Math 256B Notes

James McIvor

April 16, 2012

Abstract

These are my notes for Martin Olsson’s Math 256B: Algebraic Ge-ometry course, taught in the Spring of 2012, to be updated continuallythroughout the semester. In a few places, I have added an example orfleshed out a few details myself. I apologize for any inaccuracies this mayhave introduced. Please email any corrections, questions, or suggestionsto mcivor at math.berkeley.edu.

Contents

0 Plan for the course 4

1 Wednesday, January 18: Differentials 41.1 An Algebraic Description of the Cotangent Space . . . . . . 41.2 Relative Spec . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Dual Numbers . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Friday, January 20th: More on Differentials 72.1 Relationship with Leibniz Rule . . . . . . . . . . . . . . . . 72.2 Relation with the Diagonal . . . . . . . . . . . . . . . . . . 82.3 Kahler Differentials . . . . . . . . . . . . . . . . . . . . . . . 10

3 Monday, Jan. 23rd: Globalizing Differentials 123.1 Definition of the Sheaf of Differentials . . . . . . . . . . . . 123.2 Relating the Global and Affine Constructions . . . . . . . . 133.3 Relation with Infinitesimal Thickenings . . . . . . . . . . . 15

4 Wednesday, Jan. 25th: Exact Sequences Involving theCotangent Sheaf 164.1 Functoriality of Ω1

X/Y . . . . . . . . . . . . . . . . . . . . . 164.2 First Exact Sequence . . . . . . . . . . . . . . . . . . . . . . 184.3 Second Exact Sequence . . . . . . . . . . . . . . . . . . . . 20

5 Friday, Jan. 27th: Differentials on Pn 225.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . 225.2 A Concrete Example . . . . . . . . . . . . . . . . . . . . . . 225.3 Functorial Characterization of Ω1

X/Y . . . . . . . . . . . . . 23

1

6 Monday, January 30th: Proof of the Computation of Ω1PnS/S 24

7 Wednesday, February 1st: Conclusion of the Computationof Ω1

PnS/S 28

7.1 Review of Last Lecture’s Construction . . . . . . . . . . . . 287.2 Conclusion of the Proof . . . . . . . . . . . . . . . . . . . . 297.3 An Example . . . . . . . . . . . . . . . . . . . . . . . . . . . 317.4 Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . 31

8 Friday, February 3rd: Curves 328.1 Separating Points and Tangent Vectors . . . . . . . . . . . . 328.2 Motivating the Terminology . . . . . . . . . . . . . . . . . . 328.3 A Preparatory Lemma . . . . . . . . . . . . . . . . . . . . . 33

9 Monday, February 6th: Proof of the Closed EmbeddingCriteria 349.1 Proof of Proposition 8.1 . . . . . . . . . . . . . . . . . . . . 349.2 Towards Cohomology . . . . . . . . . . . . . . . . . . . . . . 36

10 Wednesday, Feb. 8th: Cohomology of Coherent Sheaveson a Curve 3710.1 Black Box: Cohomology of Coherent Sheaves on Curves . . 3710.2 Further Comments; Euler Characteristic . . . . . . . . . . . 3910.3 Statement of Riemann-Roch; Reduction to the More Fa-

miliar Version . . . . . . . . . . . . . . . . . . . . . . . . . . 40

11 Friday, Feb. 10th: Proof and Consequences of Riemann-Roch 4111.1 Preparatory Remarks . . . . . . . . . . . . . . . . . . . . . 4111.2 Proof of the Riemann-Roch Theorem . . . . . . . . . . . . . 4211.3 Consequences . . . . . . . . . . . . . . . . . . . . . . . . . . 43

12 Monday, Feb. 13th: Applying Riemann-Roch to the Studyof Curves 4412.1 Using Riemann-Roch to Detect Closed Immersions to Pn . . 4412.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

13 Wednesday, Feb. 15th: “Where is the Equation?” 46

14 Friday, Feb. 17th: Ramification and Degree of a Mor-phism 49

15 Wednesday, Feb. 22nd: Hyperelliptic Curves 5115.1 Canonical Embeddings . . . . . . . . . . . . . . . . . . . . . 5115.2 Hyperelliptic Curves . . . . . . . . . . . . . . . . . . . . . . 52

16 Friday, Feb. 24th: Riemann-Hurwitz Formula 5316.1 Preparatory Lemmata . . . . . . . . . . . . . . . . . . . . . 5316.2 Riemann-Hurwitz Formula . . . . . . . . . . . . . . . . . . . 55

2

17 Monday, Feb. 27th: Homological Algebra I 5617.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

17.1.1 Algebraic topology . . . . . . . . . . . . . . . . . . . 5617.1.2 Algebraic geometry – a.k.a. the Black Box . . . . . 56

17.2 Complexes, cohomology and snakes . . . . . . . . . . . . . . 5717.3 Homotopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

18 Wednesday, Feb. 29th: Homological Algebra II 6118.1 Additive and abelian categories . . . . . . . . . . . . . . . . 6118.2 δ-functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6318.3 Left- and right-exact functors . . . . . . . . . . . . . . . . . 6518.4 Universality of cohomology . . . . . . . . . . . . . . . . . . 66

19 Friday, March 2nd: Homological Algebra III 6719.1 Derived functors – a construction proposal . . . . . . . . . . 67

19.1.1 Derived functors – resolution of technical issues . . . 6919.2 Derived functors – conclusion . . . . . . . . . . . . . . . . . 7119.3 Two useful lemmas . . . . . . . . . . . . . . . . . . . . . . . 72

20 Monday, March 5th: Categories of Sheaves Have EnoughInjectives 72

21 Wednesday, March 7th: Flasque Sheaves 75

22 Friday, March 9th: Grothendieck’s Vanishing Theorem 7722.1 Statement and Preliminary Comments . . . . . . . . . . . . 7722.2 Idea of the Proof; Some lemmas . . . . . . . . . . . . . . . . 78

23 Monday, March 12th: More Lemmas 79

24 Wednesday, March 14th: Proof of the Vanishing Theorem 8024.1 Base step . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8024.2 Reduction to X irreducible . . . . . . . . . . . . . . . . . . 8124.3 Reduction to the case where F = ZU . . . . . . . . . . . . . 8124.4 Conclusion of the Proof . . . . . . . . . . . . . . . . . . . . 83

25 Friday, March 16th: Spectral Sequences - Introduction 8325.1 Motivating Examples . . . . . . . . . . . . . . . . . . . . . . 8325.2 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8425.3 Spectral Sequence of a Filtered Complex . . . . . . . . . . . 85

26 Monday, March 19th: Spectral Sequence of a Double Com-plex 8626.1 Double Complex and Total Complex . . . . . . . . . . . . . 8626.2 Application: Derived Functors on the Category of Complexes 88

27 Wednesday, March 21st: Spectral Sequence of an OpenCover 89

28 Friday, March 23rd: Cech Cohomology Agrees with De-rived Functor Cohomology in Good Cases 92

3

29 Monday, April 2nd: Serre’s Affineness Criterion 96

30 Wednesday, April 4th: Cohomology of Projective Space 99

31 Friday, April 6th: Conclusion of Proof and Applications 101

32 Monday, April 9th: Finite Generation of Cohomology 103

33 Wednesday, April 11th: The Hilbert Polynomial 106

0 Plan for the course

1. Differentials

2. Curve Theory (Assuming Cohomology)

3. Build Cohomology Theory

4. Surfaces (Time Permitting)

Professor Olsson’s office hours are Wednesdays 9-11AM in 879 EvansHall.

1 Wednesday, January 18: Differentials

1.1 An Algebraic Description of the Cotangent Space

We want to construct the relative cotangent sheaf associated to a mor-phism f : X → Y . The motivation is as follows. A differential, or du-ally, a tangent vector, should be something like the data of a point in ascheme, together with an “infinitesimal direction vector” at that point.Algebraically, this can be described by taking the morphism Spec k →Spec k[ε]/(ε2) and mapping it into our scheme. These two schemes havethe same topological space, and the nilpotency in k[ε]/(ε2) is meant tocapture our intuition of the “infinitesimal”. More generally, we can takeany “infinitesimal thickening”, i.e., a closed immersion T0 → T definedby a square-zero ideal sheaf J ⊂ OT and map that into X. Of course,we want this notion to be relative, so we should really map this thicken-ing into the morphism f , which amounts to filling in the dotted arrow infollowing diagram

T0//

_

X

T //

>>

Y

which should be thought of as representing the following picture in thecase when Y = Spec k

picture omitted/pending

Example 1.1. Take Y = Spec k, T = Spec k and T0 = Spec k[ε]/(ε2).Consider the set of morphisms filling in following diagram:

4

Spec kx∈X

//

X

Spec k[ε]/(ε2) //

88

Spec k

or, equivalently, k OX,xoo

zz

k[ε]/(ε2)

ε7→0

OO

koo

OO

Commutativity of the square forces the composite k → k to be theidentity, so the maximal ideal m ⊂ OX,x maps to zero in k, hence alongthe dotted arrow, m gets mapped into (ε), giving a new diagram

k OX,x/m2aoo

yy

k[ε]/(ε2)

ε7→0

OO

koo

b

OO

But k ⊕ m/m2 ∼= OX,x/m2 via the map (α, β) 7→ b(α) + β. More-over, k ⊕ m/m2 can be given the structure of a k-algebra by definingthe multiplication (α, β) · (α′, β′) = (αα′, αβ′ + α′β). Also, as k-algebras,k[ε]/(ε2) ∼= k ⊕ k · ε, so in fact we are looking for dotted arrows filling inthe following diagram of k-algebras:

k k ⊕m/m2aoo

yy

k[ε]/(ε2)

ε7→0

OO

koo

b

OO

Such arrows are determined by the map m/m2 → k. Thus the set ofarrows filling in diagram 1 is in bijection with (m/m2)∨. This justifies thedefinition of the tangent space of X over k at x as (m/m2)∨.

Greater generality can be achieved by working with X as a scheme oversome arbitrary scheme Y , and also by replacing the ring of dual numbersk[ε]/(ε2) by a more general construction. Before doing so, we need thenotion of relative Spec.

1.2 Relative Spec

Given a scheme X and a quasicoherent sheaf of algebras A on X, we showhow to construct a scheme Spec

X(A). Loosely speaking, the given sheaf

A includes both the data of a ring over each affine open of X, and thegluing data for these rings. We take Spec of each ring A(U) for opens Uof X, and glue them according to the specifications of the sheaf A.

Alternatively, SpecX

(A) is the representing object for the functor

F : (Schemes/X)op → Set

which sends Tf→ X to the set of OT -algebra morphisms f∗A → OT . Two

things need to be checked, namely that this functor is representable at all,

5

and that the scheme obtained by gluing described above actually is therepresenting object. Here’s why it’s locally representable:

If X and T are affine, say X = SpecB, and T = SpecR, with A = Afor someB-algebra A, then f∗A = (A⊗BR)∼, by definition of the pullbacksheaf. By the equivalence of categories between quasicoherent sheaves ofOT -modules and R-modules, the set of OT -algebra maps f∗A → OT is innatural bijection with the set of R-algebra maps A⊗B R→ R, so we arelooking for dotted maps in the following diagram

A⊗B R // R

R

OO ;;wwwwwwwwww

But the vertical map on the left fits into a fibered square as follows

A // A⊗B R // R

B

OO

// R

OO ;;wwwwwwwwww

Moreover, to give a map of R-algebras A⊗B R→ R is equivalent, by theuniversal property of the fibered square, to giving a map A → R alongthe top:

A((// A⊗B R // R

B

OO

// R

OO ;;wwwwwwwwww

This shows that in this case, the set of maps of OT -algebras f∗A → OTis in bijection with the maps of R-algebras A→ R, and hence that SpecArepresents the functor F . The remaining details of representability areomitted.

In fact, the relative Spec functor defines an equivalence of categories

affine morphisms Y → X33

qcoh OX -algebras

SpecX

ss

where the map in the other direction sends the affine morphism Yf→ X

to f∗OY .

1.3 Dual Numbers

In the example of 1.1 we took a scheme X over k and mapped the “in-finitesimal thickening” Spec k → Spec k[ε]/(ε2) into X → Spec k. In factthis thickening can be globalized as follows. If X is a scheme and F aquasicoherent sheaf of OX -algebras, define a quasicoherent sheaf of OX -algebras by

OX [F ] = OX ⊕F

6

with the multiplicative structure given open-by-open by the rule

(a, b) · (a′, b′) = (aa′, ab′ + a′b)

just like before. We then have a diagram of OX -algebras

OX OX ⊕Fpr1oo

OX

::uuuuuuuuuId

OO

where pr1 denotes projection onto the first factor and Id is the identitymap. We apply the relative Spec functor to this diagram, and define thescheme X[F ] to be Spec

XOX [F ], which by the construction comes with

a diagram

X

Id

// X[F ]

||zzzz

zzzz

X

Example 1.2. Consider the affine case, where Bf→ A is a ring map, and

M an A-module. Here we are thinking of X in the above discussion asthe scheme SpecA over the scheme SpecB, and the dual numbers are justA[M ] ∼= A⊕M , with multiplication defined as above. Thus the diagramof OX -algebras above, when working over B, looks like

A A[M ]oo

A

Id

OO ==zzzzzzzzB

OO

oo

A natural question to ask is: what maps along the diagonal, A → A[M ],make this diagram commute? We take up this question in the next lecture.

2 Friday, January 20th: More on Differentials

Given a ring map B → A and A-module M , we form A[M ] as last time,with the “dual multiplication rule”. In the case A = M = k, we recoverthe initial example: A[M ] ∼= k[ε]/(ε2).

2.1 Relationship with Leibniz Rule

The module A[M ] comes with maps of A-algebras

Aσ0 // A[M ]

π // A

whose composition is the identity. We would like to study sections of themap A[M ]→ A. First we need a definition.

7

Definition 2.1. If B → A is a ring map and M is an A-module, a B-linear derivation of A into M is a map ∂ : A→M satisfying ∂(b) = 0 forall b in B, and ∂(aa′) = a∂(a′) + a′∂(a) for al a, a′ in A.

The second property is referred to as the Leibniz rule. Note also thatby ∂(b), we really mean map b into A using the given ring map, andthen apply ∂. The set of B-linear derivations of A into M is denotedDerB(A,M), or often just Der(M) if the ring map B → A is clear fromcontext. Note also that Der(M) itself has the structure of an A-module(even though its elements are not A-linear maps!): if a is in A and ∂ is inDer(M), define (a · ∂)(x) = a · ∂(x), the dot on the right being the actionof A on M .

For an example, let B = R, A = C∞(R), and M = Ω1(R), the A-module of smooth 1-forms on R. Then take ∂ = d, the usual differentialoperator. The conditions in this case say simply that d kills the constantsand satisfies the usual product rule for differentiation.

Proposition 2.1. The set of A-algebra maps φ : A → A[M ] such thatπ φ = Id is in bijection with the set DerB(A,M).

Proof. The bijection is defined by sending a derivation ∂ to the map A→A[M ] given by a 7→ a+ ∂(a). The resulting map is a section of π since itis the identity on the first component; it is A-linear since it is additive inboth factors A and M , and for a, a′ in A,

aa′ 7→ aa′ + ∂(aa′) = aa′ + a∂(a′) + a′∂(a) = (a+ ∂(a))(a′ + ∂(a′)),

the last equality as a result of the definition of the multiplication in A[M ].The map in the other direction sends an A-linear map φ : A → A[M ]

to ∂(a) = φ(a) − a. Note that the image of ∂ actually lies in M sinceπ φ = Id.

Moreover, ∂ is a derivation since, writing φ(a) = a+m for some m inM , we have

∂(aa′) = φ(aa′)− aa′ = φ(a)φ(a′)− aa′ = (a+m)(a′ +m′)− aa′

= am′ + a′m = a(φ(a′)− a′) + a′(φ(a)− a)

= a∂(a′) + a′∂(a)

2.2 Relation with the Diagonal

Now, to study these maps φ from a slightly different point of view, wethink of M above as varying over all A-modules, and construct a universalderivation. To begin with, take Spec of the maps above, giving a diagram

SpecAπ //

Id&&LLLLLLLLLL SpecA[M ]

σ0

SpecA

Now, putting the diagram over B, we are looking for φ such that thefollowing diagram commutes:

8

SpecAπ //

Id&&LLLLLLLLLL

Id

%%

SpecA[M ]

σ0

φ// SpecA

SpecA // SpecB

Let ∆: SpecA → SpecA ⊗B A denote the diagonal morphism, cor-responding to multiplication A ⊗B A → A. Then to give a morphism φabove is equivalent to giving a map φ such that

SpecAπ //

∆&&MMMMMMMMMMM SpecA[M ]

σ0⊗φ

σ0

%%

φ

&&MMMMMMMMMM

SpecA⊗B A

pr1

pr2 // SpecA

SpecA // SpecB

commutes, by the universal property of the fiber product.If the appearance of the diagonal is surprising, you should think of it

in the following way. The composite

X → X ×Y X → X

can be interpretted as expressing X as a retract of the diagonal, sort oflike a tubular neighborhood of X in X ×Y X. We think of the image ofa point x in X in X ×Y X as “two nearby points”, which then map backdown along the second map to the original point in X.

In any case, the above situation leads to the following algebraic prob-lem. We are interested in φ : A⊗BA→ A[M ] (note that this is not exactlythe same φ as above) such that the following commutes:

A[M ]

π

zzzz

zzzz

A A⊗B Aoo

φdd

A

a7→a⊗1

OOσ0

VV

Let I be the kernel of the multiplication A ⊗B A → A. Then thecomposite

I → A⊗B Aφ→ A[M ]

π→ A

is zero, by commutativity of the triangle above; hence the image of I liesin M ⊂ A[M ], and there is an induced dotted map

9

A[M ]

π

zzzz

zzzz

Moo

A A⊗B Aoo

φddJJJJJJJJJ

Ioo

gg

A

a7→a⊗1

OOσ0

VV

Now since M2 = 0 in A[M ], I2 ⊂ A ⊗B A maps to 0 in A[M ] underany such φ, hence the desired φ must factor through A ⊗B A/I2. ThusA⊗B A/I2 is the universal object of the form A[M ], as M varies. In fact,A⊗B A ∼= A[I/I2], since the map A[I/I2]→ A⊗B A/I2 sending a+ γ to(a⊗ 1) + γ fits into the following map of exact sequences

0 // I/I2 //

∼=

A[I/I2]π //

A //

∼=

0

0 // I/I2 // A⊗A/I2 mult // A // 0

so it is an isomorphism by the five lemma.Summarizing our results, we have established

Theorem 2.1. The following four sets are in natural bijection:

1. DerB(A,M)

2. A-algebra maps φ : A→ A[M ] |π φ = Id

3. A-algebra maps | A[I/I2] //

%%KKK

A[M ]ww

A

4. HomA(I/I2,M)

The bijection between 3 and 4 is given by sending φ to its restrictionto the second factor, I/I2. We can express the equivalence of 1 and 4 assaying that the functor

Der : ModA → ModA

as described above is represented by I/I2.

2.3 Kahler Differentials

We have seen that every A-linear map A→ A[M ] corresponds to a deriva-tion ∂ : A → M ; the map A → A[I/I2] is associated to the universalderivation d: A→ I/I2, defined by

d(a) = a⊗ 1− 1⊗ a

The fact that this lies in I is implied by the following:

Lemma 2.1. The kernel I of the multiplication map A ⊗B A → A isgenerated by elements of the form 1⊗ a− a⊗ 1.

10

Proof. Clearly these elements are in I. Conversely, let Σx ⊗ y be anyelement of I. Then Σxy = 0, so Σ(xy ⊗ 1) = 0, giving Σx ⊗ y = Σx(1 ⊗y − y ⊗ 1).

Theorem 2.2. The map d defined above is a derivation, and is universalin the following sense: for any B-linear derivation ∂ : A→M , there is aunique A-linear map φ : I/I2 →M such that the diagram

Ad //

∂!!B

BBBB

BBB I/I2

φ

M

commutes.

Proof. The universal property follows from the preceding discussion ofI/I2. It remains only to check that d is actually a derivation. For b in B,we have

d(b) = 1⊗ b− b⊗ 1 = b⊗ 1− b⊗ 1 = 0

using the fact that the tensor product is taken over B. For the Leibnizrule, first note that I is generated by elements of the form 1⊗ a− a⊗ 1.Then let a and a′ be elements of A, and compute

d(aa′)− ad(a′)− a′ d(a) = aa′ ⊗ 1− 1⊗ aa′ − a(a′ ⊗ 1− 1⊗ a′)− a′(a⊗ 1− 1⊗ a)

= aa′ ⊗ 1− 1⊗ aa′ − aa′ ⊗ 1 + a⊗ a′ − a′a⊗ 1 + a′ ⊗ a= −1⊗ aa′ − aa′ ⊗ 1 + a⊗ a′ + a′ ⊗ a= (a⊗ 1− 1⊗ a)(a′ ⊗ 1− 1⊗ a′),

and this is zero in I/I2.

Remark. The above computation used the A-module structure on A⊗BAgiven by a · (x ⊗ y) = ax ⊗ y. But there is also the action given bya · (x⊗ y) = x⊗ ay. Both of these actions, of course, restrict to I, but infact when working modulo I2, these two restrictions agree. This followsfrom Lemma 1: if a is an element of A, then the difference of the twoactions of a on I can be described as multiplication by (1 ⊗ a − a ⊗ 1)(using the usual multiplication for the tensor product of rings). But thelemma shows that this is in I, and hence its action on I is zero modulo I2.Professor Olsson used this in his computation that d satisfies the Leibnizrule.

Definition 2.2. The A-module I/I2 is called the module of (relative)Kahler differentials of B → A, and denoted ΩA/B . Note that it comestogether with the derivation d: A→ ΩA/B .

In light of the previous discussion, one thinks of ΩA/B as being thealgebraic analogue of the cotangent bundle. Here is an example to furthersupport this terminology.

11

Example 2.1. With notation as above, let B = k, and A = k[x, y]/(y2−x3 − 5). Then

ΩA/B ∼=Adx⊕A dy

2y dy − 3x2 dx,

as you might expect from calculus. The universal derivaton d: A→ ΩA/Bis given by the usual differentiation formula: df = ∂f

∂xdx + ∂f

∂ydy; the

quotient ensures that this is well defined when we work with polynomialsmodulo y2 − x3 − 5. The universal property is expressed by the followingdiagram:

A

∂

@@@@

@@@@

d // ΩA/B

φ||

M

The map φ : ΩA/B →M guaranteed by the theorem sends dx and dy to∂(x) and ∂(y), respectively.

3 Monday, Jan. 23rd: Globalizing Differentials

Today we will globalize the construuctions of last week, obtaining thesheaf Ω1

X/Y of relative differentials associated to a morphism of schemesX → Y . We could do this by gluing together the modules ΩA/BB oflast time, but instead we give a global definition, and then check that itreduces to the original definition in the affine case.

Recall that for a ring map B → I, letting I be the kernel of A⊗BA→A, we defined ΩA/B = I/I2. To relate this to today’s construction, it willbe helpful to bear in mind the isomorphism

I/I2 ∼= I ⊗A⊗BA A

which emphasizes the fact that this object is a pullback along the diagonal.

3.1 Definition of the Sheaf of Differentials

Let f : X → Y be a morphism of schemes. It induces a map ∆X/Y : X →X ×Y X, from the following diagram:

X∆X/Y

%%

Id

Id

))

X ×Y X //

Xf

Xf// Y

The map ∆X/Y comes with a map of sheaves ∆#X/Y : OX×Y X →

(∆X/Y )∗OX on X ×Y X, which we will just abbreviate by ∆. Althoughreally a map of sheaves of rings, we will regard it as a map of sheaves ofOX×X -modules. Let J be its kernel, again regarded as an OX×X -module.

Definition 3.1. The sheaf of relative differentials of X → Y is the OX -module Ω1

X/Y = ∆∗J .

12

Remark. Note that ∆X/Y is not necessarily a closed immersion, so apriori J is just a sheaf of OX×Y X -modules.

Now from the diagram

X∆X/Y

//

Id$$H

HHHH

HHHH

H X ×Y X

pr1

pr2

X

we get maps of sheaves δi : pr−1i OX → OX×Y X for i = 1, 2. Applying

the functor ∆−1X/Y to these morphisms gives maps

∆−1 pr−1i OX

∆−1δi−→ ∆−1OX×Y X ,

but by the commutativity of the diagram of schemes, we have ∆−1 pr−1i OX ∼=

OX . Thus we have maps of sheaves on X, ∆−1δi : OX → ∆−1OX×Y X .Both maps (for i = 1, 2) agree when followed by the map ∆−1OX×Y X →OX , so their difference maps into the kernel of ∆−1OX×Y X → OX , giving

∆−1δ1 −∆−1δ2 : OX → ker(∆−1OX×Y X → OX)

Recall that J was defined as the kernel of OX×Y X → ∆∗OX . Applying∆−1, we have

∆−1J = ∆−1 ker(OX×Y X → ∆∗OX)

Using the left-exactness of ∆−1, we can push it across the kernel and thenapply the adjunction between ∆−1 and ∆∗ to get

∆−1J = ker(∆−1OX×Y X → OX)

Putting this together we now have a map

∆−1δ1 −∆−1δ2 : OX → ∆−1J

Now follow this with the map ∆−1J → ∆∗J = Ω1X/Y . We have now, at

great length, constructed the universal derivation d = ∆−1δ1−∆−1δ2 : OX →Ω1X/Y , which is the global version of the map dA → Ω1

A/B in the affinecase.

3.2 Relating the Global and Affine Constructions

We now verify that our new definition of the cotangent sheaf agrees withthe affine definition locally, in the process showing further that Ω1

X/Y isquasicoherent.

Proposition 3.1. 1. Ω1X/Y is quasicoherent.

2. If f : X → Y is locally of finite type and Y is locally Noetherian,then Ω1

X/Y is coherent.

13

Proof. Pick x in X, let y = f(x), take an open affine neighborhood SpecBof y, and choose an open SpecA of X which lies in the preimage of SpecBand contains x. This gives a commutative diagram

SpecAopen//

X

f

SpecB

open// Y

mapping the inclusion SpecA→ X into the fiber product gives a commu-tative diagram

SpecA∆A/B

//

open

SpecA×SpecB SpecA

X∆X/Y

// X ×Y X

Let J = ker(A⊗B A→ A). Our goal is to show that

Ω1X/Y

∣∣∣SpecA

∼= (J/J2)∼,

which will prove simultaneously that Ω1X/Y is quasicoherent, and that it

agrees with our old definition in the affine case. To do so we make thefollowing computation

Ω1X/Y

∣∣∣SpecA

∼=(

∆−1X/Y J ⊗∆−1OX×X OX

)∣∣∣SpecA

(1)

∼=(

∆−1J)∣∣∣

SpecA⊗

(∆−1OX×X )

∣∣SpecA

OX∣∣∣SpecA

(2)

∼= ker(∆−1OX×Y X → OX)∣∣∣SpecA

⊗∆−1(A⊗BA)∼ A∼ (3)

∼= ∆−1A/B

(ker((A⊗B A)∼ → A∼

))⊗∆−1(A⊗BA)∼ A

∼ (4)

∼= ∆−1J∼ ⊗∆−1(A⊗BA)∼ A∼ (5)

∼= ∆∗(J∼) (6)

∼=(J ⊗A⊗BA A

)∼(7)

∼=(J/J2)∼ (8)

The justifications for the steps are as follows:

(1) Definition of Ω1X/Y (and definition of pullback sheaf)

(2) Tensor product commutes with restriction

(3) Definition of J(4) Left-exactness of ∆−1 and commutativity of the square immediately

above this computation, and for the subscript on the ⊗, the factthat ∆−1 commutes with restriction

14

(5) Most importantly, the fact that ker((A⊗BA)∼ → A∼

)is quasicoher-

ent, which follows from the fact that in the affine case, the diagonalis a closed immersion (note that J , on the other hand, may not bequasicoherent, precisely because in general, the diagonal map neednot be a closed immersion) Thanks to Chang-Yeon for pointing thisout to me

(6) Definition of pullback

(7) Pullback corresponds to tensor product over affines

(8) An observation made at the very beginning of this lecture

This proves part 1 of the theorm. Part 2 was an exercise on HW 1.

Note: At this point Prof. Olsson made some remark about “the keyfact here is a certain property of the diagonal map ...” I didn’t write thisdown, but if anyone remembers what he said here, please let me know.

Corollary 3.1. If B → A is a ring map and f is an element of A, thenthe unique map

Ω1A/B ⊗A Af → Ω1

A′/B

is an isomorphism.

Proof. This follows immediately from quasicoherence, but here’s an al-ternate proof. Check that the localization of d: A → ΩA/B gives anAf -linear derivation df : Af → ΩAf/B . Precomposing df with the local-ization map A → Af gives a derivation A → ΩAf/B so by the universalproperty of ΩAf/B there is a unique A-linear map ΩA/B → ΩAf/B whichcommutes with the localiztion map, d, and df . Now one checks that thismap ΩA/B → ΩAf/B satifies the universal property of the localizationΩA/B → ΩA/B ⊗Af . This also was assigned on HW 1, so I will not fill inthe details.

