Deformation Theory

Michael Kemeny∗

Lecture 3

1 Category theory and families of objects

1.1 Yoneda’s LemmaRecall that a functor F is contravariant if an arrow A → B is mapped to anarrow F (B)→ F (A); in other words, if F is an ‘arrow reversing functor’.

Let C be a category. Yoneda’s lemma lets you view an object X ∈ C as the“same thing” as a contravariant functor

hX : C → (Sets),

where hX is defined byhX(Y ) = homC(Y,X).

If f : X → Y is a morphism and U ∈ C an object, we can define a function

hf (U) : hX(U) → hY (U)‖ ‖

hom(U,X) hom(U, Y ),

via pre-composition with f . In fact, hf is a morphism of functors hX → hY .We may now state the weak version of Yoneda’s lemma.

Lemma 1.1.1 (Yoneda’s Lemma). With X,Y ∈ C the function just defined:

homC(X,Y )→ hom(hX , hY ) : f 7→ hf

is bijective.

Exercise. Prove the lemma.

1.2 RepresentabilityNow suppose one wants to study families of some object, such as schemes or linebundles. One starts by defining an appropriate contravariant functor

F : Schk → (Sets).

One considers, in a loose sense, F as parametrizing the objects in the setΛ := F (Spec(k)).

∗TEX by Emre Sertöz. Please email suggestions or corrections to emresertoz@gmail.com

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Indeed, for any scheme S the set F (S) may be regarded as all possible familiesover S of objects Λ. Therefore F is a machinery telling how the elements in Λcan vary.

Now We want to have a notion of what it means for a scheme Z to parametrizethe objects in the set Λ.

The strongest interpretation of parametrizing F (Spec(k)) is via the notionof representability.

Definition 1.2.1. Let Z be a scheme. We say the functor F is represented byZ if the functors F and hZ are isomorphic.

Important Exercise. Show that F is represented by Z iff there exists auniversal object ξ ∈ F (Z) (necessarily unique) such that for any ξ′ ∈ F (Y ) thereexists a unique morphism fξ′ : Y → Z satisfying F (fξ′)(ξ) = ξ′. This object ξ isoften referred to as the universal family over Z.

In the following section we illustrate these concepts by considering familiesof line bundles on a fixed scheme.

1.3 Families of line bundlesLet X be a smooth, projective variety over a field k. Let S be a scheme over k.Denote by XS the fiber product X×Spec(k) S and let πS : XS → S be the secondprojection. The set Pic(XS) is the set of isomorphism classes of line bundles onXS .

Note that a line bundle L on XS can be regarded as a family of line bundleson X parametrized by S. Indeed, if x ∈ S is a closed point then the fiber ofXS → S over x is isomorphic to X and by pulling back L to this fiber we get aline bundle on X.

Exercise. Check that if N is a line bundle on S then L ⊗ π∗S(N ) and L givethe same family of line bundles on X. That is, on each fiber the pullbacks ofthese line bundles coincide.

To correct this problem observe that Pic(XS) is an abelian group and thatthe pullback by πS gives a group morphism Pic(S)→ Pic(XS) so that we maydefine a new functor Pic as follows:

Pic : Schk → (Sets) : S 7→ Pic(XS)/Pic(S).

Still Pic(Spec(k)) is equal to the set of isomorphism classes of line bundles onX. Moreover, Pic appears to be a better functor in parametrizing line bundleson X.

If Pic is representable then there exists a scheme T over k, together with a‘universal’ line bundle L ∈ Pic(X × T ) in the following sense. For any scheme Sover k and line bundleM∈ Pic(XS) there exists a unique morphism ϕ : S → Twith (ϕ× IdX)∗(L) isomorphic toM modulo Pic(S).

Whether some functor of families of geometric objects is representable is ingeneral a very difficult question to answer. The first step in this direction isoften to restrict the functor to Artin rings.

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2 Artin Rings

2.1 MotivationIn the previous lecture we saw that the deformation theory of associative algebrasis done in a step-by step fashion. To be precise, we started with a differentialF1 ∈ H2(A,A) and we wanted to find a 1-parameter family of deformations ofA of the form:

ft(a, b) = ab+ tF1(a, b) + t2F2(a, b) + . . .

The procedure went like this:We used F1 to produce an appropriate F2, assuming the associator of F1 is

equivalent to 0 in H3(A,A).When we have F1, . . . , Fn−1 we produced Fn under a certain condition on

the Fi’s (this time [Gn] ∈ H3(A,A) must be 0).Then by induction we get all the terms. We now want to formalize this

procedure.

2.2 Definition of an Artin ringLet k be a fixed field. Let A be a commutative algebra over k (with unit). Wesay A is Artinian if it satisfies the descending chain condition on ideals, ie. ifany descending sequence of ideals

· · · ⊂ I4 ⊂ I3 ⊂ I2 ⊂ I1 ⊂ A

eventually stabilizes. Stability here means that there is an n0 such that for alln ≥ n0 we have In = In0

.

Examples. The k-algebra k[t]/(tn) is Artinian. An integral domain is Artinianif and only if it is a field.

Recall that a commutative ring R is local if it has a unique maximal ideal. IfR is a local ring with maximal ideal m, then the field R/m is the residue field.

If A and B are local rings with maximal ideals mA and mB respectively thena morphism ϕ : A→ B is said to be a local morphism if ϕ(mA) ⊂ mB .

Notation. Let Artk be the category whose objects are local Artinian k-algebraswith residue field k and whose arrows are local morphisms.

Given A ∈ Artk we define rA : Spec(k)→ A to be the map associated to thequotient A→ A/mA ' k. Note the isomorphism of the quotient field with theresidue field is canonically induced via the structure map k → A.

For the next few weeks, the main objects of study are (covariant/ordinary)functors

F : Artk → (Sets)

such that F (k) is a set with a single element.

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2.3 Back to line bundlesLet Pic : Schk → (Sets) be the contravariant functor defined in Section 1.3. PickM∈ Pic(Spec(k)), ie. M is a line bundle on X.

