On the Smoothness of the Quot Functor

S EBASTIAN W IEGANDT

Master of Science Thesis Stockholm, Sweden 2013

S EBASTIAN W IEGANDT

Master’s Thesis in Mathematics (30 ECTS credits) Master Programme in Mathematics (120 credits) Royal Institute of Technology year 2013 Supervisor at KTH was Roy Skjelnes Examiner was Roy Skjelnes

TRITA-MAT-E 2013:29 ISRN-KTH/MAT/E--13/29--SE

Royal Institute of Technology School of Engineering Sciences

KTH SCI SE-100 44 Stockholm, Sweden

URL: www.kth.se/sci

Abstract

For a commutative ring k, we consider free k-modules E, endowing them with k[x1, . . . , xm]- module structures through a ring homomorphism k[x1, . . . , xm] → EndZ(E). These structures are then inspected by encoding the actions of the unknowns xi in matrices X1,...,Xm. We further introduce the concepts of lifts and formal smoothness for functors, and deﬁne the n QuotF/A/k functor acting on the category of k-algebras, taking some k-algebra B to the set of quotients of the form (F ⊗k B)/N, which are locally free as B-modules. Lastly, we ﬁnd con- 4 2 crete examples of modules showing that the functors Hilbk[x,y,z]/k and QuotL2 k[x,y]/k[x,y]/k are not formally smooth.

Contents

1 Introduction 3 1.1 Background ...... 3 1.2 Overview ...... 4 1.3 Acknowledgments ...... 5

2 Preliminaries 6 2.1 Categories ...... 6 2.1.1 Yoneda’s Lemma ...... 6 2.1.2 Representable Functors ...... 8

3 Free Quotients 10 3.1 Quotients in Rank One ...... 10 3.1.1 One Variable ...... 11 3.1.2 Several Variables ...... 12 3.2 Quotients in Rank Two ...... 13 3.2.1 One Variable ...... 14 3.2.2 Two Variables ...... 14

4 Lifts and Formal Smoothness 16

5 The Quot and Hilb Functors 19 5.1 The Quot Functor and Free Quotients ...... 19 5.2 Representing Simple Cases ...... 20 5.3 Smoothness of the Hilb Functor ...... 23 5.4 Smoothness of the Quot Functor ...... 27

References 31

1 INTRODUCTION

1 Introduction

n The Hilbert scheme HA of n points on a k-algebra A parametrizes ideals I ⊆ A such that A/I is locally free of rank n as a k-module. These parametrizing objects are complicated, but one has that the tangent space of H at the point corresponding to I is given by Hom(I/I2, A/I). This fact has often been used to extract information about the Hilbert scheme. Fogarty, for instance, used this fact to show that the Hilbert scheme with A = k[x, y] is smooth [2]. There is then an open set that has dimension 2n, and Fogarty showed that any point has tangent space of dimension 2n, which implied smoothness. For A = k[x, y, z, w], the Hilbert scheme has an open set of dimension 4n, and for particular choices of I one can compute and ﬁnd a tangent space of diﬀerent dimension. This is for example the case with n = 8 and

I = (x2, xy, y2, z2, zw, w2, xz − yw), which has tangent space of dimension 25 [3, p. 42]. So, here the Hilbert scheme is not even irreducible. In this work, we will take a different approach using matrix representations. Instead of starting with an ideal I ⊆ k[x1, . . . , xm] such that the quotient ring is free of rank n as a k- module, we start with an arbitrary free k-module E. Then an k[x1, . . . , xm]-module structure on E is given by pairwise commuting endomorphisms X1,...,Xm. Thereafter we ask if these actions can be infinitesimally lifted, which is an equivalent way of describing smoothness of n the moduli space HA. By doing so, we find very small examples where the Hilbert and Quot functors, which are representable by the Hilbert and Quot schemes respectively ([5]), are not smooth. In fact, these are the smallest that can exist.

1.1 Background

Given a free k-module E with basis {e1, . . . , en}, we can endow E with a k[x]-module structure through a ring homomorphism k[x] → EndZ(E), which is uniquely deﬁned by the endomorphism X it maps the element x to. Consider for example the matrix   0 0 ··· 0 a0 1 0 ··· 0 a1    0 1 ··· 0 a2  X =   , ......  . . . . .  0 0 ··· 1 an−1 encoding the action of x on E. This in fact gives us E as the quotient

n n−1 k[x]/(x − (a0 + a1x + ... + an−1x )), by considering the action of x on {1, x, . . . , xn−1}. In fact, by introducing m commuting matrices X1,...,Xm, we can endow E with a k[x1, . . . , xm]-module structure in the same way. Now, given we have a k[x1, . . . , xm]-module structure on E, suppose we have a k[x1, . . . , xm]- module homomorphism ϕ: k[x1, . . . , xm] E

3 1.2 Overview 1 INTRODUCTION onto E. By the first isomorphism theorem for modules, we then have that E = k[x1, . . . , xm]/ ker(ϕ) is a k[x1, . . . , xm]-quotient of rank n as a free k-module. Generalizing this to arbitrary k-algebras n A, the functor QuotF/A/k is then defined as taking a k-algebra B to the set of F ⊗k B-module quotients F ⊗k B E which are locally free of rank n as B-modules, with F some fixed n A-module. In particular, if F = A we get the functor HilbA/k as a special case. The concept of formal smoothness for these functors arise naturally from the corresponding concept of formal smoothness for a ring homomorphism f : A → B, which is in this sense an n infinitesimal lifting property. We will then say that the functor QuotF/A/k is smooth if it fulfills n a certain infinitesimal lifting property itself, namely that any element E in QuotF/A/k(B/N), n for some N ⊆ B with certain properties, admits a lift to some E in QuotF/A/k(B), that reduces to E modulo N.

2 Example 1.1. As a descriptive example of this formal smoothness criterion, let D2 = k[ε]/(ε ) and consider the free D2-module

2 2 3 E = D2[x, y]/(x − εy, y − εx, xy) ∈ Hilbk[x,y]/k(D2).

3 With D3 = k[ε]/(ε ), this admits a lift to

2 2 2 3 E = D3[x, y]/(x − εy, y − εx, xy − ε ) ∈ Hilbk[x,y]/k(D3), and we can identify B = k[ε]/(ε3) and N = (ε2) in the above discussion. We see that reducing E modulo ε2 indeed gives us back E, and note that such a lift was bound to exist, since the Hilbert scheme on k[x, y] is smooth.

The aim of this text is to expand on this construction of free module structures, with the ultimate goal of ﬁnding elements of certain Hilb and Quot functors that do not admit lifts in the preceding sense. With this in mind, we will develop additional ways of interpreting the smoothness condition given, primarily concerning ourselves with previously mentioned matrix representations of module structures. A brief aside will also be given on the representability of the Hilb and Quot functors in some simple cases.

1.2 Overview In Section 2, we introduce some category theoretic language, which we will use in the later stages of this paper. In particular, the concepts of categories and functors are introduced, followed by an aside on the Yoneda lemma and representable functors. In Section 3, we consider free k-modules E, and endow them with k[x1, . . . , xm]-module structures through a ring homomorphism k[x1, . . . , xm] → EndZ(E), taking xi to an endomorphism Xi : E → E. Given a basis {e1, . . . , en} for E, these can be considered as n × n matrices acting on E, encoding the actions of the unknown variables. For example the actions of x and y on k[x, y]/(x2, y2, xy) can be encoded in the matrices

0 0 0 0 0 0 X = 1 0 0 ,Y = 0 0 0 , 0 0 0 1 0 0

4 1 INTRODUCTION 1.3 Acknowledgments by considering the actions of x and y on the elements 1, x and y. Further, given r elements e1, . . . , er ∈ E, we induce a k[x1, . . . , xm]-module homomorphism

r M ϕ: k[x1, . . . , xm] → E by mapping (f1, . . . , fr) to f1(X1,...,Xm)e1 +...+fr(X1,...,Xm)er. If this mapping is surjec- Lr tive, we get E as a quotient ( k[x1, . . . , xm])/ ker(ϕ), which is free of rank n as a k-module. In Section 4, we introduce the concepts of lifts and formal smoothness, in this sense an infinitesimal lifting property concerning the ability of lifting certain B/N-modules for some k-algebra B and nilpotent N ⊆ B to the larger algebra B. Lastly, we will in Section 5 revisit many of the examples given in Section 3, translated into n a category theoretic language. We will introduce the Quot functor QuotF/A/k : k-alg → Set for a commutative ring k, a k-algebra A and an A-module F. This functor maps a k-algebra algebra B to quotients of the form (F ⊗k B)/N, which are locally free as B-modules. We will inspect some specific cases, followed by a discussion on the formal smoothness of this functor. This discussion concludes with showing the non-smoothness of two specific cases of the Quot functor, using concrete examples of non-liftable elements.