3.3 Relation with Infinitesimal Thickenings

We motivated our discussion of differentials with talk of mapping infinites-imal thickenings into a morphism X → Y . Having now made our explicitdefinition of the cotangent sheaf, it’s time we related it back to that initialdiscussion. The following theorem says loosely that if there is a way todo this at all, the number of ways to do so is controlled by the cotangentsheaf.

First fix some notation. Let T0 → T be a closed immersion defined bya square-zero quasicoherent (follows from closed immersion) ideal sheafJ . Although J is a sheaf on T , the fact that J 2 = 0 means the actionof OT on J factors through that of OT0 , so in fact we may regard J as asheaf on T0 (recall that their topologial spaces are the same). Let x0 bea point of X, and let S be the set of morphisms filling in the followingdiagram

T0

i

x0 // X

f

Ty//

>>

Y

15

Theorem 3.1. Either S = ∅ or else there is a simply transitive action ofHomOT0 (x∗0Ω1

X/Y ,J ) on S.

Intuitively, this says that if we look at a point x in our scheme X,although there is no preferred choice of tangent direction at X, once oneis chosen, we can obtain all the other by a rotation. In other words, thedifference of two tangent directions is given by a vector in the cotangentspace.

We illustrate the theorem with a concrete example.

Example 3.1. Let k be an algebraically closed field and X/k a finite-typek-scheme, with X regular, connected, and of dimension 1. Thus if x is aclosed point, the local ring OX,x is a DVR.

Proposition 3.2. In the situation above, Ω1X/k is locally free of rank 1.

Proof. We know that Ω1X/Y is coherent by part 2 of Proposition 2. We

first show that for any closed point x of X, the k-vector space Ω1X/k(x) =

Ω1X/k,x

/mxΩ1

X/k,x (equality since k is algebraically closed) is one-dimensional.

For this we calculate the dimension of the dual vector space Homk(Ω1X/Y (x), k),

which by the pevious theorem is equal to the set of dotted arrows

Spec kx //

X

Spec k[ε]/ε2 //

88

// Spec k

and we calculated in Example 1 of Lecture 1 that this is equal to Homk(mx/m2x, k),

which is one-dimensional since OX,x is a DVR. Now, applying “geometricNakayama” (see Vakil 14.7D), we can lift a generator of Ω1

X/k to some

affine neighborhood of x. Thus Ω1X,k is locally free of rank 1.

4 Wednesday, Jan. 25th: Exact Sequences In-volving the Cotangent Sheaf

Before proving two sequences which give a further geometric perspectiveon all we’ve done so far, it would be nice if the construction of Ω1

X/Y werefunctorial in some appropriate sense. Since this sheaf is associated to amorphism X → Y , we should consider a “morphism of morphisms”, i.e.,a commutative diagram.

4.1 Functoriality of Ω1X/Y

Theorem 4.1. Given a commutative diagram

X ′g//

X

Y ′ // Y

there is an associated morphism g∗Ω1X/Y → Ω1

X′/Y ′ of sheaves on X ′.

16

Proof. In the following diagram, the two outer “trapezoids” are the com-mutative square given to us, and the upper left square is the usual fibersquare

X ′ ×Y ′ X ′ //

X ′ //

X

X ′ //

Y ′

AAA

AAAA

A

X // Y

We get a map g × g : X ′ ×Y ′ X ′ → X ×Y X by the universal property ofX ×Y X, which also commutes with the relevant diagonal maps, givingus the square

X ′∆X′/Y ′

//

g

X ′ ×Y ′ X ′

g×g

X∆X/Y

// X ×Y X

From the definition of J we have a sequence

JX/Y → OX×Y X → (∆X/Y )∗OX

of sheaves on X×Y X, and an analogous sequence of sheaves on X ′×Y ′X ′.Applying (g × g)∗ to the first sequence gives

(g × g)∗JX/Y → (g × g)∗OX′×Y ′X′ → (g × g)∗(∆X/Y )∗OX

on X ′×Y ′X ′. Now commutativity of the square above shows that (g×g)∆X′/Y ′ = ∆X/Y g, so (g× g)∗(∆X/Y )∗OX = (∆X′/Y ′)∗ g

∗OX . But thepullback of the structure sheaf along any morphism is always the structuresheaf, so g∗OX ∼= OX′ , and also (g× g)∗OX×Y X ∼= OX′×Y ′X′ . This givesus a diagram of sheaves on X ′ ×Y ′ X ′:

(g × g)∗JX/Y // OX′×Y ′X′ //

∼=

(g × g)∗(∆X/Y )∗OX

JX′/Y ′ // OX′×Y ′X′ // (∆X′/Y ′)∗OX′

Even after applying (g×g)∗, the composite along the top is still zero, hence

the map (g × g)∗JX/Y → OX′×Y ′X′∼=→ OX′×Y ′X′ → (∆X′/Y ′)∗OX′ is

zero, so it factors through JX′/Y ′ , giving us a map (g × g)∗JX/Y →JX′/Y ′ . Now if we apply ∆∗X′/Y ′ to this morphism we get a map

g∗Ω1X/Y = g∗∆∗X/Y JX/Y = ∆∗X′/Y ′(g×g)∗JX/Y → ∆∗X′/Y ′JX′/Y ′ = Ω1

X′/Y ′

which is what we were after.

17

In the affine case, this construction goes as follows. We are given adiagram

SpecA′

// SpecA

SpecB′ // SpecB

which yields maps of rings

I ′ // A′ ⊗B′ A′ // A′

I

α

OO

// A⊗B A

β

OO

// A

OO

Here α is just the restriction of β to I; commutativity of the square onthe right ensures that it factors through I ′.

First we form the A′ ⊗B′ A′-module I ⊗A⊗BA(A′ ⊗B′ A′

), which is

the analogue of the pullback (g × g)∗J . Now we have an A ⊗B A-linearmap α : I → I ′, and by the change of base adjunction

HomA′⊗B′A′(I ⊗A⊗BA

(A′ ⊗B′ A′

), I ′) ∼= HomA⊗BA(I, I ′)

this gives a map I ⊗A⊗BA(A′ ⊗B′ A′

)→ I ′ of A′ ⊗B′ A′-modules, which

corresponds to the map (g× g)∗JX/Y → JX′/Y ′ in the above. Finally weapply the functor −⊗A′⊗B′A′ A

′ to this map, which gives a map(I ⊗A⊗BA

(A′ ⊗B′ A′

))⊗A′⊗B′A′ A

′ → I ′ ⊗A′⊗B′A′ A′

But I ′⊗A′⊗B′A′A′ = Ω1

A′/B′ by definition, and by “changing base all overthe place,”(I⊗A⊗BA

(A′⊗B′ A′

))⊗A′⊗B′A′ A

′ = I⊗A⊗BAA′ =

(I⊗A⊗BAA

)⊗AA′

and this is Ω1A/B ⊗A A′, so we have produced the desired map. If you

work through the construction, you find that this map is just da⊗ a′ 7→a′ · d(g(a)), where g is the given map A→ A′. In particular, this map issurjective if g is.

4.2 First Exact Sequence

The following construction is what Vakil calls the “relative cotangent exactsequence” (Thm 23.2.24).

Theorem 4.2. Given morphisms of schemes Xf→ Y

g→ Z, there is anexact sequence of sheaves on X:

f∗Ω1X/Y → Ω1

X/Z → Ω1Y/Z

18

Proof. The result essentially follows from the following diagram

X∆X/Y//

∆X/Z

$$

f##H

HHHH

HHHH

H X ×Y X //

X ×Z X

Y∆Y/Z

// Y ×Z Y

in which the square is cartesian - it is a magic square (see Vakil 2.3R andtake X1 = X2 = X). Let’s see what this says in the affine case. Then thegiven morphisms correspond to ring maps A← B ← C, and the diagramabove looks like

A A⊗B Aoo A⊗C Aoooo

B

OOccGGGGGGGGGGB ⊗C B

OO

oo

We have a right exact sequence

ker(B ⊗C B → B)→ B ⊗C B → B → 0

to which we apply the right exact functor −⊗B⊗CB A⊗C A, giving us(A⊗CA

)⊗B⊗CB

(ker(B⊗CB → B)

)→(A⊗CA

)→(A⊗CA

)⊗B⊗CBB → 0

Here we have changed base at the middle term:(A⊗C A

)⊗B⊗CB

(B⊗C

B) ∼= A ⊗C A. This is where the magic square comes in: it says that

A ⊗B A, which sits at the top left of that square, satisfies the universalproperty of the tensor product of B and A ⊗C A over the base B ⊗C B,so our sequence becomes(

A⊗C A)⊗B⊗CB

(ker(B ⊗C B → B)

)→(A⊗C A

)→ A⊗B A→ 0

Now the term on the left is already what we want; the other two termssit in their own sequences

A A

(A⊗C A

)⊗(

ker(B ⊗B → B))

// A⊗C A //

OO

A⊗B A //

OO

0

ker(A⊗C A→ A

)OO

ker(A⊗B A→ A

)OO

There is an induced map ker(A ⊗C A → A

)→ ker

(A ⊗B A → A

).

We would like to also find a map(A ⊗C A

)⊗(

ker(B ⊗ B → B))→

ker(A⊗C A→ A

). For this we have to show that the composite(

A⊗C A)⊗(

ker(B ⊗B → B))→ A⊗C A→ A

19

is zero. But this is clear if you remember that we just changed thebase on the middle term. So the map on the left is just Id⊗i, wherei :(

ker(B⊗B → B))→ B⊗B is the inclusion. Then we follow this with

multiplcation, mapping us into A. But the magic square above show thatanything in the kernel of B ⊗ B → B maps to 0 in A. So the leftmostmap factors through the kernel of the center column, and we obtain oursequence(A⊗CA

)⊗B⊗CB

(ker(B⊗CB → B)

)→ ker

(A⊗CA→ A

)→ ker

(A⊗BA→ A

)→ 0

which by definition is

f∗Ω1C/B → Ω1

C/A → Ω1B/A → 0

4.3 Second Exact Sequence

The next Theorem describes how differentials behave under a closed im-mersion. Of course, since the construction is relative, we should look at aclosed immersion X → Y over some base scheme Z.

Theorem 4.3. Let

Xi //

@@

@@@@

@@Y

Z

be a diagram of schemes, where i is a closed immersion defined by the ideal

sheaf I ⊂ OY . Then the map I → OYd→ Ω1

Y/Z induces an OX-linear

map i : i∗I → i∗Ω1Y/Z which fits into an exact sequence

i∗I d→ i∗Ω1Y/Z → Ω1

X/Z → 0

Proof. (This is the corrected version, actually given in the following lec-ture) We’ll prove it only in the affine case. In this setting, we have adiagram

SpecA/Ii //

%%KKKKKKKKKKSpecA

f

SpecB

and we wish to exhibit an exact sequence

I/I2 d→ Ω1A/B ⊗A A/I → Ω1

A/I/B → 0

We claim that the sequence can be found buried in the belly of thisbeast:

20

I ⊗A⊗BA Aa⊗f 7→a⊗f−f⊗a

ker(A⊗B A→ A)

++VVVVVVVVVVVV

I ⊗A⊕A⊗ I //

Σ**UUUUUUUUUUUUUU ker(A⊗B A→ A/I)

// // ker(A/I ⊗B A/I → A/I)

I

0

Here Σ is the map defined by a ⊗ f + a′ ⊗ f ′ 7→ af + a′f ′, extended bylinearity.

I haven’t worked through this yet. I’ll come back to it. In case I forget,the argument in Matsumura is nice and short. Read that instead.

Example 4.1 (Jacobian Matrix Calculation). Let B → A be a finite typering map, with A and B Noetherian. We compute Ω1

A/B as follows. Firstwrite

A = B[x1, . . . , xn]/(f1, . . . , fs)

Then we have morphisms

SpecA // AnB

SpecB

We have computed already that Ω1B[xi]/B

=∼= B dx1 ⊕ · · · ⊕ B dxn, andpulling this back to SpecA is the same as tensoring up with A, giving

Ω1B[xi]/B

∣∣∣A

∼= Adx1 ⊕ · · · ⊕A dxn

Now by the previous proposition, this surjects onto Ω1A/B with kernel

I/I2, where I = (f1, . . . , fs). Furthermore, there is an obvious surjectionAs → I/I2, sending the i-th basis element of As to fi. Since precomposingwith a surjection does not change the cokernel, the sequence

As → Adx1 ⊕ · · · ⊕A dxn → Ω1A/B

expresses Ω1A/B as the cokernel of the map As → A dx1 ⊕ · · · ⊕ Adxn.

But consider the image of a basis element ei under this map. It mapsfirst to fi mod I2. Recall that the map I/I2 sends the class of f todf =

∑ ∂f∂xi

dxi. Thus in terms of the bases ei for As and dxj for

Adx1 ⊕ · · · ⊕ A dxn, this map is represented by the matrix(∂fi∂xj

)i,j

, the

familiar Jacobian matrix. This is the key to computing Ω1A/B in many

applications.

21

5 Friday, Jan. 27th: Differentials on Pn

Remark. The first problem on HW 1 should have convinced you that atleast in algebraic geometry, inseparable field extensions are not patholo-gies, despite what your Galois Theory course may have led you to believe.One common situation in which they arise is the following. Say we have afinite type k-scheme X, where k is algebraically closed, and a morphismX → Y whose fibers are curves. Then one is led to consider the map offields k(x)→ k(y), where k(y) is typically not a perfect field. It may be,for example, of the form k(t1, . . . , tn), such as appeared on the assignment.

5.1 Preliminaries

Anyway, today will will compute the cotangent sheaf of PnS . There aremany more concrete proofs (e.g., in Hartshorne), but we will take thefunctorial approach. Recall that PnS represents the functor(

Schemes/S)op

→ Set

sending(T → S

)to the set of all surjections On+1

T → L, where L is alocally free sheaf of rank one on T , up to isomorphism. In particular,plugging PnS itself in to this functor, we get an isomorphism

hPnS

(PnS) ∼= F (PnS),

and the image of IdPnS

under this isomorphism is the surjection

On+1PnS→ OPn

S(1)

where OPnS

(−1) is the so-called “twisting sheaf”. A HW problem from lastterm computed the abelian group structure of Pic(PnS) under tensor prod-uct. Applying − ⊗OPn

SOPn

S(−1) (which is right exact) to this surjection

gives another surjection

OPnS

(−1)n+1 → OPnS

The following theorem asserts that the cotangent sheaf of PnS is the kernelof this surjection.

Theorem 5.1. Ω1PnS/S is locally free of rank n, and there is an exact

sequence0→ Ω1

PnS/S → OPn

S(−1)n+1 → OPn

S→ 0

5.2 A Concrete Example

To warm up, let’s take S = Spec k, with k algebraically closed. We haveseen that cotangent vectors can be regarded as dotted maps filling in thefollowing diagram

Spec k //

Pnk

Spec k[ε]/ε2 //

88

Spec k

22

Since k is algebraically closed, the map along the top is given by choos-ing a (closed) point [a0 : . . . : an] of Pnk .

To give a map from k[ε]/ε2 into Pnk is to give a point [a0 + εb0 : . . . :an + εbn] which agrees with the first point modulo ε (since the verticalmap on the left is reduction mod ε. So since the ais have already beenchosen, filling in the diagram entail choosing the bis. But we could rescaleour coordinates by an element of k[ε]/ε2 and possibly still get the samemap. Since (ai + εbi)(x+ εy) = aix+ ε(bix+ aiy), we are forced to takex = 1 in order to get the diagram to commute. So the map is determinedby choosing the bis, and the ambiguity in doing so can be expressed asthe corkernel of the map sending y 7→ (a0y, . . . , any). Thus we conclude,loosely speaking, that the (co)tangent space is the cokernel of a mapk → kn+1.

To relate this to the theorem, take the dual of the sequence whichappears there:

0→ OPnS→ OPn

S(1)n+1 →

(Ω1

PnS

)∨ → 0

where the final term is by definition the tangent sheaf of PnS over S. Thisis the coordinate-free expression of our computation above.

Compare this with the topological desription of complex projectivespace. There we have a C∗-bundle

Cn+1 \ 0

f

CPn

and if p is a nonzero vector in Cn+1, the tangent space to Cn+1 at p is justCn+1, whereas the tangent space of CPn at f(p) is isomorphic to Cn, andwe think of this as being the tangent space of Cn+1 module the tangentspace of the fiber over f(p), which is just a line:

0→ Tf−1f(p)(p)→ TCn+1\0(p)→ TCPn(f(p))→ 0

0→ C→ Cn+1 → Cn → 0

5.3 Functorial Characterization of Ω1X/Y

Before we tackle the proof, let’s ask what functor Ω1X/Y represents. We’ve

basically already done this. Given X → Y , a morphism of schemes, define

TX/Y : QCoh(X)→ Set

by sending F to HomOX (Ω1X/Y ,F). Given a quasicoherent sheaf F , we

saw the other day how to construct the following diagram

XId //

X

X[F ]

s0

<<zzzzzzzz// Y

23

Recall that, by analogy with the usual dual numbers, the scheme X[F ]comes equipped with morphisms of sheaves OX → OX ⊕ F → OX . Inparticular, there is a distinguished arrow s0 filling in the diagram, thoughof course there may be others. Since we have a distinguished choice ofarrow, Theorem 3.1 says that the set of all such arrows is isomorphicto the set of morphisms Ω1

X/Y ,F). In other words, the content of that

theorem is really that Ω1X/Y represents this functor TX/Y .

6 Monday, January 30th: Proof of the Computa-tion of Ω1

PnS/S

Today we aim to prove Theorem 5.1. We will do this in three steps, thesecond of which is the most involved. Remember that we have to showthat Ω1

PnS/S fits into an exact sequence

0→ Ω1PnS/S → OPn

S(−1)n+1 → OPn

S/S → 0.

Following the discussion of last time, all that remains is to check thatΩ1

PnS/S is the kernel of the map on the right. Let K be this kernel. The

following is an outline of the argument:

1. For any quasicoherent sheaf F on PnS ,

coker(Hom(OPn

S/S ,F)→ Hom(OPn

S/S(−1)n+1,F)

) ∼= Hom(K,F)

2. For any quasicoherent sheaf F on PnS ,

coker(Hom(OPn

S/S ,F)→ Hom(OPn

S/S(−1)n+1,F)

) ∼= Hom(Ω1PnS/S ,F)

3. By 1 and 2,Hom(K,F) ∼= Hom(Ω1PnS/S ,F), and therefore K ∼= Ω1

PnS/S

Step 3 is basically just an application of Yoneda: once we’ve shownthat Hom(K,F) ∼= Hom(Ω1

PnS/S ,F), then taking global sections gives

HomOPn(K,F

) ∼= HomOPn(Ω1

PnS/S ,F

), which implies Ω1

PnS/S∼= K, fin-

ishing the proof.Now we prove the first two steps.

Proof of Step 1. We already have an exact sequence

0→ K → OPnS

(−1)n+1 → OPnS/S → 0

We claim that application of the functor Hom gives another exactsequence of sheaves on PnS :

0→ Hom(OPnS,F)→ Hom(OPn

S(−1)n+1,F)→ Hom(K,F)→ 0

The exactness follows from the following lemma

Lemma 6.1. Let0→ K → E → E ′ → 0

be an exact sequence of quasicoherent sheaves on X, where E ′ is locallyfree of finite rank. Then for all F in QCoh(X), the sequence of sheaveson X

0→ Hom(E ′,F)→ Hom(E ,F)→ Hom(K,F)→ 0

is exact in the category of OX-modules.

24

Proof of Lemma. Exactness can be checked on stalks. Since E ′ is locallyfree of rank r for some r, every point of X has a neighborhood U onwhich E ′

∣∣U∼= OrU . Therefore replacing X by U we may assume E ′ ∼= OrX ,

generated by basis elements ei in Γ(X, E ′), i = 1, . . . , r. Then definea section by lifting each ei to one of its preimages in Γ(X, E), giving asplitting

E ∼= K ⊕ E ′

Then

Hom(E ,F) ∼= Hom(K ⊕ E ′,F)

∼= Hom(K,F)⊕Hom(E ′,F)

which implies the exactness of the sequence in the statement of the lemma.

Now, taking E = OPnS

(−1)n+1 and E ′ = OPnS

(which is locally free ofrank one), we have proved step 1.

Step 2 is the heart of the proof, and we will not finish it today. Beforewe begin, here are a few facts which we will be useful to bear in mind.

Our approach involves the fact that PnS represents the functor

F :(

Schemes/S)op

→ Set

which sends an S-scheme T → S to the set of surjections of OT -modulesOn+1T → L where L is a line bundle on T , up to some equivalence relation

that we won’t worry about here. But recall the way in which this F isactually a (contravariant) functor: given a morphism

T ′g

//

@@

@@@@

@ T

S

of S-schemes, the induced morphism F (f) sends the surjection On+1T

π→ Lin F (T ) to the surjection On+1

T ′g∗π−→ g∗L in F (T ′).

Why is g∗π still a surjection? Because g∗ is right exact, being a leftadjoint to the pushforward g∗. Why is g∗L a line bundle? Becausepullback sends vector bundles to vector bundles: this can be checkedlocally, and if T ′ = SpecA, T = SpecB, and L ∼= (B∼)n, we haveg∗L ∼= (B∼)n ⊗(B∼) A

∼ ∼= (Bn ⊗B A)∼ ∼= (A∼)n as B∼-modules. Nowwe beign the proof.

Proof of Step 2. Let U be any open of PnS , and F any quasicoherent sheafon PnS . Then Hom

(Ω1

PnS,F)(U) = HomOU

(Ω1

PnS

∣∣U,F∣∣U

), and by Theo-

rem 3.1, this set is equal to the set of dotted maps filling in the followingdiagram:

Uπ //

PnS

U [F ] //

==

S

25

Notice that in applying the theorem, we have used the fact that this set isnon-empty: we have a distinguished “retract” r : U [F ]→ U , which whencomposed with the inclusion U → PnS , fills in the diagram.

Now we consider such diagrams from the point of view of the functor Fwhich is represented by PnS . As noted above, a map T → PnS correspondsto a surjection of OT -modules On+1

T → L, with L a line bundle on T . Sothe inclusion U → PnS corresponds to a section On+1

U LU , and since theinclusion commutes with the identity map PnS → PnS , this line bundle LUis actually just OPn

S(1)∣∣U

, but we’ll continue to denote it by LU for now.We also have our distinguished dotted arrow, the retract r (followed

by inclusion into PnS), which corresponds to a surjection On+1U [F] → L

∼.

Notice that OU [F]∼= r∗OU since pullback sends the structure sheaf to the

structure sheaf. So our second surjection really looks like r∗On+1U → L∼.

But the fact that the square above commutes imposes some constraints onthese two surjections. Namely, in light of our review of the functorialityof F , we must have a commuting square

r∗On+1U

r∗π // //

r∗LU

On+1U

// // LU

Now we are asking whether there are other maps which fill in the square ofschemes above, or equivalently, fill in this square of sheaves, and the func-toriality therefore requires us to look for not just any surjection r∗On+1 →L∼, but specifically a surjection onto r∗LU . So what can be said aboutthis r∗LU? Firstly, it fits into an exact sequence of OU -modules

0→ LU ⊗F → r∗LU → LU → 0

The reason is that r∗LU can be rewritten as follows:

r∗LU = OU [F] ⊗OU LU = LU ⊗OU (OU ⊕F∣∣U

) = LU ⊕ (LU ⊗OU F),

and this fits into the (split) exact sequence above. But the map on theright is just the right hand vertical arrow in our square of sheaves twoparagraphs up, so we can fit the two diagrams together, giving

0 // LU ⊗F // r∗LU // LU // 0

On+1U ⊕Fn+1 ∼= r∗On+1

U

r∗π

OOOO

// On+1U

OOOO

All the maps here are OU -linear, and in addition, r∗π is OU [F]-linear.We ask: what other OU [F]-linear maps are there that fill in this diagram?Observe first that by commutativity of the square any two such mustagree when composed with the map r∗LU → LU killing F . Thus bythe exactness of the horizontal sequence their difference factors throughLU ⊗OU F , i.e., is a map On+1

U [F] → LU ⊗F . So we can produce the desired

26

maps by adding to r∗π an OU [F]-linear map φ : On+1U [F] → LU ⊗ F . This

associates to each element of

HomOU[F]

(On+1U [F],LU ⊗OU F

)a dotted arrow filling in our diagram. But by changing base to the categoryof OU -modules, this Hom set is in natural bijection with

HomOU(On+1U ,LU ⊗OU F

),

where LU ⊗OU F is regarded simply as an OU -module. Thus given anyelement of this set, we have produced a surjection On+1

U [F] → r∗LU of

sheaves on U [F ].Now recall that the line bundle LU is actually OU (1), and that maps

filling in the diagram correspond to elements of HomOU(Ω1

PnS

∣∣U,F), so

we have produced a map

HomOU(On+1U ,OU (1)⊗OU F

)→ HomOU

(Ω1

PnS

∣∣U,F)

But with a little calculation, we can rewrite the domain as follows:

HomOU(On+1U ,OU (1)⊗OU F

)∼= HomOU

(OU (−1)n+1 ⊗OU OU (1),OU (1)⊗OU F

)∼= HomOU

(OU (−1)n+1,HomOU

(OU (1),OU (1)⊗OU F

))∼= HomOU

(OU (−1)n+1,F

)The final isomorphism being because OU (1) is locally isomorphic to OU .For more detail, see the lemma below.

Summarizing, we now have a map

HomOU(OU (−1)n+1,F

)→ HomOU

(Ω1

PnS

∣∣U,F)

for every open U ⊆ PnS , which is compatible with restriction maps byconstruction. Thus it determines a morphism of sheaf-Homs

HomOPn(OPn(−1)n+1,F

)→ HomOPn

(Ω1

PnS,F)

The rest of the proof of Part 2 will be finished in the next lecture.

Here’s a lemma which was referred to above, and which gets used in thenext lecture also, but was neither explicitly stated nor proved in lecture.

Lemma 6.2. Let L be a line bundle on X, and F a quasicoherent sheafon X. Then there is an isomorphism of sheaves

σ : Hom(L,L ⊗OX F)∼→ F

Proof. Choose a covering of X on which L is trivial and F is the sheafassociated to some module, i.e., consisting of open affines U = SpecA, onwhich we have isomorphisms

L∣∣U∼= OU ∼= A and F

∣∣U∼= F ,

27

for some A-module F . Then we have, by the equivalence of categoriesQCoh(U) ∼= ModA,

Hom(L,L ⊗OX F)(U) ∼= HomA(A,A⊗A F )

using which σ is given explicitly by the following: if φ ∈ HomA(A,A⊗AF ),and φ(1) = 1⊗f (we can always write it like this since the tensor productis taken over A), we define σ(φ) = f . We need to check that this definitionof σ agrees on the intersection W of two affines U and V . But in this case,we may restrict further if necessary to ensure that (F

∣∣U

)∣∣W∼= (F

∣∣V

)∣∣W

.

Moreover, the isomorphisms L∣∣U∼= OU and L

∣∣V∼= OV , when restricted

to W , differ by multiplication by a unit in OW . Since this unit appearsin both sides of the equation defining σ (specifically, we multiply the 1 inboth sides by this unit), it may be cancelled, so that the equations definingσ on U and on V are the same when restricted down to W . Thus thedefinition of σ on open affines glues to give a morphism of sheaves, whichis an isomorphism of sheaves since it is so locally: HomA(A,A ⊗A F ) ∼=HomA(A,F ) ∼= F .

7 Wednesday, February 1st: Conclusion of theComputation of Ω1

PnS/S

7.1 Review of Last Lecture’s Construction

By the end of the last lecture, we claimed to have produced, for each Fin QCoh(PnS), a morphism of sheaves

γ : Hom(OPnS

(−1)n+1,F)→ Hom(Ω1PnS/S ,F).

Let’s first clarify the construction of this map γ. Consider an affineopen SpecR ⊆ PnS , with a sheaf F ∼= F , for an R-module F . We areinterested in dotted arrows

SpecR //

Id

PnS

SpecR[F ] //

r

::

S

SpecR

The inclusion SpecR → PnS corresponds to a surjection On+1R

π→ L,where L is an invertible sheaf on SpecR. 1 We observed that in factL = O(1), and that r∗L ∼= L ⊕ (F ⊗OR L) as OR-modules. This impliesthat there is an exact sequence

0→ F ⊗OR L → r∗L → L → 0,

1In the following, we will abbreviate OSpecR by OR, and similarly for R[F ].

