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Gradings of the affine line and its Quot scheme

Gustav Sædén Stå[email protected]

2011

Abstract

For a field k and a grading of the polynomial ring k[t] with Hilbert function h, weconsider the Quot functor Quot

h

V, where V =

�d

i=1 k[t] is a finitely generated andfree k[t]-module. The Quot functor parametrizes, for any k-algebra B, homogeneousB[t]-submodules N ⊆ B ⊗k V such that the graded components of the quotient(B ⊗k V )/N are locally free B-modules of rank given by h. We find that it is locallyrepresentable by a polynomial ring over k in a finite number of variables. Finally, weshow that there is a scheme that represents the Quot functor that is both smoothand irreducible.

CONTENTS G. Sædén Ståhl

Contents

1 Introduction 3

1.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 What we will do . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 The Hilbert scheme of the affine line 6

2.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Grading by G = Z . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.3 Grading by G = Z/nZ . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3 The Grassmann scheme 17

3.1 The Grassmann functor . . . . . . . . . . . . . . . . . . . . . . . . . 173.2 The graded Grassmannian . . . . . . . . . . . . . . . . . . . . . . . . 20

4 The Quot scheme of the affine line 21

4.1 Structure theorem of finitely generated graded modules over a gradedprincipal ideal domain . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4.2 The relative Quot scheme . . . . . . . . . . . . . . . . . . . . . . . . 244.2.1 Grading by G = Z . . . . . . . . . . . . . . . . . . . . . . . . 294.2.2 Grading by G = Zn . . . . . . . . . . . . . . . . . . . . . . . . 30

4.3 The Quot scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

References 38

2

G. Sædén Ståhl 1 INTRODUCTION

1 Introduction

In [6] the authors, Haiman and Sturmfels, introduce the multigraded Hilbert schemeby a general construction that, for a ring k, takes a k-algebra B to a set of certainsubmodules of a graded polynomial ring over B. This construction is also applicableto a generalization of the Hilbert scheme, namely the Quot scheme, which was firstintroduced by Grothendieck in the context of so called parameter spaces.

1.1 BackgroundFirst we will give some motivation to why we study these objects. When we parametrizea line segment in the real plane we assign, to each coordinate on the line, a real num-ber t in some interval [a, b]. To generalize this concept we see that we, basically, havea set S containing the structures that we want to parametrize (the set of coordinates)and then we want to find a set M (the interval [a, b]) such that any element in M

corresponds to an element in S.The sets M might be a bit complicated and hard to find however. There are two

main solutions to this problem. The first is to find local parametrizations, wherewe only consider some specific parts of the structure, one at a time, that might beeasier to parametrize, such that they together cover our whole structure. The otherpossibility is simply to find this more complicated set M . Both these possibilities canbe applied in the following example.Example 1.1. Consider the unit circle x

2+ y

2= 1 in the real plane R2. This can, if

we fix a point P on the circle, be parametrized in two similar ways by looking at linesthrough P .

(i) The first gives a local parametrization where we remove the point P from ourstructure. For simplicity we choose P = (−1, 0) and consider lines y = tx + t.These lines will intersect the circle in two points, the point P and a pointPt =

�1−t

2

1+t2,

2t1+t2

�. Conversely, any point Q �= P on the circle will correspond

to some line of this form for some t. Thus, we have a local parametrizationof the circle given by the bijection above between R and the circle minus thepoint P . If we instead were to remove another point, say P

�= (1, 0) we would

get another local parametrization of the circle which together with our previousparametrization covers the entire circle.

(ii) To get a global parametrization directly we would need to be able to handle thepoint P as well. Any line through a fixed point is uniquely defined by its slope

3

1 INTRODUCTION G. Sædén Ståhl

and the line that would correspond to the point P would be the line where theslope is infinite, t = ∞. It is precisely for these reasons that we have the projec-tive line that parametrizes the lines in the plane that pass through the origin,even the one with infinite slope. Therefore, we have a, global, parametrization ofthe unit circle, given by the bijection above, between the projective line over R,P1R, and the unit circle.

Here we got an example of why we introduce the projective line, and more gener-ally the projective space Pn that parametrizes lines in (n + 1)-dimensions that passthrough the origin. We can then go one step further and generalize the projectivespace by introducing the Grassmannian, often denoted Gr(r, n). This is a space thatparametrizes, for any n-dimensional vector space, the r-dimensional subspaces, simi-larly to the projective space that parametrizes the 1-dimensional subspaces. Indeed,we have that Gr(1, n) = Pn−1.Example 1.2. In an n-dimensional vector space we have that the (n− 1)-dimensionalsubspaces are uniquely defined by their normal vectors, and vice versa, from which itfollows that Gr(n− 1, n) = Gr(1, n) = Pn−1. Thus, the first example of a Grassman-nian that is not some projective n-space is Gr(2, 4) which consists of the 2-dimensionalsubspaces, i.e. the planes containing the origin, in a 4-dimensional vector space.

The Grassmannian is usually considered over vector spaces but can also be definedover arbitrary modules where we consider consider submodules such that the inducedquotient module is locally free of rank r. Then we can consider the case when themodule that we are interested in has a grading given by some abelian group and wethen get the notion of the graded Grassmannian.

The Hilbert and Quot schemes are further generalizations of this graded Grass-mannian where we also require the submodules to be modules over a graded polyno-mial ring.

The concept of parametrization leads us to the theory of representability. Acovariant functor F , from a locally small category C to the category of sets, is saidto be representable if it is naturally isomorphic to a functor of the form Hom(A,−)

for some A ∈ C. Similarly, a contravariant functor F is called representable if it isnaturally isomorphic to a functor Hom(−, A). We will in this paper consider the casewhere we have functors from the category of k-algebras for some field k to the categoryof sets. The aim is then to find some k-algebra A that gives us a representation ofour functor. Similarly to what we saw in Example 1.1 it is sometimes hard to find theA that represents the functor, sometimes it does not even exist, but we may then beable to find local representations by considering simpler subfunctors of F . Also, byinstead considering F as a contravariant functor from the category of schemes over k

4

G. Sædén Ståhl 1 INTRODUCTION

to sets it is, sometimes, possible to find a scheme that represents our functor. Thatis because the category of schemes over k contains the category of affine schemes overk that are equivalent to the category of k-algebras. All the parameter spaces that wementioned above can be defined as structures that represents some specific functors.

1.2 What we will doWe will in Section 2 start by considering the Hilbert scheme of the affine line Spec(k[t])for some field k and then, in Section 4, continue by considering a similar constructionfor the Quot scheme of the affine line. Since the functor Spec gives a one-to-onecorrespondence between the category of commutative rings and the category of affineschemes, we will be able to work with commutative rings and the theory involvingthem instead of the more theory demanding scheme theory. For a field k we willwith a k-algebra mean what is usually referred to as a commutative unital associativealgebra over k. Our method will be to define a specific functor and then show that itis representable, at least locally, by some k-algebra or, equivalently, an affine schemeover k. There are two categories that we will mostly consider, so we give these thefollowing notation.

Notation. The category of sets, where the morphisms are maps between the sets,will be denoted Set and the category of k-algebras, where the morphisms are ringhomomorphisms, will be denoted Algk.

First we will consider a grading on the polynomial ring k[t] =�

g∈G Sg by someabelian group G along with a Hilbert function h : G → N. For any k-algebra B,the Hilbert functor, Hilb

h: Algk → Set, parametrizes all homogeneous ideals I ⊆

B ⊗k k[t] = B[t] such that the graded components of the quotient B[t]/I are locallyfree of finite rank h(g) for all g ∈ G, c.f. Section 2. When G = Z/nZ then thisfunctor will be represented by the k-algebra k[a1, ..., as] together with the universalelement �

tm(t

sn+ a1t

(s−1)n+ ...+ as)

�

where the integers m and s are determined by the Hilbert function h.In Section 3 we will then review the theory of the Grassmannian via the Grassmann

functor. For some k-vector space V it is defined, similarly to the Hilbert functor, bytaking a k-algebra B to the set of submodules N ⊆ B ⊗k V such that the quotient(B⊗kV )/N is locally free of some given rank. We will show that this functor is locallyrepresentable by a polynomial ring. We will also look at the graded Grassmannianwhere we add a graded structure on the k-vector space V by some abelian group G.

5

2 THE HILBERT SCHEME OF THE AFFINE LINE G. Sædén Ståhl

Finally, we look at a generalization of the Hilbert functor by considering, forV =

�d

i=1 k[t], the Quot functor, Quoth

V: Algk → Set, that parametrizes the ho-

mogeneous B[t]-submodules N ⊆ B ⊗k V such that the graded components of thequotient (B ⊗k V )/N are locally free of rank h(g) for all g ∈ G. We will show, inSection 4, that the Quot functor is locally representable by polynomial rings overk and universal elements of the same form as in the case with the Hilbert functor.Finally, we show that the Quot functor is, globally, represented by a scheme that isboth smooth and irreducible.

I would here like to express my deepest gratitude to my two supervisors, in noparticular order other than the alphabetical one, Mats Boij and Roy Skjelnes for alltheir guidance and support.

2 The Hilbert scheme of the affine line

2.1 PreliminariesLet k be a field. When one determines a grading of the ring S = k[t] by an abeliangroup G, i.e. a decomposition k[t] =

�g∈G Sg where Sg · Sh ⊆ Sg+h for any g, h ∈ G,

this is equivalent to consider a semi-group homomorphism deg : {1, t, t2, ...} → G.This will be determined by deg(t) and we can therefore assume, without loss ofgenerality, that G is cyclic, which we will do from now on. We then have two cases,G = Z or G = Z/nZ for some n ∈ Z+. The vector space decomposition

k[t] =

∞�

r=0

k · tr

has the property that (k · tr1) ·(k · tr2) ⊆ k · tr1+r2 so this decomposition gives a gradingof k[t] with G = Z by letting each direct summand be a graded component. Anothervector space decomposistion of k[t] is

k[t] =

n−1�

r=0

trk[t

n],

and since this also has the property that (tr1k[t

n]) · (tr2k[tn]) ⊆ t

r1+r2k[tn] we get a

grading of k[t] with G = Z/nZ by the vector space decomposition when letting eachdirect summand be a graded component.

6

G. Sædén Ståhl 2 THE HILBERT SCHEME OF THE AFFINE LINE

Definition 2.1. For any graded ring S =�

g∈G Sg we define an element s ∈ S to bea homogeneous element if s ∈ Sg for some g ∈ G. A homogeneous ideal is an idealthat is generated by homogeneous elements.

If I ⊆�

g∈G Sg is a homogeneous ideal, then S/I is graded by G with

S/I =

�

g∈G

(Sg + I)/I ∼=�

g∈G

Sg/(Sg ∩ I) =

�

g∈G

Sg/Ig.