1.3 Acknowledgments I would like to give my deepest thanks to my supervisor Roy Skjelnes for his invaluable support and our many rewarding discussions.

5 2 PRELIMINARIES

2 Preliminaries

In this section, we will introduce some basic category theoretical concepts, and explore their relation to k-algebras for commutative rings k.

2.1 Categories We will begin with the fairly abstract general definition of a category, and then swiftly move on to our specific objects of interest, namely k-algebras. However, the general formulation will come in handy for some of the statements of the theory. Definition 2.1. A category C consists of a class of objects and a class of morphisms (arrows) between objects. Further, letting the class of morphisms from object a to object b be denoted hom(a, b), there is a binary operation hom(a, b) × hom(b, c) → hom(a, c) for any three objects a, b, c called composition of morphisms, written g ◦f for morphisms f : a → b, g : b → c. Further, morphism composition is associative, and for every object x there is a morphism 1x : x → x such that for any morphism f : a → b, we have 1b ◦ f = f = f ◦ 1a. To get a sense of this rather abstract definition, let us specifically look at the category k-alg of k-algebras for a commutative ring k. Here, the objects are k-algebras, while the morphisms (or arrows) f : A → B for two k-algebras A and B are the k-algebra homomorphisms from A to B, denoted Homk-alg(A, B), consisting of the set of k-linear maps f between vector spaces A and B such that f(xy) = f(x)f(y) for all x, y ∈ A and f(1) = 1. Another important and simple category is that of sets, which we will denote by Set. Its objects are, as one would assume, sets, and its morphisms are functions between sets. Now that categories can be viewed as an extension of the concept of sets, we want to introduce a concept analogous to functions, defined on general categories. Therefore, we introduce the following concept of functors: Definition 2.2. Let C and D be two categories. Then a (covariant) functor F : C → D is a mapping such that • for any object X ∈ ob(C), F (X) is an element of ob(D), and • for any morphism f : X → Y in hom(C) we get a morphism F (f): F (X) → F (Y ) in hom(D), such that F (1X ) = 1F (X) for any X ∈ ob(C), and for any two morphisms f : X → Y and g : Y → Z in hom(C) we have F (g ◦ f) = F (g) ◦ F (f).

2.1.1 Yoneda’s Lemma There is one result of category theory that will be of special interest and importance to us, called the Yoneda lemma. In order to state the result, however, we will need to introduce a couple of more concepts. Having formulated the deﬁnition of a functor, we would now like to have some way of transforming a functor F : C → D into some other functor G: C → D in a natural way. Indeed, the below deﬁnition gives us just that.

6 2 PRELIMINARIES 2.1 Categories

Deﬁnition 2.3. Given two functors F and G between categories C and D, a natural transformation Φ from F to G associates to every object X ∈ ob(C) a morphism ΦX : F (X) → G(X) in hom(D), such that for every morphism f : X → Y in hom(C), the following diagram commutes:

F (f) F (X) - F (Y )

ΦX ◦ ΦY ? ? G(X) - G(Y ). G(f)

Further, if ΦX is an isomorphism for every X ∈ C, then Φ is called a natural isomorphism, and F and G are said to be naturally isomorphic.

Now, the last step before formulating the Yoneda lemma is introducing the hom-functor for small enough categories (in the sense that hom(a, b) form sets for all a, b in the category). However, in order to not get bogged down in details, we will restrict the deﬁnition to the category k-alg. This functor will allow us to ﬁnd a correspondence between objects of our category and its set of morphisms.

Deﬁnition 2.4. Given a k-algebra A, the hom-functor hA : k-alg → Set is the functor Hom(A, −), taking a k-algebra B to the set of k-algebra homomorphisms Homk-alg(A, B), and a homomorphism f : B → B0 to the morphism f ◦ −, denoting composition.

We are now ready to state the theorem, which actually allows us to transform the study of the category k-alg to the study of the category of functors F : k-alg → Set, which might be easier, considering Set is a very well understood category.

Theorem 2.1 (Yoneda’s lemma). Let F be a functor from k-alg to Set. Then, for any k- algebra A, there is a one-to-one correspondence between the natural transformations from hA to F and the elements of F (A).

Proof. Consider a natural transformation Φ: hA → F . Then, the diagram

hA(f-) hA(A) hA(B)

ΦA ΦB ? ? F (A) - F (B) F (f) commutes for every k-algebra B and every k-algebra homomorphism f : A → B. Speciﬁcally, we have that

ΦB((hA(f))(idA)) = ΦB(f ◦ idA) = ΦB(f) = (F (f))(ΦA(idA))

7 2.1 Categories 2 PRELIMINARIES for any B. Therefore, ΦB is uniquely determined by the element ΦA(idA) ∈ F (A), and the mapping Φ 7→ ΦA(idA) is an injection Nat(hA,F ) → F (A). For any x ∈ F (A), consider now the induced map ΨB : f 7→ (F (f))(x), for any morphism f : A → B. If this deﬁnes a natural transformation Ψ: hA → F , then we speciﬁcally have

ΨA(idA) = (F (idA))(x) = idF (A)(x) = x, showing that Φ 7→ ΦA(idA) is in fact also surjective, establishing a bijection between Nat(hA,F ) and F (A). For this to be the case, we want the diagram

hA(f-) hA(B) hA(C)

ΨB ΨC ? ? F (B) - F (C) F (f) to commute for any k-algebras B and C, and for any k-algebra homomorphism f : B → C. But this is indeed the case, since for any morphism g : A → B, we have

ΨC ((hA(f))(g)) = ΨC (f ◦ g) = (F (f ◦ g))(x) = (F (f) ◦ F (g))(x) = (F (f))(ΨB(g)), showing that the diagram commutes. Thus, we have shown that the mapping Φ 7→ ΦA(idA) is a bijection, proving the theorem.

In particular, this tells us that when F = hB for some k-algebra B, we have a one-to-one correspondence between Homk-alg(B,A) and the natural transformations from hA to hB.

2.1.2 Representable Functors

We may now ask ourselves, what can be said about a general functor F in relation to hA? In particular, in the case that F is naturally isomorphic to hA, we give the following deﬁnition. Deﬁnition 2.5. A functor F : k-alg → Set is said to be representable if it is naturally isomorphic to hA for some k-algebra A. Further, a representation of F is a pair (A, Φ) such that Φ: hA → F is a natural isomorphism. Lastly, the element ξ = ΦA(idA) is called the universal element of the representation.

Now, in light of the Yoneda lemma, we know that natural transformations Φ: hA → F is in one-to-one correspondence with elements of F (A), and that ξ = ΦA(idA) corresponds to a given Φ. Further, given any element ξ ∈ F (A), we can deﬁne a natural transformation Φ: hA → F by

ΦX (f) = (F (f))(ξ) for f ∈ Hom(A, X). In order for (A, Φ) to be a representation of F , we need Φ to be an isomorphism, which is clearly dependent on ξ. Therefore, we see that by deﬁnition, ξ is a universal element if and only if the natural transformation induced by it is an isomorphism. Thus, the

8 2 PRELIMINARIES 2.1 Categories universal elements are in one-to-one correspondence with representations of the functor F . In other words, we can identify any representation of F using its corresponding universal element. All of the previous is true for the category of commutative rings, CRing, as well, allowing us to end the section by giving this fairly elementary example of a representable functor:

Example 2.1. Let A be a commutative ring and S ⊆ A be a multiplicative subset of A. Consider the functor F : CRing → Set, deﬁned by

F (B) = {ϕ: A → B | ϕ a ring homomorphism such that ϕ(s) ∈ B∗ for all s ∈ S} for any commutative ring B, and F (f) = f ◦ − for any ring homomorphism f : X → Y . Then the localization of A by S, S−1A, is a member of CRing, and i: A → S−1A, taking a to (a, 1) is a member of F (S−1A). Now, consider the natural transformation Φ: Hom(S−1A, −) → F induced by i. For any commutative ring X and ring homomorphism f : S−1A → X, we have that

ΦX (f) = (F (f))(i) = f ◦ i.