28

in which the map r∗L ∼= L ⊕ (F ⊗ L) → L is projection onto the firstfactor; we fit this into the following diagram involving π and a potentiallifting π:

0 // F ⊗OR L // r∗L // L // 0

On+1R[F ]

π

OOOO

// // On+1R

π

OOOO

If ei are basis elements of On+1R , then these lift to basis elements of

OR[F ] [Reason: elements of F ⊆ R[F ] are nilpotent, hence contained inthe nilradical, hence the radical of R[F ]. Thus Nakayama says we can liftgenerators of R[F ]/F = R to generators of R[F ]]. Abusing notation anddenoting these lifts also by ei, the commutativity of the square means thatfor any π filling in the diagram, we must have π(ei) = (π(ei), αi) for someαi in L ⊗ F . Thus π is determined by these αi, so the set of maps fillingin the diagram is in bijection with n-tuples of sections of L ⊗ F , whichis in turn in bijection with Hom

((L−1)n+1,F

),according to the following

computation:

(L ⊗ F)n+1 ∼= Hom(On+1,L ⊗ F) ∼= Hom((L−1)n+1 ⊗ L,L ⊗ F

)∼= Hom

((L−1)n+1,F

),

where we used Lemma 6.2 at the last step.This construction is compatible with restriction, so this construction

of π gives a map of sheaves

Hom(OPn(−1)n+1,F)→ maps U → Pn filling in the diagram above.

(We used L−1 ∼= OPn(−1) here). Now the sheaf on the right is a priori justa sheaf of sets, but one can check that there is an OPn module structureon it, in such a way that it is isomorphic as sheaves to Hom(Ω1

Pn ,F).Now we consider this map

γ : Hom(OPnS

(−1)n+1,F)→ Hom(Ω1PnS/S ,F).

7.2 Conclusion of the Proof

To finish the proof of Step 2, it suffices to check that γ is surjective andthat its kernel is Hom(OPn ,F). This shows that Hom(Ω1

PnS/S ,F) is the

cokernel of Hom(OPn ,F) → Hom(OPnS

(−1)n+1,F) because sheaves ofOPn -modules form an abelian category, so every surjection (here, γ) is thecokernel of its kernel.

For surjectivity, we start with an open U ⊆ Pn, and consider a surjec-tion π′ in the following diagram:

OU [F]n+1π′ // //

L′

On+1U

// // L

29

Remember that our construction began with the “canonical surjectionOn+1U → OU (1) and lifted it to a surjection On+1

U [F] → r∗L. Thus a map

π′ as above comes from our construction if and only if there exists anisomorphism ρ : L′ → r∗L over L, i.e., a diagram

L′∼= //

@@

@@@@

@ r∗L

||||

||||

L

which must furthermore fit into the following:

On+1U [F]

π′ // //

φ+π′

$$

L′ρ//

r∗L = L ⊕ (L ⊗ F)

xxqqqqqqqqqqqqq

On+1U

// // L

We exhibit this ρ locally: first find an open affine on which all three ofL,L′, and r∗L are trivial. Choose a basis for L, lift it via the surjectionsL → L and r∗L in such a way that the triangle above commutes (this canalways be done, since locally, all three are isomorphic to some ring R).Then our choice of bases for L′ and r∗L determines ρ locally. This provessurjectivity.

To determine the kernel, we ask: when do two maps φ1, φ2 : OPn(−1)n+1 →F determine the same π : On+1

U [F] → r∗L?. In other words, for which φ1, φ2

is there an isomorphism σ : r∗L → r∗L filling in the following diagram

r∗L

σ

%%KKKKKK

On+1U [F]

π+φ277oooooo

π+φ1 ''OOOOOO L

r∗L

99ssssss

Any isomorphism σ is given by multiplication by some unit in O×U [F]; therequirement that it be an isomorphism over L forces this unit to map to1 in OU after collapsing F . In other words, such units have the form1 + x ∈ OU [F], where x ∈ F ; in more otherer words, the sections of F

∣∣U

inject into the kernel of the map of sheaves of groups O×U [F] → O×U .

Now for any section x of F∣∣U

, we consider the maps

On+1U [F]

π+φ2 //

π+φ1&&LLLLLLLLLL

LU ⊕ (L ⊗ F)

·(1+x)

L ⊕ (L ⊗ F)

30

Commutativity requires that for each basis element ei of OU [F],

(π(ei), φ1(ei)) = (π(ei), π(ei) · x+ φ2(ei)).

Thus the two maps agree if φ1(ei) − φ2(ei) = π(ei) · x. So the kernel ofγ consists of these sections x of F ; that is, ker γ ∼= F ∼= Hom(OU ,F).We regard this latter as a subsheaf of Hom(Ou(−1)n+1,F) by applyingthe functor Hom(−,F) to the surjection OU (−1)n+1 → OU , giving aninclusion Hom(O,F) → Hom(OU (−1)n+1,F). Since our choice of Uwas arbitrary and the constructions are compatible with restriction, theconclusions hold with U replaced by PnS , concluding the proof.

7.3 An Example

We’d like to compute Ω1P1k/k for a field k. We know it’s a line bundle, and

we know what the line bundles on P1 are. But which line bundle is it?Theorem 5.1 gives us the exact sequence

0→ Ω1P1k/k → OP1

k(−1)2 → OP1

k/k → 0.

Now since O is free, the sequence splits, so the middle term is isomor-phic to the direct sum of the outer two. Hence the second exterior powerof the middle term is isomorphic to the tensor product of the outer twoterms. This means that

Λ2O(−1)2 ∼= Ω1P1k/k ⊗O ∼= Ω1

P1k/k,

but Λ2O(−1)2 ∼= O(−2), so we have calculated

Ω1P1k/k∼= O(−2)

More concretely, if we look on the affine open A1 ⊂ P1, we have Ω1P1∼=

Ω1A1∼= OA1 , generated by, say, dx. Similarly on the other chart A1 ∼=

D(y): there Ω1P1 is generated by dy. How are the two related? At ∞ ∈

D(y) \ D(x), the local ring OP1,∞ is a DVR with uniformizer x = 1/y,giving dx = −1/y2 dy. Thus the transition map for Ω1 is multiplicationby x2 = 1/y2. The transition map for O(1) is multiplication by y. ThusΩ1

Pn∼= O(−2).

7.4 Concluding Remarks

Before moving on to the study of curves, let’s review the techniques wehave available for calculating differentials on a scheme:

1. Try to use the calculation of Ω1Pn

2. Use the closed immersion exact sequence.

3. Use the relative cotangent exact sequence.

These ideas lead naturally into curve theory, where one of our maintasks will be to understand embeddings of curves into Pn using Ω1.

31

8 Friday, February 3rd: Curves

Now we begin our study of curves by embedding them into projectivespace. An equivalent approach which we exploit is to study line bundles(and sections thereof) on a curve (especially Ω1).

8.1 Separating Points and Tangent Vectors

Let k be an algebraically closed field, and X a proper k-scheme. Given asurjection On+1

X → L, when is the corresponding map φ : X → Pnk a closedimmersion? First we fix some notation. Let V = kn+1, with a fixed choiceof basis ei. We write On+1

X∼= V ⊗k OX , where in the tensor product

we are really tensoring with the constant sheaf associated to V .Let L be a line bundle on X, and On+1

X → L a surjection. For p apoint of X, recall our notation L(p) = Lp/mpLp. For any vector s in V ,we obtain an element sp of L(p) by the composite map

Vv 7→v⊗1−→ Γ(X,V ⊗k OX)→ Γ(X,L)→ Lp → L(p)

Recall also that by the Nullstellensatz, the residue field k(p) at anyclosed point p of X is isomorphic to k (remember we assume k is alge-braically closed).

With these observations in place, we answer the question above, viz.when do surjections onto line bundles give embeddings into projectivespace, by the following.

Proposition 8.1. If X/k is proper over an algebraically closed field, andφ : X → Pn the map associated to the surjection On+1

X

π→ L, then φ is aclosed immersion if and only if the following two conditions hold

1. (Separating Points) For every pair of distinct point p, q in X, thereis an element s of V such that sp (as defined above) is nonzero inL(p), whilst sq is zero in L(q).

2. (Separating Tangent Vectors) For every closed point p of X, the map

ker(V → L(p)

)→ mpLp/mpL2

p

is surjective.

8.2 Motivating the Terminology

We’ll prove this in the next lecture. First we motivate the conditions. Asa warmup, write V = kx0 ⊕ . . .⊕ kxn, and consider the xi as coordinateson Pn. If we take, for example, s = x0, then what is the set of closedpoints p in X such that s goes to zero in L(p)?

First recall that we can write the element φ(p) ∈ Pnk as [a0 : . . . : an] asfollows. Choose a basis element for L(p), i.e., an isomorphism L(p)

σ→ k.Then we get a map

kn+1 = V → Lpσ→ k

which sends each basis element xi to some ai in k. Of course, it is notwell-defined, since we had to choose a basis for L(p); but any other basisdiffers from our chosen one by a nonzero element of k, so our coordinates

32

ai are well-defined up to nonzero scalars, giving a usual point of projectivespace (in the classical sense). Thus we see that the set of closed pointsof X for which x0 maps to zero is just the preimage under φ of the set[0 : a1 : . . . : an] ⊂ Pnk .

More generally, if s is any linear form in the xi (i.e., an element of V ),then the points of X for which s maps to zero in L(p) are the preimageof the zero locus of s in Pnk . The zero locus of such a linear form isusually called a hyperplane in Pnk . So what condition 1 says is that forany p, q ∈ X, there exists a hyperplane passing through p but not q;succinctly, the map φ is injective on points.

Of course, this does not guarantee that φ is a closed immersion. Foran example where this fails, let Y be the cusp Spec k[x, y]/(y2 − x3) andX its normalization Spec k[t]. Then there is a morphism φ : X → Ycorresponding to the ring map x 7→ t2, y 7→ t3. This is a homeomorphismof topological spaces but not a closed immersion: the tangent space at theorigin on Y is “too big”.

To see why, look at the map on tangent spaces at the origin. On Y ,we have mO/m

2O∼= k · x ⊕ k · y, and on X, we have mO/m

2O∼= k · t. But

the induced map between them is zero, since the images of x and y areboth in m2

O ⊂ k[t]. In order for this map φ to be a closed immersion, thismap would need to be surjective.

Condition two says exactly this: the map TPn(φ(p)

)∗ → TX(p)∗ issurjective, or dually, the map on tangent spaces is injective.

8.3 A Preparatory Lemma

We’ll need the following in our proof next time.

Lemma 8.1. A map φ : X → Pnk is a closed immersion if and only if, forevery i = 0, . . . , n,

1. The subscheme Xi = p ∈ X∣∣xi 67→ 0 in L(p) is affine, and

2. the map k[y0, . . . , yn] → Γ(Xi,OXi) sending yj to π(xj)/π(xi) issurjective.

We explain the second condition: π is our surjection On+1X = V ⊗k

OX → L, which we think of as determining the homogeneous coordinatesof a point p ∈ X. Restricting to Xi, we get a map π

∣∣Xi

: Γ(Xi,On+1Xi

) →Γ(Xi,L

∣∣Xi

). For each p ∈ Xi, if we restrict to the stalk at p, we find

that π∣∣Xi

(xi) is nonzero (it’s nonzero in L(p), hence nonzero in Lp by

Nakayama). But this holds for any p ∈ Xi, therefore π(xi) must havebeen a unit in Γ(Xi,L

∣∣Xi

) (reason: it’s a unit in every stalk, hence not

contained in any prime ideal of Γ(Xi,L∣∣Xi

)).

Now since L is a line bundle, we are guaranteed an open cover of Xon which it trivializes. The above remarks show that in fact the Xi formsuch a cover, with the trivialization

OXi → L∣∣Xi

given by 1 7→ π(xi), since we just saw that π(xi) generates L over Xi.Therefore each global section π(xj) of L over Xi (some or all of which may

33

be zero), maps back under this isomorphism to an element π(xj)/π(xi).Then of course since k[y0, . . . , yn] is free, we may define a map to Γ(Xi,OXi)by sending each yj to π(xj)/π(xi). This is the map which appears in thesecond condition of Lemma 8.1, which we now prove.

Proof of Lemma 8.1. Since the condition of being a closed immersion isaffine-local on the target, the map φ : X → Pnk is a closed immersion ifand only if, for each i = 0, . . . , n, the restriction

φ−1(Pnk \ V (xi))→ Pnk \ V (xi)

is a closed immersion. By definition, this morphism is just

Xi → Spec k[y0

yi, . . . ,

ynyi

].

Moreover, since the target now is affine, this morphism is a closed immer-sion if and only if Xi also is affine (a closed immersion must be an affinemorphism) and the corresponding map of sheaves, which when Xi is affinecorresponds to a ring map k[ y0

yi, . . . , yn

yi]→ Γ(Xi,OXi), is surjective.

9 Monday, February 6th: Proof of the ClosedEmbedding Criteria

9.1 Proof of Proposition 8.1

We need yet another lemma:

Lemma 9.1. Let f : A→ B be a local homomorphism of local Noetherianrings, such that the induced map on residue fields A/mA → B/mB is anisomorphism, and the map on cotangent spaces

mA/m2A → mB/m

2B

is surjective. Then f itself is surjective.

The intuition behind the lemma is that if, say, A = k[[x1, . . . , xn]]and B = k[[x1, . . . , xn]]/(f1, . . . , fr), then surjectivity on cotangent spacesmeans that in each equation fi = 0, we can solve for xi in terms of elementsof A.

Proof. The ideal mAB ⊂ B is contained in mB since f is a local homomor-phism. Since mA/m

2A → mB/m

2B is surjective, the images of generators of

mA/m2A generate mB/m

2B , and we can lift these by Nakayama to genera-

tors of mB . This implies that the images of generators of mA generate mB ,so we have mAB = mB . Thus by our first assumption, A/mA ∼= B/mAB.In particular we have a surjection A→ B/mAB, so the image of 1 gener-ates B/mAB as an A-module. Applying Nakayama again, this time to Bas an A-module, we can lift the image of 1 in B/mAB to a generator ofB as an A-module, which means f is surjective.

34

Proof of Proposition 8.1. For the “if” direction, first observe that sinceX is proper over k, X → Pnk is also proper. Therefore the image Z ⊂ |Pnk |of |X| under φ is a closed set. The same holds with |X| replaced by anyclosed subset of X and Z replaced by φ(V ) (note closed immersions areproper and compositions of proper morphisms are still proper), so φ is aclosed map.

Next, condition (1) implies that the map φ of topological spaces is in-jective on closed points. We claim that this implies φ is actually injective.I’ll come back to this at some point.

Now a closed map that is also injective is a homeomorphism onto itsimage. Therefore (check!) φ is quasi-finite. But a proper and quasifinitemap is in fact finite (check!) and a finite map is always affine, so Xi =φ−1(Ui) is affine.

It remains to show that if p ∈ X(k), then the map

mPn,φ(p)/m2Pn,φ(p) → mX,p/m

2X,p

is surjective. [Recall our notation: X(k) is the set of k-valued points,i.e., maps Spec k → X, which correspond to closed points of X since k isalgebraically closed. So we’re saying the above should be surjective for allclosed points p]. This will suffice, because the condition thatOPn → φ∗OXbe surjective can be checked locally, since the global sections functor isexact on affines.

For notational convenience, we make a linear change of coordinates sothat φ(p) = [0 : . . . : 0 : 1] ∈ Pn. Then

mPn,φ(p)/m2Pn,φ(p)

∼= k · x0

xn⊕ . . .⊕ k · xn−1

xn,

the linear forms on the open affine Un = Spec k[ x0xn, . . . ,

xn−1

xn]. Since

p ∈ Xn, the global section sn = xn maps to a basis of Lp, i.e., gives atrivialization of Lp (the section sn canonically trivializes L over Xn). SomX,p/m

2X,p∼= mpLp/m2

pLp, and this isomorphism of OX -modules fits intothe following diagram

mPn,φ(p)/m2Pn,φ(p)

∼= //

k · x0xn⊕ . . .⊕ k · xn−1

xn

mX,p/m2X,p

∼= // mpLp/m2pLp

The map on the left is the one we want to show is surjective, and the mapon the right is the map of condition (2). This proves surjectivity of themap of sheaves locally at closed points. It is left as an exercise to checkat the non-closed points (possible hint: use fiber products somehow).

Note: we hand-waved the other implication (closed immersion impliesseparates points and tangent vectors) in the statement of the Propositionin class last time. I’ll try to come back and write it up clearly at somepoint.

35

9.2 Towards Cohomology

As before, consider a scheme X, proper over an algebraically closed field,of dimension one, normal, and connected. Such a curve has only two typesof points: a generic point, and some closed points X(k), at each of whichthe local ring is a DVR.

Remark. If P is a closed point of X, then IP , the (quasicoherent) idealsheaf associated to the embedding P → X, is invertible. In general, if Lis an invertible sheaf on X, we may write L(P ) = L ⊗OX I

−1P .

Now, when we are given a rank one surjection On+1X → L, how do we

check whether conditions (1) and (2) of Proposition 8.1 hold? First weconsider condition (1). Suppose given two distinct (closed) points P andQ of X. We hope to produce a section s ∈ V such that s(P ) = 0 buts(Q) 6= 0.

Consider the case when V = Γ(X,L). Then the s we desire is,in particular, an element of the kernel of the map of k-vector spacesΓ(X,L) = V → LP ⊗OX,P k(P ), and this kernel is just Γ(X,L(−P )).To see this, note that the map in queston is obtained by tensoring theexact sequence

0→ IP → OX → k(P )→ 0

with L, and then taking global sections (which is left-exact on the left, sorespects kernels). Then note that Γ(X,L ⊗OX k(P )) ∼= LP ⊗OX,P k(P )

since k(P ) is a skyscraper sheaf 2.In order to determine whether we can find an s such that, also, s(Q) 6=

0, we ask whether the map

Γ(X,L(−P ))→ LQ ⊗ k(Q)

is surjective. Here we regard LQ ⊗ k(Q) ∼= LQ/mQLQ simply as a one-dimensional vector space; but we may also regard LQ⊗k(Q) as a skyscrapersheaf at Q, which fits into the following exact sequence of sheaves of OX -modules:

0→ L(−P −Q)→ L(−P )→ LQ ⊗ k(Q)→ 0

Now the global sections functor is not right-exact, so the map

Γ(X,L(−P ))→ Γ(X,LQ ⊗ k(Q)) = LQ ⊗ k(Q)

is not necessarily surjective. But since LQ ⊗ k(Q) is a one-dimensionalk-vector space, there are only two possibilities: the map is zero, or else it’ssurjective. Condition (1) holds exactly when the map is not zero. Similarconsiderations apply to condition (2) by taking P = Q in the above. So weare led to the following general problem. Given the short exact sequence

0→ L(−P −Q)→ L(−P )→ LQ ⊗ k(Q)→ 0,

when is the sequence

0→ Γ(X,L(−P −Q))→ Γ(X,L(−P ))→ Γ(X,LQ ⊗ k(Q))

2This means that the sections over any open containing P are all the same, so taking thelimit does nothing.

36

obtained by applying the global sections functor also exact? This is thequestion of cohomology. Once we have defined cohomology, we will seethat there is a long exact sequence

0 // Γ(X,L(−P −Q)) // Γ(X,L(−P )) // Γ(X,LQ ⊗ k(Q))

rrdddddddddddddddddddd

H1(X,L(−P −Q)) // H1(X,L(−P )) // . . .

and thus the vanishing of H1(X,L(−P −Q)) for all P and Q (not neces-sarily distinct) is sufficient to ensure that the associated map X → Pn isa closed immersion.

10 Wednesday, Feb. 8th: Cohomology of Coher-ent Sheaves on a Curve

Before proving Riemann-Roch, we introduce, without proof of its exis-tence, the cohomology theory we need, together with its relevant prop-erties. The first section is a reproduction of a handout Professor Olssondistributed in class. The second and third contain my notes on his com-ments about the handout.

10.1 Black Box: Cohomology of Coherent Sheaves on Curves

For the results on curves that we will prove in this class, we will assumethe existence of a cohomology theory satisfying the following properties.The existence of such a cohomology theory will be a consequence of thegeneral development of cohomology which we will study later.

Fix an algebraically closed field k. In the following, a curve means aproper normal scheme X/k of dimension 1.

If X is a curve and F is a coherent sheaf on X, we write H0(X,F ) forΓ(X,F ). As part of our black box we will include the following result:

(0) For every coherent sheaf F on a curve X, H0(X,F ) is finite di-mensional over k.

We assume given for every curve X a functor

H1(X,−) : (coherent sheaves on X)→ (finite dimensional k-vector spaces)

and for every short exact sequence of coherent sheaves on X

E : 0→ F ′ → F → F ′′ → 0

a mapδE : Γ(X,F ′′)→ H1(X,F ′)

such that the following conditions hold.(i) The functor H1(X,−) is k-linear in the following sense. If a ∈ k andF is a coherent sheaf, and if ma : F → F denotes the multiplication by amap on F (which is an endomorphism in the category of coherent sheaveson X) then the induced map

H1(ma) : H1(X,F )→ H1(X,F )

37

is multiplication by a on the vector space H1(X,F ).(ii) For every short exact sequence E as above, the sequence

0 // H0(X,F ′) // H0(X,F ) // H0(X,F ′′)

δE

tthhhhhhhhhhhhhhhhhhhh

H1(X,F ′) // H1(X,F ) // H1(X,F ′′) // 0

is exact.(iii) The maps δE is functorial in the exact sequence E in the followingsense. If

0 // F ′1 //

a

F1

b

// F ′′1

c

// 0

0 // F ′2 // F2// F ′′2 // 0

(9)

is a commutative diagram of coherent sheaves on a curve X with exactrows, then the diagram

H0(X,F ′′1 )

H0(c)

δE1 // H1(X,F ′1)

H1(a)

H0(X,F ′′2 )δE2 // H1(X,F ′2)

commutes, where we write E1 (resp. E2) for the top (resp. bottom) shortexact sequence in (9).(iv) If F is a coherent sheaf on a curve X whose support is a finite set ofpoints, then H1(X,F ) = 0.(v) If X is a curve then there is an isomorphism tr : H1(X,Ω1

X)→ k suchthat for any locally free sheaf E on X with dual E∨ the pairing

H0(X,E)×H1(X,E∨ ⊗ Ω1X) // H1(X,Ω1

X)tr // k

is perfect.

Remark. Axiom (i) implies that if F and G are coherent sheaves onX and if z : F → G is the zero map then the induced map H1(z) :H1(X,F ) → H1(X,G) is also the zero map. Indeed z = m0 z, wherem0 : G→ G is multiplication by 0, and thereforeH1(z) = H1(m0)H1(z).Since H1(m0) is multiplication by 0 on H1(X,G) by (i) this implies thatH1(z) = 0.

Remark. Axiom (ii) implies that H1(X,−) commutes with finite directsums. Indeed to show this it suffices by induction to consider two coherentsheaves F and G. In this case we have a split exact sequence

0→ F → F ⊕G→ G→ 0.

Since the projection map F ⊕ G → G admits a section, then inducedmap H0(X,F ⊕ G) → H0(X,G) is surjective, whence (ii) gives an exactsequence

0→ H1(X,F )→ H1(X,F ⊕G)→ H1(X,G)→ 0

38

split by the standard inclusion G → F ⊕ G. Since finite direct sumsare isomorphic to finite direct products in both the category of coherentsheaves on X as well as the category of finite dimensional vector spaces, itfollows that H1(X,−) also commutes with finite products. This has manyconsequences. For example, if F is a coherent sheaves and σ : F ⊕F → Fis the summation map, then the induced map

H1(X,F )⊕H1(X,F ) ' H1(X,F ⊕ F )H1(σ)

// H1(X,F )

is the summation map on the vector space H1(X,F ). Similarly if f, g :F → G are two maps of coherent sheaves, then the two maps

H1(f + g), H1(f) +H1(g) : H1(X,F )→ H1(X,G)

are equal.

10.2 Further Comments; Euler Characteristic

Axiom (v) is known as Serre duality. We explain how to obtain the firstmap, leaving the existence of the trace map tr as part of the axiom. Fora perfect pairing, we would like to associate to each e ∈ H0(X,E) anoperator 〈e,−〉 on H1(X,E∨⊗Ω1

X), in such a way that this isomorphismidentifies H0(X,E) with the dual space of H1(X,E∨⊗Ω1

X). First observethat for each e ∈ H0(X,E) = Γ(X,E), we get a map OX → E, which wetensor with E∨ to obtain maps of sheaves

E∨∼= //

//

OX ⊗OX E∨e⊗1// E ⊗ E∨

Ox

where the vertical map is just evaluation. The dotted composite we nowtensor with Ω1 to give a map E∨ ⊗ Ω1 → Ω1, to which we apply thefunctor H1(X−). So for each such global section e, we have a mapH1(X,E∨ ⊗ Ω1

X) → H1(X,Ω1), and this defines the pairing. Note alsothat the existence of the trace map in the axiom should be expected ifyou are familiar with Hodge theory on complex manifolds.

We will state Riemann-Roch using Euler characteristic, and then red-erive the more familiar form. The theorem is about calculating the di-mensions of the k-vector spaces H0(X,L) for various line bundles L. Thiscan be quite hard, and it turns out to be easier to look instead at theEuler characteristic χ defined by, for F ∈ Coh(X),

χ(X,F) = dimkH0(X,F)− dimkH

1(X,F)

One useful fact is that χ is additive on short exact sequences. Moreprecisely, we have the following:

Lemma 10.1. If the sequence 0 → F ′ → F → F ′′ → 0 is exact, thenχ(X,F) = χ(X,F ′) + χ(X,F ′′).

39

Proof. From the given sequence we get a long exact sequence

0

0&&LL

LLL 0

I1

88qqqqqI3

OO

''OOOO

0 // H0(F ′) // H0(F) //

77oooooH0(F ′′) //

%%KKK

H1(F ′) //OO

H1(F) // H1(F ′′) // 0

I3

::ttt

&&LLLLL

0

88qqqqq0

Here the Ik are the kernels and images of the appropriate maps. Thisgives us four short exact sequences involving the cohomology groups andthe Ik. Each short exact sequence relates the dimensions of three of thevector spaces. If you combine the four equations appropriately, you getthe result. Note: This diagram differs slightly from the one used in class,though the argument is essentially the same. This more symmetric formof the diagram was pointed out to me by Alex Kruckman.

Remark. Let’s briefly recall the relationship between divisors and linebundles. On a curve X, we can define a divisor as a formal sum of closedpoints, D =

∑nP · P . To such an object, we associate a line bundle, de-

noted by either OX(D) or L(D), as follows. Each closed point P maybe regarded as a closed subscheme of X, and thus has an associatedideal sheaf IP [Recall further that a closed point is the same as a mapi : Spec k → X, which induces a map i# : OX → i∗OSpec k, where thelatter is most usefully regarded as the skyscraper sheaf of k(P ) = k at thepoint P . The ideal sheaf IP is defined to be the kernel of i#]. Then weset

L(D) = L(∑

nP · P ) =⊗P

I−nPP

where here the superscript means tensor power.Conversely, every invertible sheaf on X is of this form. Suppose given

a line bundle L on X. Take an open subset U ⊂ X on which we havea trivialization σ : OU

∼→ L∣∣U

. Let s = σ(1), a section of L over U .The complement of U is then a finite set of points P1, . . . , Pr (possiblyempty, if L itself happened to be trivial). Since X is a curve, the local ringat each Pi is a DVR. We want to use the valuations in each of these DVRsto produce the integers nPi . Now we proved last semester that there isan inclusion L → K , where K is the constant sheaf associated to thefunction field k(X). So we take s, and map it into K (U) ∼= k(X). Thenat each Pi we have a valuation νPi on k(X), and we define nPi = νPi(s).These nPi , for each Pi, thus define a divisor D =

∑nPi · Pi on X. Of

course, we still have to check that L ∼= L(D). For this, you can refer tothe notes from last semester.

10.3 Statement of Riemann-Roch; Reduction to the More Fa-miliar Version

From now on, we will use the notation hi(X,L) = dimkHi(X,L).

40

Theorem 10.1 (Riemann-Roch Theorem). If X is a curve, and L a linebundle on X, then

χ(X,L) ∼= degL+ χ(X,OX).

We will prove this next lecture. First notice that

H1(X,L) = H1(X, (L∨ ⊗ Ω1X)∨ ⊗ Ω1

X

),

and by Serre Duality, this is dual to H0(X,L∨ ⊗Ω1X). Hence h1(X,L) =

h0(X,L∨⊗Ω1X). This allows us to rewrite the left hand side of the equation

in the theorem as h0(X,L) − h0(X,L∨ ⊗ Ω1X). Moreover, applying the

same trick to H1(X,OX), we get h1(X,OX) = h0(X,Ω1X), which is by

definition the genus, g, of X. Finally, we have h0(X,OX) = 1, by eitherof two arguments presented at the beginning of the next lecture.

Thus we can rewrite Riemann-Roch as

h0(X,L)− h0(X,L∨ ⊗ Ω1X) = degL+ 1− g.

Remark. If H0(X,L) 6= 0, then degL ≥ 0. For if s is a nonzero globalsection, then as an element of the fraction field, its valuation at each pointP is non-negative (i.e., it’s in OX,P ). Loosely, it’s everywhere regular, sohas no poles. In fact, you can calculate the degree of L explicitly bylooking at the exact sequence

0→ OX → L → L/OX → 0,

where the term on the right is supported on a finite set of points.