We define a covariant functor

PicM : Artk → (Sets)

byPicM(A) := N ∈ Pic(Spec(A)) | r∗AN 'M

where rA : Xk → XA is induced from rA : Spec(k) → Spec(A) in the obviousway. Note that PicM(k) = M. Loosely speaking, PicM is the functor ofdeformations of the line bundleM.

In a similar fashion, from a contravariant functor F : Schk → (Sets) andan object a ∈ F (k) we can define a covariant functor Fa : Artk → (Sets) byrestriction.

2.4 Tangent spaceDefine k[ε] := k[x]/(x2) to be the dual numbers. Recall, if X is a variety overk with a chosen point x ∈ X (that is, x ∈ hom(Spec(k), X)), then the Zariskitangent space of X at x could be identified with the subset of hom(Spec(k[ε]), X)such that the unique point in the spectrum of the dual numbers hits x.

Generalizing this definition we get:

Definition 2.4.1 (Schlessinger). Let F : Artk → (Sets) be a functor such thatF (k) is a one-element set. Then F (k[ε]) is called the tangent space of F .

3 Completions of local rings

3.1 DefinitionsWe will recall the definition of inverse limits. Let Ann∈N be a sequence ofgroups together with homomorphisms θn+1 : An+1 → An. The inverse limitlim←−An is the group of all sequences an | an ∈ An, θn+1(an+1) = an.

The inverse limit comes with obvious projections:

πi : lim←−An → Ai.

The pair (lim←−An, πi | i ∈ N) is universal amongst such objects. See Section3.2 for a discussion of this notion.

If R is a commutative ring and I ⊂ R is a proper ideal, we get a commutativering

RI := lim←−(R/In).

If R is a local commutative ring with maximal ideal m we set

R := Rm.

For example:

• If R = k[x1, . . . , xn] and I = (x1, . . . , xn) then RI = k[[x1, . . . , xn]] is thering of formal power series.

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• If R = Z and I = (p) for some prime p ∈ Z then RI is the ring of p-adicintegers.

More generally, if M is an R-module and I ⊂ R is a proper ideal then wedefine

MI := lim←−M/InM,

which is a RI -module.From a ring to its completion there is a natural map taking a ∈ R to

(a ∈ R/In)n∈N. With this in mind we make the following definition.

Definition 3.1.1. Let R be a commutative ring and I ⊂ R a proper ideal. Wesay R is complete with respect to I if the natural morphism ϕ : R→ RI is anisomorphism.

Exercise. Let R be a local ring, I ⊂ R a proper ideal. Then RI is completewith respect to the ideal generated by ϕ(I).

Notation. We let Ârtk denote the category of complete, noetherian local k-algebras R such that

Rn := R/mnR

is in Artk for all n ∈ N.

Exercise. Show Artk is a subcategory of Ârtk.

Exercise. Let A ∈ Artk with mnX = 0 for some n. Pick R ∈ Ârtk and showthat hom(R,A) = hom(Rn, A).

Definition 3.1.2. (Main definition of this lecture) A functor F : Artk → (Sets)is called pro-representable if for some R ∈ Ârtk the functor F is isomorphic tothe functor hom(R,_) restricted to Artk.

3.2 Digression on the universal propertyIt is often said that an object X is universal with respect to a property P . Whatthis means is that the property P defines a category of objects C and the objectX is either the terminal or the initial object in that category (to be understoodfrom context). This interpretation makes it obvious that if a universal objectexists (with respect to a property P ) then it must be unique up to a uniqueisomorphism.

In defining the inverse limit of a sequence of groups let us see how thisinterpretation works. Fix Ann∈N and the sequence of morphisms θ = θn :An → An−1. A careful reading of the universal property of the inverse limitsuggests that we should build the category C whose objects are pairs (B,ϕ)where B is a group and ϕ = ϕn : B → An | n ∈ N, ϕn−1 = θn ϕn is asequence of maps compatible with θ.

We are claiming that (lim←−An, π) is the terminal object of this category C.This is left as an exercise.

A nicer interpretation of the inverse limit in this case is via the theory ofsheaves. Consider N with the following topology:

T = ∅ ∪ N ∪ Un := 0, . . . , n | n ∈ N.

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Let C′ be the subcategory of presheaves of groups on (N, T ) consisting ofpresheaves F such that F(Un) = An and F(Un−1 → Un) = θn. We havean equivalence of categories γ : C′ → C defined as follows:

γ(F) = (F(N), F(Un → N) | n ∈ N).

The terminal object in C′ is the one and only sheaf in C′, and it is the sheafificationof any F in C′.

Exercise. Go to the ‘Important Exercise’ and identify the category of whichthe universal property identifies the terminal object.

6

Deformation Theory

Michael Kemeny∗

Lecture 4

We follow closely the following 2 papers:

• Fantechi, Manetti - “Obstruction Calculus for Functors of Artin Rings, I”

• Schlessinger - “Functors of Artin rings”

1 ExtensionsRecall that Artk is the category of local Artinian k-algebras with residue field kand Ârtk is the category of complete Noetherian local k-algebras such that forevery n ∈ N, Rn = R/mn

R is in Artk.

Definition 1.1. A small extension in Artk (respectively Ârtk) is a short exactsequence

e : 0→M → B → A→ 0

where B → A is in Artk (resp. in Ârtk) and mBM = 0. If in addition M is aprincipal ideal of B then we say e is principal.

Exercise. Suppose e is a small extension. Show that M is a k-vector space.Show further that e is principal iff dimkM = 1.

Notation. Let e be a small extension as in Definition 1.1. Then M , B and Aare referred to as the kernel, source and target. We will also refer to them asK(e), S(e) and T (e) respectively.

Let Fvsp denote the category of finite dimensional k-vector spaces. If A ∈Ârtk and M ∈ Fvsp then we define Ex(A,M) to be the set of isomorphismclasses if extensions e with K(e) = M and T (e) = A. This set Ex(A,M) carriesa natural k-vector space structure. (This is because extensions are identifiedwith H2(A,M) where M has the A module structure induced by A→ k. Thiscohomology group obviously has a k-vector space structure.)