Suppose now that g : A → X ∈ F (X). The universal property of the localization then tells us that there is a unique ring homomorphism fb: S−1A → X such that fb◦ i = g. Therefore, Φ is a natural isomorphism, and the functor F is indeed representable by S−1A, with the universal element i: A → S−1A. ♦

9 3 FREE QUOTIENTS

3 Free Quotients

Let k be a commutative ring. We will in this section introduce and study some properties of Lr free k-modules on the form ( k[x1, x2, . . . , xm]) /I, and will speciﬁcally take an in-depth look at some cases with the rank 1 ≤ r ≤ 2. We start by proving the following proposition, showing how we can extend a k-module structure to a k[x1, . . . , xm]-module structure.

Proposition 3.1. Let k be a commutative ring, E be a k-module and A = k[x1, . . . , xm]. Then, an A-module structure on E, compatible with the given k-module structure, is equivalent to having m commuting k-linear endomorphisms Xi : E → E (i = 1, . . . , m).

Proof. The fact that E is a k-module can be expressed as (E, +) being an abelian group, and having a ring homomorphism ϕ: k → EndZ(E), where (ϕ(r))(e) := r · e. We can extend ϕ to ϕe: A → EndZ(E) by letting ϕe(r) = ϕ(r) for all r ∈ k and then letting ϕe(xi) = Xi for some commuting endomorphisms Xi (since xixj = xjxi implies Xi ◦ Xj = Xj ◦ Xi). Lastly, deﬁning ϕe on k and each variable separately deﬁnes it on all of k[x1, . . . , xm], since it is a ring homomorphism.

Corollary 3.1. With A and E as in the proposition, m endomorphisms Xi : E → E and an element e ∈ E determines an A-module homomorphism A → E, given by the composition

ev ϕ: A → End(E) −−→e E.

Further, if ϕ: A → E is surjective, we have that E is a quotient A/I, with I the kernel of the map.

Proof. We begin by noting that HomA(A, E) is isomorphic to E as A-modules, since if ϕ: A → E is an A-module homomorphism, it is uniquely determined by ϕ(1), by noting that ϕ(a) = a·ϕ(1) for all a ∈ A. Therefore, we have a one-to-one correspondence between homomorphisms ϕe and elements e = ϕe(1) in E. The map taking ϕe to e is further easily seen to be an A-module homomorphism, by the deﬁnition of the map. Now, suppose ϕe(1) = e for some e ∈ E. Then, for arbitrary a ∈ A, we have that ϕe(a) = a·e, which by deﬁnition equals (ψ(a))(e), where ψ : A → EndZ(E) is our A-module structure on E. We therefore see that the composition

ev ϕ: A → End(E) −−→e E does indeed give us back ϕe.

3.1 Quotients in Rank One

Letting A = k[x1, . . . , xm], we will now be inspecting k-modules of the form A/I that are free of rank, say, n. In fact, we will simplify it even further, by looking at just a single variable. In this case, the categorization is exceedingly easy to do.

10 3 FREE QUOTIENTS 3.1 Quotients in Rank One

3.1.1 One Variable Suppose E is a free k-module of rank n. We can endow E with a k[x]-module structure by mapping x to some endomorphism X : E → E, which, given we make sure the map ev k[x] → End(E) −−→e E is surjective, gives us that E = k[x]/I for some ideal I ⊆ k[x]. We will inspect these structures in more detail in the below example, following a lemma simplifying the discussion. Lemma 3.1. With k a ﬁeld, any k-vector space E of dimension n of the form k[x]/I has a basis {1, x, . . . , xn−1}. Proof. Since k is a ﬁeld, k[x] is a principal ideal domain, and so I = (F (x)) for some (without loss of generality, monic) F ∈ k[x]. Further, I contains some polynomial G of least degree, say m, for which F (x) | G(x). Therefore, deg F ≤ deg G, and so deg F = m. m m−1 Suppose now that m < n. Then F (x) = x − (a0 + a1x + ... + am−1x ) = 0 in E, and so {1, x, . . . , xm−1} spans E, which therefore cannot be of dimension n. However, if m = n, then {1, x, . . . , xn−1} spans E by the same reasoning, and further,

n−1 H(x) = b0 + b1x + ... + bn−1x = 0 in E implies that F (x) | H(x), and therefore deg F ≤ deg H, which is clearly untrue. Therefore, {1, x, . . . , xn−1} is a basis for E.

Example 3.1. Let k be a ﬁeld, and E a free k-module of rank n. We can, as previously mentioned, endow E with a k[x]-module structure by introducing a matrix X acting on E, and given consideration to its construction, we have that E = k[x]/I for some ideal I ⊆ k[x]. Further, by the lemma we then have that {1, x, . . . , xn−1} is a basis for E. Consider the endomorphism given by   0 0 ··· 0 a0 1 0 ··· 0 a1    0 1 ··· 0 a2  X =   . ......  . . . . .  0 0 ··· 1 an−1

ev(1) n−1 This induces a surjective map k[x] → End(E) −−−→ E, since b0 + b1x + ... + bn−1x maps n n−1 n to itself in E. Further, x 7→ a0 + a1x + ... + an−1x , showing that F (x) = x − (a0 + a1x + n−1 ... + an−1x ) = 0 in E. Moreover, the (not necessarily monic) characteristic polynomial of x is given by

−t 0 ··· 0 a0

1 −t ··· 0 a1

0 1 ··· 0 a2 , ......

0 0 ··· 1 an−1 − t

11 3.1 Quotients in Rank One 3 FREE QUOTIENTS which after expanding along the last column comes out to

n−1 ! n−1 X 2i i 2n n−1 n (−1) (−1) bit + (−1) (bn−1 − t) · t = (−1) F (t). i=1 Therefore, the monic characteristic polynomial of x is given by F (t). Then, by Cayley-Hamilton, in fact F (X) = 0 as well. ♦

3.1.2 Several Variables We will now move forward by looking at free quotients of polynomial rings in several variables, and speciﬁcally at some examples in two and three variables. Generally, such a free quotient is given by k[x1, x2, . . . , xm]/I for some ideal I deﬁning relations so that the quotient is of some rank n.

Example 3.2. Let E be a free k-module of rank 3, and let {e1, e2, e3} be a basis. Consider the endomorphisms     0 a0 c0 0 c0 b0 X = 1 a1 c1 ,Y = 0 c1 b1 0 a2 c2 1 c2 b2 on E, with ai, bi, ci some elements in k. If the nine equations given by XY − YX = 0 are satisﬁed, Proposition 3.1 tells us that they endow E with a k[x, y]-module structure. Further, letting e = e1 in Corollary 3.1, the mapping

ev(e1) k[x, y] → Endk(E) −−−−→ E is surjective (since a + bx + cy maps to ae1 + be2 + ce3 for any a, b, c ∈ k), showing that E = k[x, y]/I for some ideal I ⊆ k[x, y]. I is then given by

2 2 I = (x − (a0 + a1x + a2y), y − (b0 + b1x + b2y), xy − (c0 + c1x + c2y)), taken from the actions of x and y induced by the endomorphisms. ♦ This example illustrates how we will use Proposition 3.1 and Corollary 3.1 in the remainder of this text. Given a free k-module E, we consider a commuting set of endomorphisms on E, and if the composition k[x1, . . . , xm] → Endk(E) → E is surjective, E is given by a k[x1, . . . , xm]-quotient. Example 3.3. We will again look at a speciﬁc example, this time in three variables, and with 0 2 0 k = k [ε]/(ε ) for some commutative ring k . Let E be a free k-module with basis {e1, e2, e3, e4}, and consider the matrices 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 ε X =   ,Y =   ,Z =   , 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 ε 0 1 0 0 0