11 Friday, Feb. 10th: Proof and Consequences ofRiemann-Roch

11.1 Preparatory Remarks

Remark. Since H0(X,OX) is a k-algebra, h0(X,OX) is at least one.Here are two different reasons why it’s equal to one.

Proof 1. Let f ∈ H0(X,OX). Then f defines a map X → A1k → P1

k.Since P1

k is proper, f(X) is a closed subset of P1k, hence a finite set of

points in A1k. Since X is connected, so is f(X), which must therefore

consist of a single point α ∈ A1k.3 So the map to A1

k defined by f agreeswith the map defined by α ∈ k ⊂ H0(X,OX). Therefore the globalsections f and α differ by nilpotence, and since X is reduced, they arethe same global section. Hence H0(X,OX) = k.

3Recall that the map to A1k defined by f sends x to the residue of f in k(x) for each x ∈ X.

For closed points, k(x) = k, so in our case, we have that f maps to the same α ∈ k forevery closed point, and we think of this α as living in A1

k (i.e., as corresponding to the ideal(t− α) ⊂ k[t].

41

Proof 2. Since H0(X,OX) is finite-dimensional over k, it’s a finitely gen-erated integral (commutative) algebra over k. This implies it’s a finitelygenerated field extension of k: to find the inverse to a nonzero α, use themonic polynomial cancelling α. That is, if αn + bn−1α

n−1 + . . .+ b0 = 0,then α(αn−1 + bn−1α

n−1 + . . .+ b1) = −b0, which is nonzero since it’s adomain, so α is a unit.

Then since k is algebraically closed, there are no nontrivial finitelygenerated extensions, hence H0(X,OX) = k.

Remark. Recall from last time that if L is an invertible sheaf on X,and there is a nonzero global section s ∈ H0(X,L), then degL ≥ 0.Also, if degL = 0, then L ∼= OX (so s ∈ k). To see this, note that thesection s defines a map OX → L. For over each open U , we can defineO(U)→ L

∣∣U

by sending 1 (i.e., the restriction of 1 ∈ H0(X,OX)) to s∣∣U

.This obviously commutes with restriction maps, so it defines a morphismof sheaves. Of course, there’s no reason for this to be an isomorphism,but it is injective since s is nonzero over every open U (using that X isintegral).

We are interested in the cokernel of this map s. To measure the failureof s to be an isomorphism, we begin with an open U on which L is trivial;s itself furnishes a trivialization. The complement of U is a finite pointset P1, . . . , Pr. The cokernel is trivial over U ; looking at germs nearPi, LPi is a free rank one OX,Pi -module, and the cokernel of s locally hasthe form LPi/m

niPiLPi . Each ni is non-negative since s is a global section.

Thus the cokernel of s is the sheaf

coker(s) =⊗

P1,...,Pr

LPi/mniPiLPi

where each factor in the tensor product is a skyscraper sheaf at the ap-propriate Pi. The global sections of this sheaf forms a finite-dimensionalk-vector space, whose dimension is the degree of L, hence degL =

∑ni.

It follows, then, that if degL = 0, each ni is zero, and the cokernel istrivial, so s defines an isomorphism L ∼= OX .

11.2 Proof of the Riemann-Roch Theorem

Proof. We will prove the relation

χ(X,L) = degL+ χ(X,OX)

by induction on the degree of L. For degL = 0, it’s a consequence of theabove remarks, since we just saw that in this case, L ∼= OX .

For the inductive step, we prove that the equation holds for L if andonly if it holds for L(P ) = L ⊗ I−1

P . To see this, tensor the sequence

0→ IP → OX → k(P )→ 0

with L(P ), giving0→ L → L(P )→ k(P )→ 0

42

(note that the term on the right is a skyscraper sheaf, so tensoring withL(P ) does nothing). Since the Euler characteristic is additive on exactsequences, this yields χ(X,L(P ) = χ(X,L)+1 (note that h1(X, k(P ) = 0by one of our axioms for cohomology). Since degL(P ) = degL + 1, thisproves the equivalence.

This proves the theorem, for we have seen above that any line bundlecan be obtained from OX by adding or subtracting a finite number ofpoints

11.3 Consequences

Corollary 11.1. deg Ω1k = 2g − 2.

This comes from setting L = Ω1k in the statement of Theorem 10.1. It

suggests that the study of curves should divide into three cases: g = 0,g = 1, or g ≥ 2, according as the degree of Ω1

k is negative, zero, or positive,respectively.

Corollary 11.2. If degL > 2g − 2,

h0(X,L) = degL+ 1− g

Corollary 11.3. If degL > 2g−1, then L is generated by global sections.

Remark. We could also express this by saying that the map H0(X,L)⊗kOX → L is surjective. What map is this? Let f : X → Spec k be thestructure morphism. Since f∗ is a left adjoint to f∗, there is a naturalmap f∗f∗L → L. But f∗ is just H0(X,L) sitting over the unique pointof Spec k. This pulls back to f−1H0(X,L) ⊗f−1OSpec k

OX , and since f

is actually an open map here, and H0(X,L) and OSpec k are constantsheaves f−1 just pulls them back to the corresponding constant sheaveson X, i.e., f−1H0(X,L)⊗f−1OSpec k

OX ∼= H0(X,L)⊗k OX .

Proof of Corollary 11.3. We prove that the cokernel of the mapH0(X,L)→L described in the previous remark is zero. Since the support of the coker-nel (and indeed any sheaf) is closed, it’s enough to check that the cokernelis zero at closed points. Thus we have to show that for any closed pointP , there is a global section f whose image in L ⊗ k(P ) ∼= LP /mPLPis nonzero (we are using Nakayama here). This is equivalent to finding aglobal section f which is in Γ(X,L) but not in the subspace Γ(X,L(−P )).This we can verify simply by comparing dimensions. Our degree assump-tion ensures that the h1 term in Riemann-Roch vanishes for both sheaves(Corollary 11.2), so we have

h0(X,L) = degL+1−g and h0(X,L(−P )) = degL−1+1−g = degL−g

This Γ(X,L(−P )) is a proper subspace of Γ(X,L), so we can find thedesired global section f .

Remark. (In response to a question) Recall that L(−P ) has stalks equalto L away from P , and equal to mP ⊗ L at P .

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Of course, it is not really necessary to require that degL be greaterthan 2g − 1, for if L has any positive degree, then we can take a hightensor power to obtain a sheaf whose degree is at least 2g − 1.

Definition 11.1. A line bundle L onX is very ample if the map π : H0(X,L)⊗kL → L defines a closed immersion into Pnk . L is ample if there exists ann > 0 such that π : H0(X,L⊗n)⊗k L → L⊗n defines a closed immersion.

We will prove next time that if degL > 0, then L is ample.

12 Monday, Feb. 13th: Applying Riemann-Rochto the Study of Curves

First we introduce some new terminology. Recall that we have constructeda map H0(X,L) ⊗k OX → L of sheaves, for a line budle L on a curveX. We call the set of points at whose stalks this map is not surjectivethe base locus of L, and if it surjective everywhere (i.e., L is generated byglobal sections), we say L is base point-free.

In this language, we saw last time that if the genus of X is g anddegL > 2g−1, then L is base point-free, in which case we get a morphismof schemes X → PH0(X,L).

12.1 Using Riemann-Roch to Detect Closed Immersions to Pn

Any rank one quotient of On+1X corresponds to some morphism X → Pnk .

When L is generated by global sections, we get a rank one quotient ofthe form H0(X,L) ⊗k OX → L, with corresponding morphism X →PH0(X,L). When is this a closed immersion?

By our earlier criteria, L must separate points and tangent vectors.We now rephrase these conditions in such terms as allow us to applyRiemann-Roch.

The condition of separating points is equivalent to requiring that forany two distinct closed poitns P and Q in X, there is a section s inH0(X,L(−P )) \H0(X,L(−P −Q)). The condition of separating tangentvectors is equivalent to requiring that for any closed point P , there isa section s in H0(X,L(−P )) \ H0(X,L(−2P )). We can guarantee thesatisfaction of both conditions at once by requiring that, for any P andQ in X(k)4, not necessarily distinct, there is a section in H0(X,L(−P )) \H0(X,L(−P −Q)). This can be checked with a simple dimension count,and Riemann-Roch gives the following criterion:

Proposition 12.1. If degL > 2g, then L is very ample.

Proof. For any closed points P,Q (not necessarily distinct), our assump-tion on the degree of L ensures that h1 is zero for both L(−P ) andL(−P −Q), so Riemann-Roch computes the following:

h0(X,L(−P )) = (degL − 1) + 1− g

h0(X,L(−P −Q)) = (degL − 2) + 1− g,

4Recall that this is the set of k-valued points of X, namely the closed points.

44

This means the latter is a proper subspace of the former, and we can findthe section we desire, as described in the preceding discussion.

We also have the following:

Corollary 12.1. If degL > 0, L is ample.

12.2 Examples

Genus zeroLet X be a curve with g = 0. We will show that X is isomorphic to

P1k. For any P ∈ X(k), let, as usual, OX(P ) = I−1

P , where IP is the idealsheaf of P → X. It has degree 1 > 2g − 2 = −2, so

h0(X,OX(P )) = degL+ 1− g = 2

By choosing an isomorphism H0(X,OX(P ))∼−→ k2, we get a closed

immersion (by Proposition 12.1, since degL = 1 > 2g = 0) i : X →P1k. We claim this is actually an isomorphism of schemes. Since it’s

a closed immersion, i is a closed map. Therefore its image is a closedset, hence all of P1. Since it’s an immersion, it’s injective, and sinceproper also, it’s a homeomorphism (cf. a related statement in point-settopology - a continuous bijection from a compact space to a Hausdorffspace is automatically a homeomorphism.) Now we check the map ofsheaves is an isomorphism. Since i is a closed immersion, it induces asurjection of sheaves, and hence a surjection of stalks OP1,P → OX,P(here we identify X with its image, namely P 1, and thus speak of thesame P on both sides). The map i is a surjective morphism of schemes,hence in particular a dominant rational map, and so it induces a map offuntion fields k(P1)→ k(X). This map is not the zero map, so it must beinjective. Thus we obtain the following diagram

OP1,P // //

OX,P

k(P1) // k(X)

where the top arrow is surjective and the other three are injective. Thisforces the top arrow to be an isomorphism, like we wanted. This showsthat X is isomorphic to P1

k.

Genus oneNow let X be a genus one curve5 over k. Since 2g = 2, we need a line

bundle of degree at least three to embed in projective space. Pick a pointP ∈ X(k) and consider OX(3P ). After choosing a basis for the three-dimensional space H0(X,OX(3P )), we get a closed immersion i : X → P2

k.We will show next time that i identifies X with a cubic curve in P2

k, whichcan be given in characteristics different from 2 or 3 by an equation of theform y2 = x3 +Ax+B.

5This is not the same as saying that X is an elliptic curve - these are given by a genus onecurve together with a chosen point.

45

Remark. We chose a degree 3 line bundle to satisfy the hypothesis of thetheorem, but this is not necessarily the best we can do. For an example,take X = P1

k, and the line bundle OX(2). Then h0(P1x,O(2)), and we get

an embedding P1k → P2

k. Exercise: find the equation of this embedding(hint: it’s the Veronese map).

13 Wednesday, Feb. 15th: “Where is the Equa-tion?”

Today we prove:

Theorem 13.1. Any curve of genus one (over k) is isomorphic to a cubicin P2

k.

Recall that a homogeneous polynomial F of degree d in the variables,say, x0, . . . , xn, defines a closed subscheme of Pnk , which we denote V (F ).We ask, given a morphism g : X → Pnk corresponding to a rank one quo-tient On+1

X → L, how can we detect whether g factors through V (F )?To answer this, let the generating global sections from the surjectionOn+1X → L be s0, . . . , sn ∈ Γ(X,L). Recall that the identity map Pn → Pn

gives a surjection On+1Pn → OPn(1), with corresponding global sections

x0, . . . , xn, which we think of as the homogeneous coordinates on Pnk . Nowif we have a morphism g : On+1

Pn → OPn(1), we can pull back the surjectionOn+1

Pn → OPn(1) over Pnk to the surjection On+1X → g∗OPn(1) (recall that

the pullback of the structure sheaf is again the structure sheaf). We willabuse notation slightly and refer to the global sections of this pulled-backsurjection also as x0, . . . , xn. The surjections fit into a diagram

On+1X

(s0,...,sn)// //

(x0,...,xn)(( ((PPPPPPPPPPPPP L

∼=

g∗OPn(1)

Here the vertical map is an isomorphism because both surjections corre-spond to the same morphism X → Pnk .

Proposition 13.1. The map g factors through V (F ) if and only if F (s0, . . . , sn)is zero in Γ(X,L⊗d).

Coarse Proof. The expression of a closed point P of Pn in terms of homo-geneous coordinates [x0 : . . . : xn] is obtained by taking each of the globalsections x0, . . . , xn of OPn(1) and evaluating them in O(1)P /mPO(1)P ∼=k. A point P is in V (F ) if and only if these evaluations satisfy F = 0,or put another way, if and only if F (x0, . . . , xn), regarded as an elementof the sheaf O(1)⊗d, maps to zero at P . This makes sense, since locally,O(1)⊗d ∼= O(1) via the multiplication map a0 ⊗ . . . ⊗ an 7→ a1 · · · an (ofcourse, they’re both isomorphic to OPn locally). So the evaluation of Fat P is obtained by taking F , restricting it to some open containing P onwhich O(1) is trivial, multiplying the sections occuring in F 6, and then

6It doesn’t make sense to multiply, say, x0x1 over an arbitrary open since O(1) may notbe a ring over any open - it only is once we trivialize the line bundle.

46

passing to O(1)P /mPO(1)P . This gives an element of k, which we requireto be zero in order for P to be in V (F ).

To ask whether a point of X maps into V (F ), note that the image ofa point P ∈ X under g is given by evaluating the sections s0, . . . , sn atP . So the discussion in the previous paragraph applies to (the images of)points of X, simply replacing xi by si.

Let’s look at some examples.

Example 13.1. Let X have genus zero, and pick a point P ∈ X(k). Thespace H0(X,OX) is a proper subspace of H0(X,OX(P )). We’ve alreadycomputed dimH0(X,OX(P )) = 2, so we can write H0(X,OX(P )) as thespan of two global sections 1 and x. Here 1 spans the constant sections,namely H0(X,OX), while x necessarily has a pole at P , or else it wouldbe in H0(X,OX).

We’ve seen last time that a basis of global sections of OX(P ) deter-mines a map X → P1 (an isomorphism, in fact). Identifying X with P1

via this isomorphism, one thinks of the global sections x and 1 as de-homogenizations of the global sections (i.e., homogeneous coordinates) xand y on P1.

Example 13.2. Let X have genus one, and fix a point P ∈ X(k). The in-vertible sheaf L = O(3P ) defines an embedding X → P2

k, since h0(X,L) =3 > 2g = 2. We have a chain of subspaces (writing H0(X,OX) =H0(OX), etc., for short)

H0(O) ⊆ H0(O(P )) ⊆ H0(O(2P )) ⊆ H0(O(3P )) ⊆ H0(O(4P ))

⊆ H0(O(5P )) ⊆ H0(O(6P )) ⊆ . . .

Riemann-Roch allows us to compute the dimensions: 1,1,2,3,4,5,6,. . . .Pick a basis 1, x for H0(O(2P )). Then x has a pole of order two at P , orelse the previous space would have been two-dimensional. Similarly, wehave a basis 1, x, y for H0(O(3P )), where y has a triple pole at P . Thenx2 extends this to a basis for H0(O(4P )), and xy extends it further toa basis for H0(O(5P )). But in H0(O(6P )), we have the seven elements1, x, x2, x3, xy, y, y2, all of which have poles or order at most 6 at P . Sincethis space is six-dimensional, there must be a dependence relation amongstthem, and this relations must involve both x3 and y2, or else one of thesetwo would live in a previous space, but they both have poles of orderexactly six. Thus, after rescaling the coordinates if necessary, we obtaina relation of the form

q(x, y) = y2 + bxy + cy − x3 − ex2 − fx− g = 0

This implies that X ⊆ Y = V (q(x, y)) ⊆ P2. Now one argues thatsince X is a curve, q(x, y) is irreducible, and Y is one-dimensional, X mustbe homeomorphic to Y . To check the map X → Y is an isomorphism, it’s

47

enough to look at the stalks. For the stalk at P , we have a diagram

OY,P // //

OX,P

k(Y )∼= // k(X)

where the vertical maps are inclusions, and the bottom arrow is an iso-morphism since Y is integral (we mentioned above it’s irreducible. Checkit’s reduced). Thus the map on stalks is an isomorphism at P .

In general, given X and L, with corresponding embedding X →PH0(X,L), it possible to express

X ∼= Proj⊕n≥0

Γ(X,L⊗n).

We also havePnk = Proj

⊕n≥0

SnΓ(X,L),

where SnV denotes the nth symmetric power. Then the embedding X →Pnk is given by a map of graded rings given in degree n by SnΓ(X,L) →Γ(X,L⊗n). The equations defining X come from the kernel of this mapof graded rings.

Going back to Example 2, the equation q(x, y) = 0 defines an opensubset U of X inside the affine chart A2

x,y of P2k (the point P is in the

complement of U). We can regard this inclusion Y ⊂ A2 as being given bythe coordinate functions x and y (this is basically the Spec−Γ adjunctionfor morphisms to affine targets). But the coordinate function x also definesa map A2 → A1. Precomposing this with the inclusion U ⊆ A2 and thenfollowing with the inclusion A1 → P1 gives a map U → P1. We have seenthat such a map extends to a morphism X → P1.

Now we ask: what is the line bundle and corresponding global sectionsassociated to this map X → P1? Since we mapped A2 to A1 by “forget-ting” the y-coordinate, it is perhaps not surprising that the map to P1

comes from the space H0(X,O(2P )), which had as basis only 1 and x.

Remark. It is no accident that the degree of the line bundle OP1(1),when pulled back to X, is two. This is also the degree of

k(x)[y]/(y2 − (x− α)(x− β)(x− γ)).

Note also that, letting f be the map X → P1, we have

deg f∗OP1(1) = deg If−1(∞) = deg( ⊗P

∣∣f(P )=∞

IeP)

= −∣∣f−1(∞)

∣∣ =∑

eP

Here eP denotes the ramification index of f at P . We will discussdegree and ramification in the next lecture.

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14 Friday, Feb. 17th: Ramification and Degreeof a Morphism

Definition 14.1. Let f : X → Y be a nonconstant morphism of curves.The degree of f is the degree of the field extension [k(Y ) : k(X)]. If Pis a closed point of X, with f(P ) = Q, the ramification index eP of fat P is defined to be the ramification index of the induced map of DVRsOY,Q → OX,P .

This isn’t much of a definition until we have defined ramification in-dices for maps of DVRs. Let πP , πQ uniformizers of OX,P and OY,Q,respectively. Then πQ maps to a unit times πePP , where eP ≥ 1 since thisis a local homomorphism. Then eP is the degree of this map of DVRs.We say f is unramified at P if eP = 1, and ramified at P if eP > 1,in which case we also call Q a branch point of f .

The degree of the morphism f is, intuitively, the cardinality of thepreimage of any point Q ∈ Y , but we have to be careful at the branchpoints.

Proposition 14.1. For any closed point Q of Y ,∑P∈f−1(Q)

eP = deg f.

Proof. Let A = f∗OX . Then X = SpecY

A . The idea of the proof is tofirst prove that A is a locally free sheaf of finite rank on Y . Then we cancompute the rank at any point of Y . Over the generic point, we show therank is [k(X) : k(Y )] = deg f , while over a closed point Q, we will showthe rank is

∑P∈f−1(Q) eP , giving the desired equality.

First let’s prove A is locally free of finite rank. Since f is proper andquasifinite, f is finite, so we will have finite rank if we can show A islocally free. For this, it is enough to check at closed points Q ∈ Y (k),since the generic point is a localization of any closed point. First weobserve that AQ is a flat OY,Q-module. Flatness is equivalent to the mapI ⊗ AQ → A being injective for any ideal I of OY,Q. Since OY,Q is aDVR, it’s enough to check that the action of the uniformizer πQ on AQ

is nonzero. But over an open U , A (U) = OX(f−1(U)), and the actionof πQ is given by mapping πQ into OX(f−1(U)) and multiplying. ButOX(f−1(U)) is a domain, so this action is nonzero. Now the stalks ofA (which are stalks of OX), are finitely generated over the stalks of OYsince f is finite. But flat and finitely generated impies free, so A has freestalks. We conclude that A is locally free by the following.

Lemma 14.1. If F is a coherent sheaf on Y such that FQ is a free OQ-module for all Q, then F is locally free.

Proof. At each Q we have an isomorphism OrY,Q → FQ. Let f ′1, . . . , f′r ∈

FQ be the image of the generators under this isomorphism, and f1, . . . , frtheir representatives on some neighborhood U containing Q. Then thesefi define a map of sheaves on U (in fact, of OU -modules) σU : OrU → F

∣∣U

which is an isomorphism at the stalk at Q. Let K and G be the kerneland cokernel, respectively, of σU .

49

Since OrU and F are coherent, so are K and G. Thus their supportsare closed subsets of U not containing Q. In particular, there is an openneighborhood V ⊆ U containing Q such that KP and GP are both zerofor all P ∈ V . Since isomorphisms can be checked at stalks, this meansthat σU

∣∣V

is an isomorphism over V , and hence F∣∣V

is free.

Returning to the proof, we have now that A is locally free, and wantto compute its rank at both a closed point and the generic point η of Y .

First, letting ξ be the generic point of X, we note that

Aη = limη∈UOX(f−1(U)) = lim

ξ∈VOX(V ) = k(X),

the equality in the middle coming from the fact that for any open Vof X, we can find U ⊆ Y such that f−1(U) ⊆ V (“preimages of opensin Y are cofinal amongst opens in X”). Namely, if V = X \ Pi, setU = Y \ f(Pi). We use this to compute the rank of A at η: it’s[k(X) : k(Y )], i.e., deg f .

Now let Q be a closed point of Y . We will compute the rank of A atQ. This is by definition the rank of AQ as an OY,Q-module, which is alsoequal to the dimension of AQ ⊗OY,Q k(Q) as a k(Q)-vector space. Firstwe claim that

AQ ⊗OY,Q k(Q) ∼= Γ(X ×Y Spec k(Q),OX×Spec k(Q))

To see this, we can replace X and Y by affines SpecR and SpecS, respec-tively, where Q is contained in SpecS and f(SpecR) = SpecS since f isaffine. The fiber product in question corresponds to the diagram

R⊗S k(Q) k(Q)oo

R

OO

Soo

OO

which we can expand into a double fibered diagram

R⊗S k(Q) k(Q)oo

R⊗S OY,Q

OO

OY,Qoo

OO

R

OO

Sf#

oo

OO

Thus the claim follows from the observation that in fact AQ∼= R⊗SOY,Q,

since

R⊗S OY,Q = limQ∈D(g)

R⊗S Sg = limQ∈D(g)

Rf#g = limQ∈D(g)

OX(f−1(D(g))),

which is (f∗OX)Q = AQ.We must therefore compute the dimension over k(Q) of the global

sections of X ×Y k(Q). Since f is finite, X ×Y Spec k(Q) is a disjoint

50

union of points: f−1(Q) = P1, . . . , Ps, and therefore it is an affinescheme:

X×Y Spec k(Q) ∼=s∐

1=1

SpecOX,Pi⊗OY,Qk(Q) ∼= Spec

s∏i=1

OX,Pi⊗OY,Qk(Q).

Now, OX,Pi⊗OY,Qk(Q) ∼= OX,Pi/πeii , since the OY,Q-algebra structure

on OX,Pi is given by sending the uniformizer of OY,Q to πeii , where πi isthe uniformizer of OX,Pi , and ei the ramification index of f at Pi. Thuswe conclude that

AQ ⊗OY,Q k(Q) ∼=s∏i=1

OX,Pi/πeii ,

which has dimension

s∑i=1

ei. This is therefore the rank of A at Q, and

equating it with the rank at the generic point (deg f) gives the result.

Example 14.1. If X is a curve of genus 1, and P a closed point, thenh0(X,OX(2P )) = 2, so we get a map f : X → P1. On the complement ofP , we have a map to A1

k = Spec k[t] given by f# : k[t] → Γ(X \ P,OX).The section f#(t) extends to a global section of Ox(2P ) with a pole oforder 1 or 2 at P . Thus deg f ≤ 2. But the degree cannot be one, or elsef would be an isomorphism, which is impossible since X has genus 1. Sothe divisor 2P defines a degree two map to P1.

15 Wednesday, Feb. 22nd: Hyperelliptic Curves

So far in our study of curves, we have seen that genus one curves areisomorphic to P0 while genus one curves are cut out by cubic equationsin P2. Today we study curves of genus greater than one. We will see thatthey split into two camps: those that are “canonically embedded” in somePn, and hyperelliptic curves.

15.1 Canonical Embeddings

For a curve X of genus g ≥ 2, we have deg Ω1X > 0, so we can ask whether

the line bundle Ω1X defines a morphism, or better yet, an embedding, into

Pn. Since h0(X,Ω1X) = g, we will get a morphism to Pg−1; in the case

that it’s an embedding, we will call this the canonical embedding.To get a morphism at all, we need Ω1

X to be base point free, whichyou recall means that there is no point of X at which all global sectionsof Ω1

X vanish. Since the dimension h0(X,Ω1X(−P )) is always one less

than or equal to h0(X,Ω1X), it is equivalent to require that that for all

P ∈ X(k), h0(X,Ω1X(−P )) < h0(X,Ω1

X), i.e., there is some global sectionnot vanishing at P .

If furthermore we ask that this morphism be an embedding, Ω1X must

separate points and tangent vectors, which we have seen earlier is equiva-lent to requiring that for all P,Q ∈ X(k), not necessarily distinct h0(X,Ω1

X(−P−Q)) < h0(X,Ω1

X(−P )). Thus we obtain easily the following criterion.

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Lemma 15.1. Ω1X defines a closed embedding X → Pg−1 if and only if,

for all closed points P,Q of X (not necessarily distinct), h0(X,O1X(−P −

Q)) = g − 2.

Proof. This follows from the above discussion: the dimension ofH0(X,O1X)

is g. Being base point free is equivalent to requiring that the dimensiondrop by one when we subtract a point from O1

X ; separating points andtangent vectors happens exactly when the dimension drops one furtherafter subtraction of another point.

When is the condition of the lemma satisfied? We apply Riemann-Roch to the line bundle Ω1

X(−P −Q):

h0(X,Ω1X(−P −Q))− h0(X,OX(P +Q)) = g − 3.

So we get a closed embedding exactly when, for all P,Q inX, h0(X,OX(P+Q)) = 1. Of course, there is always an inclusion k → H0(X,OX(P +Q)),so we are really asking that there be no nonconstant functions with simplepoles at one or both of P and Q, for any choice of P and Q. But the casewhen there is such a function is also interesting, for such functions givedegree two morphisms to P1. We now turn our attention to this lattersituation.

15.2 Hyperelliptic Curves

Definition 15.1. A curve X is hyperelliptic if there exists a degree twomap X → P1.

With this terminology, we summarize our results thus far by sayingthat any curve of genus g ≥ 2 can either be (canonically) embedded inPg−1 or else is hyperelliptic. We now ask for a classification of hyperellipticcurves. This will be accomplished by explicitly writing down an equationfor such curves.

So suppose X → P1 is a degree two map. Then we have a separateddegree two, hence Galois, field extension k(P1) → k(X). This gives a Z/2action on k(X), which lifts to an action of Z/2 on X over P1. For example,if X were defined by an equation of the form y2 = (x − a1) · · · (x − an),then the action would be given by sending y to −y.

How can we obtain the Z/2 action on X from that on k(X)? In general,since the map f : X → P1 is affine, A = f∗OX is a locally free sheaf ofOP1 -algebras of rank two. For an open affine U of P1, we have maps

k(X) A (U)oo

k(P1)

OO

OP1(U)

OO

oo

The maps OP1(U) → k(P1) → k(X) are inclusions, and hence identifyOP1(U) with a subring of k(X). We claim that A (U) is the integralclosure of OP1(U) in k(X). Since the map f is finite, A (U) is a finiteOP1(U)-module, and hence A (U) is integral over OP1(U). This shows

52

that we can recover X over P1 from the inclusion of fields k(P1) → k(X)(since X = Spec A ). By the functoriality of this description of A , we geta Z/2-action on A over OP1 . This breaks A into eigenspaces; we writeA = A+ ⊕ A−, where the generator σ of Z/2 acts trivially on A+ andby −1 on A−. Both “eigensheaves” are locally free of rank one, for thiscan be checked locally, say at the generic point, where it’s true becausek(X)/k(P1) is Galois.

Also, OP1 injects into A+, and in fact this is an isomorphism (check), sothere is some line bundle L on P1 (so L ∼= OP1(d)) such that A ∼= OP1⊕Las OP1 -modules. How does the algebra structure of A look under thisisomorphism? We already know how to multiply two elements of OP1 ,or an element of OP1 and an element of L, so the algebra structure onOP1 ⊕ L is determined by a map ρ : L⊗2 → OP1 .

