343

C category of complexes 46, 47D derived category 230, 268

Da-com subcategory of D 415Da-tor subcategory of D 408

Df subcategory of D 290D@ subcategory of D 289DA subcategory of D 289D@A subcategory of D 289F subcategory of D@A 329I subcategory of D@A 329K homotopy category 200, 257M category of modules 3, 9

Mgr category of graded modules 40P subcategory of D@A 329

U(·,·) hom-set in category U 5

L(·) left derived functor 261LΛa derived a-adic completion 415⊗L derived tensor product 275

R(·) right derived functor 261RΓa derived a-torsion 407

RHom derived Hom 267

L(·) left derived functor 261R(·) right derived functor 261

xxxu sequence of uth powers 401

1M identity morphism 5

B(·) boundary subcomplex 54C(·) cokernel quotient complex 54

Coker cokernel 41Cone mapping cone 125

Cyl mapping cylinder 139D(·) disc complex 80E(·) injective envelope 443


List of Symbols 521

Ext homology of RHom 269FFD finitistic flat dimension 325FID finitistic injective dimension 325FPD finitistic projective dimension 325Gfd Gorenstein flat dimension 381Gid Gorenstein injective dimension 369Gpd Gorenstein projective dimension 354

H homology functoron category of complexes 57on derived category 248on homotopy category 216

H(·) homology subquotient complex 54Ha local homology 415Ha local cohomology 407

Hom homomorphismsin category of modules 4in category of complexes 62

I(·) injective resolution functor 223Id identity functor xxiii

Im image 41Ker kernel 41

Mm×n m×n matrices 8P(·) projective resolution functor 220Ra a-adic completion of ring R 403Tor homology of ⊗L 278

Z(·) cycle subcomplex 54

amp amplitude 75card cardinality xxiii

colim colimit 94fd flat dimension 309

gldim global dimension 322id injective dimension 301

inf infimum 75lim limit 107pd projective dimension 294

rank rank of free module 19sfli supremum of flat lenghts

of injective modules 327silf/silp supremum of injective lenghts

of flat/projective modules 327sup supremum 75

w.gldim weak global dimension 323


Index

Page numbers in italics refer to definitions and those in (parentheses) to exercises.

Symbolsℵ-generated module 487a-adic completion see completiona-torsion see torsionk-algebra xxiii, 513\-functor 51Σ-functor 127

way-out 433, 435Σ-transformation 127, 128, 433, 435

AAbelian category 513absolutely flat ring see von Neumann regular

ringacyclic complex 54, 233additive category 6additive functor 6

preserves direct sums 11adjoint functor 513adjunction isomorphism

in category of complexes 153in category of modules 17, (18)in derived category 288in homotopy category 259

amalgamated product see pushoutamplitude of complex 75, 289Artin algebra 334, 334(left/right) Artinian ring 305, 331, 452, 454,

502, 507, 513associativity isomorphism

in category of complexes 149in category of modules 16in derived category 284in homotopy category 259

Auslander Category 343, 345, 392

BBaer’s criterion 24, (307)basis 19

dual 32graded 80unique extension property 19

Bass Category 343, 345, 394biduality

in category of complexes 155in category of modules 32in derived category 318, 334in homotopy category 259

bimodule 8complex of ∼s 47graded 42symmetric 8

biproduct see also direct sum, 5, 6, 201boundary 54, 56bounded (above/below) complex 74

Ccandidate triangle 475

in homotopy category 207isomorphism of ∼s 475

cardinality xxiiiCartesian square see pullbackcategorical product see productcategorical sum see coproductcategorically flat complex 471categorically injective complex (188)categorically projective complex (177)category

Abelian 513(pre-)additive 6equivalence of ∼ies 515

523

524 Index

isomorphism of ∼ies 515(pre-)linear 5of bimodules 8, 9of complexes 46of graded modules 40of modules 3, 47, 200, 231opposite xxiii, 515triangulated 476