Let f : M → N be a morphism in Fvsp. Then there is a linear map

f∗ : Ex(A,M)→ Ex(A,N)

defined as follows: Let e ∈ Ex(A,M) then we have

0 → M → B → A → 0↓N

∗TEX by Emre Sertöz. Please email suggestions or corrections to emresertoz@gmail.com

1

and we will push forward the first exact sequence into another one as follows.Set B′ = B ⊕N . We make B′ into a ring by defining:

(b1, n1) · (b2, n2) = (b1b2, b1n2 + b2n1)

where by bi we mean the residue of bi mod mB , which is an element of k.Then B ∈ Ârtk implies B′ ∈ Ârtk. Let J ⊆ B′ be the set

(m, f(m)) | m ∈M ⊂ B.

It is easy to show that J is an ideal of B.Let B denote B′/J . This gives us the pushforward f∗(e) as follows:

f∗(e) : 0→ N → B → A→ 0

which lies in Ex(A,N).

Exercise. Let g : A → B be a morphism in Ârtk. Show that this induces alinear map g∗ : Ex(A,M)→ Ex(B,M) satisfying the following identity

f∗g∗ = g∗f∗

for any f : M → N in Fvsp.

Definition 1.2. A small extension e is called trivial if it splits (or, equivalently,if it corresponds to 0 ∈ Ex(A,M)).

We will define a morphism between two small extensions, thereby making acategory out of small extensions. Let

ei : 0→Mi → Bi → Ai → 0

for i = 1, 2 be two small extensions.A morphism f : e1 → e2 is given by the data of a commutative diagram:

0 → M1 → B1 → A1 → 0↓ ↓ ↓

0 → M2 → B2 → A2 → 0.

2 Schlessinger’s ConditionsSetup. Throughout the section F : Artk → (Sets) will be a functor such thatF (k) consists of a single element. We will let Fun∗(Artk, (Sets)) denote thecategory of such functors.

Recall that F is called pro-representable if there exists a complete local k-algebra R ∈ Ârtk such that F is equivalent to hR|Artk . From now on we will notmention when hR has to be restricted to Artk, it is assumed to be understoodfrom context.

Exercise. With F as in setup, show that for any R ∈ Ârtk there is a bijection

lim←−F (R/mnR) ' hom(hR, F )

where the hom on the RHS is the morphism of functors on Artk. Note alsothat inverse limits of sets are perfectly well defined, as in abelian groups. [Hint:To construct the bijection notice that if A ∈ Artk then any morphism R→ Afactors through R/mn

R for some n.]

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Definition 2.1. Let R ∈ Ârtk and choose ξ ∈ lim←−F (R/mnR). By the exercise

above ξ corresponds to a morphism hR → F . We call such a pair (R, ξ) apro-couple.

If F is pro-representable and (R, ξ) is a pro-couple corresponding to theisomorphism hR → F then we say the pro-couple (R, ξ) pro-represents F .

Let F → G be a morphism in Fun∗(Artk, (Sets)) and let A → B be amorphism in Artk. We have a commutative diagram:

F (A) → F (B)↓ ↓

G(A) → G(B).

Thus we get, by the universal property of fiber products, a morphism

F (A)→ F (B)×G(B) G(A).

Definition. A morphism F → G in Fun∗(Artk, (Sets)) is called smooth if forany surjective morphism A→ B in Artk, the map

F (A)→ F (B)×G(B) G(A)

is surjective.

Proposition 2.2. Let R→ S be a morphism in Ârtk. Then hS → hR is smoothif and only if there exists an n ∈ N such that

S ' R[[x1, . . . , xn]].

Proof. For any A ∈ Ârtk the map hS(A)→ hR(A) is induced by precompositionwith R→ S. Suppose hS → hR is smooth and consider

t∗S/R := mS/(m2S + mRS).

The k-vector space t∗S/R is called the Zariski cotangent space of S over R, notethat multiplication by k is induced from the action of S/mS ' k. The Zariskicotangent space over the base field k will be refered to as just the Zariskicotangent space.

Set x1, . . . , xn ∈ S inducing a basis of t∗S/R. And define T = R[[X1, . . . , Xn]];we are going to show that S ' T .

There is a morphism of local R algebras

u1 : S → T/(m2T + mRT ) : xi 7→ Xi.

To show that this morphism is well defined, observe that S/(m2S + mRS) '

k ⊕ mS/(m2S + mRS) = t∗S/R. So we may define an isomorphism between the

two vector spaces t∗S/R and t∗T/R by mapping the basis xi to Xi. Adding on thevector space k to both sides and composing with the quotient map from S givesus u1.

Now we apply smoothness of hS → hR to the surjection T/m2T → T/(m2

T +mRT ).

Let g denote the corresponding sujective morphism:

hom(S, T/m2T )→ hom(S, T/(m2

T + mRT )×hom(R,T/m2T+mRT ) hom(R, T/m2

T ).

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There is a natural morphism Ri→ T/m2

T , simply as the inclusion of the coefficientring. Via the smoothness hypothesis there exists a map

u2 : S → T/m2T

such that g(u2) = (u1, i). In particular, u2 lifts u1.By repeating this argument we may find

u3 : S → T/m3T

lifting u2. Continuing in this manner and using the universal property of theinverse limit we get:

u : S → T = lim←−(T/mnT ).

By definition of u1, the morphism u induces an isomorphism t∗S/R → t∗T/R.We wish to show that the induced map on the Zariski cotangent spaces,

ie. the morphism mS/m2S → mT /m

2T , is surjective. To that end note the exact

sequence:

0 → mRT/(m2T ∩mRT ) → mT /m

2T → t∗T/R → 0

↑ ↑ ↑0 → mRS/(m

2S ∩mRS) → mS/m

2S → t∗S/R → 0

.

The rightmost vertical arrow is a surjection, therefore if we show that the leftmostvertical arrow is also a surjection it will follow that the middle arrow too is asurjection.