Pick an open Spec k[t] = A1 ⊂ P1; then OA1 ∼= k[t]. Since PicA1 is

trivial, L∣∣A1∼= OA1 ∼= k[y]. The map ρ

∣∣A1 : O⊗2

P1∼= k[y2] → k[t] ∼= OA1

sends y2 to an element g(t) ∈ k[t]. Thus

A∣∣A1∼= OA1 ⊕ L

∣∣A1∼= k[t, y]/(y2 − g(t)),

with the Z/2 action given by y 7→ −y. In conclusion, hyperelliptic curvesX are given (in an open affine A2 ⊂ P2) by equations of the form y2 = g(t).We will see next lecture how to (almost) obtain the degree of g(t) solelyfrom the genus of X.

16 Friday, Feb. 24th: Riemann-Hurwitz Formula

Today we show how to use a morphism f : X → Y of curves to relatethe genus gX of X to the genus gY of Y . The key ingredient in thiscomputation is the familiar exact sequence

f∗Ω1Y/k → Ω1

X/k → Ω1X/Y → 0.

We will use the following terminology: the morphism f is separable ifthe corresponding field extension k(Y ) → k(X) is separable.

16.1 Preparatory Lemmata

Lemma 16.1. The map f∗Ω1Y/k

α→ Ω1X/k is injective if and only if f is

separable.

Proof. Injectivity of α can be checked at stalks. BothX and Y are integralschemes, so the vertical maps in the following diagram are injective:(

f∗Ω1Y

)x

//

Ω1X,x

(f∗Ω1

Y

)η

// Ω1X,η

where η is the generic point of X. Thus the top arrow is injective if andonly if the bottom one is, so it suffices to check injectivity at the generic

53

point. The inclusions into X and Y of the respective generic points givesa fibered diagram

Spec k(X) //

X

Spec k(Y ) // Y

This implies that(Ω1X/Y

)η∼= Ω1

k(X)/k(Y ), and this is zero if and only if f

is separable, as we checked on a previous HW assignment.

Lemma 16.2. The diagram

Pic(X)degX // Z

Pic(Y )

f∗

OO

degY // Z

· deg f

OO

commutes.

Proof. Recall that any line bundle on Y has the form OY (∑nPP ), and

the degree of such a line bundle is just∑nP . For the proof, it is enough

to check it on the generators of Pic(Y ), namely those line bundles of theform OY (P ) for P ∈ Y . Since such a line bundle has degree one, weneed to show that deg f = degX(f∗OY (P )). Let IP be the ideal sheaf ofthe inclusion P → Y . Then f∗IP is an ideal sheaf on X, isomorphic to⊗

f(Q)=P IeQQ . Therefore

degX(f∗OY (P )) = − degX(f∗IP ) = − degX( ⊗f(Q)=P

IeQQ)

= −( ∑f(Q)=P

−eQ)

= deg f,

where the last equality is from Proposition 14.1.

Our final lemma concerns the local structure of the module of differ-entials.

Lemma 16.3. Let πX be a uniformizer of the local ring OX,P . ThenΩ1X,P∼= OX,P · dπX .

Proof. Ω1X,P is generated by elements of the form df , for f ∈ OX,P . But

df = d(f − f(P )), since d(f(P )) = 0. Since f − f(P ) is in the maximalideal m of OX,P , we can write it as gπX . Then

df = d(gπX) = g dπX + πX dg.

Thus the image of dπX generates Ω1X,P /mΩ1

X,P , and therefore dπX gen-erates Ω1

X,P by Nakayama.

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16.2 Riemann-Hurwitz Formula

We now assume that f is separable, so by Lemma 16.1 the sequence

0→ f∗Ω1Y/k

α→ Ω1X/k → Ω1

X/Y → 0.

is exact. In particular, α is injective; we want to understand its image.For this we look at the stalk, where the sequence looks like

0→ Ω1Y,f(P ) ⊗OY,f(P )

OX,PαP→ Ω1

X,P → Ω1X/Y,P → 0.

Let πY be a uniformizer for OY,f(P ) and πX a uniformizer for OX,P .We can write the image of πY in OX,P as uπePX for some unit u andinteger eP ≥ 1. Since Ω1

Y,f(P ) is generated as an OY,P -module by dπY ,

f∗Ω1Y,f(P )

∼= dπYOX,P , so using this isomorphism, αP sends dπY to

d(uπePX ) = πePX du+ epuπeP−1X dπX .

If it happens that the characteristic of k divides eP , the situation isconsiderably more complicated, and gets a name, but we will not treatthis case.

Definition 16.1. The morphism f has wild ramification at P if thecharacteristic of k divides eP .

So assume that the characteristic of k does not divide eP . Then fol-lowing αP with the quotient map to Ω1

X,P /meP Ω1

X,P , the term πePX du

in the above vanishes, so the image is generated by πeP−1X dπX . By

Nakayama, then, the image of αP itself is generated by πeP−1X dπX . Thus

f∗Ω1Y,f(P )

∼= πeP−1X Ω1

X,P . Allowing P to vary, we obtain the main theoremof the lecture.

Theorem 16.1. If f is separable and has no wild ramification, then

f∗Ω1Y

( ∑P∈X

(eP − 1) · P) ∼= Ω1

X

Theorem 16.2 (Riemann-Hurwitz). If f is separable and has no wildramification, then

2gX − 2 = deg f(2gY − 2) +∑P∈X

(eP − 1)

Proof. Take degrees in Theorem 16.1.

Example 16.1. Let f : X → P1 be a degree two map, i.e., X be ahyperelliptic curve. By the previous lecture, we can write the equationfor X as y2 = (x − a1) · · · (x − ar), so f is ramified at the points ak andpossibly at ∞. For those points at which f is ramified, its ramificationindex is two, so using 16.2 we find

2gX − 2 = 2(−2) + r + ε,

where ε comes from the ramification at ∞: it is 1 if f is ramified there,and 0 if not. By parity, we see that if r is odd, ε = 1, so f is ramified at∞and gX = (r−1)/2. If r is even, f is unramified at∞ and gX = (r−2)/2.This makes good on the promise made at the end of the previous lecture.

Example 16.2. As a reality check, consider the case where r = 3, i.e., Xis an elliptic curve. It’s ramified at infinity, and the computation aboveagrees with our knowledge that gX = 1. Phew!

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17 Monday, Feb. 27th: Homological Algebra I

Lecture and typesetting by Piotr Achinger

17.1 Motivation

17.1.1 Algebraic topology

was the main motivation for the concepts of homological algebra. Theterm homology itself comes from Poincare’s first attempts on algebraictopology. Given a nice topological space7 X and a commutative ring R,one can associate to it the following sequence of maps between R-modules:

0→ C0(X)d−→ C1(X)

d−→ C2(X)d−→ . . .

where Ci(X) is the dual of the free R-module spanned by the set of allcontinuous maps f : [0, 1]n → X (,,singular cubes in X”) and the maps dare dual to the maps taking such ,,singular cubes” to their ,,boundaries”.It is intuitively clear that the boundary of the boundary is zero, i.e., dd =0. We can therefore look at the cohomology groups8 :

Hi(X,R) :=ker(d : Ci(X)→ Ci+1(X))

im(d : Ci−1(X)→ Ci(X)).

These encode important topological information of X, and have manyuseful properties, to note a few:

(a) H0(X,R) is a free R-module spanned by the (path-)connected com-ponents of X.

(b) The Hi(X,R) are functorial in both X and R – contravariantly inX and covariantly in R (the map Hi(f) induced by f : X → Y isusually denoted by f∗),

(c) given a nice inclusion9 i : Y → X, one gets a long cohomology exactsequence

→ Hi−1(Y,R)δ−→ Hi(Z,R)

p∗−→ Hi(X,R)i∗−→ Hi(Y,R)

δ−→ Hi+1(Z,R)→ . . .(10)

where Z = X/Y is X with Y contracted to a point, and p : X → Zis the projection.

17.1.2 Algebraic geometry – a.k.a. the Black Box

Our main goal is to define and study the sheaf cohomology functorsHi(X,F ),where X is a topological space and F a sheaf of abelian groups on X.We already know some of their properties, having used them as a ,,blackbox” to study algebraic curves:

(a) H0(X,F ) = Γ(X,F ),

7e.g. a CW-complex8If we didn’t take duals here, we would get homology groups – but cohomology serves a

better motivation for our purposes.9e.g. a cofibration

56

(b) The H0(X,F ) are functorial in both X and F – contravariantly inX and covariantly in F (contravariance in X is to be understood asfollows: if f : X → Y is a continuous map and G is a sheaf on Y ,we get a map Hi(Y,G )→ Hi(X, f−1G )).

(c) Given a short exact sequence 0→ F → G → H → 0 of sheaves onX, one gets a long cohomology exact sequence

→ Hi−1(X,H )δ−→ Hi(X,F )→ Hi(X,G )→ Hi(X,H )

δ−→ Hi+1(X,F )→(11)

Propaganda 17.1. Our goal, motivated by algebraic topology, is to de-velop a general notion of a ,,cohomology theory”, encompassing (17.1.1)and (17.1.2) above10 and many other examples in a uniform fashion. Oursecond goal is to develop a theory which would handle the non-exactnessof the functor Γ(X,−) (and other non-exact functors).

17.2 Complexes, cohomology and snakes

We start with a rather lengthy definition. I hope most readers will befamiliar with most of the notions presented here.

Definition 17.2. Let R be a commutative ring.

1. A complex of R-modules is a pair C• = (Cn, dn) where Cn (n ∈ Z)are R-modules and dn : Cn → Cn+1 are maps (called differentials)satisfying the condition dn+1 dn = 0. (In practice, one usuallywrites d instead of dn, so our condition can be written succintly asd2 = 0).

2. A morphism of complexes f• : C• → D• is a family of maps fn :Cn → Dn commuting with the differentials, i.e., dnD fn = fn+1 dnC . One therefore gets a category Ch(R − mod) of complexes ofR-modules.

3. The cohomology groups of a complex C• are the groups Hi(C•) :=ker dn/ im dn−1. One easily checks that they are covariant functorsHi : Ch(R−mod)→ R−mod. Elements of ker dn are called cyclesand denoted by Zn(C) and elements of im dn−1 are called boundariesand denoted by Bi(C). Therefore Hi(C) = Zi(C)/Bi(C).

4. The kernel, image and cokernel of a morphism of complexes aredefined in the usual way. One can then talk about subcomplexes,quotient complexes, exact sequences of complexes, complexes of com-plexes, cohomology complexes of complexes of complexes and so on.

Propaganda 17.3. Here is one principle of our approach to homologicalalgebra: to study an object A, associate to it a complex C•(A). Thiscomplex will depend on the ,,cohomology theory”. In other words – toconstruct a cohomology theory H ∗, one first describes a way to attach acomplex C•(A) to A and define H i(A) as Hi(C•(A)). Usually C•(A) isnot functorial in A (but its cohomology is).

10In fact (again for ,,nice” spaces), (1) is a special case of (2): Hi(X,R) = Hi(X,R) whereR is the constant sheaf associated to R.

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How do we get long exact sequences? We extend our principle asfollows: if an object X fits into a ,,short exact sequence” Y → X Z,then there should be an exact sequence of complexes 0 → C•(Y ) →C•(X)→ C•(Z)→ 0. Then we get a long exact sequence from the SnakeLemma (below). This encompasses both long exact sequences (10) and(11).

Lemma 17.1 (Snake Lemma11). (a) Given a commutative diagram ofR-modules with exact rows

. C D E 0

0 C′ D′ E′ .

f g h

there exists an arrow δ : ker g → coker f such that the sequence

ker f → ker g → kerhδ−→ coker f → coker g → cokerh

is exact. Here is the whole picture:

ker f ker g kerh

C D E 0

0 C′ D′ E′

coker f coker g cokerh

f g h

δ

(b) The arrow δ is functorial, that is, given two diagrams as above and acommutative system of arrows between them, the two δ-arrows keepthe system commutative.

(c) Given a short exact sequence of complexes 0→ C•i−→ D•

p−→ E• → 0,there is a long exact sequence of cohomology

. . .→ Hi−1(D•)δ−→ Hi(C•)

i∗−→ Hi(D•)p∗−→ Hi(E•)

δ−→ Hi+1(C•)→ . . .

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(d) The arrows δ above are functorial, that is, given two short exact se-quences as above and a commutative system of arrows between them,one gets a commutative ladder between the two long cohomology ex-act sequences.

Proof. You should prove (a) and (b) yourself. For (c), apply (a) to thediagram

. Ci/Bi(C) Di/Bi(D) Ei/Bi(E) 0

0 Zi+1(C) Zi+1(D) Zi+1(E) .

diC diD diE

(it is straightforward to check that the rows are exact) obtaining

Hi(C) Hi(D) Hi(E)

Ci/Bi(C) Di/Bi(D) Ei/Bi(E) 0

0 Zi+1(C) Zi+1(D) Zi+1(E)

Hi+1(C) Hi+1(D) Hi+1(E)

diC diD diE

δ

Then (d) follows from (b) and (c).

17.3 Homotopy

In Section 1, we discussed cohomology groups Hi(X,R) of a topologicalspace X. In fact, these are homotopy invariants of X: that two mapsf, g : X → Y are homotopic if there exists a continuous map H : X ×[0, 1]→ Y such that H(x, 0) = f(x) and H(x, 1) = g(x). One proves thathomotopic maps induce the same maps on cohomology groups. To provethis, one first constructs a family of maps si : Ci(Y ) → Ci−1(X) such

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that ds+ sd = f∗ − g∗, as in the diagram below.

. . . Ci−1(Y ) Ci(Y ) Ci+1(Y ) . . .

. . . Ci−1(X) Ci(X) Ci+1(X) . . .

di−1Y diY

di−1X

diX

g∗f∗ g∗f∗ g∗f∗si+1 si+1

Then, the statement follows formally from the equation ds+sd = f∗−g∗12.

Definition 17.4. Let C•, D• be complexes and let f, g : C• → D• betwo maps of complexes. We say that f and g are homotopic (denotedf ∼ g) if there exists a family of maps si : Ci → Di−1 such that

f i − gi = di−1D si + si+1 diC .

Here is the diagram for your convenience:

. . . Ci−1 Ci Ci+1 . . .

. . . Di−1 Di Di+1 . . .

d

d

gfs s

Lemma 17.2. If f ∼ g, then Hi(f) = Hi(g) for all i.

Proof. Let z ∈ Zi(C). We want to prove that there is an x ∈ Di−1 suchthat f(z)− g(z) = dx. But

f(z)− g(z) = d(s(z)) + s(d(z)) = d(s(z)) + 0

since dz = 0 by assumption. Therefore we can take x = s(z).

Some more vocabulary:

Definition 17.5. The family of maps si is called a homotopy betweenf and g. A map which is homotopic to zero is called nullhomotopic. Wecall an f : C• → D• a homotopy equivalence if there exists a g : D• → C•

such that f g is homotopic to idD and g f is homotopic to idC .

Remark. By the Lemma, a homotopy equivalence induces an isomor-phism on cohomology groups (such a map is called a quasi-isomorphism).The problem with quasi-isomorphisms is that they are not preservedunder additive functors, for example a short exact sequence of sheaves

0 → F ′α−→ F

β−→ F ′′ → 0 can be seen as a quasi-isomorphism as in thediagram below:

. . . 0 F ′ 0 0 . . .

. . . 0 F F ′′ 0 . . .β

α

12For the construction of si, see e.g. A. Hatcher Algebraic Topology, Theorem 2.10

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This map of complexes will not however remain a quasi-isomorphism af-ter applying Γ(X,−). On the other hand, homotopy equivalences arepreserved by additive functors, by the nature of their definition. Thismeans that homotopy equivalence is a much better behaved notion. Infact, they will play a key role in our construction of ,,cohomology theories”(i.e., derived functors).

18 Wednesday, Feb. 29th: Homological AlgebraII

Lecture and typesetting by Piotr AchingerHere is what we know after Lecture 1:

• we know what is a complex (of abelian groups or R-modules), whatare its cohomology groups, what it means for it to be exact,

• we know the Snake Lemma and that a short exact sequence of com-plexes gives us a long cohomology exact sequence,

• we know the notion of homotopy of maps between complexes andthat homotopic maps induce the same map on cohomology.

18.1 Additive and abelian categories

A model example of a category in which one can make homological con-siderations is the category R- Mod of (left) modules over a ring R.

Observation 18.1. In the category R- Mod there exist:

(a) a zero (initial and terminal) object 0.

(b) a structure of an abelian group on the sets of morphisms Hom(M,N)satisfying the distributivity laws of left and right composition withrepect to addition (i.e. for f : M → N the functions

x 7→ x f : Hom(N,N ′)→ Hom(M,N ′),

x 7→ f x : Hom(M ′,M)→ Hom(M ′, N)

are group homomorphisms).

(c) direct sums M ⊕N which are simultaneously products (i.e. we haveboth the projections pM : M ⊕ N → M , pN : M ⊕ N → N andthe inclusions iM : M → M ⊕ N , iN : N → M ⊕ N satisfyingpM iM = idM , pN iN = idN , iM pM + iN pN = idM⊕N ).

Definition 18.2. A category satisfying (a)–(c) above is called additive13.An additive functor is a functor F : C → D between additive categoriestaking 0 to 0, direct sums to direct sums and such that the induced mapsHom(M,N)→ Hom(FM,FN) are group homomorphisms.

In an additive category, it makes sense to talk about complexes, butnot cohomology. Therefore we need additional properties.

13One may think that ,,additive” is an extra structure on a category. In fact, whenevera category is additive, the group structure on the sets Hom(M,N) is uniquely determined.Therefore ,,additive” is a property of a category

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Observation 18.3. In the category R- Mod there also exist:

(d) kernels, i.e. equalizers of any morphisms with the zero morphism(that is, for any f : M → N there is an object ker f = K togetherwith a map if = i : K → M with the following universal property:given g : M ′ → N such that f g = 0, there is a unique g : M ′ → Ksatisfying g = i g),

(e) cokernels, i.e. coequalizers of any morphisms with the zero morphism(that is, for any f : M → N there is an object coker f = C togetherwith a map pf = p : N → C with the following universal property:given g : N → N ′ such that g f = 0, there is a unique g : C → N ′

satisfying g = g p),

(f) canonical isomorphims ker(pf ) = coker(if ), that is ,,the cokernel ofthe kernel is the kernel of the cokernel” (by the universal property ofkernel and cokernel, there is a canonical map ker(pf )→ coker(if ) –we require it to be an isomorphism).

Definition 18.4. An additive category satisfying additionally (d)–(f) iscalled abelian. In an abelian category, the usual notions of cohomology of acomplex or an exact sequence make sense. An additive functor F : C → Dbetween two abelian categories is called exact if it takes kernels to kernelsand cokernels to cokernels (or equivalently – takes exact sequences toexact sequences).

Observation 18.5. In the category R- Mod we have the Snake Lemma,which gives us long cohomology exact sequences for short exact sequencesof complexes.

At the first glance, it is not obvious how the proof of the Snake Lemmashould go in any abelian category – recall that our proof used ,,elements”of the objects in the diagram. However, the Snake Lemma and its con-clusions hold in any abelian category, thanks to the following result:

Theorem 18.1 (Freyd-Mitchell, 1964). Let A be a small14 abelian cat-egory. Then A is a full abelian subcategory of the category R- Mod ofmodules over a certain ring R. More precisely, there exists a ring R anda fully faithful exact additive functor F : A → S −Mod.

Remark. In fact, the Freyd-Mitchell theorem allows us to perform dia-gram chasing in any abelian category – given a diagram (whose objectsform a set) we can pass to the smalles abelian subcategory containing theobjects of the diagram (which should be small). I will ignore set theoreticissues of this type from now on.

Remark. The category Ch(A ) of complexes of objects of an abeliancategory A is abelian (we take sums, kernels and cokernels ,,levelwise”).The functors

Hi : Ch(A )→ A , Hi(A•) = ker(di)/im(di−1)

are additive, but obviously not exact.

Our considerations concerning R- Mod and the Freyd-Mitchell theoremshow that

14A category is small if its objects form a set.

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Theorem 18.2. If A is an abelian category and ξ : 0 → A•f−→ B•

g−→C• → 0 is a short exact sequence of complexes in A , then there exists thelong cohomology exact sequence

. . .→ Hi−1C•δi−1ξ−−−→ HiA•

Hif−−−→ HiB•Hig−−→ HiC• → . . .

which is natural in the following sense: if ξ′ : 0→ A′•f−→ B′•

g−→ C′• → 0is another such sequence and we are given a morphism of short exactsequences15 φ : ξ → ξ′, then the diagrams

HiC• Hi+1A•

HiC′• Hi+1A′•

δiξ

φ∗ φ∗

δiξ′

commute.

Examples 18.6. The following additive categories are abelian:

• abelian groups, R-modules,

• sheaves and presheaves of abelian groups on a topological space X,sheaves of OX -modules on a ringed space (X,OX),

• coherent and quasi-coherent OX -modules on a scheme X,

• complexes in an abelian category A .

and the following are not:

• finitely generated free abelian groups,

• vector bundles (i.e., locally free OX -modules of finite rank) on ascheme X.

18.2 δ-functors

Propaganda 18.7. Recall that our slogan was ,,replace the objects westudy by complexes”. One can find manifestations of this idea on everylevel of homological algebra, the most refined of them being the notion ofa derived category, developed by J.-L. Verdier in 1960s.

Trying to realize our motto in a naive way, imagine that to a givenobject A of an abelian category A there is assigned a complex X• insome other abelian category B. For simplicity (and better analogy withour motivational examples coming from topology) we will assume thatXi = 0 for i < 0.

With this picture there should be an associated cohomology theory,that is, a sequence of functors Hi(A) = Hi(X•) and for every short exactsequence 0 → A → B → C → 0 a sequence of arrows δi : Hi(C) →Hi+1(A) (as if 0 → A → B → C → 0 gave a short exact sequence ofcomplexes), giving a long cohomology exact sequence.

The notion realizing the above picture is called a δ-functor.

15This is simply a commutative ,,ladder” between ξ and ξ′.

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Definition 18.8. Let A and B be abelian categories. By a (covariant,cohomological) δ-functor T • from A to B we mean the following data

• a sequence of additive functors T i : A → B (i ≥ 0),

• a sequence of morphisms δiξ : T iC → T i+1A (i ≥ 0) for any shortexact sequence ξ : 0→ A→ B → C → 0 in A ,

satisfying the following conditions

1. for any exact sequence ξ : 0→ Af−→ B

g−→ C → 0 in A the sequence

0→ T 0AT0f−−→ T 0B

T0g−−→ T 0Cδ0ξ−→ . . .→ T iA

T if−−→ T iBT ig−−→ T iC

δiξ−→ T i+1A→ . . .

is exact in B,

2. for any two exact sequences ξ : 0 → A → B → C → 0 and ξ′ : 0 →A′ → B′ → C′ → 0 and a morphism φ : ξ → ξ′ the diagram

T iC T i+1A

T iC′ T i+1A′

δiξ

φ∗ φ∗

δiξ′

commutes

Remark. There also exist notions of a contravariant δ-functor and of ahomological δ-functor.

Propaganda 18.9. What we would also like to have (this condition maynot sound very well motivated to a person without background from sayalgebraic topology) is for this ,,cohomology theory” to be ,,determined byits coefficients” – where by ,,coefficients” we mean the functor T 0. Indeed,the cohomology functors Hi(X,R) satisfy

1. H0(X,R) = R for X connected,

2. if Hi(X) is an ordinary cohomology theory, then H0(pt) determinesthe whole theory (at least for CW-complexes),

3. a morphism R→ R′ of coefficient rings extends uniquely to a ,,mor-phism of theories”, i.e., a system of natural transformationsHi(−, R)→Hi(−, R′), compatible with the connecting homomorphisms δ in thelong exact sequences.

Another motivation comes from our Black Box – we want universal func-tors Hi satisfying the properties we need.

To understand, what 3. above should mean in the context of δ-functors, we should introduce the notion of a morphism of δ-functors(,,natural transformation of cohomology theories” – in fact the motivatingexample for Eilenberg and MacLane’s definition of a natural transforma-tion of functors).

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Definition 18.10. Let T •, T ′• be two δ-functors from an abelian categoryA to an abelian category B. A morphism of δ-functors τ : T • → T ′• isa sequence of natural transformations τ i : T i → T ′i which commute withδi, δ′i, that is, for every short exact sequence ξ : 0 → A → B → C → 0the diagram

T iC T i+1A

T ′iC T ′i+1A

δiξ

τ iC τ i+1A

δ′iξ

commutes.

We can now say what it means that a δ-functor T • is determined byits coefficients T 0.

Definition 18.11. Let T • be a δ-functor from an abelian category Ato an abelian category B. We say that T • is universal if for every otherδ-functor S• from A to B and a natural transformation σ : T 0 → S0

there is a unique morphism of δ-functors τ : T • → S• with τ0 = σ.

As usual, the universal property implies that whenever a universalδ-functor with a fixed T 0 exists, it is unique.

18.3 Left- and right-exact functors

What additive functors F : A → B does there exist a (not necessarilyuniversal) δ-functor T • with T 0 = F? From the ,,long exact sequence”property it follows that for every short exact sequence 0 → A → B →C → 0 the sequence

0→ FA→ FB → FC

(with no 0 on the right!) is exact.

Definition 18.12. Let A , B be abelian categories and let F : A → Bbe an additive functor. We call F left-exact if for any short exact sequence0→ A→ B → C → 0 in A , the sequence

0→ FA→ FB → FC

is exact; F is right-exact if for any short exact sequence 0 → A → B →C → 0 in A , the sequence

FA→ FB → FC → 0

is exact16.

Remark. There also is a notion of a contravariant left- or right-exactfunctor. We just consider such a functor as a covariant functor A op → B.

16That is, F is left/right exact iff it preserves kernels/cokernels – or equalizers/coequalizers– or (because F is additive) finite limits/colimits.

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Examples 18.13. For any object A ∈ A , the functors Hom(A,−) andHom(−, A) are left-exact. If A ∈ R- Mod, then A⊗R − is right-exact. Iff : X → Y is a map of topological spaces, then f∗ : Sh(X)→ Sh(Y ) is left-exact and f−1 : Sh(Y )→ Sh(X) is right exact (where Sh are the categoriesof sheaves of abelian groups) and similarly for f∗ and f∗ for sheaves ofmodules if X and Y are ringed spaces. In particular, Γ : Sh(X)→ A b isleft-exact. More generally, if F : A → B is left adjoint to G : B → Athen F is right exact and G is left exact (Problem 4).

A few observations are in order. First, a functor is exact if and only ifit is both left- and right-exact. Secondly, a functor extends to a δ-functoronly if it is left-exact. Finally, if F : A → B is exact, then T 0 = F ,T i = 0 for i > 0 defines a universal δ-functor. Therefore F can give aninteresting universal δ-functor only if it is left-, but not right-exact.

Propaganda 18.14. We now change our paradigm a little: we consider(universal) δ-functors as a tool to study not (or not only) objects, but thenon-exactness of left-exact functors. That is, we focus on the second goalmentioned in the first lecture: to develop a theory which would handle thenon-exactness of the functor Γ(X,−) (and other non-exact functors).

Our main goal is now the construction of a universal δ-functor extend-ing a given left-exact functor (the construction for right-exact functors isanalogous) – we shall call them (left) derived functors.

18.4 Universality of cohomology

To make sure that our approach may be right and imagine how the generalconstruction should work, we solve the following ,,toy” problem: since themain prototype of the notion of a δ-functor are the cohomology functorsHi : Ch≥(A ) → A (Ch≥(A ) is the category of complexes concentratedin degrees ≥ 0), do they form a universal δ-functor?

Proposition 18.1. The functors Hi : Ch≥(A ) → A together with themorphisms δ from Theorem 2 form a universal δ-functor.

Proof (sketch). Let there be given a δ-functor S• from Ch≥(A ) to Aand a natural transformation τ0 : H0 → S0. We shall construct theτ i : Hi → Si inductively.

Suppose that τi−1 has already been constructed. From the Lemmabelow it follows that for any complex A• there is an injection A• →I• into an exact complex I•. This gives us a short exact sequence ofcomplexes 0 → A• → I• → B• → 0. Looking at the long cohomologyexact sequences gives us the diagram

Hi−1A• Hi−1I• Hi−1B• HiA• HiI• = 0

Si−1A• Si−1I• Si−1B• SiA• SiI•

τ i−1 τ i−1 τ i−1 τ i

By some easy diagram chasing we convince ourselves that this determinesτ i : HiA → SiA. We then have to check that τ does not depend on thechoice of A• → I•, that it is a natural transformation etc.

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Lemma 18.1 (Problem 3). For any A• ∈ Ch(A ) there is an injectionA• → I• where I• ∈ Ch(A ) and Hi(I•) = 0 for all i.

Remark for algebraic topologists. We can look at the above proofin the following way: we embed a given space A into the cone I = CA,which is contractible. Then B = I/A = ΣA is the suspension of A. By thesuspension axiom we get Hi(A) = Hi−1(B) for any cohomology theoryH•. But on Hi−1 everything is already defined, hence the inductive step.