Cech complex 410, 413, 421homology of 411limit of dual Koszul complexes 410semi-free resolution of 417

center xxiii, 513central element 513chain map 46, 53, 65, 71

cycle in Hom complex 62null-homotopic 127

character complex 178, 184, 313, 314character module 26, (36), 178, (498)Cocartesian square see pushoutcocontinuous functor 96cofinal subset 100(left/right) coherent ring 316, (317), 514colimit 94

filtered 99in category of (graded) modules (105)in opposite category (121)of telescope 104right exactness 95uniqueness (105)universal property 94

colon submodule 514commutativity isomorphism

in category of complexes 148in category of modules 15in derived category 282, 283in homotopy category 258

completely reducible module see semi-simplemodule

completion functor 403not half-exact 404

complex 43acyclic 54, 233bounded (above/below) 74categorically flat 471concentrated in certain degrees 43contractible 144, 200degreewise finitely generated 74degreewise finitely presented 74derived complete 415derived reflexive 396derived torsion 408dualizing 330, 331homomorphism of ∼es 46

invertible 337minimal 440morphism of ∼es 46of bimodules 47pure-acyclic (498)semi-flat 191, 191, 461semi-free 165semi-injective 180, 184semi-projective 172, 173sub- and quotient 46totally acyclic 351, 367, 378

connecting homomorphism 4connecting morphism 49

in homology 58, 136continuous chain of modules 492continuous functor 110contractible complex 144, (177), (188), 200contraction 144contravariant functor xxiiicoproduct 514

in category of complexes 84, 94, 101, (106)uniqueness (91)universal property 84

in category of graded modules (91)in category of modules 3, 7in derived category 234in homotopy category 201in opposite category (93)

coresolution see resolutioncotorsion module 274, (473), (498)counit of adjunction 513counitor


countably generated module 488countably related module 489covariant functor xxiiicycle 54, 56cyclic module 19

Ddegree

of graded homomorphism 40of homogeneous element 40

degreewise finitely generated complex 74, 99degreewise finitely presented complex 74,

464degreewise split exact sequence 51, (218)derived category 230

bounded subcategories of 289has (co)products 234isomorphism in 231


Index 525

triangulated 243universal property 236, 238, 239, 245–247,

260zero object in 233

derived complete complex 415, 424derived completion functor 415, 421

idempotence of 422right adjoint for RΓa 424

derived functor 262, 263universal property 265, 266

derived Hom 267additional ring actions 268, 272boundedness 290degreewise finitely generated homology

291infimum of 270supremum of 270, 290way-out 435, 437

derived reflexive complex 396, 396derived tensor product 275

additional ring actions 277, 280boundedness 290degreewise finitely generated homology

292infimum of 278, 290supremum of 278way-out 435

derived torsion complex 408, 414, 424derived torsion functor 407, 413

idempotence of 422left adjoint for LΛa 424

derived transformation 262DG-flat see semi-flatDG-injective see semi-injectiveDG-projective see semi-projectivedifferential 43differential graded module see complexdirect limit see colimitdirect product see productdirect sum see also coproduct

in category of complexes 89in category of graded modules 41in category of modules 7internal see independent family of

submodulesdirect summand

in category of complexes 89in category of graded modules 41in category of modules 7

direct system 93category of ∼s (105)colimit of 94in opposite category (121)morphism of ∼s 94

directed set see filtered setdisc complex 81, 144distinguished triangle 476

in derived category 242, 247in homotopy category 208split 481, 481

divisible module 25, (407)division ring 514Dold complex 43, 54, 165, (177), (188), 352,

367domain 515dual basis 32dual numbers 440dualizing complex 330, 331, 332

existence 333parametrization 349

EEilenberg’s swindle (31), 352enveloping algebra 9, (13, 14)equivalence of categories 515

triangulated 479essential submodule 441exact complex see acyclic complexexact functor 12exact sequence see also short exact sequence

in category of complexes 48split 49

in category of modules 3pure 487split 10

Ext 269(long) exact sequence 271

extension property see unique extensionproperty

exterior power 42

Ffaithful functor 12, 515faithfully exact functor 12fibered coproduct/sum see pushoutfibered product see pullbackfield of fractions

is Σ-injective (507)is flat 27is injective (407), (458)

filtered colimit 99commutes with homology 102exactness 101

filtered set 515finitely generated module 18finitely presented module 26finitely related module (31)