Definition 2.2. A homogeneous ideal I of a graded k-algebra S, graded by G, S =�g∈G Sg, is called admissible if Sg/Ig is of finite dimension over k for all g ∈ G. An

admissible ideal has Hilbert function hI : G → N defined by hI(g) = dimk(Sg/Ig).

When S = k[t] and h : G → N is any function then we want to parametrizeall admissible ideals I with Hilbert function hI = h. Since the tensor product isdistributive over direct sums we have, for any k-algebra B, an induced grading B ⊗k

S = B ⊗k

��g∈G Sg

�=

�g∈G (B ⊗k Sg), where each graded component B ⊗k Sg is

considered as a B-module. Note that B ⊗k k[t] = B[t].

Definition 2.3. If A is a ring, then the A-module M is locally free of rank m if thereexist elements r1, ..., rd ∈ A for some d with (r1, ..., rd) = A such that Mri is a freeAri-module of rank m for every i.

Remark 2.4. 1. Note that Definition 2.3 is equivalent to saying that, for any p ∈Spec(A), there is some f ∈ A \ p such that Mf is a free Af -module of rank m.

2. Definition 2.3 implies that an A-module M , that is locally free of rank m, hasthe property that, for any p ∈ Spec(A), Mp is a locally free Ap-module of rank m.However, we do not in general have an equivalence here. If we have, for any p ∈Spec(A), that Mp is a locally free Ap-module this does not imply that M is locallyfree, unless we pose some extra requirements, e.g. that M is finitely presented or thatthe rank is constant for all prime ideals p, see [2, Section II.5.2, Theorem 1].

3. A free A-module M is always locally free since (1) = A and when localizing atthe element 1 we get that M1 = M is a free A1-module, and A1 = A.

We will now with the following results prove some properties of locally free mod-ules.

Lemma 2.5. Let G be an abelian group and h : G → N a function such that�g∈G h(g) < ∞. If M =

�g∈G Mg is graded module over a ring A and if each graded

component Mg is locally free with rank h(g), then M is a locally free A-module of rank�g∈G h(g).

7


Proof. Take some p ∈ Spec(A) and consider M ⊗A Ap. Since any graded componentof M is locally free of rank h(g) we have that Mg ⊗A Ap = A

h(g)p . This implies that

M ⊗A Ap =

��

g∈G

Mg

�⊗A Ap =

�

g∈G

(Mg ⊗A Ap) =

�

g∈G

Ah(g)p =

�g∈G h(g)�

i=1

Ap.

Hence, the localization of M at any prime ideal p ∈ Spec(A) is a locally free Ap-module of rank

�g∈G h(g) and since that sum is finite and independent of the prime

p we have, from Remark 2.4, that M is a locally free A-module of rank�

g∈G h(g).

Lemma 2.6. Let A,B be rings and ϕ : A → B a ring homomorphism. If M is a

locally free A-module of rank m then M ⊗A B is a locally free B-module of rank m.

Proof. Take a prime ideal p ⊂ B. Then q = ϕ−1(p) ⊂ A is a prime ideal. Therefore,

since M is locally free as an A-module it follows that there is some f ∈ A \ q suchthat Mf = M ⊗A Af is a free Af -module of rank m. Letting g = ϕ(f) we get aninduced homomorphism Af → Bg. Then

(M ⊗A B)g = (M ⊗A B)⊗B Bg = M ⊗A (B ⊗B Bg) = M ⊗A Bg.

Furthermore, we trivially have Bg = Af ⊗AfBg which gives us that

(M ⊗A B)g = M ⊗A Bg = M ⊗A (Af ⊗AfBg) = (M ⊗A Af )⊗Af

Bg = Mf ⊗AfBg

which is a free Bg-module of rank m, since

Mf ⊗AfBg =

�m�

i=1

Af

�⊗Af

Bg =

m�

i=1

(Af ⊗AfBg) =

m�

i=1

Bg.

Hence, for any prime ideal p ⊂ B there exists some g ∈ B \ p such that (M ⊗A B)g

is a free Bg-module of rank m. Thus M ⊗A B is a locally free B-module.

Lemma 2.7. If R is a ring, M,N are R-modules and L ⊆ M a submodule, then

(M/L)⊗R N = (M ⊗R N)/im(L⊗R N),

where im(L⊗R N) denotes the image of the map incl⊗ id : L⊗R N → M ⊗R N .

8


Proof. This is just the right exactness property of the tensor product. Indeed, considerthe short exact sequence

0 → L → M → M/L → 0.

This turns into a right exact sequence when we tensor it with N ,

L⊗R N → M ⊗R N → (M/L)⊗R N → 0

and then we can consider the short exact sequence

0 → im(L⊗R N) → M ⊗R N → (M/L)⊗R N → 0.

Hence we get an isomorphism

(M/L)⊗R N ∼= (M ⊗R N)/im(L⊗R N).

Fix a grading of S = k[t] by a group G and consider a Hilbert function h : G → N.Then, for any k-algebra B, we define the set

Hilbh(B) =

I ⊆ B[x] :

I is homogeneous and theB-module (B ⊗k Sg)/Ig is locallyfree of finite rank h(g) for all g ∈ G

. (1)

Proposition 2.8. Let B1 and B2 be k-algebras and ϕ : B1 → B2 a ring homomor-

phism. Then there is an induced homomorphism (ϕ ⊗ id) : B1 ⊗k k[t] → B2 ⊗k k[t].

For any I ∈ Hilbh(B1) we have that

im(I ⊗B1 B2) = (incl⊗ id)(I ⊗B1 B2) ∈ Hilbh(B2).

Proof. Take an I ∈ Hilbh(B1), I ⊆ B1[t], and consider the ideal im(I ⊗B1 B2) ⊆

B2 ⊗k k[t] = B2[t]. Since the ideal is generated by homogeneous elements it followsthat the ideal is homogeneous. By Lemma 2.7 we have that

(B1[t]/I)⊗B1 B2 = (B1[t]⊗B1 B2)/im(I ⊗B1 B2) = B2[t]/im(I ⊗B1 B2)

and that each graded component of this module is a locally free B2-module of rankh(g) follows from Lemma 2.6. Hence im(I⊗B1B2) = (incl⊗ id)(I⊗B1B2) ∈ Hilb

h(B2).

9


Definition 2.9. The Hilbert functor Hilbh: Algk → Set is defined, for any k-algebra

B, by Hilbh(B) given by (1), and for any ring homomorphism ϕ : B1 → B2, for some

k-algebras B1, B2, by Hilbh(ϕ)(I) = im(I ⊗B1 B2) = (incl⊗ id)(I ⊗B1 B2).

Definition 2.10. If A and B are k-algebras, a k-algebra homomorphism ϕ : A → B

is called an B-valued point of A. The set of all such points is, naturally, denotedHomk(A,B).

Our aim is to describe the functor Hilbh and we will do this by finding a corre-

spondence between the elements of Hilbh(B) and the B-valued points of A for some

k-algebra A. This gives a natural explanation for the following definition.Definition 2.11. A functor F from the category of k-algebras to the category of sets,F : Algk → Set, is representable if there is a natural isomorphism Φ : Homk(A,−) →F from the functor Homk(A,−) : Algk → Set for some k-algebra A. The pair (A,Φ)

is called the representation of the functor F .Remark 2.12. By Yoneda’s Lemma, see e.g. [4, Lemma VI-1], we have that the setof natural transformations Φ : Homk(A,−) → F are in a one-to-one correspondencewith the elements of F (A). The correspondence is given, for any natural transforma-tion Φ : Homk(A,−) → F , by ΦA(idA) = a ∈ F (A) and conversely, for any a ∈ F (A)

we get a natural transformation Φ : Homk(A,−) → F by ΦB(ϕ) = (Fϕ)(a) for anyϕ ∈ Homk(A,B). Note that the natural transformation Φ that is induced by anelement a ∈ F (A) need not be a natural isomorphism. That is the case if and onlyif a has the property that, for any B and any b ∈ F (B), there is a unique morphismϕ : A → B such that (Fϕ)(a) = b. If a ∈ F (A) has this property then it is called auniversal element of F . Thus, if we have a representation of F given by (A,Φ) wecan exchange Φ for a = ΦA(idA) and call (A, a) a representation of F .

2.2 Grading by G = ZA homogeneous admissible ideal I ⊆

�∞r=0 k · tr = k[t] is, since k[t] is a P.I.D.,

generated by tm for some m, i.e. I = (t

m). We have then that

Ir = (tm) ∩ k · tr =

�0 0 ≤ r < m,

k · tr r ≥ m,

which implies that k[t]/I =�∞

r=0 k · tr/Ir = k⊕ k · t⊕ ...⊕ k · tm−1 and thus, the onlypossible Hilbert functions when one has a grading by G = Z is of the form

hI(r) =

�1 0 ≤ r < m,

0 r ≥ m, r < 0.

10


Let h : Z → N be the Hilbert function of some homogeneous admissible ideal I ⊆ k[t].With the grading from G = Z we have that the graded components of k[t] are of theform k · tr for r ∈ N and therefore the induced grading on B[t], for any k-algebra B,is B[t] =

�∞r=0 (B ⊗k (k · tr)) =

�∞r=0 B · tr. Thus, we want to study the set

Hilbh(B) =

I ⊆ B[t] :

I is a homogeneous ideal andB · tr/Ir is locally free over B andof finite rank h(r) for all r ∈ Z

.

Proposition 2.13. A homogeneous ideal I ⊆�∞

r=0 B · tr such that B · tr/Ir is locally

free of rank

h(r) =

�1 0 ≤ r < m,

0 r ≥ m, r < 0,

for all r ∈ Z is generated by the element tm.

Proof. For all r ≥ m we must have, since h(r) = 0, that B · tr/Ir = 0, i.e. Ir =

I ∩ (B · tr) = B · tr, which implies that tm ·B[t] ⊆ I. Suppose that I is also generatedby some other homogenous element b·tr for some r < m and b �= 0. Then b·B ·tr ⊆ Ir.If b is a unit in B then B · tr/Ir ⊆ B · tr/(b ·B · tr) = B · tr/(B · tr) = 0 and since 0 isnot a B-module of rank h(r) = 1 we have a contradiction. On the other hand, if b isa non-unit, then we get that the B-module B · tr/Ir is not locally free since it is notfaithful, for any possible basis element e we always have be = 0 even though b �= 0,also a contradiction. Hence I = t

m · B[t].

Thus, we have that Hilbh(B) = {tmB[t]}. We can now prove the important result

of this section.