Notice that mR/m2R → mRT/(m

2T ∩mRT ) is an isomorphism. Since the map

u : S → T is such that the inclusion R→ T factors through as R→ Su→ T , the

isomorphism above factors through as:

mR/m2R → mRS/(m

2S ∩mRS)→ mRT/(m

2T ∩mRT ).

This implies that the leftmost vertical arrow has to be surjective.If A is a ring and a ⊆ A is an ideal then we define the graded ring of A with

respect to a as

G(A) :=

∞⊕n=0

an/an+1.

Since the map between the cotangent spaces of S and T is surjective, we get asurjective map G(S)→ G(T ) = T . Applying Lemma 10.23 of Atiyah-MacDonald(page 112) we conclude u : S → T is surjective.

Constructv : T = R[[X1, . . . , Xn]]→ S

by picking yi ∈ u−1(Xi) for each i and setting v(Xi) = yi. (To show that sucha map can be defined you may start with the polynomial ring over R and usethe universal property of the localization at the origin followed by the universalproperty of completions.)

Since u v = IdT , the morphism v must be injective. Notice also that v islocal, R-linear and that it induces an isomorphism between the cotangent spaces.Using the graded rings again we may conclude that v is surjective. Hence, v isan isomorphism.

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Exercise. We established the “only if” part of the proof. The converse is easyand is left as an exercise.

Definition 2.3. Let (R, ξ) be a pro-couple for F ∈ Fun∗(Artk, (Sets)) corre-sponding to a morphism hR → F . Then (R, ξ) is called a hull of F if thecorresponding map hR → F is smooth and the induced map

hom(R, k[ε])→ F (k[ε])

on tangent spaces is bijective.

We now aim to give sufficient conditions for a functor F ∈ Fun∗(Artk, (Sets))to

i) have a hull

ii) be pro-representable.

To this end we introduce Schlessinger’s Conditions H1, H2, H3 and H4.Throughout the definitions, f : A→ A′ and g : A′′ → A will be in Artk. Such acouple induce a map

F (A′ ×A A′′)→ F (A′)×F (A) F (A′′)

which we will denote by f ∗ g.

Property H1. F is said to satisfy property H1 if for any f and any surjectiveg : A′′ → A yielding a principal small extension, the induced map f ∗ g issurjective.

Property H2. F satisfies H2 if we take g : A[ε] → A to be the quotient mapand pick any f : A→ A′ then the induced map f ∗ g is bijective.

We will show later that if properties H1 and H2 hold then F (k[ε]) is canoni-cally a k-vector space. So we may introduce the next property.

Property H3. If F satisfies H1, H2 and if in addition dimk F (k[ε]) <∞ thenF is said to satisfy H3.

Property H4. If g : A′′ → A is a principal small extension then for any f theinduced map f ∗ g is bijective.

In the new few weeks we intend to prove the following theorem:

Schlessinger’s Theorem 2.4. Let F ∈ Fun∗(Artk, (Sets)). Then F has a hullif and only if F satisfies H1, H2 and H3. Furthermore, F is pro-representable ifand only if in addition F satisfies H4.

Next week. We will study the Schlessinger conditions on specific examplessuch as the Picard functor.

5

Deformation Theory

Michael Kemeny∗

Lecture 5

1 RecallLast week we saw Schlessinger conditions H1, . . . ,H4 on a functor F : Artk →(Sets) with F (k) a point.

Schlessinger’s theorem which we will start proving next week says that F ispro-representable when H1, . . . ,H4 are satisfied.

The goal of this week is to get a feel for Schlessinger conditions. We are firstgoing to prove that the Picard functor

PicM : Artk → (Sets)

satisfies the Schlessinger conditions.

2 Picard FunctorLet X be a scheme over k and M ∈ Pic(X) ' H1(X,O∗X). For A ∈ Artk wehave

PicM (A) := L ∈ Pic(XA) | L⊗A k =M.Recall that XA := X ×Spec(k) Spec(A).

For any morphism A→ B in Artk the induced morphism XB → XA give apullback map

Pic(XA)→ Pic(XB).

For L ∈ Pic(XA) we denote the pullback of L as L⊗A B. In the definition ofPicM (A) above k is the residue field A→ A/mA ' k.

From now on we make the following two assumptions about X:

1. H0(X,OX) ' k

2. dimkH1(X,OX) <∞.

For instance, if X is a proper algebraic variety then the assumptions aresatisfied. We will first show that PicM satisfies conditions H1 and H2. We willrecall their definitions from Lecture 4.

F satisfies condition H1 if for any f : A′ → A, g : A′′ → A ∈ Artk with g asurjection giving a small principal extension the following map is surjective:

f ∗ g : F (A′ ×A A′′)→ F (A′)×F (A) F (A′′).

∗TEX by Emre Sertöz. Please email suggestions or corrections to emresertoz@gmail.com

1

F satisfies condition H2 if f ∗ g is bijective when we take g to be the quotientmap g : k[ε]→ k.

3 PreparationBefore we can prove that PicM satisfies the first two Schlessinger conditions, wehave to prepare two algebraic results.

Lemma 3.1. Let A be a ring, J ⊆ A a nilpotent ideal and u : M → N amorphism of A-modules such that N is flat over A.

If u :M/J → N/J is an isomorphism then so is u :M → N .

Proof. Let Q be the cokernel of u and consider the exact sequence

Mu→ N → Q→ 0.

Tensor this sequence of A-modules with A/J . Since tensoring is right exact westill have an exact sequence

M/JMu→ N/JN → K/JK → 0.

By assumption u is an isomorphism and thus K/JK = 0. Therefore K = JK =JnK for any n ≥ 0. As J is nilpotent we conclude K = 0, ie. u is surjective.

Therefore we have a short exact sequence:

0 H M N 0u

Since N is flat over A, after tensoring with A/J the first derived term on theleft is 0 and we get an exact sequence:

0 H/JH M/JM N/JN 0u

Once again we conclude H/JH and thus H is 0. Therefore u is injective.

Exercise. Using this lemma prove that if M is flat over A ∈ Artk then M is infact free.