An important observation here is that the proof of Proposition 18.1essentially shows the following

Lemma 18.2 (Effaceable δ-functors are universal. See Problem 5). LetT • : A → B be a δ-functor which is effaceable, that is, for every A ∈ Athere exists an injection A → J into a T •-acyclic object J (J is calledacyclic for T • if T i(J) = 0 for i > 0). Then T • is universal.

19 Friday, March 2nd: Homological Algebra III

Lecture and Typesetting by Piotr AchingerHere is what we know after Lecture 2:

• what is an additive or abelian category, what is an additive or exactfunctor,

• what is a δ-functor and what it means for it to be universal,

• what is a left- or right-exact functor,

• that an effaceable17 δ-functor is universal.

Our goal is to extend a given left-exact functor to a universal δ-functor.

19.1 Derived functors – a construction proposal

Let us think about what was crucial in our proof of ,,universality of co-homology”. The key idea was that any object A could be embedded intoan object I on which the given functor was ,,exact” (that is, the ,,coordi-nates” of the δ functor for i > 0 vanished on I).

Lemma 19.1 (Effaceable δ-functors are universal). Let T • : A → B bea δ-functor which is effaceable, that is, for every A ∈ A there exists aninjection A → J into a T •-acyclic object J (J is called acyclic for T • ifT i(J) = 0 for i > 0). Then T • is universal.

Let A and B be abelian categories and let F : A → B be a left-exact functor. We are looking for a universal δ-functor T • with T 0 = F .Motivated by the above Lemma, we want to find a good class I of objectsof A , such that

0. every object A ∈ A injects into an object I(A) ∈ I ,

1. objects of I are acyclic for F .

17A δ-functor T • is effaceable if any object injects into an object on which T 1, T 2, . . . vanish.

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It is not clear what it should mean for an I ∈ I to be acyclic, since wedon’t have a δ-functor yet, but only its zero component T 0 = F . But inany case, the following should hold: whenever 0→ I →M → N → 0 is ashort exact sequence, 0 → FI → FM → FN → 0 is also exact (becausethe next term after FN should be T 1I = 0). This of course holds when0→ I →M → N → 0 splits, since a split short exact sequence stays split(therefore exact) after applying F . Let us be greedy here and ask for theclass of objects I for which all such sequences split.

Definition 19.1. An object I ∈ A is called injective if any short exactsequence 0 → I → M → N → 0 splits. An object P ∈ A is calledprojective if any short exact sequence 0→M → N → P → 0 splits (thatis, if P is injective in A op).

Problem 6. What are the injective and projective objects in the categoryof abelian groups?

The class of projective objects plays the same role as injective objectswhen one wants to study right-exact functors (or contravariant left-exactfunctors). Then one has to consider surjections P A and the wholeprogram carries over.

With these assumptions in place, set I to be the class of injective ob-jects in A and let us try to construct T i using the principle of ,,inductionby dimension shifting” as in the proof of Lemma 19.1. Given A ∈ A , findA → I, with I ∈ I and let B be the cokernel:

0→ Aα−→ I

β−→ B → 0.

After applying F , we get

0→ FAFα−−→ FI

Fβ−−→ FB

which we want to continue to the right

0→ FAFα−−→ FI

Fβ−−→ FB → T 1A→ T 1I → . . . ,

and since we wish that T iI = 0 for i > 0, we should put T 1A = cokerFβand T i+1A = T iB for i ≥ 1.

Thus to compute say T 2A, we should find an short exact sequence0 → B → J

γ−→ C → 0 and compute cokerFγ. This seems alreadycumbersome, but we can make a long exact complex 0 → A → I →J → . . . (an ,,injective resolution” of A) by ,,splicing” all the short exactsequences.

Construction. Given A ∈ A , construct an injective resolution A→

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I• of I as follows

0 0

B2

0 A I0 I1 I2 . . .

B1 B3

0 0 0 0

iA

iB1

iB2

iB3

(where I0 = I(A), B1 = coker iA, Ii = I(Bi) for i > 0, Bi+1 =coker(Bi → Ii)). Then apply F to I•:

F (I•) : F (I0)→ F (I1)→ F (I2)→ . . .

and defineT iA = Hi(F (I•)).

By left-exactness of F , since 0 → A → I0 → I1 is exact, 0 → FA →FI0 → FI1 is exact, hence T 0A = FA. Obviously this agrees with ourprevious approach.

Remark. It is convenient to view 0 → A → I• → . . . as a quasi-isomorphism A → I• (where A is identified with the complex . . . →0 → A → 0 → . . .). This means that in the derived category of A (thecategory of complexes where quasi-isomorphisms are made to be actualisomorphisms), A and I• are the same.

There are several obvious problems here:

2. Why should T iA be well-defined (i.e., independent of the choice ofA→ I•)?

3. Why should it be functorial in A?

4. And how to define the connecting homomorphisms δ to get a δ-functor?

Suppose that (2)-(4) were resolved. Then (1) follows, since we canchoose 0→ I → I → 0 to be the injective resolution of I, which gives usT iI = 0 for i > 0. If in additions (0) would hold, then by Lemma 19.1 wehave constructed a universal δ-functor.

19.1.1 Derived functors – resolution of technical issues

Regarding (2) – T iA are well-defined

Note that T i constructed as above will not depend on the choice of 0 →A → I• if any two such resolutions will be homotopic (since homotopicmaps

1. remain homotopic after applying any additive functor,

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2. induce the same map on cohomology).

What we need is the following

Theorem 19.1. 1. for any two resolutions 0 → A → I• amd 0 →A → J• there exist φ, ψ – morphisms between 0 → A → I• i0→ A→ J• (i.e. dφ = φd oraz dψ = ψd) giving the identity on A:

0 A I0 I1 I2 . . .

0 A J0 J1 J2 . . .

φψ φψ φψ

2. any two such φ, ψ be the homotopy inverses of each other – i.e. thereexist si : Ii → Ii−1 and ti : J i → J i−1 (i ≥ 0, we put I−1 = A =J−1) such that ψφ− id = ds+ sd and φψ − id = dt+ td:

0 A I0 I1 I2 . . .

0 A I0 I1 I2 . . .

ψφ− id ψφ− id ψφ− ids s s s

0 A J0 J1 J2 . . .

0 A J0 J1 J2 . . .

φψ − id φψ − id φψ − idt t t t

Let us try to construct φ inductively – the first step looks as follows:

0 A I0 I1 I2 . . .

0 A J0 J1 J2 . . .

We would like to extend the dotted arrow A → J0 to I0. This followsfrom

Problem 7. An object I ∈ A is injective iff any morphism A → Idefined on A ⊆ B can be extended to B (or, equivalently, iff the functorHom(−, I) is exact).

Regarding (3) – Functoriality

In fact, a more general version of Theorem 19.1 holds:

Theorem 19.2. Let f : A′ → A be a map in A and let A→ I•, A′ → I ′•

be resolutions of A and A′, respectively, with Ii injective (the objects I ′i

do not have to be injective). Then there exists a chain map f : I• → I ′•,inducing f on H0, which is unique up to homotopy.

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Problem 8. Prove Theorem 19.2 and deduce Theorem 19.1.

Thus, the exercise above resolves both our issues with (2) and (3).

Regarding (4) – Construction of δi

This is called the Horseshoe Lemma:

Lemma 19.2. Given a short exact sequence 0 → A → B → C → 0 andalready chosen injective resolutions A→ I•A and C → I•C one can find aninjective resolution B → I•B so that one obtains a short exact sequence ofcomplexes 0 → I•A → I•B → I•C → 0 which is 0 → A → B → C → 0 onH0.

Problem 9. Prove the above statement. Hint: set I0B = I0

A ⊕ I0C and

construct an arrow B → I0B . Use the Snake Lemma, obtaining a short

exact sequence 0→ I0A/A→ I0

B/B → I0C/C → 0. Proceed by induction.

Thus, given a short exact sequence 0 → A → B → C → 0, weobtain the morphisms δi : T iC → T i+1A from the long cohomology exactsequence associated to the short exact (why?) sequence of complexes 0→F (I•A) → F (I•B) → F (I•C) → 0 where I•A, I

•B , I

•C are as in the Horseshoe

Lemma.We need to check that the morphisms δ are well defined and natural,

which is also left to the reader.

Regarding (0) – Sufficiently many injective objects

The only problem may happen with (0). For example, if A is the categoryof all finitely generated abelian groups, then A has no nonzero injectiveobjects!

Definition 19.2. An abelian category A is said to have sufficiently manyinjective objects if for any object there exists an injection into an injectiveone. We say that A has sufficiently many projective objects if for anyobject there is a surjection from a projective one.

Theorem 19.3. For any ring R, the category R- Mod has sufficientlymany injective and projective objects.

19.2 Derived functors – conclusion

Theorem 19.4. Let A and B be abelian categories and let T 0 : A → Bbe a left-exact functor. If A has sufficiently many injective objects, thenT 0 extends to a universal δ-functor T • with T 0 = F .

Definition 19.3. We denote the functors T i by RiT 0 and call the rightderived functors of T 0.

A similar Theorem holds for right-exact functors and projective objects(then one has to use homological δ-functor). The left-derived functors ofa right-exact functor T0 are denoted by LiT0.

Examples 19.4. Let A ∈ A , then Exti(A,−) := Ri Hom(A,−) andExti(−, A) = Ri Hom(−, A) (these two coincide, so one can computeExti(A,B) from an injective resolution of B or a projective resolutionof A). For A ∈ R −Mod, → ri(A,B) := Li(A ⊗R B) (here one proves

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that → ri(A,B) '→ ri(B,A)). For a topological space X, Hi(X,−) :=RiΓ : Sh(X) → A b are called the sheaf cohomology functors (one checksfirst that the category of abelian sheaves on X has sufficiently many in-jective objects). For a continuous map f : X → Y , the derived functorsRif∗ : Sh(X)→ Sh(Y ) are called the higher direct image functors. For agroup G, the functor of G-invariants (−)G : G−Mod→ A b is left-exactand we call its derived functors Hi(G,−) = Ri(−)G the group cohomologyfunctors. One checks easily that MG = Hom(Z, G) where Z has the triv-ial Z-action. Therefore Hi(G,M) = Exti(Z,M) can be computed from aprojective resolution of Z, which is useful since we do not have to pick anew resolution whenever M changes, and also because projective objectsare generally easier to work with than injective objects.

19.3 Two useful lemmas

Lemma 19.3 (Acyclic resolutions suffice). Let F : A → B be left-exact,where A has sufficiently many injective objects. Let A ∈ A and let 0 →A → J0 → J1 → . . . be an exact sequence with all Jn acyclic for F(that is, RiF (Jn) = 0 for i > 0 and all n ≥ 0). Then there are naturalisomorphisms Hi(F (I•)) ' RiF (A).

Lemma 19.4. Let A , B and C be abelian categories, F : A → B andG : B → C be left-exact functors. Suppose that A and B have enoughinjectives (so that the derived functors RiF , RiG and Ri(G F ) exist)and that F maps injective objects of A to G-acyclic objects of B (that is,if I ∈ A is injective, then (RiG)(F (A)) = 0 for i > 0). Let A ∈ A . Then

(a) if (RiF )(A) = 0 for i > 0, then Ri(G F )(A) ' (RiG)(F (A)),

(b) if (RiG)(F (A)) = 0 for i > 0, then Ri(G F )(A) ' G(RiF (A)).

In general, under the assumptions of the lemma, there is a spectralsequence

Epq2 = (RpG)(RqF (A)) ⇒ Rp+q(G F )(A).

20 Monday, March 5th: Categories of SheavesHave Enough Injectives

We have seen last week how to construct the right derived functors ofan additive left-exact functor, so long as the categories we are interestedin have enough injectives. Today we verify that this is the case for thecategories we are interested in, namely sheaves of OX -modules ModOX ,or perhaps just sheaves of abelian groups Sh(X). The functor whose rightderived functors we construct is f∗, where f : X → Y is a morphism ofschemes. We will be especially interested in the case where Y = Spec k; inthis case f∗ is the just the global sections functor Γ(X,−) : Sh(X)→ Ab.

Remark. In some sense, it won’t really matter what categories betweenwhich we regard f∗ as a functor. In fact this is reflected in the notation:Rif∗ does not specify Sh(X) or ModOX , etc. Of course, some care needsto be taken. For instance, we will not consider categories of coherentsheaves, because over SpecZ, there aren’t any (except for the zero sheaf):

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the reason is that injective Z-modules are things like Q and Q/Z, whichare not finitely-generated.

The goal of today’s lecture is to establish the following, which holdseven for ringed spaces with noncommutative rings:

Theorem 20.1. If (X,OX) is a ringed space, then ModOX has enoughinjectives.

Remark. Note that by taking OX = Z, we obtain as a corollary thatSh(X) has enough injectives.

Proof. We first show that categories of modules over a ring have enoughinjectives, and then globalize. The following simple lemma is at the heartof both steps of the proof.

Lemma 20.1. If F : A → B is an additive functor which has an exact18

left adjoint L : B → A , then F takes injectives to injectives.

Proof. Let I ∈ A be an injective object. To show that F (I) is an injectiveobject of B we must show there exists a dotted arrow in the followingcommutative diagram in B

0 // M //

N

FI

By the adjunction, this is equivalent to filling in the dotted arrow in

0 // LM //

LN

||I

which can be done because I is injective. Note that the map LM → LNis still injective because L is left exact.

Now let R be a ring (commutative for simplicity). Then ModR hasenough injectives. This follows easily from the following lemma.

Lemma 20.2. Let R be a ring. Then the forgetful functor L : ModR →Ab has a right adjoint F : Ab→ ModR, and if M ∈ ModR, and LM → Iis a monomorphism in Ab, then the composite M → FL(M) → FI is amonomorphism.

Proof. We define, for A any abelian group, FA = HomAb(R,A) whichhas an R-module structure given by (r · f)(x) = f(rx). To see this givesan adjunction we need to establish a natural bijection, for M ∈ ModR,A ∈ Ab,

HomAb(LM,A) ' HomR(M,HomAb(R,A))

18We actually only use left-exactness of L, but right exactness comes for free because left-adjoints preserve colimits

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The map going to the right sends f : M → A to the map m 7→ (r 7→f(rm)). The map going to the left sends φ : M → HomAb(R,A) to themap m 7→ φm(1), where φm is the image of m under φ. We omit theverification that these constructions are inverse and functorial in M andA.

Now we must check that given an inclusion LM → I, the map M →FL(M) → FI is also an inclusion. But this map is exactly the mapcoming from the adjunction HomAb(LM, I) ' HomR(M,FI). We needto check that given φ such that

Kφ//

0**

M // FL(M)

then actually φ is zero. Applying L and using the adjunction gives

LKLφ//

0

))LM

Id// LM

Thus Lφ = 0, hence φ = 0, since L is faithful (Lφ is the same map as φ,it just ignores the R-module structure). This finishes the proof.

To complete the proof of the theorem, we now consider OX -modules.Fix x ∈ X, i.e., a morphism j : x → X, and for ease of notation, letA = ModOX , the category of OX -modules, and B = ModOX,x , thecategory of modules over the local ring OX,x at x. Then we have anadjoint pair of functors

A

j−1

))B

j∗

ii

Now j∗ is a right adjoint, so if I ∈ B is injective, then so is j∗I, by thelemma19.

For the remainder of the discussion, fix F ∈ A , and since B has enoughinjectives, pick for each x an inclusion Fx → I(x) for some injective OX,x-module I(x). Then we have a map

F α−→∏x∈X

jx∗I(x),

where each map jx is the inclusion of the point x. A product of injectivesis still injective, coarsely speaking, because both products and injectivesare defined by maps in. Moreover, α is a monomorphism. We check thisat the stalk at an arbitrary point z ∈ X:

Fzαz−−→

( ∏x∈X

jx∗I(x))

z→ jz∗I

(z) ∼= I(z),

19Note that j−1 is exact: in general, for f : X → Y , and F a sheaf on Y , the stalk of f−1Fat x ∈ X is the same as the stalk of F at f(x) ∈ Y .

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where the isomorphism on the right is because jz∗ and j−1z are an adjoint

pair and j−1 is exact. The composite map here is just the chosen inclusionFz → I(z). Since this is injective, so is αz. Thus we have take an arbitrarysheaf F of OX -modules, and embedded it into an injective object of A ,which shows that A has enough injectives.

21 Wednesday, March 7th: Flasque Sheaves

Having shown that ModOX has enough injectives, we can define , for Fan OX -module

Hi(X,F) = Hi(Γ(X, I0)→ Γ(X, I1)→ Γ(X, I2)→ · · ·),

where F → I• is an injective resolution of F . However, in practice itis easier to work with flasque sheaves, which we now define, rather thaninjectives.

Definition 21.1. A sheaf F on a topological space is flasque if for everyinclusion, U ⊂ V , F(V )→ F(U) is surjective.

This notion is useful because, in consideration of whether a sheaf isflasque, it does not matter whether we regard it as a sheaf of O-modulesor abelian groups. Moreover, every injective sheaf is flasque. In the proof,we will use the following.

Definition 21.2. Let j : U → X be an open subset, and F be a sheaf ofOU -modules. Then the extension by zero of F to X is defined to bethe sheafification of the presheaf

W 7→

F(W ) if W ⊂ U0 if not.

It is denoted j!F .

Note that j!F has stalks equal to those of F inside U , and zero else-where. Also, j!F is a left adjoint to j−120.

Lemma 21.1. Let (X,OX) be a ringed space. If I ∈ ModOX is injective,then I is flasque.

Proof. Let V ⊂ U be two opens in X, with inclusions i : U → X, j : V →U . By the adjunction, we have for any sheaf F of OX -modules,

HomOX (i!OU ,F) ' HomOU (OU , i−1F) ' F(U)

Now there is an inclusion of OU -modules j!j−1OV → OU , which gives a

mapHomOX (i!OU ,F)→ HomOX (i!(j!j

−1OV ),F)

by restricting along the above inclusion. But HomOX (i!OU ,F) ' F(U),and similarly HomOU (OU , i−1F) ' F(V ). So we have a natural mapF(U) → F(V ). This is for any F , but now take F = I an injective

20Remember that j−1 is itself a left adjoint to j∗.

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OX -module. Then since i! is exact21, the inclusion j!j−1OV → OU gives

an inclusion i!j!j−1OV → i!OU , and then injectivity implies the map

HomOX (i!OU , I) → HomOX (i!(j!j−1OV ), I) is surjective, i.e., I(V ) →

I(U) is surjective.

Our next goal is to show that flasque sheaves are acyclic. For this weneed a(nother) lemma.

Lemma 21.2. Let F be a flasque sheaf of OX-modules, and embed F → Iinto an injective sheaf I, with cokernel G. Then for all opens U of X,I(U)→ G(U) is surjective.

Proof. We have an exact sequence 0 → F → I → G → 0. Fix a sectiong ∈ G(U). By exactness at the stalk, for each point of U , there is aneighborhood V over which we can lift g

∣∣V

to some gV ∈ I(V ). Suppose

we have also some other neighborhood W , and a lifting gW of g∣∣W

to

I(W ). We would like to glue these to a lift in I(V ∪W ). Since gV∣∣W∩V

and gW∣∣W∩V both map to g

∣∣W∩V in G(W ∩ V ), the exactness of the

above sequence implies that their difference gV∣∣W∩V − gW

∣∣W∩V lies in

F(W ∩ V )22. Since F is flasque, F(W ) surjects onto F(W ∩ V ), so thereis some ε ∈ F(W ) which maps to gV

∣∣W∩V −gW

∣∣W∩V in F(W∩V ). Since ε

maps to zero in G(W ), gW + ε ∈ I(W ) is a lift of g∣∣W

to I(W ). Moreover,

gV∣∣W∩V − (gW + ε)|W∩V = 0 in I(W ∩ V ), so they glue to give a lifting

of g over W ∪ V .Thus we have shown that we can lift sections of G “one open at a

time”. We now apply Zorn’s lemma to the set

(V, gV | gV lifts g over V ,

partially ordered by (V, gV ) ≤ (V ′, gV ′) if V ⊂ V ′ and gV ′∣∣V

= gV . Anyincreasing chain of such objects has an upper bound by taking the unionof the V . Thus we have a maximal such pair, which must be a lifting ofg over all of U : if it were not, we could extend further by the above andviolate maximality.23

We can now prove

Proposition 21.1. Let F be a flasque sheaf of OX-modules. Then Hi(X,F) =0 for i > 0.

Proof. First we pick an embedding F → I of F into an injective I. ThenI is flasque by Lemma 21.1. The quotient G is also flasque, because ifV ⊂ U is an inclusion, then we have a diagram

I(U) // //

G(U)

I(V ) // // G(V )

21This follows from the description of the stalks.22We are really using the left-exactness of the functor Γ(W ∩ V,−) here.23Cf. also Hartshorne Ex II.1.16

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in which the horizontal arrows are surjective by Lemma 21.2. This impliesthat the vertical arrow on the right is surjective also, hence G is flasque.Now H1(X,F) = 0 since it is the cokernel of the map Γ(X, I)→ Γ(X,G),which is surjective by the previous lemma. The short exact sequence0 → F → I → G → 0 gives rise to a long exact sequence in cohomology,in which the middle terms Hr(X, I) are zero because I is injective. Thuswe obtain isomorphisms Hi(X,G) ' Hi+1(X,F). Since G is flasque, theresult follows by induction.

22 Friday, March 9th: Grothendieck’s VanishingTheorem

22.1 Statement and Preliminary Comments

Ther next few lectures will be devoted to the proof of the following

Theorem 22.1. If X is a Noetherian topological space of dimension n,and F a sheaf of abelian groups on X, then Hi(X,F) = 0 for i > 0.

Our primary application of this theorem will be when (X,OX) is aringed space and F an OX -module. In order to do so, we will needthe fact that computations of cohomology in the categories of abeliangroups give the same result. More precisely, we have a universal δ-functorHi

M (X,−) from ModOX → Ab (the subscript “M” is for module), andalso a δ-functor Hi

A(X,−) from Sh(X)→ Ab (“A” is for abelian group),which is defined by regarding a sheaf of abelian groups F ∈ Sh(X) as asheaf of Z-modules, where Z is the constant sheaf24 There is also theforgetful functor ε : ModOX → Sh(X). Since ε is exact, the compositionHi

A(X,−) ε is a δ-functor. We ask whether it is isomorphic, as δ-functors, to Hi

M (X,−).Theorem 22.2. If (X,OX) is a ringed space and F an OX-module, thenHiA(X, εF) ' Hi

M (X,F).

Proof. We have an isomorphism

H0M (X,−) ' H0

A(X, ε(−))

which, by the universality of HiM , induces a morphism of δ-functors

HiM (X,−) → Hi

A(X,−) ε.

This is an isomorphism because HiA(X,−) ε is also universal. This

follows because it is effaceable: for F ∈ ModOX , pick an embeddinginto an injective OX -module I. This injective is also flasque, and so εI isflasque in Sh(X). Flasque sheaves in Sh(X) are acyclic for Hi

A by applyingLemma 21.1 with OX = Z.

24Note that the same argument as in the case of ModOX shows that Sh(X) has enoughinjectives, by taking OX = Z.

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22.2 Idea of the Proof; Some lemmas

The proof of the vanishing theorem begins by reducing first to the casewhen X is irreducible, and then reducing to the case where F = j! Z forj : U → X and open. Then we use induction on the dimension of X. Tomake the reduction on F , we will need to know how flasque sheaves andcohomology interact with direct limits.

Lemma 22.1. On a Noetherian topological space, a direct limit of flasquesheaves is flasque.

Proof. HW exercise.

The next lemma says that “cohomology commutes with direct limits”.More precisely, if (Fα) is a directed system in Sh(X), then for each αwe have a map Fα → lim−→Fα. Applying Hi we get maps Hi(X,Fα) →Hi(X, lim−→Fα) for each α. By the universal property of direct limit, theseinduce a map

lim−→Hi(X,Fα)→ Hi(X, lim−→Fα)

Lemma 22.2. If X is a Noetherian topological space, the map lim−→Hi(X,Fα)→Hi(X, lim−→Fα) is an isomorphism.

Proof. Let A be a directed set, which we can regard as a category. Wedenote by Sh(X)A the category of functors A → Sh(X). Note that it isan abelian category; kernels, cokernels, etc., are defined “index by index”.It also has arbitrary products, and enough injectives. From the followingdiagram

Sh(X)Alim−→ //

Γ

Sh(X)

Γ

AbAlim−→

// Ab

we obtain two δ-functors Sh(X)A → Ab, namely the right-derived functorsof the two different compositions in the square. More precisely, the twoδ-functors in question are

1. (Fα) 7→ Hi(X, lim−→Fα)

2. (Fα) 7→ lim−→Hi(X,Fα)

We will be done if we can show that these are both universal, for then theuniversality gives a canonical isomorphism between them. We show thatthey are both effaceable. The reason is that for any sheaf F , there is afunctorial embedding F → F of F into a flasque sheaf, namely take F tobe the so-called sheaf of discontinuous sections of F , given by

F(U) = s : U⋃p∈U

FP∣∣ s(p) ∈ Fp for each p.

Then the first δ-functor is effaceable since this is functorial, i.e., given anobject (Fα) ∈ Sh(X)A, the construction produces a system Fα of flasquesheaves, whose direct limit is flasque by Lemma 22.1, henceHi(X, lim−→Fα) =

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0 for i > 0. The second is effaceable since each Hi(X, F) = 0 for i > 0,so lim−→Hi(X,Fα) = 0.

23 Monday, March 12th: More Lemmas

We stated without proof last time that the functor lim−→ : Sh(X) → Ab isexact. Let’s prove that now.

Proposition 23.1. For A a directed set, the functor lim−→ : Sh(X)A →Sh(X) is exact.

Proof. Let0→ (F ′α)→ (Fα)→ (F ′′α)→ 0

be a short exact sequence in Sh(X)A. For each x ∈ X, and each α ∈ A,the sequence

0→ F ′α,x → Fα,x → F ′′α,x → 0

is exact, and the functor lim−→ : ModAOX,x → ModOX,x is exact, so we haveshort exact sequences

0→ lim−→F′α,x → lim−→Fα,x → lim−→F

′′α,x → 0

for each x ∈ X. Hence we will be done if we can show that lim−→ commuteswith stalks, i.e.,

lim−→(Fα)x = lim−→Fα,xBut stalk is a special case of pullback, so it suffices to prove more generallythat for f : Y → X, the functor f−1 : Sh(X) → Sh(Y ) commutes withdirect limits. For this we use Yoneda: let (Fα) ∈ Sh(X) and G ∈ Sh(Y ).Then

HomY (f−1 lim−→Fα,G) = HomX(lim−→Fα, f∗G)

= lim−→HomX(Fα, f∗G)

= lim−→HomY (f−1Fα,G)

= HomY (lim−→Fα,G),

so f−1 lim−→Fα = lim−→Fα.25

The next lemma says that if we have a sheaf F on a closed subsetZ ⊂ X, we can compute the cohomology of F directly on Z, or push itforward to X and compute there, and we will get the same result.

Lemma 23.1. Let i : Z → X be a closed subset; then for any F ∈ Sh(Z),

Hi(Z,F) ' Hi(X, i∗F)

25The argument is a common one: “colimits commute with left adjoints”. Also note thatthis isomorphism holds for OX -modules, although our proof only works in Sh(X), becausef−1 is not well-defined for OX -modules.

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Proof. First note that since i is a closed immersion, i∗ is exact.26 Also,i∗ has an exact left-adjoint, so by Lemma 20.1, it takes injectives to in-jectives. Thus taking an injective resolution F → I• of F in Sh(Z), weget an injective resolution i∗F → i∗I• of i∗F in Sh(X). But for any sheafG on Z, Γ(X, i∗G) = Γ(Z,G), so applying the global sections functor tothese two injective resolutions gives the same sequence of abelian groups,hence the result.

Remarks. 1. To see that i∗ is exact when i is a closed immersion, notethat the stalk of i∗F is

(i∗F)x =

Fx if x ∈ Z0 if not

.

Since exactness can be checked at stalks, it is easy to see in this casethat i∗ is exact. However, this fails when i is not a closed immersion:consider the open immersion

i : Y = A2k \ (0, 0) → A2

k = X

Then the stalk (i∗OY )(0,0) is just OX,(0,0). Intuitively, an openneighborhood U of the origin, and the punctured neighborhood U \(0, 0) have the same sections, because the point has codimensiontwo.

2. In applying the lemma, we will frequently use the following fact. IfF is a sheaf on a space X, whose support is contained in a closedsubset i : Z → X, then F = i∗i

−1F .

At this point we began the proof of Theorem 22.1, but I’ll consolidatethe whole argument into the notes for the next lecture.

24 Wednesday, March 14th: Proof of the Vanish-ing Theorem

We recall first the statement: If X is a Noetherian topological space ofdimension n, and F a sheaf of abelian groups on X, then Hi(X,F) = 0for i > 0.