Theorem 2.14. Let S = k[t], graded by the abelian group G = Z, i.e. k[t] =�∞r=0 k · tr, and fix a Hilbert function h : Z → N defined by

h(r) =

�1 0 ≤ r < m,

0 r ≥ m, r < 0.

Then the functor Hilbh

is represented by the k-algebra H along with the universal

element tmH[t] where H = k.

Proof. Take a k-algebra B. Then it follows directly from Proposition 2.13 that thereis a one-to-one correspondence between the sets Hilbh

(B) and Homk(H,B) since both

11


sets only contains one element, Homk(k,B) = {inclusion : k → B} and Hilbh(B) =

{tmB[t]}. The one-to-one correspondence, given by, say, ΦB : Homk(H,B) → Hilbh(B),

is clearly functorial in B which means that we have a representation of the functorHilb

h given by (H,Φ) with H = k and Φ defined by Φ(B) = ΦB. By Remark 2.12we have that Φ gives us an element ΦH(idH) = t

mH[t] ∈ Hilb

h(H) and we thus have

a representation of Hilbh given by the pair�H, t

mH[t]

�=

�k, (t

m)�.

2.3 Grading by G = Z/nZNow we consider the case G = Z/nZ = Zn. A homogeneous admissible ideal I ⊆�

n−1r0

trk[t

n] = k[t] is then generated by homogeneous elements of the form t

rp(t

n)

for different r ∈ Zn and polynomials p(t) ∈ k[t]. Since k[t] is Noetherian it followsfrom [9, Corollary 1] that I is generated by a finite number of homogeneous elements,tr1p1(t

n), ..., t

rdpd(tn). Then, since k[t] has the Euclidean algorithm, it follows that I

is generated by the greatest common divisor of the homogeneous generators whichmust be a polynomial of the form q(t) = t

m(t

sn+a1t

(s−1)n+ ...+as) for some m ∈ Zn,

s ∈ N and elements a1, ..., as ∈ k. Hence any homogeneous ideal I ⊆ k[t] is generatedby a single homogeneous element. We will now show that there is only one possibleform that the Hilbert function of an admissible homogeneous ideal can take.

Proposition 2.15. If I ⊆�

n−1r=0 t

rk[t

n] is a homogeneous admissible ideal, I =

(tmp0(t

n)) where p0(t

n) = t

sn+ a1t

(s−1)n+ ... + as, and hI : Zn → N its associated

Hilbert function, then

hI(r) =

�s+ 1 0 ≤ r < m,

s m ≤ r < n.

Proof. We want to find the dimension of the k-vector space trk[t

n]/Ir for all r, where

Ir = (trk[t

n])∩ I. In order to do this we start by describing Ir for all r ∈ Z/nZ. Take

some trq(t

n) ∈ t

rk[t

n]. Then t

rq(t

n) ∈ I if and only if trq(tn) = α(t)t

mp0(t

n) for some

α(t) ∈ k[t]. We now get two cases.

Case r < m. Then we get q(tn) = α(t)t

m−rp0(t

n), i.e. α(t)t

m−r ∈ k[tn] which

implies that α(t) must be of the form α(t) = tn−(m−r)

α0(tn) for some α0(t

n) ∈ k[t

n].

Thus we have that q(tn) = α0(tn)t

np0(t

n). On the other hand, if q(tn) = β(t

n)t

np0(t

n)

for some β(tn) ∈ k[t

n] then we have that

trq(t

n) = t

r(β(t

n)t

np0(t

n)) =

�tn+r−m

β(tn)�tmp0(t

n) ∈ t

mk[t

n].

Hence Ir = {trq(tn) ∈ Sr : tnp0(t

n) | q(tn)} = t

np0(t

n)Sr.

12


Case r ≥ m. Then we have that tr−m

q(tn) = α(t)p0(t

n), i.e. α(t) ∈ t

r−mk[t

n].

Thus we have that α(t) = tr−m

α0(tn) for some α0(t

n) ∈ k[t

n]. Then we have that

q(tn) = α0(t

n)p0(t

n). On the other hand, if q(tn) = β(t

n)p0(t

n) for some β(t

n) ∈ k[t

n]

thentrq(t

n) = t

r�β(t

n)p0(t

n)�=

�tr−m

β(tn)�tmp0(t

n) ∈ t

mk[t

n].

Hence Ir = {trq(tn) ∈ Sr : p0(tn) | q(tn)} = p0(t

n)Sr.

Thus we have that

hI(r) = dimk(Sr/Ir) =

�dimk(Sr/(t

np0(t

n)Sr)) = s+ 1 0 ≤ r < m,

dimk(Sr/(p0(tn)Sr)) = s m ≤ r < n.

When k[t] is graded by Z/nZ we have that its graded components are of the formtrk[t

n] from which it follows, for any k-algebra B, that we have the induced grading

B[t] =�

n−1r=0 (B ⊗k t

rk[t

n]) =

�n−1r=0 t

rB[t

n]. Hence, we want to consider the set

Hilbh(B) =

I ⊆ B[t] :

I is an homogeneous ideal such thatthe B-module t

rB[t

n]/Ir is locally free

of finite rank h(r) for all r ∈ Zn

.

First of all, we show that if the graded components of B[t]/I are locally free of finiterank then they are actually free.

Proposition 2.16. For any r ∈ Zn, if the B-module trB[t

n]/Ir is locally free of rank

h(r) = s, then trB[t

n]/Ir is free of rank s (globally) with basis t

r, t

r+n, ..., t

r+(s−1)n.

Proof. Since M = trB[t

n]/Ir is locally free we have, for any p ∈ Spec(B), that the

Bp-module Mp = (trB[t

n]/Ir)p is free of rank s. We will first determine a basis for

this module. Since Bp is a local ring, with maximal ideal m = pBp, we can constructits fraction field κ = Bp/m, and we have that Mp/mMp is a vector space of dimensions over κ. By construction we have that the quotient classes of the elements {tr+ni}i∈Nin the vector space Mp/mMp generates the vector space over κ.

If the quotient classes of the elements tr, tr+n, ..., t

r+(s−1)n would not be linearly in-dependent over κ this would imply that we could reduce our generating set {tr+ni}i∈Nto a set consisting of less than s elements, a contradiction. Hence it is clear thatthe s elements t

r, t

r+n, ..., t

r+(s−1)n are linearly independent over κ and thus forma basis. From Nakayama’s lemma [1, Proposition 2.8] we get that the elements

13


tr, t

r+n, ..., t

r+(s−1)n form a minimal generating set of Mp, and since the module isfree it follows that they are a basis. This is true for any prime p and thus it follows,[8, Corollary 1.2], that tr, tr+n

, ..., tr+(s−1)n is a generating set for M , and thus form a

basis with s elements.

In the following two propositions we will first show, in Proposition 2.17, that ifthe graded components of B[t]/I are locally free of finite rank then Ir is principal forany r ∈ Z/nZ. After that we will use this result to show, in Proposition 2.18, that Iis, in fact, principal.

Proposition 2.17. For any r ∈ Zn, if the B-module trB[t

n]/Ir is free of rank s, then

there is a unique monic polynomial G(t) = tr(t

sn+ b1t

(s−1)n+ ...+ bs) ∈ t

rB[t

n] such

that Ir = (G(t)).

Proof. Since trB[t

n]/Ir is free of rank s we have a basis consisting of the images of

the elements tr, tr+n, ..., t

r+(s−1)n by the quotient map, otherwise we could reduce thegenerating set {tr+ni}i∈N to a set consisting of less than s elements, a contradiction.We use the notation that we write the image of an element x by the quotient map as xitself. This means that tr+sn can be written, in a unique way, as a linear combinationof these elements, i.e. tr+sn

= b1tr+(s−1)n

+ ...+ bstr.

Hence the polynomial tr+sn − b1t

r+(s−1)n − ... − bstr ∈ t

rB[t

n] is contained in

Ir, i.e. (G(t)) ⊆ Ir. The uniqueness of G(t) follows from the uniqueness of thelinear combination. To show the other inclusion we note that the canonical maptrB[t

n]/(G(t)) → t

rB[t

n]/Ir is a surjection between two free modules of the same

rank, so it must be an isomorphism. Hence Ir = (G(t)).

Proposition 2.18. For any k-algebra B we have that the set Hilbh(B), where

h(r) =

�s+ 1 0 ≤ r < m

s m ≤ r < n,

is in a one-to-one correspondence with the set of monic polynomials of the form

tm(t

sn+ b1t

(s−1)n+ ...+ bs).

Proof. Take an I ∈ Hilbh(B). Since the graded components of B[t]/I are locally

free B-modules it follows from Proposition 2.16 that the graded components are freeB-modules. Furthermore, from Proposition 2.17 we have that each Ir is generated bya unique polynomial pr(t) = t

r(t

sn+ b1,rt

(s−1)n+ ... + bs,r) for each r. Now consider

the case r = m. Then we have that

pm(t) = tm(t

sn+ b1,mt

(s−1)n+ ...+ bs,m) = t

mp(t

n)

14


generates Im. It follows that td · pm(t) ∈ t

m+dk[t

n] for any d ∈ Zn so we have that

pm+1(t) = tpm(t) = tm+1

p(tn) ∈ (t

mp(t

n))

...pn−1(t) = t

n−1−mpm(t) = t

n−1p(t

n) ∈ (t

mp(t

n))

p0(t) = tn−m

pm(t) = tnp(t

n) ∈ (t

mp(t

n))

...pm−1(t) = t

n−1pm(t) = t

m−1tnp(t

n) ∈ (t

mp(t

n)).

It follows that I = (pm(t)) = (tmp(t

n)).

Conversely, if we have a monic polynomial on the form tm(t

sn+ b1t

(s−1)n+ ...+ bs)

then it is clear that the ideal generated by it will be an element in Hilbh(B).

We are now ready to prove the final, and important, results of this section.

Theorem 2.19. Let S = k[t], graded by the abelian group G = Z/nZ and take a

Hilbert function

h(r) =

�s+ 1 0 ≤ r < m,

s m ≤ r < n.

Then, for any k-algebra B, there is a one-to-one correspondence between the sets

Hilbh(B) and Homk(H,B), where H = k[a1, ..., as], given by the map ΦB : Homk(H,B) →

Hilbh(B) defined, for any ϕ ∈ Homk(H,B), by

ΦB(ϕ) =�tr(t

sn+ ϕ(a1)t

(s−1)n+ ...+ ϕ(as))

�.

Proof. Take a k-algebra B. From Proposition 2.18 we have that

Hilbh(B) = {I ⊆ B[t] : I = (t

mp(t

n)) for some p(t

n) = t

sn+ a1t

(s−1)n+ ...+ as}.