12 3 FREE QUOTIENTS 3.2 Quotients in Rank Two along with the 4 × 4 identity matrix. We have that XY = YX = XZ = XZ = YZ = 0 and

0 0 0 0 0 0 ε2 0 ZY =   , 0 0 0 0 0 0 0 0 and so the map into the endomorphism ring is commutative (since ε2 = 0 in k). Further, the composition ev1 k[x, y, z] → Endk(E) −−→ E is surjective, since each endomorphism maps e1 to a diﬀerent basis element among {e1, e2, e3, e4}. Therefore, these endomorphisms endow E with a k[x, y, z]-module structure, and further tells us that E = k[x, y, z]/I for some I ⊆ k[x, y, z]. The induced actions of x, y and z then give us that I = (xy, xz, yz, x2, y2 − εz, z2 − εx). ♦

3.2 Quotients in Rank Two

We will now focus our attention to quotients of the form F/I with F = k[x1, . . . , xm] ⊕ k[x1, . . . , xm], and more speciﬁcally the case with one and two variables. Before we begin, we will have to alter Corollary 3.1 slightly. While a ring homomorphism A → End(E) still endows E with an A-module structure, we now want to ﬁnd A-module homomorphisms F → E. Therefore, we generalize the corollary to the following (while the proof remains essentially the same): Lr Corollary 3.2. Let k be a commutative ring and A = k[x1, . . . , xm]. With F = A and E a k-module, having m commuting endomorphisms Xi : E → E and r elements ei ∈ E (i = 1, . . . , r) is equivalent to an A-module homomorphism F → E, given by the composition

r M ev ϕ: F → End(E) −→ E, where the last arrow denotes the map (F1,...,Fr) 7→ F1(e1) + ... + Fr(er). Lr Proof. We note that EndA(F,E) = E, again since an A-module homomorphism F → E is uniquely determined by where it maps the r elements (0, 0,..., 1, 0,..., 0), with a 1 in the ith place (i = 1, . . . , r). Having chosen e1, . . . , er, the corresponding homomorphism ϕe : F → E is given by

ϕe(a1, . . . , ar) = ϕe((a1, 0,..., 0) + ... + (0, 0, . . . , ar)) = a1 · e1 + ... + ar · er, which by deﬁnition equals (ψ(a1))(e1) + ... + (ψ(ar))(er), where ψ : A → EndZ(E) is the A-module structure on E. This ﬁnally shows that ϕe is indeed the map ϕ as given above.

13 3.2 Quotients in Rank Two 3 FREE QUOTIENTS

3.2.1 One Variable Suppose E is a free k-module of rank n, and we want to ﬁnd a k[x]-module homomorphism k[x] ⊕ k[x] → E. Then a ring homomorphism k[x] → EndZ(E) endows E with a k[x]-module structure as previously. The same machinery as in the previous subsection can therefore be used in higher ranks. Speciﬁcally, if

2 2 M M eve k[x] → End(E) −−→ E is a surjective k[x]-module homomorphism, then E is given by a quotient (k[x] ⊕ k[x])/I. We will inspect a speciﬁc example below.

Example 3.4. Let E be a free k-module of rank 2 with basis {e1, e2}. Consider the endomorphism on E given by a b X = 1 1 . a2 b2 This again endows E with a k[x]-module structure, and if the composition

2 M k[x] ⊕ k[x] → Endk(E) → E is surjective, we have that E = (k[x] ⊕ k[x])/N for some submodule N ⊆ k[x] ⊕ k[x]. Choosing to evaluate in e1 and e2, the map will be surjective if X itself is. Considering the action of x on E, we get that N = ((x, 0) − (a1, a2), (0, x) − (b1, b2)). ♦

3.2.2 Two Variables Given a basis for a free module in two variables, we can describe the actions of x and y in the same manner as above. The fully general situation is a bit more tedious to write down, so we will restrict the scope and analyze the following example instead.

Example 3.5. Let E be a free A-module with basis {e1, e2}, and let the actions of x and y on E be described by the matrices

1 0 0 ε X = ,Y = , ε 1 0 0 with A = k[ε]/(ε2) for some commutative ring k. Again, for these to give rise to an A[x, y]- module structure, the endomorphisms need to commute. We have that

0 ε ε2 ε XY = ,YX = , 0 ε2 0 0

2 and so they do over A, where ε = 0. Further, evaluating in e1 and e2, the map k[x, y] → EndA(E) → E is surjective, since

X(ce1 + (d − εc)e2) = ce1 + de2,

14 3 FREE QUOTIENTS 3.2 Quotients in Rank Two showing that the image spans all of E. Thus, the structure they generate is a free A-module, given by the quotient

(A[x, y] ⊕ A[x, y])/((x − 1, −ε), (0, x − 1), (y, 0), (−ε, y)).

♦

The reason for these somewhat obscure examples using the dual numbers will become clear in the next section, where we will deﬁne and discuss lifts and formal smoothness of maps.

15 4 LIFTS AND FORMAL SMOOTHNESS

4 Lifts and Formal Smoothness

We will now introduce the concept of formally smooth maps, at ﬁrst for k-algebras, but then later extend this to formally smooth transformations between functors. Formal smoothness in this sense is essentially an inﬁnitesimal lifting property, in that a mapping into a small neighborhood can be lifted to a larger space.

Deﬁnition 4.1. Let f : A → B be a k-algebra homomorphism. The map f is then said to be formally smooth if for every k-algebra homomorphism A → A0, every nilpotent ideal N ⊆ A0 2 0 such that N = (0) and every A-algebra homomorphism gb: B → A /N, there exists an A-algebra homomorphism (called a lift) g : B → A0 so that the diagram

g B b - A0/N 6 6 g f

- A - A0 is commutative.

Example 4.1. Consider the diagram

p B = k[x, y]/(xy) -b k[ε]/(ε2) = A0/N 6 6 p

- A = k - k[ε]/(ε3) = A0 having identiﬁed A, B and A0 in the above deﬁnition, with N = (ε2) begin a nilpotent ideal in 3 0 k[ε]/(ε ), and the map pb: B → A /N taking both x and y to ε. We note that since pb(xy) = ε2 = 0 in k[ε]/(ε2), this is a k-algebra homomorphism. Due to the singularity at the origin, we would not expect f : A → B to be smooth, and we will see that this is indeed the case. 0 Suppose that the lift p: B → A of pb exists. Then, since the above diagram should commute, we must have p(x) = ε + aε2 and p(y) = ε + bε2 for some a, b ∈ k. However, then we have that p(xy) = p(x)p(y) = (ε + aε2)(ε + bε2) = ε2 6= 0 in k[ε]/(ε3), and so no such homomorphism p exists. Therefore, k → k[x, y]/(xy) is not formally smooth. ♦

Drawing inspiration from this deﬁnition, we note that by Yoneda, every k-algebra homomorphism A → B corresponds to exactly one natural transformation hB → hA. Further, suppose we have k-algebra homomorphisms f : A → B and g : B → C. Then the functor B 7→ hB will take the morphism g to the precomposition by it, g ◦ f, and so this functor will invert the arrow f to the transformation f∗ : hB → hA. Therefore, the following deﬁnition is an intuitive analog of the previous one in the case of representable functors that also extends to arbitrary functors.

16 4 LIFTS AND FORMAL SMOOTHNESS

Deﬁnition 4.2. Let F,G: k-alg → Set be two functors, and ϕ: F → G be a natural transformation. We say that the map ϕ is formally smooth if for every k-algebra B and every nilpotent 2 ideal N ⊆ B such that N = (0) and every transformation ψb: hB/N → F there exists a lift ψ : hB → F so that the diagram

ψb F hB/N

ψ ϕ

? ? G hB is commutative. Further, we call a functor F smooth if for every transformation ψb: hB/N → F there exists a lift ψ : hB → F so that

ψb F hB/N

? hB commutes.