Proof. 24.1 Base step

We proceed by induction on n = dimX. For the case n = 0, the theoremsays exactly that the global sections functor is exact. But here X is adisjoint union of points, finite by the Noetherian hypothesis. A sheaf ona point is just an abelian group, so a sheaf on X is just a finite productof abelian groups. Thus Γ(X,−) is exact - it sends a finite product ofabelian groups to ... that same finite product of abelian groups.

Now we fix n > 0, and assume the theorem holds for any space ofdimension less than n. Fix a sheaf F of abelian groups on X.

26It’s a right adjoint, so it’s always left exact (even if i is not a closed immersion). Forright-exactness in the case of a closed immersion, see the remark following the proof.

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24.2 Reduction to X irreducible

Here we argue that it is enough, under our inductive hypothesis, to assumethat X is irreducible. Suppose it were not. Then pick an irreducible closedsubset X ′ of X, let U be the complement in X ′ of the other irreduciblecomponents of X, and let Y = X \ U . Then we have inclusions

Uj→ X

i← Y

with j open and i closed. Let FU = j!(F∣∣U

) and FY = i∗i−1F . We

obtain27 an exact sequence of sheaves on X:

0→ FU → F → FY → 0,

which gives rise to a long exact sequence of cohomology

H0(X,FU )→ H0(X,F)→ H0(X,FY )→ H1(X,FU )→ H1(X,F)→ · · ·

By Lemma 23.1, we have Hi(X,FY ) = Hi(Y, i−1F). Also, since FUis supported on X ′, we may regard it as a sheaf on X ′, and we haveHi(X,FU ) = Hi(X ′,FU ), again by Lemma 23.1 28. So the long exactsequence looks like

· · · → Hi−1(Y, i−1F)→ Hi(X ′,FU )→ Hi(X,F)→ Hi(Y, i−1F)→ · · ·

Since X ′ is irreducible, and Y ′ is a closed subset of X with less irreduciblecomponents than X, this sequence will prove that Hi(X,F) = 0 for i > nby induction on the number of irreducible components, so long as we canestablish this in the case where X is irreducible (i.e., the base step for theinduction on the irreducible components).

24.3 Reduction to the case where F = ZU .

Now we assume X is irreducible. For j : U → X any open subset, we letas above ZU = j! Z, where Z is the constant sheaf Z on U . We will simplywrite ZU from now on. Define

S =∐U⊂X

F(U),

and let A be the set of finite subsets of S , partially ordered by inclusion.A typical element of A looks like

α = (U1, f1), . . . , (Ur, fr), fk ∈ F(Uk)

Such a thing defines a map

r⊕i=1

ZUi(fi)→ F

27Cf. Hartshorne, Ex II.1.19(c).28Abusing notation slightly: should be Hi(X,FU ) = Hi(X′, s∗s−1FU ), where s : X′ → X

is the inclusion. See remark 2 following Lemma 23.1.

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For each α as above, we let Fα be the image of this map. Then theinclusions Fα → F induce a map lim−→Fα → F which is an isomorphism.It’s injective since the Fα are subsheaves of F . It’s surjective on eachopen, since if f ∈ F(U), then take α = (U, f); the image of 1 underZU (U)→ Fα(U)→ (lim−→Fα)(U) is a preimage for f over U .

Since Hi(X, lim−→Fα) ' lim−→Hi(X,Fα), it is now enough to prove that

when i > n, Hi(X,Fα) = 0. With α = (U1, f1), . . . , (Ur, fr) as above,and setting α′ = (U1, f1), . . . , (Ur−1, fr−1), we have a diagram

0 // Fα′ // Fα // Q // 0

0 //⊕r−1

i=1 ZUi

OOOO

//⊕r

i=1 ZUi

OOOO

// ZUr

OO

// 0

where the horizontal rows are exact by construction (here Q is just thecokernel of Fα′ → Fα), and the left two vertical arrows are surjective bydefinition. It follows that the rightmost vertical arrow is also surjective.Now from the long exact sequence in cohomology induced by the topsequence here, and using induction on the cardinality of α, it suffices toshow that Hi(X,Q) = 0 for i > n, where Q is any quotient of a sheaf ofthe form ZU .

For such a sheaf Q, we have an exact sequence

0→ K → ZU → Q → 0

and again using the long exact sequence, we see it will be enough to showthat Hi(X,K ) = 0 for i > n, where K is any subsheaf of ZU (includingZU itself). So fix some subsheaf K of ZU ; then the stalk Kx at x ∈ Xis a principal ideal (dx) ⊂ Z, where we may assume that dx ≥ 0. If allintegers dx are zero, then K is the zero sheaf, whose higher cohomologyvanishes, so there’s nothing to show. Otherwise, set d = infdx, wherethe infimum is taken over all x such that dx 6= 0. Then we will have dx = dfor at least one point of X. Take some neighborhood V ⊂ U containingthis x, and a section s ∈ K (V ) representing d on this neighborhood. Thesection s defines an injective map ZV → K sending 1 to d. Composingwith the inclusion K → ZU gives a map

ZV → K → ZU

of sheaves on U , which is multiplication by d at stalks of V , and the zeromap elsewhere. Now we claim that in fact KV is isomorphic to ZV . Wecheck this at the stalk at a point y of V . There the inclusions above looklike

ZV,y → Ky → ZU,y,where ZV,y ' ZU,y ' Z and the composite map is multiplication by d. Butthis multiplication must factor through the image of Ky → ZU,y, whichis just the subgroup generated by dy. So (d) ⊆ (dy); but d is minimalamongst the dx as x varies, therefore dy = d and Ky ' (d) ' ZV,y. Inparticular, Z

∣∣V' K

∣∣V

.

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This isomorphism extends to an inclusion ZV → KV → K . Letthe cokernel of ZV → K be D . Then since ZV ' K over V , D issupported on U \V , and hence on U \ V , which has lower dimension thanX. For if Z0 ⊂ · · ·Zr is a chain of irreducible closed subsets in U \ V ,then the inclusion Zr ⊂ X extends this chain, since X has been assumedirreducible. Thus the long exact sequence in cohomology coming from

0→ ZV → K → D → 0

shows that Hi(X,ZV ) ' Hi(X,K ) for i > n, so it will be enough toprove Hi(X,ZV ) = 0 for i > n (note that the long exact sequence evenshows Hn+1(X,ZV ) ' Hn+1(X,K ), since the cohomology of D vanishesin degrees greater than or equal to n).

24.4 Conclusion of the Proof

Now we have finished our reduction. To show that Hi(X,ZV ) = 0 indegrees greater than n, consider the inclusion ZV → Z (the constant sheafon X), and let Q be the cokernel. Since X is irreducible, Z is flasque,so its cohomology vanishes in positive degree. The long exact sequencecoming from

0→ ZV → Z→ Q → 0

thus shows that Hi(X,Q) ' Hi+1(X,ZV ) for all i > 0. But Q is sup-ported on X \ V , which as above has lower dimension than X sinceX is irreducible29. So Hi(X,Q) = 0 for i ≥ n by induction30. ThusHi(X,ZV ) = 0 for all i ≥ n+ 1, finishing the proof.

25 Friday, March 16th: Spectral Sequences - In-troduction

In the following few lectures, we will state without proof the results aboutspectral sequences which we will use, along with some common applica-tions. Our aim is to obtain the following two results

1. If X is affine and F a quasicoherent sheaf on X, then the higher co-homology of F vanishes (we saw the vanishing of H1 last semester).

2. If f : X → Y is an affine morphism and F ∈ QCoh(X), then thehigher direct image sheaves Rif∗F are zero for i > 0. This general-izes the previous result by taking Y =point.

25.1 Motivating Examples

Examples 25.1. 1. Let X be a topological space and U = Ui anopen cover. If F ∈ Sh(X), we ask: how can we compute Hs(X,F)

29The V from above was non-empty by construction. If we were proving this for general V ,the case when V is empty implies that ZV is the zero sheaf, so there’s nothing to show.

30We’re using Lemma 23.1 and Remark 2 following it again.

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in terms of the Hs(Ui,F∣∣Ui

)? Consider first the case when s = 0.

Then by the sheaf property, H0(X,F) is the kernel of the map∏i

H0(Ui,F) ⇒∏i,j

H0(Uij ,F).31

For the case s = 1, let

K = ker(∏

i

H1(Ui,F) ⇒∏i,j

H1(Uij ,F)).

We also define groups (the “Cech cohomology groups”)

H1(U,F) =ker(∏

i,j H0(Uij ,F) ⇒

∏i,j,kH

0(Uijk,F))

im(∏

iH0(Ui,F) ⇒

∏i,j H

0(Uij ,F))

H2(U,F) =ker(∏

i,j,kH0(Uijk,F) ⇒

∏i,j,k,lH

1(Uijkl,F))

im(∏

ij H0(Uij ,F) ⇒

∏i,j,kH

0(Uijk,F))

At this point you should ask “what actually are all those maps ⇒?”We’ll come to that later. We will see that these groups fit into anexact sequence

0→ H1(U,F)→ H1(X,F)→ H2(U,F)

In the case mentioned above, namely when X is affine and F quasi-coherent, the

∏iH

1(Ui,F) = 0, so K = 0, and we get H1(U,F) 'H1(X,F).

2. Consider continuous maps Xf→ Y

g→ Z, and a sheaf F on X. Wewould like to relate the composite right derived functorsRqg∗(R

pf∗F)with the right derived functors of the composite Rn(fg)∗F . This isaccomplished by the Leray spectral sequence, which we will see soon.

25.2 Definition

Definition 25.2. Let A be an abelian category, and a ∈ N. A spectralsequence starting at page a consists of the following data:

1. Objects Epqr of A , for p, q ∈ Z and r ≥ a some integer. The collectionEpqr for fixed r is called the “rth page”.

2. Morphisms dpqr : Epqr → Ep+r,q−r+1r such that d2 = 0 (whenever d2

makes sense). The maps go “r to the right and r − 1 down”.

3. Isomorphisms Epqr+1 ' H∗(Epqr ).

Usually we will consider situations where p, q ≥ 0, in which case foreach p, q there exists a page r beyond which the differentials d are all zero(this r will depend on p, q), so Epqr = Epqr+1 = . . .. We denote this commonvalue by Epq∞ . We say then that the spectral sequence converges.

31As usual, Uij = Ui ∩ Uj , and in the following we will denote triple intersections byUijk = Ui ∩ Uj ∩ Uk, etc. Also, we will carelessly write F even where we should actually bewriting F

∣∣Ui

, etc.

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Definition 25.3. Let H∗ = Hnn∈Z be a collection of objects of A . Wesay that Epqr converges to H∗ if a) it converges in the sense describedabove, and b) each Hn has a filtration32 F • such that F p/F p+1 ' En,n−p∞ .We use the notation Epqr ⇒ H∗ in this case.

In other words, Epqr ⇒ H∗ means that the nth antidiagonal of E∞gives the graded pieces of a filtration on Hn.

Remark. Note that this information is not always enough to determineHn. It is, for example when A is the category of k-vector spaces. On theother hand, the two complexes of abelian groups

0→ Z/2 ·2→ Z/4→ Z/2→ 0 and

0→ Z/2→ Z/2× Z/2→ Z/2→ 0

Both have filtrations with quotients Z/2, but are not isomorphic com-plexes.

25.3 Spectral Sequence of a Filtered Complex

LetK• . . . 0→ Kr → Kr+1 → . . .

be a complex in A , with a filtration by subcomplexes F i, so that

0 = F s ⊆ F s−1 ⊆ . . . ⊆ F 1 ⊆ F 0 = K•.

We say that the filtration is finite because the F i are eventually zero,and exhaustive because F 0 = K•. We will use frequently the followingresult:

Theorem 25.1. In the situation above, there is a spectral sequence

Epq1 = Hp+q(grpK•)⇒ Hp+q(K•).

A word on notation. When we write Epq1 = Hp+q(grpK•)⇒ Hp+q(K•),we mean that the spectral sequence begins at page 1, with the first pagebeing Hp+q(grpK•). Here grpK• means the pth graded piece of the givenfiltration on K•, namely F p/F p+1 (note that this, too, is a complex, sowe can take cohomology). The notation for the spectral sequence givesno indication of what the differentials are, not even on the first page. Tounderstand what they are, you must study the proof of the theorem. SeeWeibel’s book, section 5.4, for instance.

One thing we can say is that the mapHp+q(grpK•)→ Hp+1+q(grp+1 K•)is the boundary map coming from application of the Snake Lemma to theshort exact sequence

0→ F p+1/F p+2 → F p/F p+2 → F p/F p+1 → 0.

32For us a filtration will always mean a decreasing filtration, i.e., a sequence of inclusions. . . ⊂ F i+1 ⊂ F i ⊂ . . . Hn.

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Example 25.1. The “filtration bete” on a complex

K• = K0 → K1 → K2 → . . .

is as follows:

F 2 : 0 //

0 //

K2 //

. . .

F 1 : 0 //

K1 //

K2 //

. . .

K• : K0 // K1 // K2 // . . .

In this case grpK• = F p/F p+1 is the complex with the object Kp indegree p and zeroes elsewhere, which we denote Kp[−p]33 Therefore

Hp+q(grpK•) =

Kp if q = 0

0 if q 6= 0

Thus the first page of our spectral sequence is

0 0 0 0 . . .

0 0 0 0 . . .

K0 // K1 // K2 // K3 // . . .

which converges by the second page since the differentials are zero there.But on the second page the objects are Ep02 = Hp(K•). So the Theoremtells us that the objects Hp+q(K•) have a filtration whose pth gradedpiece is 0 unless q = 0, in which case it is Hp(K•). In other words, Hn isfiltered by Hn. Bete, indeed.

26 Monday, March 19th: Spectral Sequence of aDouble Complex

26.1 Double Complex and Total Complex

Today we discuss another important example of a spectral sequence. Sup-pose we have a double complex I•,•; that is, a collection of objects Ipq

33In general, whenever we write an object as standing for a complex, we mean the complexcontaining that object in degree zero with zeroes elsewhere. Here we have shifted the degreeto make the object sit in degree p.

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and two differentials d and δ, such that the following diagram commutes

Ip,q+1δ // Ip+1,q+1,

Ip,q

d

OO

δ// Ip+1,q

d

OO

Then we can form a new complex, the total complex tot(I) of I•,•,defined as follows:

(tot I)n =⊕i+j=n

Ii,j

The differential of this complex is

D∣∣Iij

= d+ (−1)jδ

There are other sign conventions in the literature - other authors takeD = d + δ but insist that the squares in the original double complexanticommute; the important point is that in order to get a complex, thesquare in the following diagram must anticommute

Ii,j+2

Ii,j+1∓δ//

d

OO

Ii+1,j+1

Ii,j ±δ//

d

OO

Ii+1,jδ//

d

OO

Ii+2,j

D2 is the sum of all the paths from the lower left to the three objects onthe upper right antidiagonal - these should sum to zero.

There is a filtration on tot(I•,•) whose pth term is

(F p)n =⊕i+j=ni≥p

Iij

In other words, the pth term F p of the filtration is that part of the nthantidiagonal of I•,• which lies to the right of the p − 1 column. Fromthis description, we see that the quotient F p/F p+1 is a complex whoseobjects are precisely the pth colum of I•,•. But Ipq sits in degree n = p+qaccording to the above definition, so the complex is graded by q, but witha shift down in degree by p. I.e.,

F p/F p+1 = Ip,•[−p]34

Theorem 25.1 in this situation can be stated as:

Theorem 26.1. In the situation above, there is a spectral sequence

Epq1 = Hp+q(Ip,•[−p])⇒ Hp+q(tot I•,•)

34Recall our convention about grading shifts: (C[r])n = Cr+n.

87

Note that Hp+q(Ip,•[−p]) = Hq(Ip,•).

Corollary 26.1. Let f : I•,• → J•,• be a morphism of double complexessuch that for each p, Ip,• → Jp,• is a quasi-isomorphism. Then tot I•,• →tot J•,• is a quasi-isomorphism.

Proof. The morphism f induces a morphism of spectral sequences, asfollows. By assumption, the maps Hq(Ip,•) → Hq(Jp,•) induced by fare isomorphisms between the first pages of the two spectral sequences.Thus when we take cohomology we get isomorphisms between the 2ndpages, and so on. So isomorphisms between the first pages give rise toisomorphisms

Epq∞ (I)→ Epq∞ (J)

Now it is a general fact that a map of filtered complexes which inducesisomorphisms on each graded piece must be an isomorphism of complexes.So the above implies that we get an isomorphism

Hn(tot I•,•)'→ Hn(tot J•,•).

26.2 Application: Derived Functors on the Category of Com-plexes

The above will be useful for us in the following situation. Let F : A → Bbe a left exact functor between abelian categories, where A has enoughinjectives. Let K• be a complex in A with Kp = 0 for p << 0. We wantto compute RiFK• ∈ B.

First resolve the complex K• by injectives:

. . . // Kp−1 //

Kp //

Kp+1 //

. . .

. . . // Ip−1,• // Ip,• // Ip+1,• // . . .

Here the rows and columns are complexes, and the columns are also res-olutions (i.e. exact). In particular, it’s a double complex I•,•. We canthus define

RiFK• = Hi(totFI•,•)

Then of course we have to check that this does not depend on the reso-lution I•,•. Suppose given two injective resolutions I•,• and J•,• of thecomplex K•. Then one argues as in the HW a few weeks ago that it isenough to consider the case when we have a morphism I•,• → J•,• whichcommutes with the inclusions of K• into each double complex. Fixing p,we have injective resolutions Kp → Ip,• and Kp → Jp,•. We know thatthe derived functors of F on objects do not depend on the resolution, i.e.,the maps I•,• → J•,• induce isomorphisms Hn(FIp,•) ' Hn(FJp,•), andhence the conditions of the corollary are met, showing that the derivedfunctors RnF are well-defined on the complex K•.

88

27 Wednesday, March 21st: Spectral Sequence ofan Open Cover

Let (X,OX) be a ringed space and F an OX -module. Fix an open coverU = Uii∈I of X, and assume I is totally ordered. We use the notation

i = (i0, . . . , ik), i0 < i1 < . . . < ik

for a multi-index in I, with |i| = k denoting the length (length minusone, really, since our multi-indices start with i0). For such a multi-index,define Ui = Ui0 ∩ . . . ∩ Uik , and let

ji : Ui → X

be the inclusion.We define a complex C •(U,F) of sheaves of abelian groups on X, the

Cech complex of F relative to the open cover U, as follows. Foreach k ≥ 0, set

C k(U,F) =∏|i|=k

ji∗F∣∣Ui.

The first few terms are thus

C 0(U,F) =∏i∈I

ji∗F∣∣Ui

C 1(U,F) =∏i1<i2

j(i1,i2)∗F∣∣Ui1∩Ui2

The differential on this complex is obtained by first defining, for each0 ≤ j ≤ k + 1, a map

dj : C k(U,F)→ C k+1(U,F)

which sends an element

(ai)i=i0i1···ik 7→ (as0s1···sj ···sk )s0s1···sk

That is, the component of dj((ai)) on the s0s1 · · · sk factor is obtained bydeleting the jth index sj and taking the appropriate entry of (ai).

Then we define the differential on C k(U,F) by

d =

k+1∑j=0

dj

One checks that this defines a complex of sheaves.

Example 27.1. Take (X,OX) to be P1k with its structure sheaf, and

U = U0, U1 the standard open cover of P1 by affines. Then

C 0(U,OP1) = j0∗OU0 ⊕ j1∗OU1

C 1(U,OP1) = j01∗OU01

C k(U,OP1) = 0 for k > 1

Then if (a0, a1) is a local section of C 1, the two maps dj for j = 0, 1 sendit to a0 and a1, respectively, so the differential d : C 0 → C 1 maps (a0, a1)to a1 − a0.

89

The reason we care about this Cech complex is the following:

Proposition 27.1. There is a quasi-isomorphism of complexes F →C •(U,F).

Remark. Recall that when we speak of a sheaf F as a complex, we meanthe complex which has F in degree zero and zeroes elsewhere, and withdifferential zero. Then to say that this complex is quasi-isomorphic tosome other complex K• (with Kp = 0 for p ≤ 0) means that K• mustbe exact except in degree zero, since the cohomology of the “complex”F is zero in higher degrees. Moreover, the kernel of K0 → K1 must beisomorphic to F because of the isomorphism of cohomology in degree zero.In other words, “F quasi-isomorphic to K•” means “K• is a resolution ofF”.

Proof. First note that we have a map

F →∏i∈I

ji∗F∣∣Ui

Given by the product of the restriction maps. This map is actually anisomorphism onto the kernel of d0 because of the sheaf axiom for F . Itremains to show that the higher cohomology sheaves H q(C •(U,F)) arezero for q > 0. The question is local since we’re working with sheaves,so we can assume X is one of the Uis. In fact, after some fussing aroundwith indices and signs, we may assume that X = U0 for simplicity.

Now we define a homotopy operator

hk : C k+1(U,F)→ C k(U,F)

which sends a local section (ai0···ik+1) to (a0i0···ik )i0···ik . One checks thatthese maps satisfy

dhk + hk+1d = IdC• ,

that is, they define a homotopy between the identity map on C •(U,F)and the zero map. This shows that the identity map and the zero mapinduce the same map on cohomology, so it must be that (C •(U,F), d) isexact (except, of course, in degree zero).

Next we want to apply our results on spectral sequences to Cech com-plexes. Take an injective resolution F → I • of F . Then the Cech sheavesC (U,I ) are all injective sheaves, and we obtain a double complex

......

C 0(U,I 1) //

OO

C 1(U,I 1) //

OO

· · ·

C 0(U,I 0) //

OO

C 1(U,I 0) //

OO

· · ·

90

The rows are (injective) resolutions of the sheaves I k, by the proposition.We can regard the complex I • as a double complex I •,• with only onecolumn. Since the maps I k → C •(U,I k) are resolutions (i.e., quasi-isomorphisms), Corollary 26.1 from the previous lecture applies, and saysthat the map of total complexes tot I •,• → tot C •(U,I •) is a quasi-isomorphism. But the total complex of I •,• is just I •. Thus we have achain of quasi-isomorphisms

F ' I • ' tot(C p(U,I q))

In other words, the total complex of the double complex above is a reso-lution of F , also.

Now we take global sections all over the place. Let

Cq(U,F) = Γ(X,C q(U,F)) =∏|i|=q

F(Ui).

Then the quasi-isomorphisms I k → C •(U,I k) give rise to quasi-isomorphisms

Γ(X,I k)→ C•(U,I k)

of complexes of abelian groups35. Then we have a double complex

......

C0(U,I 1) //

OO

C1(U,I 1) //

OO

· · ·

C0(U,I 0) //

OO

C1(U,I 0) //

OO

· · ·

where here the rows are resolutions of Γ(X,I k). By the same argumentas above, we find that the total complex totC•(U,I •) is a resolution ofΓ(X,I •), so we can use this total complex to compute the cohomologygroups of F on X:

H∗(totC•(U,I •)) ' (H∗(X,F)).

On the other hand, our discussion of the spectral sequence of a doublecomplex showed that there is a spectral sequence

Epq1 = Hq(Cp(U,I •))⇒ Hp+q(totC•(U,I •))

But

Hq(Cp(U,I •)) = Hq( ∏|i|=p

I •(Ui))

= 36∏|i|=p

Hq(Ui,I•) =

∏|i|=p

Hq(Ui,F)

35This is only true because we began with resolutions of injective objects I k. In general,if K• and K′• are quasi-isomorphic complexes of injectives, then application of a left exactfunctor F gives quasi-isomorphic complexes FK• and FK′•, but the fact that both K• andK′• are injective is essential.

36Not sure under what circumstances we can interchange cohomology and products likethis.

91

In conclusion, we have constructed a spectral sequence

Epq1 =∏|i|=p

Hq(Ui,F)⇒ Hp+q(X,F)

This is the spectral sequence of the open cover U.

28 Friday, March 23rd: Cech Cohomology Agreeswith Derived Functor Cohomology in Good Cases

Last time, given an open cover U = Uii∈I of a topological space X, anda sheaf F of abelian groups on X, we constructed a spectral sequence

Epq1 =∏|i|=p

Hq(Ui,F)⇒ Hp+q(X,F)

Recall that the “global” Cech complex consisted of groups

Cp(U,F) =∏|i|=p

F(Ui),

with differentials defined in the same way as for the sheafy version C • lasttime. We call the cohomology groups

Hp(C•(U,F))

the Cech cohomology of F relative to the open cover U, and denotethem by Hp(U,F). Today we will use the above spectral sequence to provethat in good circmstances, the Cech cohomology groups agree with thecohomology groups defined by derived functors of Γ(X,−).

Let us write out the first few pages of the “global” spectral sequencefrom last time. The E1 page looks like

......

∏iH

1(Ui,F)γ//∏i<j H

1(Uij ,F) // · · ·

C0(U,F) // C1(U,F) // · · ·

0 0 · · ·The second page looks as follows, letting K = ker γ.

......

K∂

++XXXXXXXXXXXXXXXXX · · ·

++VVVVVVVVVVVVVVVVV · · · · · ·

H0(U,F)

++XXXXXXXXXXXXXXXXX H1(U,F)

++WWWWWWWWWWWWWW H2(U,F) · · ·

0 0 0 · · ·

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The map in to H0(U,F) is also zero, so this shows that H0(U,F) is theonly term in our filtration of H0(X,F), so

H0(U,F) ' H0(X,F)

To study the filtrations on the higher cohomology groups Hi(X,F),we look now at the third page:

......

ker ∂

((RRRRRRRRRRRRRRRRRRRRRRR · · · · · · · · ·

H0(U,F) H1(U,F) · · · · · ·

0 0 0 0

Note that now the first antidiagonal p+ q = 1 stabilizes, as the mapsinto and out of both ker ∂ and H1(U,F) are zero hereafter. So we knowthat there is a filtration on H1(X,F):

0 = F 2 ⊆ F 1 ⊆ F 0 = H1(X,F)

with

F 0/F 1 ' E0,1∞ = E0,1

3 ' ker ∂

F 1/F 2 ' E1,0∞ = E1,0

3 ' H1(U,F)

Since F 2 = 0, we have

ker ∂ ' F 0/F 1 '(F 0/F 2)/(F 1/F 2) ' H1(X,F)/H1(U,F)

So we get an exact sequence

0→ H1(U,F)→ H1(X,F)→ ker ∂ → 0.

It turns out that the map on the right, H1(X,F) → ker ∂, is compatiblewith the inclusions

ker ∂ → K →∏i

H1(Ui,F)

Thus we obtain a commutative diagram

0 // H1(U,F) // H1(X,F) //

##GGGGGGGGGG ker ∂ //

0

K

∏iH

1(Ui,F)

This is good because it relates H1(X,F) to H1(U,F), but at the expenseof the “smaller” groups

∏iH

1(Ui,F). Of course, what we want is a way to

93

get a handle on these H1(Ui,F). More generally, we would like to handlethe groups Hi(Ui,F), where i > 0 and the multi-index i has arbitrarylength. A first step in this direction is to take the cover U to consist ofopen affines. The next theorem shows why this is useful. It is also the firstof the two desired results mentioned at the beginning of our discussion ofspectral sequences.

Theorem 28.1. Let X be an affine scheme and F ∈ QCoh(X). Thenfor all q > 0, Hq(X,F) = 0.

Proof. We proceed by induction on q. The case when q = 1 was estab-lished last semester: “the global sections functor is exact on affines”. Sonow assume that for any affine scheme X and any quasicoherent sheaf Fon X, Hi(X,F) = 0 for 1 ≤ i ≤ q. We fix an affine scheme X, with acover37 by open affines X = U0 ∪ . . . ∪ Ur. We can arrange it so that allpossible intersections of the Ui are also affine (by taking them to be basicopens, say). We consider the spectral sequence from before, where by ourinductive assumption, all Hi(Ui,F) are zero, where i is a multi-index and0 < i ≤ q. The first page is thus∏

iHq+1(Ui,F) //

∏i<j H

q+1(Uij ,F) // · · ·

0 0 0

0 0 0

C0(X,F) // C1(X,F) // · · ·

We include only two rows of zeroes for simplicity, but of course there willbe more for larger q. These rows are zero by our inductive assumption.

We now proceed to page two. The first row will consist of the groupsHp(C•(U,F)), which are the right derived functors of Γ(X,−), applied tothe “Cech sheaves” C •(U,F). These sheaves are quasicoherent, so sinceX is affine, Γ(X,−) is exact and the higher right derived functors vanish.So the second page is

K · · · · · ·

0 0 0

0 0 0

H0(X,F) 0 0

Here K is just some subgroup of∏iH

q+1(Ui,F). Since this spectralsequence converges to Hp+q(X,F), we have that Hq+1(X,F) ' K →∏iH

q+1(Ui,F). It turns out that, by the construction of the spectralsequence (which we did not go into in any detail), this map

Hq+1(X,F) →∏i

Hq+1(Ui,F)

37Recall that an affine scheme is quasicompact, so we may take a finite cover here, althoughwe will not use the finiteness.

94

is induced by the restriction maps along Ui → X. We want to use this toshow that Hq+1(X,F) = 0, for which we need the following lemma.