Therefore it follows that ΦB is well defined, ΦB(ϕ) ∈ HilbB), and that ΦB has a

natural inverse ΨB : Hilbh(B) → Homk(H,B) defined by

I =�tr(t

sn+ b1t

(s−1)n+ ...+ bs)

��→

�ψ : H → B

�

where ψ is defined by ψ(ai) = bi for i = 1, ..., s. We thus have our bijection.

15


Theorem 2.20. Let S = k[t], graded by the abelian group G = Z/nZ and let

h : Z/nZ → N be a Hilbert function defined by

h(r) =

�s+ 1 0 ≤ r < m,

s m ≤ r < n.

Then the Hilbert functor Hilbh

is represented by the k-algebra H = k[a1, ..., as] with the

natural isomorphism Φ : Homk(H,−) → Hilbh

defined by Φ(B) = ΦB : Homk(B) →Hilb

h(B) that takes a homomorphism ϕ : H → B to the ideal

�tr(t

sn+ ϕ(a1)t

(s−1)n+ ...+ ϕ(as))

�.

Proof. For any k-algebra B we have, from Theorem 2.19, a one-to-one correspondencebetween the sets Hilbh

(B) and Homk(H,B). This is clearly functorial in B so this isa natural isomorphism. Thus, we have a representation of the Hilbert functor Hilb

h

by (H,Φ).

With the notation from Theorem 2.20 we have that

ΦH(idH) =�tr(t

sn+ idH(a1)t

(s−1)n+ ...+ idH(as))

�=

�tr(t

sn+ a1t

(s−1)n+ ...+ as)

�,

so by Yoneda’s Lemma it follows that we can write our representation of Hilbh as

(H, (F (t)) where F (t) = tm(t

sn+ a1t

(s−1)n+ ... + as). All this means is that for any

k-algebra B and ideal I ∈ Hilbh(B), there is a unique homomorphism ϕ : H → B,

such that we get a co-Cartesian diagram

H

◦��

ϕ�� B

��

H|t]/(F (t)) �� B[t]/I

where B ⊗H (H[t]/(F (t))) = B[t]/I.Example. If n = 1 we have a grading over the trivial group G = 0. Then we havea Hilbert function h : 0 �→ s and Theorem 2.20 tells us that the functor Hilb

h isrepresented by the ring k[a1, ..., as]. This is, if we instead work in the category ofschemes, equivalent to the fact that the Hilbert scheme of s point on the affine line,A1

= Spec(k[t]), is the scheme As= Spec(k[a1, ..., as]), i.e. the affine s-space.

Remark 2.21. Note that the case G = Z and Theorem 2.14 is just a special case ofthe grading G = Zn and Theorem 2.20 when we let n → ∞. Indeed, when n → ∞we have that

�n−1r=0 t

rk[t

n] →

�∞r=0 t

rk, since the powers tn will grow so big that they

will never be attained. Then the homogeneous ideals will be of the form (tm) for some

m and therefore Theorem 2.20 tells us that we have a representation of Hilbh givenby (k, (t

m)) which is precisely the result given by Theorem 2.14.

16

G. Sædén Ståhl 3 THE GRASSMANN SCHEME

3 The Grassmann scheme

3.1 The Grassmann functorWe will now consider the Grassmann scheme which is the scheme that represents theGrassmann functor. Since this theory can be found in several texts on the subjectwe will only go through this here in order to find methods that can be applicable forproving the representability of the Quot functor in the next section and so we referthe interested reader to e.g. [5, 8.4] for a more comprehensive study.

Definition 3.1. Let k be a field and V a k-vector space. For any r ∈ N the Grassmann

functor Grassr

V: Algk → Set from the category of k-algebras to the category of of

sets is defined, for any k-algebra B, by

Grassr

V(B) = {N ⊆ B ⊗k V : (B ⊗k V )/N is locally free over B of rank r},

and for any homomorphism ϕ : B1 → B2 of R-algebras B1, B2 we define Grassr

V(ϕ)(N) =

(incl⊗ id)(I ⊗B1 B2) = im(I ⊗B1 B2).

This is similar to the Hilbert functor in the sense that we consider quotients thatare locally free, but we do not require the submodules N ∈ Grass

r

V(B) to be B[t]-

modules which we do, implicitly, for the Hilbert functor when we require I ∈ Hilbh(B)

to be an ideal in B[t]. In the following we will consider the case when V is the finitelygenerated k-vector space V = k

n of dimension n.Example 3.2. First we will give a motivation for the definition of the Grassmannfunctor. The Grassmannian of a vector space of dimension n, often denoted Gr(r, n),consists of the subspaces of dimension r. We want our version to be the naturalextension of this. For this motivation, choose a Noetherian k-algebra B and takesome N ∈ Grass

r

V(B) and let P = (B ⊗k V )/N . Since B is Noetherian we have that

P is a finitely presented and locally free B-module, and this is equivalent to P beingprojective. Since π : B⊗k V � P is surjective it therefore splits, locally, and we havethat B ⊗k V = ker(π)⊕ P where ker(π) = N . Thus, N is locally a direct summandand this is our natural extension of the notion of subspaces of a vector space.

The Grassmann functor is not that easy to work with, instead we will considermore easily representable subfunctors of this functor. Since our k-vector space V = k

n

is finitely generated of dimension n we have that B ⊗k V = Bn. Let us denote the

set of canonical basis elements of kn by X and let Y ⊂ X consist of r elements. Welet k

Y denote the k-vector space with basis Y and B ⊗k kY

= BY denote the free

B-module with the induced basis from Y . Then we have an induced homomorphismϕY : B

Y�→ B

n.

17

3 THE GRASSMANN SCHEME G. Sædén Ståhl

Definition 3.3. Let k be a field, V = kn a k-vector space of dimension n, r ∈ N an

integer and Y a set consisting of r basis elements of kn. Then we define the relative

Grassmann functor, relGrassY

Vas the subfunctor of Grass

r

Vdefined by

relGrassY

V(B) = {N ⊆ B

n: B

n/N is free with basis Y }.

Lemma 3.4. If k is a field, V = kn

a k-vector space of dimension n, r ∈ N an

integer and Y a set consisting of r basis elements of kn, then the relative Grassmann

functor relGrassY

Vis naturally isomorphic to the functor G : Algk → Set defined by

G(B) = HomB(Bn/B

Y, B

Y).

Proof. Fix a k-algebra B. It is clear that N ∈ relGrassY

V(B) is equivalent to the

composition BY

ϕY�→ B

n � Bn/N being an isomorphism, so we have

relGrassY

V(B) = {N ⊆ B

n: B

Y�→ B

n � Bn/N is an isomorphism}.

Now, for any N ∈ relGrassY

V(B) we have an isomorphism ϕ : B

Y → Bn/N which

implies that the composition ψ : Bn � B

n/N

ϕ−1

→ BY has kernel ker(ψ) = N and

ψ ◦ ϕY = id. Conversely, if we have a homomorphism χ : Bn → B

Y with χ ◦ ϕY = id

then it is clear that ker(χ) ∈ GrassY

V(B). Hence we have that relGrass

Y

Vis naturally

isomorphic to the functor F : Algk → Set, defined by

F (B) = {ψ ∈ HomB(Bn, B

Y) : ψ ◦ ϕY = id}.

Furthermore, since Bn= B

Y ⊕ Bn/B

Y we have that maps ψ : Bn → B

Y satisfyingψ ◦ ϕY = id are in a one-to-one correspondence with maps B

n/B

Y → BY (the

element in the first coordinate must be mapped to itself and the other coordinate canbe mapped to anything). Thus we also have the that relGrass

Y

Vnaturally isomorphic

to the functor G : Algk → Set, defined by G(B) = HomB(Bn/B

Y, B

Y).

A useful result in commutative algebra is the following.

Proposition 3.5 ([3, Theorem 10.43]). If A is a ring and M,N and L are A-modules,

then there is a canonical isomorphism

HomA(M ⊗A N,L) ∼= HomA(M,HomA(N,L)).

Corollary 3.6. For A-modules M,E with E being finitely generated and free we have

that any homomorphism M → E can be canonically identified with a homomorphism

M ⊗A E∗ → A, where E

∗denotes the dual of E.

18

G. Sædén Ståhl 3 THE GRASSMANN SCHEME

Proof. Let N = E and P = A in Proposition 3.5. Then we have that

Hom(M ⊗ E∗, A) ∼= Hom(M,Hom(E

∗, A)) = Hom(M,E

∗∗) ∼= Hom(M,E),

where the last isomorphism comes from the fact that there is a canonical isomorphismE ∼= E

∗∗ when E is finitely generated and free.

If we apply Corollary 3.6 to the result of Lemma 3.4 with Bn/B

Y= M,B

Y=

E and B = A we conclude that the functor relGrassY

Vis naturally isomorphic to

H : Algk → Set defined by

H(B) = HomB(Bn/B

Y ⊗ (BY)∗, B).

Since every module homomorphism of a module M into a commutative k-algebraB extends uniquely to an algebra homomorphism of the symmetric algebra overM , denoted S(M), into B we have in particular that any B-module homomor-phism ϕ : B

n/B

Y ⊗ (BY)∗ → B extends uniquely to an k-algebra homomorphism

ψ : S(Bn/B

Y ⊗ (BY)∗) → B. Since B

n/B

Y ∼= Bn−r is free of rank n − r and (B

Y)∗

is free of rank r it follows that the tensor product Bn/B

Y ⊗ (BY)∗ is free of rank

r(n− r). The symmetric algebra of a free module of rank r(n− r) is the polynomialring in r(n − r) variables, B[x1, ..., xr(n−r)]. This must hold for any B and thus, itfollows that the functor H is represented by the polynomial ring k[x1, ..., xr(n−r)] andwe can conclude our main result of this section, for a more detailed proof see [5,Lemma 8.13].

Theorem 3.7. Let k be a field, V = kn

a finitely generated k-vector space of dimen-

sion n, r ∈ N an integer and Y a set consisting of r basis elements of kn. Then the

relative Grassmann functor relGrassY

Vis represented by k[x1, ..., xr(n−r)].

It is then possible to prove, using these results, that the original functor Grassr

V

is representable in the category of schemes. The proof uses theory involving conceptssuch as open subfunctors and open coverings of functors. This theory is postponeduntil Section 4.

Theorem 3.8 ([5, Proposition 8.14]). If k a field and V = kn

a finitely generated k-

vector space of dimension n, then the Grassmann functor Grassr

Vis representable with

a scheme that is finitely covered of the affine schemes Ar(n−r)= Spec(k[x1, ..., xr(n−r)]).