In setting up for the next result we need a definition. Note that we chose thegeometric (ie. scheme theoretic) convention for our naming.

Definition 3.2. Let B → A be a ring morphism. If N is a B module then wewill call N ⊗B A the pullback of N to A.

If M is an A-module isomorphic to the pullback of N , then there are B-module morphisms u : N →M (factoring through N → N ⊗B A) which inducean isomorphism N ⊗B A→M . We will call such morphisms restriction maps.

Lemma 3.3. Suppose A′ → A is a morphism and A′′ → A is a surjectivemorphism with nilpotent kernel J .

Let M be an A module, M ′ a free A′ module and M ′′ a flat A′′ module;such that M ′ and M ′′ pull back to M . Let u′ :M ′ →M and u′′ :M ′′ →M berestriction maps.

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With B = A′ ×A A′′ and N = M ′ ×M M ′′ we may construct the followingCartesian diagram using the projection maps p′ and p′′:

N =M ′ ×M M ′′ M ′′

M ′ M

p′′

p′ u′′

u′

.

Under these circumstances, N is free over B. Moreover, N pulls back to M ′and M ′′ while p′ and p′′ provide restriction maps.

Proof. Choose a basis (x′i)i∈I of M ′ over A′. Since u′ :M ′ →M is a restrictionmap, M is free over A with basis (u′(x′i))i∈I .

Since M ' M ′′/JM ′′ we may choose x′′i ∈ M ′′ such that u′′(x′′i ) = u′(x′i).The morphism ⊕

i∈IA′′x′′i →M ′′

pulls back to an isomorphism on A. Since J is nilpotent we may apply Lemma3.1 to conclude that M ′′ is free with basis (x′′i )i∈I . As u′′(x′′i ) = u′(x′i) the pairs(x′′i , x

′i) ∈ N form a free basis over B.

Exercise. Convince yourself of this fact.

Note also that the projection maps p′ and p′′ map the basis to the corre-sponding bases. Making them restriction maps.

Corollary 3.4. Together with the hypotheses of Lemma 3.3, assume that L is aB module and that we have a commutative diagram of B modules:

L M ′′

M ′ M

q′′

q′ u′′

u′

.

By the universal property of fiber products we get a morphism u : L → Nmaking the following diagram commutative:

L

N M ′′

M ′ M

q′′

q′

u

p′′

p′ u′′

u′

.

If q′ is a restriction map then u is an isomorphism.

Proof. First, as an easy exercise, show that B → A′ is surjective with nilpotentkernel J ′. Use the explicit construction of product of rings.

Lemma 2 assures that the projection p′ : N →M ′ is a restriction map. Thecommutativity of the diagram and our hypothesis on q′ implies u must pull backto an isomorphism on A′. This pull back can be written in the form:

u : L/J ′L→ N/J ′N.

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Lemma 2 also says that N is free and thus flat over B. Since J ′ is nilpotent wemay use Lemma 1 to conclude that u is itself an isomorphism.

4 Main resultWe are finally ready to prove the main result of this lecture.

Proposition 4.1. Let f : A′ → A and g : A′′ → A be morphisms in Artk withg a surjection. Then the natural map

f ∗ g : PicM (A′ ×A A′′)→ PicM (A′)×PicM (A) PicM (A′′)

is a bijection. In particular PicM satisfies H1 and H2.

Proof. For simplicity of notation let P denote PicM . Let (L′, L′′) ∈ P(A′)×P(A)

P(A′′) with the common image L ∈ P(A). Let B denote the product A′ ×A A′′.Let |X| denote the underlying topological space of X. Note that the spectrum

of an Artinian ring A consists of a point and so |XA| = |X|.We may thus consider the following commutative diagram of sheaves on |X|:

OXBOXA′′

OXA′ OXA

.

This induces a canonical morphism OXB→ OXA′ ×OXA

OXA′′ of sheaves on |X|.Since A′′ is Artinian, the kernel of g : A′′ → A is nilpotent. Therefore we

may apply Corollary 3.4 to conclude that the canonical map just defined is anisomorphism of sheaves of B modules:

OXB

∼−→ OXA′ ×OXAOXA′′ .

Therefore N := L′ ×L L′′ is an invertible sheaf on XB . By Lemma 3.3, N pullsback to L′ and L′′ on XA′ and XA′′ respectively. In other words f ∗ g(N) =(L′, L′′) and thus f ∗ g is surjective.

It remains to verify injectivity of f ∗ g. Let M ∈ P(B) map to (L′, L′′)via f ∗ g and pull back to L over XA. Notice that there is an automorphismθ : L→ L which makes the following diagram commute:

M L′′

L′ L L

q′′

q′ θ

.

We will now describe this automorphism θ. Recall that we assumed k →H0(X,OX) is an isomorphism. This implies that the natural map

A→ H0(XA,OXA) = H0(X,OX ⊗k A)

is an isomorphism. This is a consequence of the following exercise.

Exercise. The canonical map k → H0(X,OX) is an isomorphism if and onlyif for all k-vector spaces V the canonical map V → H0(X,OX ⊗k V ) is anisomorphism. [Hint: OX ⊗k V '

⊕v∈BOX , where B is a basis for V .]

4

Thus θ is multiplication by a unit a in A. Replacing q′′ with 1a · q

′′ we mayassume θ is the identity map. That is u′q′ = u′′q′′.

Now we may apply Corollary 3.4 to conclude that M is isomorphic to N .Proving that f ∗ g is injective.

Next week. We will show that PicM satisfies H3, H4 and we will begin provingSchlessinger’s Theorem.

5

Deformation Theory

Michael Kemeny∗

Lecture 6

The meaning of bonus exercises. While typing the notes, I (Emre) addedbonus exercises to emphasize some non-trivial points in the argument that arenot covered in lecture notes. They are occasionally very time consuming, so dothem at your own risk. In anycase, they are by no means included in the exam,should you be interested in taking one. But, any run-of-the-mill exercises andanything that is said to be ‘left to the reader’ are from the lecture notes; ignorethem at your own risk.