Lemma 28.1. IF X is any scheme, and α ∈ Hi(X,F) for i > 0. Thenthere is an open cover X =

⋃j Uj such that the image of α in each

Hi(X,Uj) is zero.

Proof. First choose an injective resolution F → I •. Our element α ∈Hi(X,F) can be represented by an element α ∈ ker(Γ(X,I i)→ Γ(X,I i+1)).Since the copmlex of sheaves I • is exact, we can find a neighborhood Uof any point in X such that α

∣∣U

is in the image of I i−1(U) → I i(U).Now choose an open cover

⋃j Uj of X by such U . Then we have [α] = 0

in Hi(Uj ,F), for each j, as desired.

Using the Lemma, we may replace, i.e., refine, our original cover to getsome new Uj such that α ∈ Hq+1(X,F) restricts to 0 in each Hq+1(Uj ,F).But the map Hq+1(X,F) →

∏iH

q+1(Ui,F) is injective, so α is zero,hence Hq+1(X,F) = 0.

Corollary 28.1. If f : X → Y is an affine morphism, and F a quasico-herent sheaf on X, then Rif∗F = 0 for all i > 0.

The proof follows immediately from the theorem and the followinglemma.

Lemma 28.2. If f : X → Y is a morphism of schemes, and F a sheafon X, then the higher pushforward Rif∗F is the sheaf associated to thepresheaf on Y

T iF : U 7→ Hi(f−1(U),F).

Proof. My notes for Professor Olsson’s argument are unclear here. Theyinclude only the cryptic phrase “sheafification commutes with quotients”.See Hartshorne III.8.1 for a different argument.

Now we make good on the title of today’s lecture

Theorem 28.2. If X is a scheme and U = Uiri=1 a cover by openaffines with all possible intersections affine38, then for any quasicoherentsheaf F on X, we have

Hn(X,F) ' Hn(U,F)

Proof. Here the spectral sequence of the open cover, namely

Epq1 =∏|i|=p

Hq(Ui,F)⇒ Hp+q(X,F),

is especially simple. On the first page, all the rows except the first arezero, by Theorem 28.1. On the second page, the sequence has alreadyconverged, and the nonzero stable groups are the Hn(U,F). By the the-orem on this spectral sequence, these groups form a one term filtration ofHn(X,F), which proves it.

38This happens, for instance, when X is separated.

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Example 28.1. We’ll use the above to compute H∗(P1k,OP1

k)39. We have

a cover of P1 by two affines Spec k[x] and Spec k[y], whose intersection isSpec k[x±], where y corresponds to 1/x on this intersection. The Cechcomplex is

C0(U,O) = k[x]× k[y]→ k[x±] = C1(U,O)

where this map is given by (p(x), q(y)) 7→ p(x)−q(1/x). The kernel of thismap consists of the constants k ' (c, c) ⊂ k[x] × k[y]. It is surjective,so the cokernel is zero. Therefore

Hn(P1k,O) '

k i = 0

0 else

29 Monday, April 2nd: Serre’s Affineness Crite-rion

Today we prove a useful result of Serre that uses cohomology of idealsheaves to test whether a Noetherian scheme is affine. The idea of theproof introduces the point of view that cohomology groups should bethought of as containing “obstructions” to various things.

We begin with a few more examples to review the concepts from lasttime. Recall that the hypotheses of Theorem 28.2 are automatically sat-isfied if X is a scheme over R which is separated. The reason is that fortwo open affines U, V ⊂ X, we have a diagram

U ∩ V //

U ×R V

X∆ // X ×R X

and when X/R is separated, ∆ is a closed immersion, hence so is the toparrow. But a closed immersion is an affine morphism and U×RV is affine,hence so is U ∩ V .

Example 29.1. Let X be the affine line with doubled origin, A1k

∐Gm A1

k.This scheme is not separated, but nonetheless we have a cover by twoaffines, with affine intersection, so the theorem applies.

Example 29.2. If X is affine, then we can take the open cover to consistof just X itself, so Theorem 28.2 applies and Hi = Hi. Moreover, theCech groups are zero in positive degree since our open cover has only oneelement. This isn’t really an application of the theorem, however, sincewe used this fact in the proof; but we see another point of view on whyaffine schemes have no higher cohomology.

Now to the main course:

Theorem 29.1 (Serre). If X is a Noetherian scheme, the following areequivalent:

1. X is affine.

39Hereafter we’ll drop the subscript P1k from the notation of the structure sheaf.

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2. Hi(X,F) = 0 for all quasicoherent sheaves F on X and all i > 0.

3. H1(X,I ) = 0 for all quasicoherent ideal sheaves I on X.40

Proof. All that needs to be shown is that 3 implies 1. For this we use thefollowing two facts

1. (Hartshorne II.2.17) X is affine if and only if there are global sectionsf1, . . . , fr of OX such that for each i, Xfi is affine and the ideal(f1, . . . , fr) = Γ(X,OX).

2. If X is a Noetherian scheme, and U an open subset which containsall the closed points of X, then U = X.

Now we produce the sections fi “one point at a time.” Specifically,pick a closed point P ∈ X, and an affine neighborhood U of P . LetY = X \ U , and give it a scheme structure (the reduced one, say). Wehave the skyscraper sheaf k(P ) at P , and it fits into a short exact sequence

0→ IY ∪P → IY → k(P )→ 0.

This comes from the fact that Y is a closed subscheme of Y ∪ P , so wehave an inclusion IY ∪P → IY . This is an isomorphism away from P ; thecokernel is then just the sheaf k(P ). Taking cohomology gives the longexact sequence

0→ Γ(X,IY ∪P )→ Γ(X,IY )→ k(P )→ 0

where the rightmost term is zero because of our assumption (3) that H1

vanishes for quasicoherent ideal sheaves. Thus the map Γ(X,IY )→ k(P )is surjective, so we can find a global section f of IY which does not vanishat P (namely a preimage of any nonzero element of k(P )). This meansthat P ∈ Xf , and moreover, since f vanishes on Y , we have that Xf ⊆ U .Therefore the points where f vanishes are exactly the points where theimage of f in OX/IY ' OU vanishes, which is just Uf . So Xf is affine.

So we can cover all the closed points of X by these affine sets Xf .Therefore their union is all of X, by fact 2. Again using the Noetherianhypothesis, we see that finitely many such Xf suffice to cover X. Thuswe have produced global sections f1, . . . , fr such that X =

⋃Xfi .

It remains to check that (f1, . . . , fr) = Γ(X,OX). The global sectionsfi define a map

OrX → OX

by sending (a0, . . . , ar) to∑aifi. This map is surjective since we can

check at stalks, and at each point of X, not all fi vanish since the Xficover X. So we obtain an exact sequence

0→ K → OrX → OX → 0

for some subsheaf K of OrX . Looking at the induced long exact sequence

in cohomology, we see that if we can show that H1(X,K ) = 0, then wewill have a surjection Γ(X,OX)r → Γ(X,OX), which will finish the proof.

40(In reply to a question) An example of an ideal sheaf which is not quasicoherent: j!I ⊂OX , where j : U → X is any proper open subset, and I ⊂ OU an ideal sheaf on U .

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The idea is to filter K in such a way that we can apply our assumption3 to the quotients of the filtration. The filtation

OX → . . .Or−2X → Or−1

X → OrX ,

where Or−j has local sections of the form (∗, . . . , ∗, 0, . . . , 0). This givesrise to a filtration of K :

Kr → . . . → K2 → K1 → K

where Ki = Or−iX ∩K . The local sections of Ki/Ki+1 have zeroes in all

but one slot, so they are isomorphic to a subsheaf of OX ' Or−iX /Or−i−1

X .More precisely, we have the following diagram

0

0

kerα

0 // Ki+1//

Ki//

Ki/Ki+1//

α

0

0 // Or−i−1X

//

Or−1X

//

OX // 0

0 // Or−i−1X /Ki+1

//

Or−1X /Ki

0 0

The map on the bottom is injective since the upper left square iscartesian. From the snake lemma, we obtain an exact sequence

0→ kerα→ Or−i−1X /Ki+1 → Or−1

X /Ki → . . .

and since the map Or−i−1X /Ki+1 → Or−1

X /Ki is injective, kerα = 0. Thisshows that Ki/Ki+1 is a subsheaf of OX . We now use this to show thatH1(X,Ki) = 0. The long exact sequence induced by 0→ Ki+1 → Ki →Ki/Ki+1 → 0 gives, in degree 1,

. . .→ H1(X,Ki+1)→ H1(X,Ki)→ H1(X,Ki/Ki+1)→ . . .

Now since Ki/Ki+1 is an ideal sheaf, H1(X,Ki/Ki+1) = 0 by our assump-tion of (3). Assume inductively that H1(X,Ki+1) = 0; then it follows thatH1(X,Ki) = 0. So we get that H1(X,K ) = 0 by decreasing inductionon i. As we mentioned above, this suffices to finish the proof that X isaffine.

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30 Wednesday, April 4th: Cohomology of Pro-jective Space

We begin with a quick warm-up, using the criterion of last time to showthat X = A2

k \ (0, 0) is not affine. By 29.1 it will be enough to finda quasicoherent sheaf F on X for which H1(X,F) 6= 0. Using the opencover

U1 = Spec k[x±, y]

U2 = Spec k[x, y±]

U1 ∩ U2 = Spec k[x±, y±],

we have

H1(X,F) = coker(F(U1)×F(U2)

(a,b)7→a−b−→ F(U1 ∩ U2))

Now we can take F = OX . Then x−1y−1 ∈ OX(U1 ∩ U2) is not in theimage of the difference map, so H1(X,OX) is non-zero (it’s not even finite-dimensional).

Now we turn to the calculation of the cohomology of the sheavesOPr (n). Let A be a ring, S = A[x0, . . . , xr], and set X = PrA = ProjS.We use the notation Γ∗(OX) for the graded ring

⊕n≥0 Γ(X,OX(n)).

Theorem 30.1. (a) There is an isomorphism S'→ Γ∗(OX) induced by

the universal sections s0, . . . , sr ∈ Γ(X,OX(1)).

(b) Hi(X,OX(n)) = 0 for 0 < i < r and i > r.

(c) Hr(X,OX(−r − 1)) ' A41.

(d) For each n, the map

H0(X,OX(n))×Hr(X,OX(−n−r−1))→ Hr(X,OX(−r−1)) ' A

is a perfect pairing42

Remarks. 1. The theorem allows us to compute all cohomology groupsof OX(n) for any n.

2. The pairing of (d) is defined as follows. If s ∈ H0(X,OX(n)), itdefines a map of sheaves OX → OX(n), or equivalently, a mapOX(−n)→ OX . Tensoring with OX(−r − 1) gives a map OX(−n−r − 1) → OX(−r − 1). Taking rth cohomology gives our mapHr(OX(−n− r − 1))→ Hr(X,OX(−r − 1)).

Proof. Let F =⊕

n∈Z OX(n)43. We will compute Hi(X,F). The stan-dard open cover U of PrA satisfies the hypotheses of Theorem 28.2, so wecan use the Cech complex to compute this. Moreover, formation of theCech complex commutes with direct sums, so we have

Hi(X,F) =⊕n∈Z

Hi(X,OX(n)),

41Not canonically.42This means it induces an isomorphism H0(X,OX(n)) ' Hr(X,OX(−n− r − 1))∨.43This sheaf is actually the pushforward of the structure sheaf of X = Ar+1

A \ O alongthe map X → PrA.

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so we will be able to recover the cohomology groups Hi(X,OX(n)) foreach n from our calculation. More precisely, the Cech complex looks like

r⊕i=1

F(Ui)→⊕i<j

F(Uij)→ . . .

Note that for each n, the sections of OX(n) over the open set Ui = D+(xi)are jsut the degree n part of Sxi . Summing over all n gives F(Ui) ' Sxi ,F(Uij) ' Sxixj , etc., so we are interested in the following graded complexof graded S-modules

r⊕i=1

Sxi →⊕i<j

Sxixj → . . .

Now the cohomology groups of this complex are by definitionHi(C•(X,F))).But because the differential in the complex is just localization, it is com-patible with the grading of the individual groups so the degree n piece ofHi(C•(X,F))) is just Hi(X,OX(n)).

We now use the observations to prove the result, in the following order:(a),(c),(d),(b).

We’ve basically already seen (a) earlier. Here’s an example in the caser = 1. then H0 is the kernel of the map

A[x±, y]×A[x, y±]→ A[x±, y±],

which is isomorphic to A[x, y]. For (c), since the Cech complex vanishesin degree r + 1 and larger, we have

Hr(X,F) = coker( r∏k=0

Sx0···xk···xr → Sx0···xr

).

But Sx0···xr is a free A-module with basis xl00 · · ·xlrr | li ∈ Z, so thiscokernel is free with basis xl00 · · ·xlrr | li < 0. Looking at the degree npiece, we see that since each li < 0,

∑li ≤ −(r + 1), so

Hr(X,OX(n)) = 0

whenever n > −r− 1, and for n = −r− 1, the only monomial in the basisis x−1

0 · · ·x−1r , so Hr(X,OX(−r−1)) is a free A-module of rank one. This

proves (c).Now for (d). First, if n < 0, then OX(n) has no global sections, and

since −n− r − 1 > −r − 1, Hr(X,OX(−n− r − 1)) = 0, as was observedin the previous paragraph. So there is nothing to show. When n = 0,H0(X,OX) ' A and Hr(X,OX(−r − 1)) ' A by (c). For each globalsection s ∈ A we get a map

A ' Hr(X,OX(−r − 1))→ Hr(X,OX(−r − 1)) ' A

and this association defines an isomorphism ofH0(X,OX) withHr(X,OX(−r−1))∨ since Hr(X,OX(−r − 1)) is free of rank one.

Now we consider the case when n > 0. Here H0(X,OX(n)) is free withbasis xm0

0 · · ·xmrr |mi ≥ 0,∑mi = n, while Hr(X,OX(−n−r−1)) has

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basis xl00 · · ·xlrr | li < 0,∑li = −n − r − 1. The pairing acts on these

bases by〈xm0

0 · · ·xmrr , xl00 · · ·xlrr 〉 = xm0+l0

0 · · ·xmr+lrr

But in Hr(X,OX(−r − 1)), xm0+l00 · · ·xmr+lr

r is zero unless each mi + liis -1, in which case it is 1. So these form dual bases with respect to thepairing, which is therefore perfect. This proves (d). We will prove (b)next time.

31 Friday, April 6th: Conclusion of Proof andApplications

We will prove (b) by induction on r. Note that the statement for i > ris true, regardless of r, simply because the Cech complex is zero after therth degree. If r = 1, the claim is vacuous.

Note that C•(U,F)xi is the Cech complex of F∣∣Ui

with respect to the

open cover Ui = Ui ∩ Ujrj=1. Therefore since Ui is affine, it is an exactcomplex except in degree 0. This means that for each j, an for all i > 0,Hi(X,F)xj = 0. But to say that this localization is zero is to say that

the S-module Hi(X,F) is annihilated by xj , so to prove (b), it will beenough to show that

·xj : Hi(X,F)→ Hi(X,F)

is injective for each 0 < i < r, 0 ≤ j ≤ r. Now xj ∈ Γ(X,OX(1)) gives amap OX(n− 1) → OX for each n, so it defines a short exact sequence ofgraded S-modules

0→ S(−1)·xj→ S → S/(xj)→ 0.

This gives rise to a short exact sequence of sheaves on Pr

0→ F(−1)→ F → FH → 0,

where H = V+(xj) ' Pr−1A ⊂ PrA. Taking cohomology gives the long exact

sequence

H0(X,F)·xj

// H0(X,F) // H0(H,FH)

ssggggggggggggggggggggggg

H1(X,F) // H1(X,F) // H1(H,FH)

sshhhhhhhhhhhhhhhhhhhhhhhhhhh

......

...

Hr−1(X,F)·xj// Hr−1(X,F)

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First, the map H0(X,F) ' S → H0(H,FH) ' S/(xj) is surjective, sothe map

H1(X,F)·xj−→ H1(X,F)

is injective. Next, we may assume by our induction hypothesis thatHi(H,FH) = 0 for 0 < i < r − 1, so the right hand column consistsof zeroes except in the top row. Therefore each map

Hi(X,F)·xj−→ Hi(X,F)

for i = 2, 3, . . . , r − 1 is injective. This finishes the proof.

Example 31.1. Let F ∈ k[x0, . . . , xr] be a homogeneous polynomial ofdegree d, and consider the hypersurface

X = V+(F )i→ Prk.

We want to calculate Hi(X,OX). The global section F ∈ Γ(Pr,OPr )defines an exact sequence

0→ OPr (−d)·F→ OPr → i∗OX → 0

Recall that since i is a closed immersion we haveHi(X,OX) = Hi(Pr, i∗OX).So the long exact sequence can be written as

H0(Pr,OPr (−d)) // H0(Pr,OPr ) // H0(X,OX)

ssgggggggggggggggggggggg

H1(Pr,OPr (−d)) // H1(Pr,OPr ) // H1(X,OX)

ssgggggggggggggggggggggggggggg

......

...

Hr(Pr,OPr (−d))α // Hr(Pr,OPr ) // Hr(X,OX)

We have already computed the first two columns; the left columnis zero except in degree r, the middle column is zero except in degrees0 and r. Moreover, in degree zero, H0(Pr,OPr ) ' k. Also, we knowHi(X,OX) = 0 in degrees greater than r−1 for dimension reasons. Hencewe have

H0(X,OX) ' k

Hi(X,OX) ' 0 for r 6= 0, r − 1

The only tricky part is to determine Hr−1(X,OX), which is the kernelof the map α : Hr(Pr,OPr (−d)) → Hr(Pr,OPr ). Recall that the globalsection F defines a map

·F : OPr → OPr (d).

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Twisting by −r − 1 and taking global sections we have

H0(Pr,OPr (−r − 1))F−→ H0(Pr,OPr (d− r − 1))

Taking duals gives a map

H0(Pr,OPr (d− r − 1))∨F−→ H0(Pr,OPr (−r − 1))∨.

Applying Serre duality, we get a map Hr(Pr,OPr (−d)) → Hr(Pr,OPr ).Unwinding the definitions, it turns out that this map is in fact α. So

the kernel of α is isomorphic to the kernel of H0(Pr,OPr (−r − 1))F−→

H0(Pr,OPr (d− r − 1)).

Example 31.2. Specializing the previous example, let F be the polyno-mial zy2 − (x3 − axz2 + bz3); this defines an elliptic curve E ⊂ P2

k. Wehave

h0(E,OE) = 1

h1(E,OE) =

(2

2

)= 1

Example 31.3. Recall that the canonical sheaf ωX is the top exteriorpower of Ω1

X ; it’s a line bundle on X. In the case of Pr, we’ve calculatedthat ωPr is the line bundle OPr (−r − 1). For P1, we have ωP1 = Ω1

P1 =OP2(−2); so H1(P1, ωP1) = H1(P1,OP1(−2)).

32 Monday, April 9th: Finite Generation of Co-homology

Today we discuss the question of whether the cohomology groups of acoherent sheaf are finitely generated modules over the base ring. Theanswer is yes in good circumstances.

Theorem 32.1. If A is a Noetherian ring, X/A projective, L is a veryample sheaf on X, and F is any coherent sheaf on X, then

(a) For all i > 0, Hi(X,F) is a finitely generated A-module.

(b) There is an integer n0 such that for all i > 0 and n ≥ n0,

Hi(X,F ⊗L⊗n) = 0.

Remark. Recall that L is very ample if there is an immersion i : X → PrAsuch that i∗OPr (1) ' L . In the following we will write L as OX(1), inorder to use the simpler notation F(n) = F ⊗L⊗n.

The theorem also admits the following generalization.

Theorem 32.2. 1. If A is a Noetherian ring and X is proper over A,then for any coherent sheaf F on X and all i > 0, Hi(X,F) is afinitely generated A-module.

2. If f : X → Y is a proper morphism of locally Noetherian schemes,then for any coherent sheaf F on X and all i > 0, Rif∗F is acoherent sheaf on Y .

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Statement 2 is the most general; this is proved by reducing it to state-ment 1, and then reducing that to the proof of the previous theorem,which as we will now see, involves a reduction to the calculation of thecohomology of projective space.

We will need the following technical lemma

Lemma 32.1. Let F be a coherent sheaf on Pr and let s ∈ Γ(D+(xi)).Then there is an integer m such that sxmi extends to a global section ofΓ(Pr,F(m)), i.e. there is a section s ∈ Γ(Pr,F(m)) such that s

∣∣D+(xi)

=

sxmi ∈ Γ(D+(xi),F(m)).

Proof. Here’s a sketch of the idea. Given coherent F , set G =⊕

n≥0 F(n).By the sheaf property, we have the following two equalizer diagrams, wherethe vertical arrows are the localizations maps

G(X) //

∏j G(D+(xj)) //

∏j<j′ G(D+(xjxj′))

G(D+(xi)) //∏j G(D+(xixj)) //

∏j<j′ G(D+(xixjxj′))

One can lift sections up along the vertical maps by multiplying by asufficiently high power of xi. Use this and a diagram chase to lift s ∈G(D+(xi)) to a section of G(X).

Corollary 32.1. If F is a coherent sheaf on PrA, then F admits a sur-jection E → F , where E is a finite direct sum of sheaves of the formOPr (q).

Proof. If we have generators s1, . . . , st for F over D+(xi), multiply eachone by an appropriate power of xi to get a section of F(m). These sectionsdefine a map Ot

Pr → F(m), where this m is the maximum of the twistsinvolved in lifting the various sj . This map is surjective over D+(xi).Tensoring with OPr (−m) gives a map of sheaves OPr → F which is sur-jective over D+(xi). Do this oover each D+(xi), and then take the directsum to prove the corollary.

Proof of Theorem 32.1. First we reduce to the case X = PrA. Let i : X →PrA be the closed immersion. Since i∗ is exact and has an exact left adjoint,it doesn’t change the cohomology. Also, it preserves coherence, since thiscan be checked locally. If on some open affine, F corresponds to the B/I-module M , then i∗F corresponds to the module MB obtained by basechange along the ring map B → B/I, which is still finitely generatedsince this is a surjective ring map. Thus Hi(X,F) ' Hi(Pr, i∗F), so wemay assume X = Pr.

The result is true for i > r: in that case Hi = 0 since Pr can becovered by r + 1 affines44. We also know the result in the case that Fis OX(n), since we have explicitly calculated Hi(Pr,OX(n)); we also get

44Note - we cannot argue that Hi = 0 for i > r because of dimension reasons, since A isarbitrary, so it is difficult to get our hands on the dimension of Pr.

104

the result when F is a finite direct sum of copies of these since Hi is anadditive functor.

Now for the inductive step: assume that for some fixed i0, and for allcoherent G, Hi(Pr,G) is finitely generated whenever i > i0. We will usethis to show that Hi0(Pr,F) is finitely generated. Now use the lemmaabove to find an E =

⊕i OPr (qi) and a morphism E → F , and let K be

the kernel. Then we have an exact sequence

0→ K → E → F → 0,

which gives a long exact sequence in cohomology in which the followingterms appear:

. . .→ Hi0(Pr, E )α→ Hi0(Pr,F)

β→ Hi0+1(Pr,K )→ . . . .

Note that Hi0(Pr, E ) is finitely generated, as we observed above, and thatHi0+1(Pr,K ) is finitely generated by induction. Since this sequence isexact, we can break off the following short exact sequence

0→ imα→ Hi(Pr,F)→ kerβ → 0,

and both imα and kerβ are finitely generated, since kernels and imagesof finitely generated modules over a Noetherian ring are also finitely gen-erated. This proves (a).

For (b), we only provide a sketch; the crux of the argument is the sameas (a). We will need the following result, which is useful enough to isolateas a separate lemma.

Lemma 32.2 (Projection Formula). If i : (X,OX) → (Y,OY ) is a mor-phism of ringed spaces, F an OX-module, and E a locally free OY -moduleof finite rank, then

i∗(F ⊗OX i∗E ) ' (i∗F)⊗OY E

A similar argument was made for this statement in the special casewhen E is a line bundle in an earlier lecture. In our case, taking Y = PrA,and E = OPr (1), we have

i∗F(n) = i∗(F ⊗OX OX(1)⊗n) = (i∗F)⊗OPr OPr (1)⊗n = (i∗F)(n),

which shows that we can assume that X = PrA. Now, the statementof (b) is true when i > r, as in (a). We also know that for F =OX(n), Hi(X,OX(n) = 0 for 0 < i < r and Hr(X,OX(n)) is dual toH0(X,OX(−n − r − 1)), which shows that the statement holds whenF = OX(n); it holds for finite direct sums of these again by additivity ofHi.

For the inductive step fix i0, and assume that for any G and any i > i0,there is an integer n0 such that Hi(X,G(n)) = 0 if n ≥ n0. Then oneshows that there is an m0 such that Hi0(X,F(m)) = 0 for m ≥ m0, byarguments similar to those of (a).

Note in conclusion that the integer n0 of (b) in principle depends on allof X, A, L , and F , but the essential dependence is on F . In particular,we can always twist F down to force larger values of n0.

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33 Wednesday, April 11th: The Hilbert Polyno-mial

Fix a projective scheme X over a Noetherian ring A. Let OX(1) be avery ample sheaf. For a coherent sheaf F on X, we define the Eulercharacteristic of F to be the integer

χ(F) =∑i

(−1)ihi(X,F).

Note that by the results of the previous lecture, each hi is finite, and thereare only finitely many nonzero terms, so the sum makes sense.

For such F , we also define the Hilbert polynomial of F to be thefunction pF : Z→ Z given by

pF (n) = χ(F(n))

Note that for sufficiently large n, we have pF (n) = h0(X,F(n)), againusing the results of the previous lecture. To justify calling this function apolynomial, we prove the following

Theorem 33.1. pF is a polynomial function.

Proof. First we reduce to the case whereX = PrA. Pick a closed immersioni : X → PrA such that i∗OPr ' OX(1). This can be done since OX(1) isvery ample. Then as we saw last time, the projection formula implies(i∗F)(n) = i∗(F(n)), so pi∗F = pF , and we may replace X by PrA.

Now pick a hyperplane H ⊂ X = Pr; this gives us an exact sequence

0→ OX(−1)→ OX → OH → 0

where we are omitting the pushforward of OH from the notation. Sincetensoring with F is right exact we get another exact sequence

0→ K → F(−1)→ F → F∣∣H→ 0,

where K is defined as the kernel of F(−1)→ F . Let I denote the imageof the map F(−1)→ F . We can break this sequence into two short exactsequences

0→ K → F(−1)→ I → 0

0→ I → F → F∣∣H→ 0

Since χ is additive on short exact sequences (we proved this for curves afew weeks ago), we get two equations

pK + pI = pF(−1)

pI + pF|H = pF .

The difference of these two equations is the equality

pF|H − pK = pF − pF(−1).

By induction, we may assume pF|H is a polynomial. Also, the sheaf K issupported on H, and a coherent sheaf supported on a closed subscheme

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can be realized as the pushforward of a coherent sheaf on that closed sub-scheme (see lemma below for details). Thus we may also assume by induc-tion that pK is a polynomial. Finally, note that pF(−1)(n) = pF (n − 1).Thus the equation above shows that pF (n) − pF (n − 1) is a polynomialfunction. In fact (check) this implies that pF itself is a polynomial func-tion.

To explain our treatment of F∣∣H

in the above, we have the followinglemma and its corollary.

Lemma 33.1. Let X be a Noetherian scheme, F a coherent sheaf on Xwith set-theoretic support Z ⊂ X. Give Z the reduced scheme structure,and let I ⊂ OX be the ideal sheaf of Z. Then there exists r such that I r

annihilates F .

Corollary 33.1. F is isomorphic to the pushforward of a coherent sheafon V (I r).

Proof. The question of whether, for given r, I r annihilates F can bechecked locally. Having produced such r locally, we may take a finitecover of X by open affines and use the maximum of all the r obtainedin our local constructions. Thus we may as well let X = SpecA. ForI ⊂ A an ideal and F a finitely generated A-module, the condition that Fis supported on V (I) means that, if g1, . . . , gs are generators for I, then

F is 0 on D(g1) ∪ . . . D(gs), i.e., Fgi = 0 for each i. Let m1, . . . ,mt begenerators for F . Since Fgi = 0, there is an ni such that gnii mj = 0 forall j. Let n be the maximum of these ni as i varies from 1 to s. Thusgni mj = 0 for all i, j.

We can choose r big enough so that, for any monomial ga11 · · · gass ofdegree r, at least one of the ai is at least n. Then for this r, any monomialof degree r annihilates F , so IrF = 0. This proves the lemma.

To see the corollary, observe that if IrF = 0, then F has the struc-ture of an A/Ir-module, so the sheaf F on SpecA can be viewed as the

pushforward of a sheaf F ′ on SpecA/Ir along the ring map A → A/Ir.Globalizing, we have a closd immersion i : V (Ir) → X. The action ofi−1OX on i−1F factors through OV (Ir), and i−1F is still coherent asan OV (Ir)-module. Moreover, the map F → i∗i

−1F is an isomorphismof OX modules, so we can view F as the pushforward of the coherentOV (Ir)-module i−1F .

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