526 Index

finitistic dimensionflat 325Gorenstein flat 390Gorenstein injective 377Gorenstein projective 366injective 325of (left/right) Noetherian ring 326, 327of (left/right) perfect ring 326projective 325, 485vs. global dimension 326

Five Lemmain category of complexes 48in category of modules 4in derived category 248in homotopy category (218)in triangulated category 480

flat dimension 309, 311, 382complexes of finite 329complexes of finite 320, (322), 325, 339,

345, 347of module 315over (left/right) Noetherian ring 314, 315over (left/right) perfect ring 315vs. projective dimension 347vs. injective dimension 313, 314vs. projective dimension 310, 315, 325

flat module 27, 195, 406, 469colimit of ∼s 195countably related 489faithfully 27finitely presented 28finitely related (31)graded- 191over principal ideal domain 406product of ∼s 316projective dimension of 489vanishing of Tor 309

flat preenvelope 498flat resolution 315fp-injective module 275, (498)free module 19

graded- 80rank of 20unique extension property 19

free presentation 26free resolution 169full functor 515functor

\- 51Σ- 127additive 6adjoint 513derived 262exact 12

faithful 12, 515faithfully exact 12full 515half exact 11homotopy invariant 202left exact 11(multi-)linear 6opposite xxiii, 516preserves colimits 96preserves coproducts 514preserves homotopy 142, 143, 255preserves limits 110, 111preserves products 516quasi-triangulated 213, 214right exact 11triangulated 478

Ggenerators of module 18

graded 80global dimension 322

of (left/right) Noetherian ring 324vs. finitistic dimension 326weak 323

Gorenstein flat dimension 381, 386complexes of finite 390, 392, 398of complex of finite injective dimension

382of module 389over Noetherian ring 389refinement of flat dimension 382vs. Gorenstein injective dimension 381vs. Gorenstein projective dimension 382,

389Gorenstein flat module 378

finitely generated 388vanishing of Tor 380

Gorenstein injective dimension 370, 374complexes of finite 392, 394, 398of complex of finite projective dimension

376of module 377refinement of injective dimension 370vs. Gorenstein flat dimension 381

Gorenstein injective module 367, 379vanishing of Ext 369

Gorenstein projective dimension 354, 359complexes of finite 392, 396, 398of complex of finite injective dimension

360of module 365, 366over (left/right) Noetherian ring 364, 366refinement of projective dimension 355vs. Gorenstein flat dimension 382, 389


Index 527

Gorenstein projective module 351finitely generated 362, 388, 395vanishing of Ext 354, 362

Gorenstein ring see also Iwanaga–Gorensteinring

Govorov and Lazard’s theorem 461, 468, 469graded basis 80, (82)

unique extension property (82)graded direct sum 41graded direct summand 41graded Hom 40graded homomorphism 40graded module 39

(homo)morphism of ∼s 40semi-simple 137

graded natural transformation see triangulatednatural transformation

graded quotient module 41graded submodule 40graded tensor product 41

universal property 41graded-flat module 190, 191, 468graded-free module 80, 80

unique extension property (82)graded-injective module 179, 180

lifting property 179graded-projective module 171, 172

degreewise finitely generated 453lifting property 171

Greenlees–May Equivalence 424gross Auslander Category 343gross Bass Category 343Grothendieck Duality 335

Hhalf exact functor 11, (14)

vanishing of 429–431(left/right) hereditary ring (31), (147), (177),

(189), (197), (240), (251), (267), (300),(308), 515

Hom 62, 65, 256\-functor 66Σ-functor 129acyclicity of 431, 432bounded above 76degreewise finitely generated 77exactness 67, 68, 272preserves homotopy 143way-out 433, 434