1 Finishing the proof from last weekLast week we showed that PicM satisfies the first two Schlessinger conditionsprovided X is such that H0(X,OX) = k and dimkH

1(X,OX) <∞. This weekwe will prove that PicM further satisfies the last two Schlessinger conditions.

Remark. Condition H4 requires that f ∗ g be bijective whenever g is a principlesmall extension. Previous week we showed that for any two morphisms f and gin Artk, with g surjective, the induced map f ∗ g for PicM is bijective. HencePicM satisfies H4.

The fact that H3 is satisfied by PicM is a consequence of the following lemma.

Lemma 1.1. If X is a scheme over k such that H1(X,OX) is finite dimensional,then PicM (k[ε]) is a finite dimensional k-vector space

Proof.

Step 1:

We fixed M , a line bundle on X, to define PicM . Now we are going to showthat we may assume M = OX for this proof.

Let L0 ∈ PicM (k[ε]) be the zero element of the vector space structure (infact L0 is the image of M with respect to the structure map k → k[ε]). Define amap:

PicM (k[ε]) → PicOX(k[ε])

L 7→ L⊗ L∨0 .

Since tensoring with a line bundle is an invertible operation, this map is abijection.

∗TEX by Emre Sertöz. Please email suggestions or corrections to emresertoz@gmail.com

1

Bonus exercise. Show that the map above is a vector space isomorphism.

Step 2:

From this point onwards we may, and will, take M = OX . Let Y be Xk[ε] =X ×Spec(k) Spec(k[ε]). Let r : Spec(k)→ Spec(k[ε]) be the map correspondingto the quotient map k[ε]→ k. Let π : X → Spec(k) be the structure morphism.The inclusion of the central fiber is given by µ := IdX ×(r π) : X → Y . Bydefinition we have

PicOX(k[ε]) = L ∈ Pic(Y ) | µ∗L ' OX = ker(H1(Y,O∗Y )

µ∗→ H1(X,O∗X)).

Step 3:

Now we will concentrate on the map between the cohomologies. First, recallthat µ is a homeomorphism of the underlying topological spaces of X and Y . Themorphism between the structure sheaves µ# : OY → OX is simply the projectionon to the first factor where we make the obvious identification OY = OX ⊕ εOX .

Further note that we have O∗Y = O∗X · (1 + εOX). This gives an exactsequence:

0 OX O∗Y O∗X 0exp µ#

where the exponential map is defined so that exp(f) = 1 + εf . Note that thesurjection indeed coincides with the projection onto the first factor. We maytake the associated long exact sequence, observing that µ# obviously surjects onthe level of global sections, we are left with:

0 H1(OX) H1(O∗Y ) H1(O∗X)µ∗

.

Consequently,PicOX

(k[ε]) ' H1(OX).

But by hypothesis the latter vector space is finite dimensional over k and thusso is PicOX

(k[ε]) once you prove the following.

Bonus exercise. The map PicOX(k[ε]) ' H1(OX) is a k-vector space isomor-

phism.

Summing up the results so far we have:

Theorem 1.2. Let X be a scheme over k such that H0(X,OX) = k anddimkH

1(X,OX) <∞. For any M ∈ Pic(X) the functor PicM satisfies the fourSchlessinger’s conditions H1, H2, H3 and H4.

2

2 Schlessinger’s theoremRecall that a natural transformation ν : F → G in Fun∗(Artk, (Sets)) is calledsmooth if for any surjection f : A→ B the induced map

fν : F (A)→ F (B)×G(B) G(A)

is surjective.Recall also that we defined the ‘tangent space’ of F to be tF := F (k[ε]) which

in certain cases will have a k-vector space structure. As for a ring R we willrefer to thR

as tR.Finally, recall that a hull of F is a couple (R, r) with R ∈ Ârtk and r ∈ F (R)

such that the induced map r : hR → F is smooth and the map between thetangent spaces tR → tF is an isomorphism.

We will now prove:

Theorem 2.1. Assume that F ∈ Fun∗(Artk, (Sets)) has a hull. Then F satisfiesthe first three Schlessinger conditions.

As always let f : A′ → A and g : A′′ → A be in Artk. Let (R, r) be a hull ofF .

To be able to distinguish the two induced maps, we use the following notation:

f ∗F g : F (A′ ×A A′′)→ F (A′)×F (A) F (A′′)

f ∗hRg : hR(A

′ ×A A′′)→ hR(A′)×hR(A) hR(A

′′).

Proof of condition 1. We will show that smoothness of r : hR → F implies thatf ∗F g is surjective, for any f and g as above. By elementary diagram chasingwe construct a commutative diagram:

F (A′ ×A A′′) F (A′)×F (A) F (A′′)

hR(A′ ×A A′′) hR(A

′)×hR(A) hR(A′′)

f∗F g

f∗hRg

Check that the smoothness condition implies, for any B ∈ Artk, that hR(B)→F (B) is surjective. (Hint: Take the map to be B → k.) Therefore, the verticalarrows are surjective by smoothness.

By showing that the bottom horizontal map is surjective we may concludethat the top horizontal map is surjective. Since hR is a hom functor, it preservesfiber products. Therefore, the bottom horizontal arrow is a bijection.

Proof of condition 2. We take g to be the map k[ε]→ k. We wish to prove thatf ∗F g is a bijection for any f . Denote A′×k k[ε] by A′[ε]. Using also the tangentspace notation, the commutative diagram in the previous proof becomes:

F (A′[ε]) F (A′)× tF

hR(A′[ε]) hR(A

′)× tR

f∗F g

∼

3

We will show that f ∗F g is injective. Take x, y ∈ F (A′[ε]) which map to thesame (α, β) ∈ F (A′)× tF . Fix a lift a ∈ hR(A′) of α ∈ F (A′). By smoothnessthe following map is surjective:

hR(A′[ε])→ hR(A

′)×F (A′) F (A′[ε]).