19

3 THE GRASSMANN SCHEME G. Sædén Ståhl

Remark 3.9. Another way of realizing the Grassmannian is by considering matrices.Let k be a field and V a finitely generated k-vector space of rank n and suppose thatwe want, for any k-algebra B, to consider the submodules N ⊆ B ⊗k V such thatthe quotients (B ⊗k V )/N are locally free of rank r say. Then we can consider theset W of r× n-matrices M over B, modulo multiplication from the left by invertibler × r-matrices. If I ⊂ {1, 2, ..., n} is a subset consisting of r elements we let MI

denote the I-th submatrix. If the determinant of MI is a unit in B, i.e. MI is non-singular, it follows that the subset of the canonical basis elements {ei}i∈I of V = k

n

is a basis for the quotient (B⊗k V )/N . Furthermore, if the matrix MI is non-singularwe may multiply M with M

−1I

to make that submatrix into the identity, and allthe other (n − r)r elements are then the coordinates of the quotient. The subsetWI ⊆ W consisting of the r × n-matrices that have a non-singular I-th submatrix(the equivalence to our relative Grassmann functor) is then realized as an affine(n− r)r-space A(n−r)r

B= Spec(B[x1, ..., x(n−r)r]).

If we let the affine nr-space Anr

Bdenote the set of our r × n-matrices we see that

it is therefore covered by affine (n− r)r-spaces A(n−r)rB

and since their intersection iswell behaved (we omit the details) these can be glued into a scheme that representsthe Grassmann functor. For a more comprehensive explanation, see [4, III.2.7].

3.2 The graded GrassmannianLet V now be a finitely generated k-vector space with a grading by an abelian groupG, i.e. V =

�g∈G Vg, and let h : G → N be any function. Then we can define the

graded Grassmann functor grGrassh

V: Algk → Set by

grGrassh

V(B) =

N ⊆ B ⊗ V :

N is a homogeneous submodule suchthat (B ⊗ Vg)/Ng is a locally freeB-module of rank h(g) for all g ∈ G

.

If N ∈ grGrassh

V(B) then N is graded so N =

�g∈G Ng. There is then a canonical

map grGrassh

V→

�g∈G Grass

h(g)Vg

given, for each k-algebra B, by

grGrassh

V(B) � N �→

�

g∈G

Ng ∈�

g∈G

Grassh(g)Vg

(B).

This is a natural isomorphism of functors so we get a decomposition of the gradedGrassmannian into a direct product of regular Grassmannians, grGrass

h

V∼=

�g∈G Grass

h(g)Vg

.

20

G. Sædén Ståhl 4 THE QUOT SCHEME OF THE AFFINE LINE

4 The Quot scheme of the affine line

4.1 Structure theorem of finitely generated graded modulesover a graded principal ideal domain

Definition 4.1. If M is a graded module, then M(d) is the graded module whereM(d)k = Mk+d.

Definition 4.2. If M and N are graded modules then a graded homomorphism

ϕ : M → N is a homomorphism that has the property that ϕ(Mk) ⊆ Nk, i.e. ϕ pre-serves the grading of the modules. We will denote the set of graded homomorphismfrom M to N by grHom(M,N). A graded isomorphism is a graded homomorphismthat is an isomorphism. Two modules M and N are isomorphic as graded modules,denoted M ∼=0 N , if there is a graded isomorphism between them.

Remark 4.3. By definition the zero-homomorphism is graded since the element 0 ishomogeneous and is a member of every component.

Lemma 4.4. If S = k[t] =�

n−1r=0 t

rk[t

n], then S(−d) ∼=0 t

dS =

�n−1r=0 t

r+dk[t

n].

Proof. The homomorphism which maps 1 ∈ S(−d) to td ∈ t

dS is clearly a graded

isomorphism.

The following comes from Proposition 5 in [9].

Proposition 4.5. If ϕ : M → N is a graded homomorphism then both ker(ϕ) and

im(ϕ) are homogeneous and the canonical bijection M/ ker(ϕ) → im(ϕ) is a graded

isomorphism.

Proof. First of all, im(ϕ) is generated by the images of the homogeneous generatorsof M and since ϕ is graded then the images are also homogeneous. If x ∈ ker(ϕ) ⊆ M

then x can be written as a sum of its graded components x =�

gxg.

Thus 0 = ϕ(x) = ϕ

��gxg

�=

�gϕ(xg), where the ϕ(xg) are the graded compo-

nents of 0, which implies that ϕ(xg) = 0 for all g, or simply xg ∈ ker(ϕ) for all g. Thatthe isomorphism M/ ker(ϕ) → im(ϕ) is graded then follows from the construction ofthe quotient module and the fact that ϕ is graded.

Definition 4.6. A graded principal ideal domain is a principal ideal domain in whichevery homogeneous ideal is generated by a homogeneous element.

21

4 THE QUOT SCHEME OF THE AFFINE LINE G. Sædén Ståhl

Remark 4.7. 1. Note that the k-algebra k[t] is a graded principal ideal domain whenit is graded by Z or Zn.2. There could also be a different interpretation of a graded principal ideal domain

than that of Definition 4.6. It could also mean that it simply is a principal idealdomain that is graded, but where every ideal does not need to be generated by ahomogeneous element.

The following lemma is inspired by Theorem 12.4 in [3].

Lemma 4.8. If D is a graded principal ideal domain and M a free graded D-module

of rank n then any homogeneous submodule N ⊆ M has the following property: There

is some basis y1, ..., yn of M , some m ∈ N and homogeneous elements a1, ..., am ∈ D

with a1 | a2 | · · · | am such that a1y1, ..., amym is a basis of N .

Proof. If N = 0 there is nothing to prove. Suppose N �= 0, then N is free of rankm say (see [3, Theorem 12.4]). For any graded homomorphism ϕ : M → D we have,since ϕ is graded, that ϕ(N) is a homogeneous ideal in D. Since D is a principalideal domain ϕ(N) is generated by a single element, which we denote by aϕ. Let usconsider the set

Σ = {(aϕ) : ϕ ∈ grHomD(M,D) so that aϕ is homogeneous},

which has a partial ordering by inclusion, ⊆. It is clear that Σ is non-empty sincethe zero-homomorphism is graded and its image is generated by the homogeneouselement 0. Since D is is a P.I.D. it is Noetherian so Σ has a maximal element whichcorresponding to some aν for some graded homomorphism ν, and we let the degreeof aν be d. The maximality implies that (aν) is not properly contained in any otherelement of Σ. Let a1 = aν . The fact that a1 ∈ ν(N) implies that there is some y ∈ N

such that ν(y) = a1 and since ν is graded it follows that y has degree d.Now we choose a basis x1, ..., xn for M and let πi : M → D be the projection

of M on the i:th coordinate corresponding to this basis. Note that πi is a gradedhomomorphism for any i. Since N �= 0 it is clear that there is some i such thatπi(N) �= 0, thus it is clear that a1 �= 0.

Take some ϕ ∈ grHomD(M,D). We will now show that a1 divides ϕ(y). Since y

has degree d and ϕ is graded we have that ϕ(y) has degree d as well. Let us considerthe ideal generated by these two elements, I = (a1,ϕ(y)), which is generated bysome element, say b. Since D was a graded principal ideal domain it follows that b

is homogeneous of degree d. This implies that b ∈ Σ and thus a1 | b and thereforea1 | ϕ(y).

22


We can now apply the fact that a1 divides ϕ(y) to the projections πi so thatπi(y) = a1bi for some bi. We define

y1 =

n�

i=1

bixi.

First we note that a1y1 = y from which it follows that a1 = ν(y) = ν(a1y1) = a1ν(y1),hence ν(y1) = 1. That is, y1 has degree 0. We now claim that M = Dy1 ⊕ ker(ν).Take some arbitrary x ∈ M and wite x = ν(x)y1 + (x− ν(x)y1). Since

ν(x− ν(x)y1) = ν(x)− ν(x)ν(y1)

= ν(x)− ν(x)

= 0,

we have that (x− ν(x)y1) ∈ ker(ν). Hence M = Dy1 +ker(ν). To show that the sumis direct we have to show that the intersection is trivial. Suppose that ry1 ∈ ker(ν)

for some r ∈ D, then 0 = ν(ry1) = rν(y1) = r, so the intersection is indeed trivial.Hence M = Dy1 ⊕ ker(ν).

We now show that N = Da1y1 ⊕ (N ∩ ker(ν)). Take some x ∈ N . Then a1

divides ν(x), by construction, and therefore we have that ν(x) = ba1 for some b ∈ D.Write x = ν(x)y1 + (x− ν(x)y1) = ba1y1 + (x− ν(x)y1) and since the second term isan element of N and, similarly to above, an element in the kernel of ν we have that(x−ν(x)y1) ∈ (N∩ker(ν)). That the sum is direct follows by the same argumentationas above. Hence N = Da1y1 ⊕ (N ∩ ker(ν)).

Now, since ker(ν) is free of rank n − 1 we have, by induction, that there is abasis of homogeneous elements of degree 0, y2, ..., yn, of ker(ν) and homogeneouselements a2, ..., am ∈ D with a1 | · · · | am such that a2y2, ..., amym is a basis forN ∩ ker(ν). Thus, we have that y1, ..., yn is a basis of M and a1y1, ..., amym is abasis of N . The only thing we have to show is that a1 divides a2. This we do byconsidering the graded homomorphism ϕ : M → D defined by ϕ(y1) = ϕ(y2) = 1 andϕ(y3) = ... = ϕ(yn) = 0. Then a1 = ϕ(a1y1), i.e. a1 ∈ ϕ(N) and by maximality of (a1)it follows that (a1) = ϕ(N). Since a2 = ϕ(a2y2) ∈ ϕ(N) it follows that a1 | a2.

We are now ready to prove our structure theorem.

Theorem 4.9. If M is a finitely generated graded module over a graded principal

ideal domain D (graded by an abelian group G), then

M ∼=0

�m�

k=1

D(−gk)/akD

��

�n�

k=m+1

D(−gk)

�

23


for some g1, ..., gm, gm+1, ..., gn ∈ G, where n,m ∈ N, and homogeneous elements

a1, ..., am ∈ D such that a1 | · · · | am.

Proof. Let {α1, ...,αn} be a minimal homogeneous generating set of M , i.e. a set ofhomogeneous generators of M such that no other homogeneous generating set of Mis properly contained in {α1, ...,αn}, where each αi is homogeneous of degree di ∈ G.That a minimal homogeneous generating set exists follows from the fact that if M isfinitely generated and graded then we can choose a finite generating set and each ofthose generators can be written as a sum of a finite number of homogeneous elements.

We have that the D-module�

n

k=1 D(−di) is free of rank n, say with basis x1, ..., xn.Then we have a surjective graded homomorphism ϕ :

�n

k=1 D(−di) → M defined byϕ(xi) = αi for i = 1, ..., n. Thus

n�

k=1

D(−di)/ ker(ϕ)∼=0 M.