Hom–tensor adjunction see also adjuctionisomorphism

(co)unit of 338Hom-vanishing Lemma 501homogeneous element 39

homological functor 216homologically trivial complex see acyclic

complexhomology 54

torsion 414homology class 57homology equivalence see quasi-isomorphismhomology functor

homotopy invariant 202on category of complexes 57on derived category 248on homotopy category 216

homology isomorphism see quasi-isomorphism

homomorphismof bimodules 8of complexes 46of graded modules 40of modules 3

homomorphism evaluationin category of complexes 160in category of modules 35in derived category 321, 321, (399)in homotopy category 259inderived category 392

homothety 40, 46homothety formation

in category of complexes 329in derived category 332in homotopy category 330

homotopic chain maps 60, 143, 255homotopically flat complex (197)homotopically injective complex (189)homotopically projective complex (177)homotopically trivial complex see con-

tractible complexhomotopy 60

congruence relation 60diagram commutative up to 61functor that preserves 142, 143, 255

homotopy category 200has (co)products 201isomorphism in 200triangulated 208universal property 203–205, 213–216, 255zero object in 200

homotopy class 200homotopy equivalence 138, 143, 145homotopy invariant functor 202homotopy inverse 138Homotopy Lemma 485Horseshoe Lemma (178), (189), 251


528 Index

IIBN 20, 515ideal xxiii

nil(potent) 515idempotent 515identity functor xxiiiidentity morphism 5indecomposable injective module 502, 504

localization of 505over (left/right) Artinian ring 502, 507over Z 504over Noetherian ring 504

indecomposable module 515injective 502, 504

independent family of submodules 8inductive limit see colimitinductively ordered set 515infimum of complex 75, 289injection 7, 514injective dimension

complexes of finite 329injective dimension 301, 302, 370

complexes of finite 321, (322), 335, 339,341, 345, 347

of module 306over (left/right) Artinian ring 305vs. flat dimension 313, 314

injective envelope 442injective hull see injective envelopeinjective limit see colimitinjective module 23, 25, 187Σ- (507)coproduct of ∼s 306faithfully 23, 501filtered colimit of ∼s 306graded- 180indecomposable 502, 504lifting property 23over (left/right) Noetherian ring 306, 502over principal (left/right) ideal domain 25product of ∼s 24vanishing of Ext 301

injective precover 503injective preenvelope 187, (458)injective resolution 188integral domain (458), 507, 515invariant basis number see IBNinverse limit see limitinverse system 107

category of ∼s (121)in opposite category (121)limit of 107morphism of ∼s 108

invertible complex 337, 350

inverse of 341irreducible module see simple moduleisomorphism

in category of complexes 146in derived category 231in homotopy category 200of candidate triangles 475of categories 515

Iwanaga–Gorenstein ring see also Gorensteinring, 333, 344, 347

Iwanaga-Gorenstein ring 397, 398

JJacobson radical 515Jensen’s theorem 325, 346

proof of 485

KK-flat complex (197), (281), (317)K-injective complex (189), (275), (308)K-projective complex (177), (275), (300)Koszul complex 44, 408, 409

as mapping cone 408Koszul homology 54

annihilator of 408Krull dimension see also dimension

Llarge submodule see essential submoduleleft adjoint functor see adjoint functorleft bounded complex see bounded above

complexleft derived functor see derived functorleft derived tensor product see derived tensor

product functorleft exact functor 11

vanishing of 430left fraction 227

composition of ∼s 229left prefraction 225

equivalence of ∼s 226lifting property

of graded-injective module 179of graded-projective module 171of injective module 187of injective module 23of projective module 22of semi-injective complex 184, 219of semi-projective complex 173, 219

limit 107in category of (graded) modules (121)in opposite category 112, (121)left exactness 109