Remark. Observe that all the arrows have been reversed in relation to the definition for smoothness of k-algebra homomorphisms. For now, we will not specifically concern ourselves that much with the concept of formal smoothness, but rather look into what a lift of quotients entails. Suppose E0 is a free A/N- module of rank n for some commutative ring A and nilpotent ideal N as above. We then say that E0 can be lifted to a free A-module E if the projection of E onto A/N is E0 and E is free of rank n. Further, identifying a quotient of a polynomial ring A/N[x1, . . . , xm] with the actions of its variables as we have done previously, a lift can also be found by finding commutative matrices with elements in A whose projections onto A/N give us the original matrices. We illustrate this in an example below. This is not yet an argument about functorial smoothness, but rather about free quotients, as noted. However, in the next section, this machinery will immensely help us show the non-smoothness of certain functors that are to be introduced.

Example 4.2. Let A be the ring k[ε]/(ε3) and N = (ε2). The element E = (A/N)[x, y]/(x2 − εy, y2 − εx, xy) is a free A/N-module with basis {1, x, y}, for which the actions of x and y are given by 0 0 0 0 0 0 X = 1 0 0 ,Y = 0 0 ε . 0 ε 0 1 0 0

17 4 LIFTS AND FORMAL SMOOTHNESS

The products of these are

0 0 0  0 0 0 XY = 0 0 0  ,YX = 0 ε2 0 , 0 0 ε2 0 0 0 showing that they commute over A/N, but that the obvious (trivial) lift is not commutative over A. This is also evident from direct computation in E, where we have

0 = (xy)x = x2y = εy2 = ε2x, meaning that the obvious lift would not be free as an A-module generated by {1, x, y}. However, lifting the matrices to 0 0 ε2 0 ε2 0 X = 1 0 0  , Y = 0 0 ε , 0 ε 0 1 0 0 we have that ε2 0 0  XY = YX =  0 ε2 0  , 0 0 ε2 rendering a commutative representation. Further, this lifts E to A[x, y]/(x2−εy, y2−εx, xy−ε2), where ε2x = (xy)x = x2y = εy2 = ε2x, voiding the previous issue. ♦

As this example shows, it is often not enough to consider the most obvious lift of a quotient, but we rather have to ﬁnd some clever way to construct a lift that is actually a free module. In some cases, which we will cover in the next section, a lift is indeed not even possible to ﬁnd, which will lead us to conclude that certain functors are not formally smooth.

18 5 THE QUOT AND HILB FUNCTORS

5 The Quot and Hilb Functors

Having given an overview of the construction of free quotients, we are now ready to abstract this concept to a more general setting involving functors acting on the category of k-algebras. However, regardless of this abstraction, we will continue looking at concrete examples, and draw several parallels to the previous sections. Deﬁnition 5.1. An A-module M is said to be locally free of rank n if there exist elements f1, . . . , fp ∈ A such that (f1, . . . , fp) = (1), and further,

n M Mfi = Afi j=1 for i = 1, . . . , p. In other words, the localization of M at any of the elements fi is free as a module over the corresponding localization of A.

Remark. Equivalent to the above definition is the condition that Mp is free of rank n as an Ap-module for every prime ideal p in A. Definition 5.2. Let k be a commutative ring, k → A a homomorphism of rings, and F an n A-module. Let Quot (F/A/k) be the set of A-module quotients F E, where E is locally 0 free of rank n as a k-module, and two quotients F E and F E are equivalent if their 0 0 0 0 0 kernels coincide. For any k → k , let A = A ⊗k k and F = F ⊗k k . The Quot functor n QuotF/A/k : k-alg → Set is then defined as

n 0 n 0 0 0 QuotF/A/k(k ) = Quot (F /A /k ) for any k0.

n Remark. If F = A, the Quot functor is also called the Hilbert functor, denoted HilbA/k. We will use this notation in the remainder of this text. There is a rather straightforward connection between this deﬁnition and our previous exposition of locally free quotients, as we will see in the next subsection.

5.1 The Quot Functor and Free Quotients Before jumping in and starting to analyze our previous examples in this new language, we will bridge the gap between these two points of view. Lr 0 Let k be a commutative ring, A = k[x1, . . . , xm], F = A and k be an arbitrary k-algebra. 0 0 Further let E be a free k -module. We can endow E with an A ⊗k k -module structure with a mapping 0 0 A ⊗k k = k [x1, . . . , xm] → End(E), 0 0 by ﬁxing m commuting matrices Xi with entries in k , as seen previously. Then, an A ⊗k k - 0 module homomorphism ϕ: F ⊗k k → E is given by the composition r 0 M ev F ⊗k k → End(E) −−→ E,

19 5.2 Representing Simple Cases 5 THE QUOT AND HILB FUNCTORS after picking r elements e1, . . . , er ∈ E to evaluate in. If this map is surjective, we have that 0 E is given by the quotient (F ⊗k k )/ ker(ϕ). This is indeed identical to the work we did on free quotients, with the main diﬀerence when working in the general setting of the Quot functor being that E is generally only locally free. However, we will continue working with our free examples given in the previous section. For some concrete examples, letting F = A = k[x], we have that any k0-valued element n 0 0 of QuotF/A/k is a locally free k -module of rank n given by a quotient k [x]/I for some ideal I. This, of course, corresponds to the examples given in Section 3.1.1. Further, if A = k[x] and F = A ⊕ A, we have that any k0-valued element is given by a quotient (k0[x] ⊕ k0[x])/N, corresponding to Section 3.2.1.

5.2 Representing Simple Cases We will begin by examining the Hilb functor in the simplest of cases, namely A = k[x]. Given any n k-algebra B, we have that k[x]⊗k B ' B[x], and so for any element E of Hilbk[x]/k(B) we have a Ln surjective map ϕ: B[x] E, or equivalently an ideal I such that, locally, B[x]/I = B. We will begin with a proposition that somewhat relaxes the conditions in the case of one variable, after proving the following lemma:

Lemma 5.1. Let A be a ring, and M a ﬁnitely generated A-module. We have that if {(x1, 1), ..., (xn, 1)} generates Mp for every p ∈ Spec(A), then {x1, . . . , xn} generates M. Ln Proof. Let ϕ: A → M denote the map taking (a1, . . . , an) to a1x1 +...+anxn. This induces an exact sequence n M 0 → K → A → M → C → 0, with K the kernel and C the cokernel of ϕ. This, in turn, induces an exact sequence n M 0 → Kp → Ap → Mp → Cp → 0 for every p ∈ Spec(A). Now, if Mp = ((x1, 1),..., (xn, 1)), we have that ϕp is surjective, and therefore that Cp = (0). Coupling this with the fact that an A-module N is trivial if and only if Np is trivial for every prime p, we see that this implies that C = (0), and so ϕ is surjective. Using this lemma, we can now show the following proposition: Proposition 5.1. Any locally free A-module M of the form A[x]/I is free.

Proof. The Nakayama lemma tells us that if {x1,..., xn} generate Mp/pM as an Ap-module, then in fact {x1, . . . , xn} generate Mp as an Ap-module. Therefore, since M = A[x]/I is locally free as an A-module, we have that n M (Mp)/pM = Ap/p is actually a vector space with basis S = {1, x, . . . , xn−1}, and therefore that S generates (A[x]/I)p as an Ap-module for every prime p ⊆ A. Then by the previous lemma, S generates A[x]/I as an A-module, and it is therefore free.

20 5 THE QUOT AND HILB FUNCTORS 5.2 Representing Simple Cases

The following proposition then somewhat formalizes the argument from Section 3.1.1.

n Proposition 5.2. The functor Hilbk[x]/k is representable by k[t1, . . . , tn]. Further, the universal element is given by , n ! n X i−1 ξ = k[t1, . . . , tn][x] x − tix . i=1 Proof. Let E be a free B-module of rank n, given a B[x]-module structure through X : E → E such that E = B[x]/I for some I ⊆ B[x]. Representing X as an n × n matrix with elements in the commutative ring B, we have that, with F the characteristic polynomial of X, F (X) = n n−1 X − (a0I + ... + an−1X ) = 0 by Cayley-Hamilton. Therefore, with ψ : B[x] → EndZ(E) being the B[x]-module structure on E, we get ψ(F (x)) = 0, and thus F (x) ∈ ker(ψ).