From Lemma 4.8 we have that the submodule ker(ϕ) is free of some rank m ≤ n, andthat we can choose a basis, y1, ..., yn, of

�n

k=1 D(−di) such that a1y1, ..., amym is abasis of ker(ϕ) for some homogeneous elements a1, ..., am of D with ai | ai+1. Hence

M ∼=0 (D(−d1)y1 ⊕ ...⊕D(−dn)yn)/(D(−d1)a1y1 ⊕ ...⊕D(−dm)amym).

Now we can consider the graded homomorphism

π : D(−d1)y1⊕...⊕D(−dn)yn → D(−d1)/(a1)⊕D(−dm)/(am)⊕D(−dm+1)⊕...⊕D(−dn)

that maps (f1y1, ..., fnyn) to (f1+(a1), ..., fm+(am), fm+1, ..., fn). Then we have thatker(π) = {(f1, ..., fm, 0, ..., 0) : fi ∈ (ai)} = D(−d1)a1y1 ⊕ ...⊕D(−dm)amym. Hence

M ∼=0

�m�

k=1

D(−gk)/akD

��

�n�

k=m+1

D(−gk)

�.

4.2 The relative Quot schemeWe will study the Quot scheme in the same way as we did with the Hilbert schemein Section 2, namely by first defining the Quot functor and then show that it is rep-resentable. However to show that it is representable is not so easy, instead we will,

24


similarly to the Grassmannian in Section 3, first consider specific subfunctors of ourQuot functor, so called relative Quot functors, and show that these are representable.That result will then be used to show the representability of the Quot functor in thenext section.

As before we let k be a field, S be the k-algebra k[t] graded by the abelian groupG and h be a Hilbert function G → N. For any graded k[t]-module V and k-algebraB we define the set

Quoth

V(B) =

N ⊆ B ⊗k V :

N is a homogeneous B[t]-submodule suchthat [(B ⊗k V )/N ]

gis locally free over B

and of finite rank h(g) for all g ∈ G

. (2)

Lemma 4.10. Let V be a k[t]-module graded by G, B1, B2 be k-algebras and h : G →N a Hilbert function. Then, for any N ∈ Quot

h

V(B1) the submodule im(N ⊗B1 B2) =

(incl⊗ id)(N ⊗B1 B2) ∈ Quoth

V(B2).

Proof. Take an N ∈ Quoth

V(B1) and consider im(N ⊗B1 B2) ⊆ B2 ⊗k V . Since

this submodule is generated by homogeneous elements it is clearly a homogeneoussubmodule. By Lemma 2.7 it follows that

((B1⊗kV )/N)⊗B1B2 = ((B1⊗kV )⊗B1B2)/im(N⊗B1B2) = (B2⊗kV )/im(N⊗B1B2)

and that each graded component of this module is a locally free B2-module of rankh(g) follows from Lemma 2.6. Hence im(N⊗B1B2) = (incl⊗ id)(N⊗B1B2) ∈ Quot

h

V(B2).

Definition 4.11. Let V be a k[t]-module graded by G and h : G → N a Hilbertfunction. The Quot functor, Quot

h

V: Algk → Set, is defined, for any k-algebra B,

by Quoth

V(B) given by (2), and for any morphism of k-algebras, ϕ : B1 → B2, by

Quoth

V(ϕ)(N) = im(N ⊗B1 B2) = (incl⊗ id)(N ⊗B1 B2).

Remark 4.12. Note that the Quot functor is really a generalization of the Hilbertfunctor. Indeed, for V = k[t] we have B ⊗k k[t] = B[t] and the requirement thatN ⊆ B[t] is a B[t]-module is equivalent to saying that N is an ideal in B[t]. Hencewe have

Quoth

k[t](B) =

I ⊆ B ⊗k k[t] :

I is a homogeneous ideal such thatthe B-module (B ⊗k Sg)/Ig is locallyfree of finite rank h(g) for all g ∈ G

= Hilb

h(B).

25


We will in the following consider the case V being the finitely generated freek[t]-module V =

�d

i=1 k[t] of rank d say.

Notation. When V is the k-vector space V =�

d

i=1 k[t], then we introduce e1 =

(1, 0, ....., 0), e2 = (0, 1, 0, ..., 0), ..., ed = (0, 0, ...., 1) to denote the canonical basis in V

as a k[t]-module.

Take some k-algebra B. If N ∈ Quoth

V(B) then we have that (B⊗k V )/N must be

a graded B[t]-module such that each graded component is a locally free B-module offinite rank. If we localize at some prime ideal p ∈ Spec(B) we have that [(B ⊗k V )/N ]p

is a Bp[t]-module such that every graded component�[(B ⊗k V )/N ]p

�g

is a free Bp-module of rank h(g) for g ∈ G. Passing to the fraction field of Bp we can use ourstructure theorem of finitely generated graded modules over graded P.I.D.s (Theo-rem 4.9) to see that it will be free with basis e1, te1, ..., t

s1e1, e2, ...., tsded for some

s1, ..., sd. As stated earlier, the Quot functor is quite hard to handle, so instead wewill, motivated by what the this calculation, consider more manageable subfunctors.Before we define them we will first add some useful notation.

Notation. Let X ⊆ V be a subset consisting of homogeneous elements of V gradedby the abelian group G. We then let Xg denote the subset of X that consists of theelements of degree g ∈ G and define x(g) = |Xg| to be the number of elements of Xof degree g. Also we will let χ

(g)1 ,χ

(g)2 , ...,χ

(g)x(g) denote the elements in X of degree g.

Since we have that the Quot functor is defined with a Hilbert function h : G → Nwe will choose an X ⊆ V such that h(g) = x(g) for all g ∈ G and say that X agreeswith h.

Definition 4.13. For the k-vector space V =�

d

i=1 k[t] with an induced gradingfrom k[t] =

�g∈G Sg, for some abelian group G, and a Hilbert function h : G → N we

define the relative Quot functor as the subfunctor of Quoth

Vdefined by

relQuotX

V(B) =

N ⊆ B ⊗k V :

N is a homogeneous B[t]-submodulesuch that [(B ⊗k V )/N ]

gis free over B

with basis Xg for all g ∈ G

.

Remark 4.14. The reason that we did not have to consider subfunctors when weshowed the representability of the Hilbert functor was that subfunctors in that casecoincided with the original Hilbert functor. We showed that the module was locally

26


free implied that the module was actually free (Proposition 2.16). Indeed, when wehad a grading by Zn and Hilbert function

h(r) =

�s+ 1 0 ≤ r < m

s m ≤ r < n

for some m ∈ Zn, s ∈ N, c.f. Section 2, the result was

Hilbh= relQuot

{1,t,...,tsn−1}k[t] .

As we remarked earlier, Remark 2.21, the case when we had a grading by Z was aspecial case of this when we let n → ∞.

Notation. If we write a set of basis elements as X = {e1, te1, ..., ts1−1e1, e2, te2, ...., t

sd−1ed}

for some s1, ..., sd then it will be implied that if sj = 0 then there are no basis elementsin the ej-coordinate.

Proposition 4.15. If V =�

d

i=1 k[t] is a free k[t]-module, graded by G = Z or Zn,

of finite rank d and X = {e1, te1, ..., ts1−1e1, e2, te2, ...., t

sd−1ed} then there is, for any

k-algebra B, a one-to-one correspondence between the set relQuotX

V(B) and the set

{p1(t), ..., pd(t)} : pj(t) = tsjej −

x(sj)�

r=1

b(j)rχ(sj)r

for some b(j)1 , ..., b

(j)x(d) ∈ B for j = 1, ..., d

.

Proof. Let us denote the set of monic polynomials above by MP . Fix a k-algebraB and take some N ∈ relQuot

X

V(B). Now, consider the image of the element

ts1e1 ∈ B⊗k V in the quotient (B⊗k V )/N . Since (B⊗k V )/N is a module over the

graded ring B[t] we must have that ts1e1 = ts1−1

e1 ·t, it must respect the grading whenwe multiply by t. Hence t

s1e1 must be written as a linear combination of the basiselements of the same degree s1, i.e. ts1e1 =

�x(s1)r=1 b

(1)r χ

(sj)r for some b

(1)1 , ..., b

(1)x(s1)

∈ B.The same argument holds for any sj. Hence N contains these elements. Furthermore,if N was generated by at least one more non-redundant polynomial then the subsetsXg would no longer be a basis for the quotients graded components [(B ⊗k V )/N ]g.Thus N is generated by precisely the polynomials t

sjej −�x(sj)

r=1 b(j)r χ

(sj)r .

Conversely, take an element {p1(t), ..., pd(t)} ∈ MP . Then it is clear that thesubmodule of B ⊗k V that they generate is an element of relQuot

X

V(B). Hence we

have a natural correspondence given by the correspondence between the submodulesand their generators.

27


Example 4.16. To illustrate we give an example of Proposition 4.15. Let V = k[t]

and G = Zn so that we are back to the setting of Section 2. Then X = {1, t, ..., ts−1}for some s. For simplicity we write s = s

�n + m for some integers s

�,m where m is

the smallest non-negative integer possible. Then χs

1, ...,χs

x(s) are simply the elementstm, t

m+n, ..., t

m+(s�−1)n, in some order. Proposition 4.15 then states that there is, forany k-algebra B, a one-to-one correspondence between the sets relQuot

X

V(B) and the

set of monic polynomials ts�n+m

+ b1t(s�−1)n+m

+ ... + bs� , which is precisely the sameas what Proposition 2.18 says.

Now we are ready for our main result of this section, namely that the relativeQuot functor is representable.Theorem 4.17. Let V =

�d

i=1 k[t] be a free k[t]-module, graded by G = Z or Zn, of

finite rank d and X = {e1, te1, ..., ts1−1e1, e2, te2, ...., t

sd−1ed}. Then the relative Quot

functor relQuotX

Vis represented by the k-algebra Q = k

�a(1)1 , ..., a

(1)x(s1)

, a(2)1 , ..., a

(d)x(sd)

�,

i.e. the polynomial ring over k in�

d

j=1 x(sj) variables, and the natural isomorphism

Φ : Homk(Q,−) → relQuotX

Vgiven, for any k-algebra B and ϕ ∈ Homk(Q,B), by

(ϕ : Q → B) �→

ts1e1 −

x(s1)�

r=1

ϕ(a(1)r)χ

(s1)r

, ..., tsded −

x(sd)�

r=1

ϕ(a(d)r)χ

(sd)r

.