Therefore we may pick lifts ϕx, ϕy ∈ hr(A′[ε]) of (a, x) and (a, y) respectively.Using the bottom horizontal arrow of our commutative diagram, we may associatethese lifts to pairs (a, ϕ′x) and (a, ϕ′y) in hR(A′)× tR, uniquely. Since (R, r) is ahull, the rightmost vertical arrow is a bijection between the second components.But both ϕ′x and ϕ′y map to the same element β so they must be equal.

It follows that x and y are equal. This proves injectivity of f ∗F g and havingestablished surjectivity in the previous proof, we are done.

Proof of condition 3. This is immediate because tF ' tR by hypothesis. ButtR ' R/mR ∈ Artk which implies that dimk tR <∞.

3 Additional LemmasFor future reference we will prove two lemmas. This first implies a fact we havebeen using so far: If F satisfies H2 then tF has a canonical vector space structure.We will be interested in the following kind of functors.

Definition 3.1. A functor satisfying the first two Schlessinger conditions is saidto be a functor with good deformation theory.

Let us first introduce a notation. For a k-vector space V we define k[V ] tobe the k-algebra where elements in V give a square zero ideal. Note for instancethat k[V ]×k k[W ] ' k[V ⊕W ].

Bonus exercise. Prove that if F satisfies H2 then it satisfies the hypothesis ofthe following lemma.

Lemma 3.2. Suppose F is a functor such that for any two finite dimensional k-vector spaces V and W the canonical map F (k[V ]×kk[W ])→ F (k[V ])×F (k[W ])is a bijection. Then F (k[V ]), and in particular tF = F (k[ε]), has a canonicalvector space structure. Moreover, there is a natural isomorphism:

tF ⊗ V → F (k[V ]).

Proof. The addition on V defines an “addition” map on k[V ] as follows:

k[V ]×k k[V ] → k[V ]

(c+ v, c+ w) → (c, v + w)

where c ∈ k and v, w ∈ V . Similarly, scalar multiplication on V defines scalarmultiplication on k[V ], e.g. a ∈ k takes (c, v) to (c, av). Using the canonicalisomorphism (induced by the projection maps)

F (k[V ]×k F [V ])→ F (k[V ])× F (k[V ])

4

we can carry the addition and scalar multiplication maps on k[V ] to the setF (k[V ]) via F . Its easy to check that these satisfy the necessary vector spaceaxioms making F (k[V ]) a vector space. (e.g. by showing that F preserves thecommutativity of necessary diagrams.)

Since homC(k[ε], k[V ]) ' V , which is a k-linear map with respect to thenatural vector space structures on both sides, given (x, v) ∈ tF we can treatv ∈ V as a map k[ε]→ k[V ] thereby define F (v)(x) ∈ F (k[V ]). This is a k-linearmap and it induces the map

tF ⊗ V → F (k[V ]).

Notice k[V ] '∏dimVi=1 k[ε] where the product is taken over k. Thus it is easy to

see that the map above is an isomorphism.

We will need the following exercise for the final lemma:

Exercise. F satisfies H1 iff for any f : A′ → A and any small extensiong : A′′ → A in Artk the map F (A′ ×A A′′)→ F (A′)×F (A) F (A

′′) is surjective.(That is, the condition of being principle may be omitted.)

Lemma 3.3. Suppose F is a functor with good deformation theory. Let B → Abe a small extension in Artk with kernel M . Then there is a canonical transitiveaction of F (k[M ]) = tF ⊗kM on the fibers of F (B)→ F (A).

Proof. First off let V := tF ⊗kM .To see why we should expect such an action, and motivate the remaining

part of the proof, we will first assume B → A is split. That is B ' A[M ]. Bythe previous lemma and using H2 we have

F (B) ' F (A)× F (k[M ]) ' F (A)× V

In this case the fiber of the map F (B)→ F (A) is just V and the action of V onthe fibers is clearly transitive as well as canonical (because the bijections aboveare canonical).

As for the general proof, consider B ×A B. It is left as an exercise to showthat B[M ] ' B ×A B, where the isomorphism is canonical with respect to theiruniversal properties (Hint: The map is defined by (b,m)→ (b, b+m)).

Using H2 again we conclude F (B[M ]) = F (B) × V . This gives us thefollowing sequence of maps:

F (B)× V ' F (B[M ]) ' F (B ×A B) F (B)×F (A) F (B)

where the last arrow is a surjection due toH1. The projection on to the first factorgives us the first factor we began with and projection on to the second factorallows us to define the V action on the set F (B). The fibers of F (B)→ F (A) areobviously preserved, simply due to the codomain of the map above. Transitivityof the action follows from the surjectivity of that map.

Bonus exercise. Check that this is indeed a group action. (Hint: Express thiscondition using diagrams or you are in trouble.)

5

Deformation Theory

Michael Kemeny∗

Lecture 7

1 Schlessinger’s theorem, the other directionNow we wish to prove that the first three Schlessinger’s conditions imply theexistence of a hull. This is the converse of what we proved last week.

Theorem 1.1. Let F ∈ Fun∗(Artk, (Sets)) satisfy H1, H2, H3. Then F admistsa hull.

Proof. Our goal is to construct R ∈ Ârtk and an element u ∈ F (R) such thatthe map hR → F induced by u is smooth and is an isomorphism on the tangentspaces. By definition of Ârtk, R ' lim←−R/m

nR. Therefore, in defining R we may

as well start with a sequence of small surjections and take their limit:

. . . Ri+1 Ri Ri−1 . . . R1 R0 = k ∈ Artk .

In constructing the element u we need to choose a sequence of elements ui ∈ F (Ri)that are compatible with respect to the maps above.

Step 1

Observe that for any R ∈ Ârtk we have a surjection R R/m2 ' k[M ] whereM = mR/m

2R. This gives an isomorphism tR → tk[M ]. Hence the discussion

below describes how we should pick M and thus R1 as well as u1 in order toensure that tR → tF is an isomorphism.

Let R1 = k[M ] be the square zero extension for some k-vector spaceM . Notethat tR1

= homArtk(k[M ], k[ε]). We have a natural pairing:

tR1× F (k[M ]) → F (k[ε]) = tF

(f, x) 7→ F (f)(x).