Further, with ϕ: EndZ(E) → E being the evaluation map granting E its quotient structure, we have that F (x) ∈ ker(ϕ ◦ ψ). Thus, {1, x, . . . , xn−1} is a minimal spanning set of E as a B-module, and hence a basis. Conversely, if G(x) is a polynomial of degree n, then B[x]/(G(x)) is free of rank n. We therefore see that there is a one-to-one correspondence between ideals I such that E = B[x]/I and polynomials of degree n. n Now, with A = k[t1, . . . , tn], ξ induces the natural transformation Φ: hA → Hilbk[x]/k given n by ΦB(f) = (Hilbk[x]/k(f))(ξ) for any k-algebra B and k-algebra homomorphism f : A → B. Any such homomorphism is uniquely determined by where it maps t1, . . . , tn, so let f(ti) = bi for some bi ∈ B. Then, we have that

, n ! n X i−1 ΦB(f) = B[x] x − bix , i=1

n which by the previous is a bijection for all k-algebras B. Therefore, Hilbk[x]/k is naturally isomorphic to hA, with aforementioned universal element.

Having translated this simple example of a quotient in rank one into our new category theoretic language, we now want to do the same for some simple case in rank two. Doing this, we move away from the Hilb functor, and will instead be inspecting the Quot functor. However, as the general Quot functor is fairly difficult to work with, we will start by defining a modification of it, taking into account a specified basis for its quotients.

S Deﬁnition 5.3. The relative Quot functor, relQuotF/A/k : k-alg → Set, is deﬁned as

S n |S| o relQuotF/A/k(B) = E ∈ QuotF/A/k(B) | E is free as a B-module with basis S , for any k-algebra B.

S Letting F = k[x] ⊕ k[x], we then have for any E ∈ relQuotF/A/k with |S| = n, that

n M E = (F ⊗k B)/N = (B[x] ⊕ B[x])/N = B

21 5.2 Representing Simple Cases 5 THE QUOT AND HILB FUNCTORS for some N ⊆ B[x] ⊕ B[x]. We can thus fix a basis of n elements, chosen (for example) among all elements of the form (xi, xj). Let us assume that, for instance, n = 5, and fix the basis {(1, 0), (x, 0), (x2, 0), (0, 1), (0, x)}. From the previous proposition, one might then S naively assume that we can therefore represent relQuotF/A/k with Hom(k[t1, . . . , t5], −). This will however not be the case. We state the following non-general proposition, regarding the relQuot functor taking into account a specified basis. For a more in depth discussion of the relative Quot functor, see [7].

S Proposition 5.3. With F = k[x] ⊕ k[x], the functor relQuotF/k[x]/k, with S = {(1, 0), (x, 0), 2 (x , 0), (0, 1), (0, x)}, is representable by k[t1, . . . , t10].

Proof. We can write (B[x] ⊕ B[x])/N = Be1 + ... + Be5, 3 where we identify e1 = (1, 0), . . . , e5 = (0, x). Upon examination, we see that (x , 0) = b1e1 + ... + b5e5 for some bi ∈ B, and that therefore,

3 2 (x , 0) − [(b1, 0) + (b2x, 0) + (b3x , 0) + (0, b4) + (0, b5x)] ∈ N.

Similarly, we have that

2 2 (0, x ) − [(c1, 0) + (c2x, 0) + (c3x , 0) + (0, c4) + (0, c5x)] ∈ N for some ci ∈ B, and so N is in fact determined by 10 coeﬃcients. By a similar argument as in the previous case, we can therefore construct a natural isomorphism between Hom(k[t1, . . . , t10], −) S and relQuotF/A/k. As a last introductory example of the functorial deﬁnitions, we will revisit Example 3.3, as this will play a major role in subsequent parts of this text. We do this mainly to give a translation of an earlier example into the language of Quot and Hilb functors, and to show the intimate connection with free quotients.

Example 5.1. Let A = k[x, y, z] and B = k[ε]/(ε2). Then the B-module

E = B[x, y, z]/(xy, xz, yz, x2, y2 − εz, z2 − εx)

4 is a member of HilbA/k(B), as A ⊗k B = B[x, y, z] and as we saw in Example 3.3, the endomorphisms 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 ε X =   ,Y =   ,Z =   0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 ε 0 1 0 0 0

ev(e ) are surjective, with the induced map A → End(E) −−−−→1 E spanning E. Further, {1, x, y, z} forms a basis of E, and so it is free of rank 4. This again emphasizes the fact that elements of the Quot and Hilb functors are intimately connected with the previous exposition of free quotients. ♦

22 5 THE QUOT AND HILB FUNCTORS 5.3 Smoothness of the Hilb Functor

5.3 Smoothness of the Hilb Functor We will now revisit some of our previous examples of quotients in rank one, in a slightly more general setting. We will go through the creation and implications of our matrix representations of elements of the Hilb functor, and eventually arrive at a discussion of when the functor is and is not smooth. Previously, we have started with a free k-module E, and given this module a k[x1, . . . , xm]- module structure through a map k[x1, . . . , xm] → EndZ(E), defining the actions of the variables xi on E. Conversely, however, suppose we are given a free k-module with a k[x1, . . . , xm]-module structure. We can then define a map Φ: k[x1, . . . , xm] → End(E) by considering the actions of xi within the given module structure. This way, we can represent any free k-module of the form Lr ( k[x1, . . . , xm])/N with its corresponding matrices. The following definition lets us use this language in the remainder of the text.

Deﬁnition 5.4. Let k be a commutative ring and E a free k-module given a k[x1, . . . , xm]- module structure. We then say that the matrices Xi encoding the action of xi on E represent the k[x1, . . . , xm]-module structure on E. With this deﬁnition, we note that we have already come across a number of representations of free k-modules, namely our maps k[x1, . . . , xm] → End(E) endowing E with k[x1, . . . , xm]- module structures, which we have shown induce surjective maps k[x1, . . . , xm] E. We will in the remainder revisit many of these examples, using our knowledge of their structures. We begin by summarizing the results from Example 3.1 and Proposition 5.2 in the following proposition: Proposition 5.4. Letting k be a commutative ring and B be a k-algebra, elements E of n Hilbk[x]/k(B) have representations given by   0 0 ... 0 a0 1 0 ... 0 a1    0 1 ... 0 a2  X =   , ......  . . . . .  0 0 ... 1 an−1

n n−1 where E = B[x]/(x − (a0 + a1x + ... + an−1x )). Before moving on to analyzing the smoothness of these functors, we will look at one more example. Suppose that instead of just one variable, we have three. In this case, a representation consists of three matrices, each corresponding to the action of one of the variables. Indeed, this is the case in our next example, where we look at one such representation. 4 Example 5.2. Let k be a commutative ring and B a k-algebra, and suppose E ∈ Hilbk[x,y,z]/k(B) has a basis {1, x, y, z} as a B-module. We then claim that E has a representation of the form       0 axx axy axz 0 ayx ayy ayz 0 azx azy azz 1 bxx bxy bxz  0 byx byy byz  0 bzx bzy bzz  X =   ,Y =   ,Z =   , 0 cxx cxy cxz  1 cyx cyy cyz  0 czx czy czz  0 dxx dxy dxz 0 dyx dyy dyz 1 dzx dzy dzz

23 5.3 Smoothness of the Hilb Functor 5 THE QUOT AND HILB FUNCTORS along with the 4 × 4 identity matrix, where the three above matrices commute pairwise. This will in fact generate a quotient k[axx, . . . , dzz]/I, where the ideal I is given by the 48 equations in 36 unknowns stemming from the relations XY = YX, XZ = ZX and YZ = ZY . Further, this 4 quotient will completely represent all elements of the functor Hilbk[x,y,z]/k with basis {1, x, y, z}. As noted previously, the fact that these matrices commute immediately forces the columns with the same mixed indices to be equal. Further, because {1, x, y, z} is a basis for E, we can ﬁnd relations  2 x = axx + bxxx + cxxy + dxxz  2  y = ayy + byyx + cyyy + dyyz  2  z = azz + bzzx + czzy + dzzz xy = axy + bxyx + cxyy + dxyz   xz = a + b x + c y + d z  xz xz xz xz  yz = ayz + byzx + cyzy + dyzz, with coeﬃcients in k, which indeed correspond to the actions of x, y and z on E and therefore also with the entries in the matrices. These identities are also respected by the matrices (the steps of which are a bit tedious to write out, but simple to verify). For example, we have that 2 X = axxI + bxxX + cxxY + dxxZ, which is most easily checked by noting that both sides act in the same way on our basis elements. This fact, however, is dependent on the commutativity of the set of endomorphisms. ♦

We can now use these matrix representation of elements of the Hilb functor to check its smoothness in some cases. We begin by stating a necessary and suﬃcient condition for the more general Quot functor to be smooth.