Proof. Fix a k-algebra B and take some N ∈ relQuotX

V(B). From Proposition 4.15 we

have that N is generated by polynomials on the form pj(t) = tsjej −

�x(sj)r=1 a

(j)r χ

(sj)r

for some a(j)1 , ..., a

(j)x(d) ∈ B for j = 1, ..., d. So we can define ΦB : Homk(Q,B) →

relQuotX

V(B) by

(ϕ : Q → B) �→

ts1e1 −

x(s1)�

r=1

ϕ(a(1)r)χ

(1)r, ..., t

sded −x(sd)�

r=1

ϕ(a(d)r)χ

(d)r

.

This has a canonical inverse ΨB : relQuotX

V(B) → Homk(Q,B) defined by

ts1e1 −

x(s1)�

r=1

b(1)rχ(s1)r

, ..., tsded −

x(sd)�

r=1

b(d)rχ(sd)r

�→ (ϕ : Q → B)

where ϕ is defined by ϕ(a(i)j) = b

(i)j

for each j = 1, ..., x(si) and i = 1, ..., d. Wethus have an isomorphism between Homk(Q,B) and relQuot

X

V(B) given by ΦB and

since this is clearly functorial in B we get a natural isomorphism Φ : Homk(Q,−) →relQuot

X

Vgiven by Φ(B) = ΦB for any k-algebra B.

28


Remark 4.18. Using the notation of Theorem 4.17 we have that

ΦQ(idQ) =

ts1e1 −

x(s1)�

r=1

idQ(a(1)r)χ

(s1)r

, ..., tsded −

x(sd)�

r=1

idQ(a(d)r)χ

(sd)r

=

ts1e1 −

x(s1)�

r=1

a(1)rχ(s1)r

� �� =F1(t)

, ..., tsded −

x(sd)�

r=1

a(d)rχ(sd)r

� �� =Fd(T )

∈ relQuot

X

V(Q).

So we can then say, by using Yoneda’s Lemma, that we can write our representationof relQuot

X

Vas the pair (Q, (F1(t), ..., Fd(t))).

We will in the following sections give some examples of the use of Theorem 4.17.

4.2.1 Grading by G = Z

Similarly to Section 2 we start by considering the case when k[t] is graded by theabelian group G = Z, k[t] =

�∞r=0 k · tr.

Example 4.19. First we consider the case when V = k[t], to show the correspondencewith the Hilbert scheme. Let X = {1, t, ..., tm−1}. Then Theorem 4.17 tells us thatrelQuot

X

Vis represented by Q = k since there are no basis elements in X of degree

(m− 1) + 1 = m. This is precisely what Theorem 2.14 says, that Hilbh= relQuot

X

k[t]

is represented by H = k, where h(r) =

�1 0 ≤ r < m,

0 m ≤ r < n

Example 4.20. Consider the case V = k[t]⊕ k[t] and X = {(1, 0), (t, 0), (0, 1), (0, t)}.Then we have from Theorem 4.17 that relQuot

X

Vis represented by Q = k, since there

are no basis elements of degree 2.

Example 4.21. Consider the case V = k[t]⊕k[t] and X = {(1, 0), (t, 0), (t2, 0), (0, 1), (0, t)}.Then we have from Theorem 4.17 that relQuot

X

Vis represented by Q = k[a], since

the element (0, t2) = t · (0, t) can be written as a linear combination of the one basis

element of degree 2, (0, t2) = a · (t2, 0) but the element (t3, 0) = t · (t2, 0) can not be

written as a linear combination of basis elements of degree 3 since there are none so(t

3, 0) = 0.

29


4.2.2 Grading by G = Zn

Example 4.22. As we did for the case G = Z, we first compare our result to what wegot with the Hilbert scheme in Section 2. Let V = k[t] and X = {1, t, ..., tsn−1}. Then,from Theorem 4.17, we have that relQuot

X

Vis represented by Q = k[a1, ..., as] since

there are s elements in X of degree (sn− 1) + 1 = sn = 0, namely 1, tn, t

2n, ..., t

sn−1.This is the same result that we would have gotten from Theorem 2.20, that says that

Hilbh= relQuot

X

Vis represented by H = k[a1, ..., as], where h(r) =

�s+ 1 0 ≤ r < m,

s m ≤ r < n,

for some m ∈ Zn.

Example 4.23. Let G = Z2 and consider the case V = k[t]⊕ k[t] andX = {(1, 0), (t, 0), (t2, 0), (0, 1), (0, t)}. Then we have from Theorem 4.17 that relQuot

X

V

is represented by the polynomial ring in five variables, Q = k[a1, a2, ..., a5]. That isbecause the element (t

3, 0) = t · (t2, 0) can be written as a linear combination of

the two basis elements of degree 3 = 1, namely (t, 0) and (0, t), and the element(0, t

2) = t · (0, t) can be written as a linear combination of the three basis elements of

degree 2 = 0, namely (1, 0), (t2, 0) and (0, 1).

4.3 The Quot schemeWhat we will do now is to show that the Quot functor, Quot

h

V, c.f. Definition 4.11,

is representable with a scheme when V is a finitely generated and free k[t]-module.To do this we will first need to review some theory. This will be more of a summary,for a full treatment of the subject see, e.g., [4, VI.1.1].

We start by widening our definition of a representable functor. Since the categoryof commutative rings are a dual of only the affine schemes, not all schemes, we wouldotherwise be too restrictive in our class of representable functors.

Definition 4.24. A functor F from the category of k-algebras to the category ofsets, F : Algk → Set, is representable by a scheme if there is a natural isomorphismΦ : Homk(−, X) → F from the functor Homk(−, X) : Algk → Set, which maps a k-algebra B to the set Homk(Spec(B), X), for some scheme X. As there is no ambiguitywith the representation of a functor by a k-algebra, we call also the pair (X,Φ) arepresentation of the functor F .

Remark 4.25. Note that any functor F : Algk → Set that is represented by a k-algebraA is also represented (by a scheme) by the affine scheme Spec(A).

30


Definition 4.26. Let F,G be functors Algk → Set, with a natural transformationα : G → F . Then G is an open subfunctor of F if

1. For any k-algebra A, the induced map G(A) → F (A) is injective, i.e. G is asubfunctor of F .

2. For any k-algebra A and each element a ∈ F (A), which by Yoneda’s lemmagives a natural transformation Φa : Homk(A,−) → F , we have that the functorGΦa , that is defined by

GΦa(B) = {(g,ϕ) ∈ G(B)× Homk(A,B) : α(g) = Φ(ϕ) in F (B)},

is a subfunctor of Homk(A,−) that is representable by an open subscheme ofSpec(A).

Remark 4.27. Note that the functor GΦa from Definition 4.26 induces a Cartesiandiagram

GΦa��

��

Homk(A,−)

��

G �� F

and G is an open subfunctor of F if the upper arrow in the diagram makes GΦa

into a subfunctor of Homk(A,−) and if GΦa is represented by an open subscheme ofSpec(A).

We also need the notion of a Zariski sheaf which is a functor that satisfies thesheaf axiom for any open covering of an affine scheme of open affine schemes but willleave the details for the interested reader to find in, e.g., [4, VI.1.1].

Now we state (without proof) the theorem that we will need for the representabil-ity of the Quot functor. It can be found, in one form or another, in several introduc-tory texts on the subject of representability of functors, e.g. [4, Theorem VI-14] or[5, Theorem 8.9].

Theorem 4.28. A functor F : Algk → Set is representable by some scheme X if and

only if

1. F is a Zariski sheaf,

2. There exists a collection of open representable subfunctors {Gi → F} such that,

for every field extension K/k, F (K) =�

iGi(K).

31


With the results from the previous section on the relative Quot functor we nowhave the theory we need to prove that the Quot functor is representable, but first weneed a lemma.

Lemma 4.29. Let V =�

d

i=1 k[t] be a free k[t]-module, graded by G = Z or Zn, of fi-

nite rank d, h : G → N a Hilbert function and X = {e1, te1, ..., ts1−1e1, e2, te2, ...., t

sd−1ed}

a set of basis elements that agree with h. Then the relative Quot functor relQuotX

Vis

an open subfunctor of Quoth

V.

Proof. That it is a subfunctor follows from the fact that any free module is alsolocally free. In order to show the other part we will use two particular facts fromcommutative algebra so we state them here for clarity.

(i) A square matrix M over a ring R is invertible if and only if its determinant isa unit in R. Hence if we consider M over Rdet(M) it is invertible.

(ii) A localization of a ring R (with respect to the multiplicatively closed set S) hasthe universal property that if ϕ : R → R

� is a ring homomorphism that mapsthe elements of S to units in R

� then there is a unique map S−1R → R

� thatfactor through ϕ.

Take a k-algebra A and some N ∈ Quoth

V(A). From Yoneda’s lemma this gives

us a natural transformation Φ : Homk(A,−) → Quoth

Vwhere Φ is defined, for any

ϕ ∈ Homk(A,B), by ΦB(ϕ) = Quoth

V(ϕ)(N) = im(N ⊗A B). We then consider the

commutative diagram(relQuot

X

V)Φ

��

��

Homk(A,−)

��

relQuotX

V�� Quot

h

V

where (relQuotX

V)Φ is the functor defined by

(relQuotX

V)Φ(B) =

�(N

�,ϕ) ∈ relQuot

X

V(B)×Homk(A,B) : N

�= ΦB(ϕ) = im(N⊗AB)

�.

Thus, we see that (relQuotX

V)Φ(B) consists of the ring homomorphisms ϕ ∈ Homk(A,B)

such that im(N⊗AB) is an element of relQuotX

V(B), i.e. that the graded components

of the quotient (B ⊗k V )/im(N ⊗A B) are free over B with basis X.Now, we note that we can, for any k-algebra B, visualize the set Quot

h

V(B) as a

set of matrices over B, similarly to what was described in Remark 3.9 at the end ofSection 3.1 for the Grassmannian. The matrices will in this case consist of infinitely

32


many columns corresponding to the fact that B⊗k V =�

d

i=1 B[t] will not be finitelygenerated over B. The number of rows of the matrix will, due to Lemma 2.5, be�

g∈G x(g). Then, relQuotX

V(B) will be equivalent to the set of matrices that have a

non-singular I-th submatrix MI , where I determines the columns of the matrix thatcorresponds to the basis set X. This means, from fact (i) of commutative algebrastated above, that the determinant of the matrix MI is a unit in B. In particular wehave for the k-algebra A that there is a matrix M

(N) that corresponds to the elementN ∈ relQuot

X

V(A) and we let detN ∈ A denote the value of the determinant of the

I-th submatrix M(N)I

where I corresponds to the set X.Since im(N ⊗A B) ∈ Quot

h

V(B) it corresponds to a matrix M

(im(N⊗AB)) over B.Then it follows that im(N ⊗AB) ∈ relQuot

X

V(B) if and only if the determinant of the

I-th submatrix M(im(N⊗AB))I

is a unit in B.Thus we have that (relQuot

X

V)Φ corresponds to the ring homomorphisms ϕ : A →

B such that the element detN ∈ A is mapped to a unit in B. From fact (ii) statedabove it follows that any such ϕ factors through a unique �ϕ : AdetN → B.