Index 529

of tower 114uniqueness (121)universal property 107

linear category 5linear functor 6local cohomology 407, 413

(long) exact sequence 407local homology 415, 421

(long) exact sequence 415local ring 453, 515

completion of 506localization

of module/ring 5, 515

Mmapping cone 125, 207

acyclic 136, 234and Hom 129, 131and tensor product 131, 132contractible 145, 201of morphism of candidate triangles 476,

483mapping cylinder 140, 212middle linear map 515minimal complex 440

of injective modules 444of projective modules 455

minimal semi-injective resolution 305, 445existence and uniqueness 445

minimal semi-projective resolution 298, 457existence 457uniqueness 457

Mittag-Leffler Condition 115module xxiii

cotorsion 274countably related 489cyclic 19divisible 25finitely generated 18finitely presented 26finitely related (31)flat 27, 195, 469fp-injective 275free 19Gorenstein flat 378Gorenstein injective 367Gorenstein projective 351graded 39homomorphism of ∼s 3in a certain degree 39indecomposable 515injective 23, 25, 187localization of 515projective 22, 22

pure-injective 487pure-projective 487reflexive 395semi-simple (13), 517simple 517torsion 517

Morita Equivalence 339, 341morphism

in category of complexes 46homotopy equivalence 138, 143mapping cone of 125mapping cylinder of 140null-homotopic 60, 143quasi-isomorphism 133, 143

in category of graded modules 40in derived category 230, 270in homotopy category 200

quasi-isomorphism 202, 260of candidate triangles 475

multilinear functor 6

NNakayama’s lemma 447natural isomorphism 515natural transformation 203, 204, 238, 515Σ- 127, 128opposite xxiii, 516quasi-triangulated 213, 214triangulated 479

nil(potent) ideal 515(left/right) Noetherian ring 77, 168, 169, 306,

457, 502, 515null-homotopic chain map 60, 127

boundary in Hom complex 62

Ooctahedral axiom 477opposite category xxiii, 515

of Abelian category 513of module category (14)of (pre)linear category 5of triangulated category 477

opposite functor 516triangulated 479

opposite natural transformation xxiii, 516opposite ring xxiii, 516

Ppartially ordered set 516perfect complex 337(left/right) perfect ring 196, (198), 298, 326,

454, 454Prüfer p-group 442


530 Index

precover 503preenvelope 497prelinear category 5preordered set 516pretriangulated category 477principal (left/right) ideal domain 20, 23, 25,

233, (274), 406, 516product 516, 517

in category of complexes 87, 108uniqueness (92)

in category of graded modules (92)in category of modules 3, 6in derived category 234in homotopy category 201in opposite category 89, (93)universal property 86

projection 6, 516projective cover 447projective dimension 294, 295, 355

complexes of finite 320, 321, (322), 325,329, 335, 339–341, 345, 347

of transfinite extension 492of flat module 325, 346of module 299over (left/right) Noetherian ring 297, 300over (left/right) perfect ring 298over semi-perfect ring 299vs. flat dimension 310, 315, 325

projective limit see limitprojective module 22, 22, (36)

coproduct of ∼s 23finitely generated 28, 453graded- 172lifting property 22over principal (left/right) ideal domain 23vanishing of Ext 293

projective precover 22, (458)projective resolution 176pullback 113pure epimorphism 487pure exact sequence 487, (498)pure monomorphism 487pure submodule 487pure-acyclic complex (498)pure-injective module 487, (498)pure-projective module 487pushout 97

Qquasi-Frobenius ring 517quasi-isomorphism 133, 136, 143, 232

in homotopy category 202, 260injective (138)surjective 134

quasi-ordered set see preordered setquasi-triangulated functor 213, 214quasi-triangulated natural transformation 213,

214quotient complex 47

Rrank of free module 20reflexive module 395regular ring 307resolution

flat 315free 169injective 188projective 176semi-free 166semi-injective 181semi-projective 174

minimal 457restriction of scalars 12, (18)right adjoint functor see adjoint functorright bounded complex see bounded below

complexright derived functor see derived functorright derived Hom see derived Hom functorright exact functor 11ring xxiii