Lemma 5.2. Let k be a commutative ring, A a k-algebra and F an A-module. The functor n QuotF/A/k is then smooth if and only if for every k-algebra B and every nilpotent ideal N ⊆ B 2 n n such that N = (0), any element E ∈ QuotF/A/k(B/N) admits a lift E ∈ QuotF/A/k(B), such that E reduces to E modulo N.

Proof. This is basically just a restatement of the deﬁnition. Suppose the diagram

n ψb QuotF/A/k hB/N

? hB

n n commutes. Then the composition hB/N → hB → Q equals the map hB/N → Q ; however, n hB/N → hB corresponds to the natural map B → B/N, and both mappings into Q corre- n n n spond to elements of Q (B/N) and Q (B), respectively. Therefore, the map hB → Q then corresponds to the lift of the given element in Qn(B/N). Conversely, if a lift E exists, then the corresponding map into Qn(B) makes the diagram commute.

24 5 THE QUOT AND HILB FUNCTORS 5.3 Smoothness of the Hilb Functor

Consider now an element E = B[x , x , . . . , x ]/I in Hilbn (B) with B = B/N 1 2 m k[x1,...,xm]/k for some nilpotent N such that N 2 = (0) and some k-algebra B, admitting a lift to E in Hilbn (B). We then have the diagram k[x1,...,xm]/k

- eve - B[x1, . . . , xm] EndB(E) E

? ? ? - eve - B[x1, . . . , xm] EndB(E) E, where the horizontal compositions are surjective, for some e ∈ E, showing that ﬁnding a lift for E is equivalent to ﬁnding a lift for its matrix representation that is still commutative. We will start by examining the simplest possible case, where we have only one variable:

n Proposition 5.5. The functor Hilbk[x]/k is formally smooth. n Proof. A representation of any element in Hilbk[x]/k(A) for any k-algebra A is given by matrices X as shown in Proposition 5.4. The composition A[x] → EndA(V ) → V is also indeed surjective, i since X e1 = ei+1 for 0 ≤ i < n, and so evaluating in the ﬁrst column will span all of V . Further, since powers of X always commute, any lift of a representation {Xi} of an ele- n i n ment E ∈ Hilbk[x]/k(B/N) to a representation {X } of E ∈ Hilbk[x]/k(B) will be commutative. Therefore, any such element E has a lift, and the functor is smooth.

n Remark. Of course, as we saw in Proposition 5.2, Hilbk[x]/k is representable by k[t1, . . . , tn], the polynomial ring in n variables. This proposition then tells us that this polynomial ring is non-singular, which is as we would expect. Essentially, this shows that in the case of one variable, there is nowhere for anything to go wrong so to speak, since any representation will automatically commute. We will see that this breaks down when introducing more variables, and that at a certain point this leads the Hilb functor to become non-smooth. However, we will continue with an example over two variables, where one could be led to believe that the functor is non-smooth, but a clever choice of lift indeed makes the lifted representation commute. In fact, this is a revisit of Example 4.2, and we will therefore omit some of the details that were given previously.

3 3 2 Example 5.3. Consider Hilbk[x,y]/k(B/N) with B = k[ε]/(ε ) and N = (ε ). The element E = (B/N)[x, y]/(x2 − εy, y2 − εx, xy) is a free B/N-module with basis {1, x, y}, and is represented by 0 0 0 0 0 0 X = 1 0 0 ,Y = 0 0 ε , 0 ε 0 1 0 0 for which we have 0 0 0  0 0 0 XY = 0 0 0  ,YX = 0 ε2 0 . 0 0 ε2 0 0 0

25 5.3 Smoothness of the Hilb Functor 5 THE QUOT AND HILB FUNCTORS

This representation therefore commutes over B/N, but not obviously over B. However, lifting the matrices to 0 0 ε2 0 ε2 0 X = 1 0 0  , Y = 0 0 ε , 0 ε 0 1 0 0 we have that ε2 0 0  XY = YX =  0 ε2 0  , 0 0 ε2 rendering a commutative representation. ♦ We are now ready to approach our ﬁrst example of a non-smooth functor. In fact, we have already come in contact with the quotient used in the proof, namely in Example 3.3. 4 Proposition 5.6. The functor Hilbk[x,y,z]/k is not formally smooth. Proof. Consider again B = k[ε]/(ε3) and N = (ε2). We then have that

(B/N)[x, y, z]/(xy, xz, yz, x2, y2 − εz, z2 − εx) is represented by 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 ε X =   ,Y =   ,Z =   , 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 ε 0 1 0 0 0 for which XY = YX = XZ = ZX = YZ = 0 and 0 0 0 0 0 0 ε2 0 ZY =   , 0 0 0 0 0 0 0 0 showing that the representation is commutative over B/N, but not over B. If Y and Z are 2 2 lifts of Y and Z, we have that Y = Y + ε C and Z = Z + ε D, with C = (cij) and D = (dij) arbitrary 4 × 4 matrices with elements in k. Then, we have that Y Z − ZY = (YZ − ZY ) + ε2(CZ + YD − DY − ZC). Carrying this out, we get   c14 − d13 0 0 0 2  c24 − d23 0 −1 0  Y Z − ZY = ε   , c34 + d11 − d33 d12 d13 d14  c44 − c11 − d43 −c12 −c13 −c14 showing that no matter our choices of C and D, the lift will never commute since the element in position (2, 3) in the product diﬀers. Thus, we have shown that there exists a B and a nilpotent ideal N such that an element of 4 4 Hilbk[x,y,z]/k(B/N) does not admit a lift to an element in Hilbk[x,y,z]/k(B), and that the functor therefore is not smooth.

26 5 THE QUOT AND HILB FUNCTORS 5.4 Smoothness of the Quot Functor

5.4 Smoothness of the Quot Functor

We would now like to do the same thing when F = Lr A, and speciﬁcally take a closer look at the case r = 2. Again, ﬁnding a mapping A ⊗k B → EndB(E) corresponding to the action of A ⊗k B on E grants us an A ⊗k B-module structure on E. We begin by revisiting Example 3.4, again to get a sense of this example set against this new language.

2 Example 5.4. Let A = k[x] and F = A ⊕ A. Then, if E ∈ QuotF/A/k(B) for some k-algebra B, we have that 2 loc M E = (B[x] ⊕ B[x])/N = B for some N ⊆ B[x] ⊕ B[x]. Suppose further that {(1, 0), (0, 1)} forms a basis for E as a free B-module. We then have that (x, 0) = a1(1, 0) + a2(0, 1) and (0, x) = b1(1, 0) + b2(0, 1) for some ai, bi ∈ B. Considering the action of x on the basis elements, we can again write this as a matrix a b X = 1 1 , a2 b2 deﬁning the submodule N = ((x, 0) − (a1, a2), (0, x) − (b1, b2)). ♦

The biggest difference between representations of elements of the Hilbert functor and those of elements of the Quot functor is that, in general, representations of elements of Hilb have matrices with very simple first columns, corresponding to the actions of xi on 1. The Quot functor, on the other hand, takes into account the module F , and as such there is generally no multiplicative identity. As such, the first column of the matrices no longer have this simple structure.