Hence (relQuotX

V)Φ(B) ∼= Homk(AdetN , B) and since this is functorial in B it

follows that (relQuotX

V)Φ is represented by the open affine subscheme Spec(AdetN )

and therefore, relQuotX

Vis an open subfunctor of Quot

X

V.

Remark 4.30. Note that we see from the proof of Lemma 4.29 that (relQuotX

V)Φ is

always representable by an affine open subscheme of Spec(A), which is a much strongerstatement than that of an open subfunctor which only requires that (relQuot

X

V)Φ is

representable by any open subscheme.

Theorem 4.31. If V =�

d

i=1 k[t] is a free k[t]-module, graded by G = Z or Zn, of

finite rank d and h is a Hilbert function G → N, then the Quot functor Quoth

Vis

representable with a scheme.

Proof. That Quoth

Vis a Zariski sheaf follows from the definition, since submodules

can ge glued together, see [7, Section II.5]. Let X1, ..., Xn for some n ∈ N, where eachXi is of the form

Xi = {e1, te1, ..., ts(i)1 −1

e1, e2, ..., ts(i)d −1

ed},

be those different possible set of basis elements given by h. From Lemma 4.29 it followsthat relQuot

XiV

is an open subfunctor of Quoth

Vfor any i and we let {relQuot

XiV

→ Quoth

V}

be our collection of open subfunctors. From Theorem 4.17 we know that relQuotXiV

isrepresentable for all Xi. That Quot

h

V(K) =

�irelQuot

XiV(K) for any field K over k,

follows directly since, for any N ∈ Quoth

V(K), the K-module (K ⊗k V )/N is a vector

33


space so it has a well defined basis which must be given by at least some Xi. Hence,by Theorem 4.28, Quot

h

Vis representable.

Definition 4.32. The scheme that represents the Quot functor is called the Quot

scheme.

Finally we prove some geometrical properties of the Quot scheme.

Theorem 4.33. When V =�

d

i=1 k[t] is a free k[t]-module, graded by G = Z or Zn,

of finite rank d and h : G → N a Hilbert function, then the Quot scheme is smooth,

connected and, therefore, irreducible.

Proof. That it is smooth follows from the fact that it is locally representable by apolynomial ring. In order to show that it is connected we need to show that theintersection of two open subfunctors, relQuot

X

Vand relQuot

Y

V, is non-empty, where

X = {e1, te1, ..., ts1−1e1, e2, te2, ...., t

sd−1ed}

andY = {e1, te1, ..., tσ1−1

e1, e2, te2, ...., tσd−1

ed}are two arbitrary sets that agree with h. We therefore have to consider the functorrelQuot

X

V∩ relQuot

Y

Vwhich is defined, for any k-algebra B, by

(relQuotX

V∩ relQuot

Y

V)(B) = relQuot

X

V(B) ∩ relQuot

Y

V(B).

That is to say, we consider the submodules N such that the quotient (B ⊗k V )/N

is free as a B-module with both basis X and Y . We have to show that the set(relQuot

X

V∩ relQuot

Y

V)(B) is non-empty. It is enough if we can do this for any k-

algebra B and we choose B = k.As before, we let x(g) denote the number of elements in X of degree g and we

introduce y(g) to denote the number of elements in Y of degree g. Clearly we musthave that x(g) = y(g) for any g since they come from the same Hilbert function.We also let χ

(g)1 , ...,χ

(g)x(g) denote the elements in X of degree g and we introduce

ξ(g)1 , ..., ξ

(g)x(g) to denote the elements in Y of degree g.

Any N ∈ relQuotX

V(B) is, from Proposition 4.15, generated, as a B[t]-module, by

the set of elements

NX =

ts1e1 −

x(s1)�

i=1

a(1)iχ(s1)i

, ..., tsded −

x(sd)�

i=1

a(d)iχ(sd)i

,

34


for some a(1)1 , ..., a

(d)x(sd)

∈ B. If N ∈ relQuotY

V(B) then it must also be generated by

the set

NY =

tσ1e1 −

x(σ1)�

i=1

b(1)iξ(σ1)i

, ..., tσded −

x(σd)�

i=1

b(d)iξ(σd)i

,

for some b(1)1 , ..., b

(d)x(σd)

∈ B. Since the elements in NY are linearly independent thisis true if and only if tσjej −

�x(σj)i=1 b

(j)iξ(σj)i

can be written as a linear combination ofthe elements in NX for j = 1, ..., d. Thus we have a system of equations

tσ1e1 −

x(σ1)�

i=1

b(1)iξ(σ1)i

= p(1)1 (t)

ts1e1 −

x(s1)�

i=1

a(1)iχ(s1)i

+ ...+ p(1)d(t)

tsded −

x(sd)�

i=1

a(d)iχ(sd)i

...

tσded −

x(σd)�

i=1

b(d)iξ(σd)i

= p(d)1 (t)

ts1e1 −

x(s1)�

i=1

a(1)iχ(s1)i

+ ...+ p(d)d(t)

tsded −

x(sd)�

i=1

a(d)iχ(sd)i

where p(1)1 (t), ..., p

(d)d(t) ∈ B[t]. Suppose that G = Zn, the case G = Z will follow

in the same way and is easier. It is enough to prove that this has a solution in thecase when the basis elements in X and Y only differ in the smallest possible way,so we can assume that we have σ1 = s1 + r, σ2 = s2 − r, σi = si for i = 3, ..., d andr = s1 − s2 − nx > 0 where x is as large as possible, i.e.

X = {e1, te1, ..., ts1−1e1, e2, te2, ..., t

s2−1e2, ..., t

sd−1ed}

andY = {e1, te1, ..., ts1+r−1

e1, e2, te2..., ts2−r−1

e2, ..., tsd−1

ed}.Then, for any g ∈ G, we have that Xg and Yg differ at most with one element. Lets1 = q1+nm1 and s2 = q2+nm2 where m1,m2, q1, q2 ∈ N and q1, q2 < n. The numberof basis elements of the form t

ue1 in X of degree s2 is equal to m1 + δ

X

1 where

δX

1 =

�0 if q1 − q2 ≥ 0,

−1 if q1 − q2 < 0.

In the same way we have that the number of basis elements of the form tue2 of degree

s1 in X is equal to m2 + δX

2 where

δX

2 =

�0 if q2 − q1 ≥ 0,

−1 if q2 − q1 < 0.

35


Furthermore, we have that σ1 = q1+ r+nm1 and σ2 = q2− r+nm2. If q1+ r < n

then there are m1 elements of degree s1 in Y and if q1 + r ≥ n then there are m1 + 1

elements of degree σ1 in Y . We therefore define

δY

1 =

�0 if q1 + r < n,

1 if q1 + r ≥ n,

so that there are m1 + δY

1 elements of degree s1 in Y . Similarly with σ2 we define

δY

2 =

�0 if q2 − r ≥ 0,

−1 if q2 − r < 0.

Let also qi + r = qi + r − nδY

ifor i = 1, 2.

We only need to find one solution to our system of equations and we will showthat we can find one if we choose, for i, j = 3, ..., d,

p(i)j(t) =

�1 if i = j,

0 if i �= j,

as well as a(j)r = 0 and b(j)r = 0 for r = 1, ..., x(si). We also let p(1)1 (t) = t

r, p(2)2 (t) = 0,

p(1)3 (t), ..., p

(1)d(t) = 0

andp(2)2 (t), p

(2)3 (t), ..., p

(2)d(t) = 0.

Finally, we let those coefficients, a(j)i, b

(j)i

, of the basis elements that are of the formtuer be zero for r = 3, ..., d, j = 1, 2. Then, our system of equations is reduced, if we

write our basis elements explicitly, to

tσ1e1 −m1+δY1�

i=1

b(1)i tq1+r+n(i−1)e1 −m2+δX2 +δY2�

i=1

b(1)i+m1+δtq1+r+n(i−1)e2

= tr

ts1e1 −m1�

i=1

a(1)i tq1+n(i−1)e1 −m2+δX1�

i=1

a(1)i+m1tq1+n(i−1)e2

tσ2e2 −m1+δX1 +δY1�

i=1

b(2)i tq2−r+n(i−1)e1 −m2+δY2�

i=1

b(2)i+m1tq2−r+n(i−1)e2

= p(2)1 (t)

ts1e1 −m1�

i=1

a(1)i tq1+n(i−1)e1 −m2+δY2�

i=1

a(1)i+m1tq1+n(i−1)e2

.

36


This might seem quite complicated but by choosing

p(2)1 (t) = t

σ2−q1−(m2−1)n= t

q2−r+n(m2+δ2)−q1−n(m2−1)= t

q2−q1−r+n(1+δ2)

we can simply read of the solution one coordinate at a time. It follows that there is asolution to our system and that solution gives us a submodule N that is a member ofboth relQuot

X

V(B) and relQuot

Y

V(B). Hence the intersection of any two relative Quot

functors is non-empty and it follows that the Quot scheme is connected.Since any smooth connected scheme is irreducible, it follows that the Quot scheme

is irreducible.

37

REFERENCES G. Sædén Ståhl

References

[1] Michael F. Atiyah and Ian G. Macdonald. Introduction to Commutative Algebra.Westview Press, 1969.

[2] Nicolas Bourbaki. Commutative algebra. Springer-Verlag Berlin and Heidelberg,1998.

[3] David S. Dummit and Richard M. Foote. Abstract Algebra. John Wiley & SonsInc., Hoboken, NJ, third edition, 2004.

[4] David Eisenbud and Joe Harris. The Geometry of Schemes. Springer-Verlag NewYork, Inc., 2000.

[5] Ulrich Görtz and Torsten Wedhorn. Algebraic geometry I : Schemes With Ex-

amples and Exercises. Vieweg+Teubner Verlag, first edition, 2010.

[6] Mark Haiman and Bernd Sturmfels. Multigraded Hilbert Schemes. ArXiv Math-

ematics e-prints, January 2002, arXiv:math/0201271.

[7] Robin Hartshorne. Algebraic geometry. Springer, 1977.

[8] Ernst Kunz. Introduction to Commutative Algebra and Algebraic Geometry.Birkhäuser, 1985.

[9] Marvin Mendelssohn. Graded rings, modules and algebras. Master’s thesis, TheMassachusetts Institute of Technology, 1970.

[10] Shigeru Mukai. An Introduction to Invariants and Moduli. Cambridge UniversityPress, 2003.

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