(left/right) Artinian 513(left/right) coherent 514(left/right) hereditary 515local 515localization of 515(left/right) Noetherian 515opposite xxiii, 516(left/right) perfect 196, 454, 454quasi-Frobenius 517semi-local 452semi-perfect 452, 452semi-simple 24, 517simple 517von Neumann regular (106), 517

SSchanuel’s lemma 297, 305, 314self-injective ring

(left/right) self-injective 517(left/right) left self-injective ring 517semi-basis 165semi-flat complex 191, 191, 461, 470, 471,

(473)of projective modules (473)

semi-flat replacement 276, 309semi-free complex 165, (170)


Index 531

is semi-projective 173semi-free filtration (170)semi-free resolution 166

existence of 166, 168, 169semi-injective complex 180, 184semi-injective replacement 301semi-injective resolution 181, (189)

existence of 185, 187minimal 445

functoriality 223minimal 445

semi-local ring 452semi-perfect module 449, 451semi-perfect ring 299, 452, 452, 457semi-projective complex 172, 173

is semi-flat 192semi-projective replacement 294semi-projective resolution 174, (178)

existence of 174minimal 457

functoriality 220semi-simple module (13), 137, 517semi-simple ring 24, (61), 137, 146, (177),

(188), (206), (251), 273, (300), (308),323, 517

shift 53short exact sequence

in category of complexes 48, 140, 247degreewise split (68)split 49

in category of modules 3pure 487split 10, (68)

simple module 517simple ring 517skew field see division ringsmall submodule see superfluous submoduleSnake Lemma

in category of complexes 49in category of modules 4

split complex see contractible complexsplit distinguished triangle 481split exact sequence

in category of complexes 49, (217)in category of modules 10, 272

stable image of tower 115stable module category (483)stalk complex 74strict triangle in homotopy category 208subcomplex 46submodule

pure 487superfluous submodule 446supremum of complex 75, 289

suspension see shiftswap isomorphism


T(left) T-nilpotency 453t-structure 290telescope 104

colimit of 104morphism of ∼s 104

tensor evaluationin category of complexes 156in category of modules 33in derived category 319, 320, 390, (398,

399)in homotopy category 259

tensor power 42tensor product 68, 71, 257\-functor 72Σ-functor 131acyclicity of 432bounded below 78degreewise finitely generated 78exactness 73, 279preserves homotopy 143way-out 433

Tor 278(long) exact sequence 279

torsion element/subcomplex/complex 402torsion element/submodule/module 517torsion functor 401torsion-free module 406, 517totally acyclic complex

of finitely generated free modules 351, 362of flat modules 378of injective modules 367of projective modules 351

totally reflexive module 395tower 114

limit of 114commutes with homology 119exactness 117

Mittag-Leffler Condition 115morphism of ∼s 114stable image of 115trump 115

transfinite extension 492transformation

derived 262translation see also shift, 476triangulated category 476


532 Index

triangulated functor 213, 214, 245–247, 478way-out 436, 437

triangulated natural transformation 436, 437triangulated natural transformation 214, 215,

245, 246, 479triangulated subcategory 480trivial Mittag-Leffler Condition 115truncation

hard (above/below) 79soft (above/below 289soft (above/below) 79

induces quasi-isomorphism 134

Uunique extension property

of free module 19of graded-free module (82)

unit of adjunction 513unitor


universal propertyof (graded) tensor product 41of colimit 94of coproduct 84of derived category 236, 238, 239, 245–247,

260of derived functor 265, 266

of homotopy category 203–205, 213–216,255

of limit 107of product 86

Vvon Neumann regular ring 28, (106), 273,

280, (317), 323, (499), 517

Wway out

(above/below) functor on C(R) 433, 434(above/below) functor on D(R) 435, 437

weak global dimension 323weakly proregular sequence 411wedge product 42Whitehead’s problem 300

YYoneda Ext 271

Zzero morphism 6

in derived category (240)in homotopy category 200

zero object 5in derived category 233in homotopy category 200

zerodivisor 54, 517Zorn’s lemma 515


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