Example 5.5. Let again A = k[x] and F = A ⊕ A. Suppose further that an element E of 4 QuotF/A/k is given by

2 2 E = (B[x] ⊕ B[x])/((x , 0) − (a1e1 + a2e2 + a3e3 + a4e4), (0, x ) − (b1e1 + b2e2 + b3e3 + b4e4)), where e1 = (1, 0), e2 = (x, 0), e3 = (0, 1) and e4 = (0, x), so that E has basis {(1, 0), (x, 0), (0, 1), (0, x)} as a B-module. The action of x on E can then be described by the matrix

  0 a1 0 b1 1 a2 0 b2 X =   , 0 a3 0 b3 0 a4 1 b4 giving a distinct appearance of being separated into two vertical components. ♦

Just as in the case with the Hilb functor, there is no concern with commutativity in one variable. We have the following proposition, mirroring Proposition 5.5.

27 5.4 Smoothness of the Quot Functor 5 THE QUOT AND HILB FUNCTORS

Lr n Proposition 5.7. Let A = k[x] and F = A. Then the functor QuotF/A/k is formally smooth. Proof. Just as in the proof of Proposition 5.5, any matrix encoding X of the action of x on an n element in QuotF/A/k is commutative (with itself) and can be lifted to a commutative matrix X. Further, if the determinant of X is nonzero, then the same is true for the lift, showing that it is also surjective.

Moving on to several variables, it is in fact much easier to ﬁnd an example of a non-smooth Quot functor than ﬁnding a similar example for the Hilb functor. We will begin by giving a complete description of a simple case in two variables.

Example 5.6. Let A = k[x, y] and F = A ⊕ A. Suppose E = (F ⊗k B)/N is an element of 2 QuotF/A/k(B) with basis {e1, e2} for some k-algebra B. We can then describe the actions of x and y on E with the matrices a b e f X = ,Y = , c d g h

L2 where the composite map F ⊗k B → EndB(E) → E is surjective, and the two matrices commute. The fact that the two matrices commute give rise to three equations in the variables a, . . . , h, as seen by bg − cf af + bh − be − df [X,Y ] = XY − YX = . ce + dg − ag − ch cf − bg Rearranging, we get the three equations   bg − cf = 0 f(a − d) + b(h − e) = 0  g(a − d) + c(h − e) = 0, showing that as long as the mapping is surjective, bg = cf, a = d and e = h, this induces a k[x, y]-module structure on E. ♦ Note that for a case of this, we can look at Example 3.5, where a = d = 1, e = h = 0 and bg = cf = 0 in the given ring. We will now expand on this example, to get a second case of a non-smooth functor.

2 Proposition 5.8. With A = k[x, y] and F = A ⊕ A, the functor QuotF/A/k is non-smooth. Proof. Let B = k[ε]/(ε3) and N = (ε2), and consider the endomorphisms

1 0 0 ε X = ,Y = ε 1 0 0

L2 in End( B/N). As seen in Example 3.5, these matrices commute, and the evaluation in e1 L2 and e2 is surjective onto B/N, giving us the quotient E = (B/N[x, y] ⊕ B/N[x, y])/((x − 1, −ε), (0, x − 1), (y, 0), (−ε, y)).

28 5 THE QUOT AND HILB FUNCTORS 5.4 Smoothness of the Quot Functor

Suppose now that there is a lift of these matrices to X and Y . Then, since the projection of the lifts to B/N must be X and Y , they take the form

c c d d X = X + ε2 11 12 = X + ε2C, Y = Y + ε2 11 12 = Y + ε2D, c21 c22 d21 d22 whereby we have that XY − Y X = XY − YX + ε2(XD + CY − DX − YC). However, we also have that ε2 0 ε2X = = ε2I ε3 ε2 and 0 ε3 ε2Y = = 0 0 0 in B, and so XD − DX = CY − YC = 0, modulo ε3. Therefore, since

1 0 XY − YX = ε2 6= 0 0 −1 in B, no lift X and Y will commute, and so the given quotient has no lift. Further, by direct computation in E, we have that

(0, 0) = x(y, 0) = y(x, 0) = y(1, ε) = ε(0, y) = (ε2, 0), which is true modulo ε2, but not modulo ε3.

4 To conclude, we have found concrete examples to show that the functors Hilbk[x,y,z]/k and Quot2 are non-smooth. Using more sophisticated arguments, one can prove the L2 k[x,y]/k[x,y]/k following, more general, theorem.

n Theorem 5.1. The functor QuotF/A/k is smooth if • A = k[x],

• A = k[x, y] and F = A, or

• A = k[x1, . . . , xm], F = A and n ≤ 3. Proof sketch. We have proved the ﬁrst point previously. For a proof of the second point, see [2], Theorem 2.4. Lastly, an outline of the proof of the third point is given by the observation 2 that a basis for a module of rank 3 is of the form {1, xi, xi } or {1, xi, xj}, bringing us back to one of the earlier two cases.

Remark. This theorem points to an interesting diﬀerence between the Hilbert and Quot schemes. Consider the aﬃne variety C of commuting pairs of 2 × 2 matrices (X,Y ). If

a b e f X = ,Y = , c d g h

29 5.4 Smoothness of the Quot Functor 5 THE QUOT AND HILB FUNCTORS we have that C = k[a, b, . . . , h]/I, where

I = (bg − cf, f(a − d) + b(h − e), g(a − d) + c(h − e)), describes all such pairs. The ideal I is in fact prime, and C is thus irreducible and singular at some point P . The aﬃne variety C ⊗k k[u1, u2] corresponds to pairs of commuting matrices and an element u = (u1, u2) ∈ E = k ⊕ k. Any point (X, Y, u) then gives a map k[x, y] → E of k[x, y]-modules. The condition that this map is surjective is Zariski open, and therefore determines an open subset U1 ⊂ Spec(C[u1, u2]). One can further check that the set of invertible 2 × 2 matrices 2 2 GL2 acts freely on U1, and that U1/GL2 gives an open subset of Hilbk[x,y]/k. Since Hilbk[x,y]/k is smooth, we have in particular that U1 avoids the singular set, which is determined by the singularities of C. 2 The Quot scheme QuotF/k[x,y]/k with F = k[x, y] ⊕ k[x, y], on the other hand, is singular. In the variety C[u1, u2, v1, v2], points correspond to k[x, y]-module maps

k[x, y] ⊕ k[x, y] → E, and subjectivity of this map is again a Zariski open condition and determines an open subset U2 ⊂ Spec(C[u1, u2, v1, v2]). Again, the group GL2 acts freely on U2, and U2/GL2 is an open 2 subset of QuotF/k[x,y]/k. Since this scheme is singular, this implies that U2 does not avoid the singularities of C. Hence, even though the variety of commuting matrices is the same for both Hilb and Quot, the surjectivity condition on the module homomorphisms determines whether the scheme is smooth or not. In particular, with the matrices given in the proof of Proposition 5.8 we cannot ﬁnd a single element e ∈ E corresponding to surjective maps into E.

30 REFERENCES REFERENCES

References

[1] Torsten Ekedahl and Roy Skjelnes. Recovering the good component of the hilbert scheme, 2000, arXiv:math/0405073.

[2] John Fogarty. Algebraic families on an algebraic surface. American Journal of Mathematics, 90(2):511–521, April 1968.

[3] Robin Hartshorne. Deformation Theory. Graduate Texts in Mathematics. Springer, 2010.

[4] Saunders Mac Lane. Categories for the Working Mathematician. Graduate Texts in Math- ematics. Springer, 2nd edition, 1998.

[5] Nitin Nitsure. Construction of hilbert and quot schemes, 2005, arXiv:math/0504590.

[6] Norbert Roby. Lois polynomes et lois formelles en th´eorie des modules. Annales scientiﬁques de l’E.N.S.´ , 80(3):213–348, 1963.

[7] Gustav Sædén St˚ahl. Gradings of the affine line and its quot scheme. Master’s thesis, Kungliga Tekniska högskolan, 2011.

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