Recall that we have an isomorphism

M → homArtk(k[ε], k[M ])

m 7→ (ε 7→ m),

which allowed us to define the isomorphism

tF ⊗M → F (k[M ])

v ⊗m 7→ F (ε 7→ m)(v).

∗TEX by Emre Sertöz. Please email suggestions or corrections to emresertoz@gmail.com

1

Similarly we have an isomorphism

M∨ → homArtk(k[M ], k[ε])

ϕ 7→ (s+ ϕ · ε)

where s : k[M ]→ k is the canonical projection.Using these identifications the pairing above becomes

M∨ × (tF ⊗M) → tF

(ϕ, v ⊗m) 7→ ϕ(m) · v.

If we choose an element u1 =∑i vi ⊗mi ∈ tF ⊗M such that the vi form

a basis of tF and the mi form a basis of M then it is easily checked that thepairing above gives an isomorphism:

tR1 =M∨ → tF .

Step 2

Let S = k[[T1, . . . , Tr]] where r = dimk tF and Ti’s form a basis for t∨F .Further let η = (T1, . . . , Tr) be the maximal ideal. We let R0 = S/η = k andR1 = S/η2 = k[t∨F ]. We pick the only element in F (R0) to be u0 and we chooseu1 ∈ F (R1) such that tR1

→ tF is an isomorphism. (In fact, for fun, we canchoose u1 so that tR1

= (t∨F )∨ → tF is the canonical isomorphism.)

We will construct the other pairs (Rn, un) by induction as follows. We wantRi = S/Ji for some ideal ηJi−1 ⊆ Ji ⊆ Ji−1 and we want F (Ri Ri−1)(ui) =ui−1. Suppose (Ri, ui) has been constructed in this way for 0 ≤ i ≤ n−1. Let Indenote the set of all ideals ηJn−1 ⊆ J ⊆ Jn−1 such that there exists u′ ∈ F (S/J)satisfying F (S/J Ri−1)(u

′) = ui−1.We claim that In is closed under arbitrary intersections. Therefore we may

define Jn to be the smallest element, or the intersection of all elements, in In.Leaving the proof of this claim to the next step, let us continue in constructingR.

Having defined Jn and thus Rn = S/Jn we pick any un ∈ F (Rn) lifting un−1.So that we may define

R = lim←−Rn, u = lim←−un ∈ F (R).

Remark. In general R/mnR 6' Rn. However the fact that R ' S/ ∩i Ji meansmnR ⊆ Jn/ ∩i Ji for all n. This is why we can define u as the limit of un’s. Wewon’t go into this here.

It is clear that tR = hR(k[ε]) = hR1(k[ε]) = tR1

and the map tR → tFinduced by u coincides with the map tR1

→ tF induced by u1. That is, hR → Finduced by u gives an isomorphism of the tangent spaces. It remains to showthat this map is smooth, which we postpone to Step 4 of this proof.

Step 3

We are going to show that In defined above is closed under arbitrary inter-sections. Notice however that the ideals in In give vector subspaces of the finitedimensional vector space Jn−1/ηJn−1. Conversely we claim that any vector

2

space V in Jn−1/ηJn−1 pulls back to an ideal containing ηJn−1 and containedin Jn−1. The pull back can be performed via the ring quotient S → S/ηJn−1hence if V ⊂ S/ηJn−1 is an ideal, its pull back will be an ideal. The exercisebelow provides the missing ingredient.

Bonus exercise. Let V ⊂ Jn−1/ηJn−1 be a vector space. Show that themultiplication action of S/ηJn−1 on V factors through S/η = k. Hence V is asquare zero ideal in S/ηJn−1.

An intersection of an arbitrary collection of subspaces within a finite dimen-sional vector space can be realized as the intersection of a finite subcollectionof these subspaces. Hence, it is enough to show that In is closed under finiteintersections, or by induction, under the intersection of any two ideals.

Pick I, J ∈ In and let K = I ∩ J . By the following exercise we will assumethat I + J = Jn−1.

Exercise. Given I, J ∈ In we may expand J such that I + J = Jn−1 withoutleaving In (Easy, but check.) Thus we have

S/(I ∩ J) ' S/I ×Rn−1S/J.

Using the exercise we get a natural map

F (S/K) ' F (S/I ×Rn−1 S/J)→ F (S/I)×F (Rn−1) F (S/J).

Moreover this map is surjective byH1 because S/J → S/Jn−1 is a small extension(as ηJn−1 ⊂ J). By definition of In we have uI ∈ F (S/I) and uJ ∈ F (S/J) bothof which map to un−1 ∈ F (Rn−1). Hence (uI , uJ) ∈ F (S/I)×F (Rn−1) F (S/J).Consequently there is a uK ∈ F (S/K) mapping to (uI , uJ) and thus to un−1.That is, K ∈ In.

Step 4

Now we will prove that the map ν : hR → F just constructed is smooth.That is, for any ϕ : B A in Artk we need to show that

νϕ : hR(B)→ hR(A)×F (A) F (B)

is surjective. Let us simplify the problem using the following lemma.

Lemma 1.2. A natural transformation ν : G → F in Fun∗(Artk, (Sets)) issmooth iff for any principal small extension ϕ : B A in Artk the map

νϕ : hR(B)→ hR(A)×F (A) F (B)

is surjective.

Proof of lemma. Since mnB = 0 for some n, we can factor the surjection ϕ asfollows. Let I = kerϕ and Ai = B/miBI so that we have

B = An → An−1 → · · · → A1 = A

where each map is a small extension. Similarly we can factor any small extensionby principal small extensions.

3

Let us outline the induction step. Suppose A3ϕ2→ A2

ϕ1→ A1 are two mapssuch that νϕ1 and νϕ2 are surjective. We will show that νϕ1ϕ2 is surjective.Consider the diagram below:

F (A3) F (A2) F (A1)

G(A3) G(A2) G(A1)

Pick g3 ∈ G(A3) and f1 ∈ F (A1) lying over g1 ∈ G(A1). Let g2 ∈ G(A2) be theimage of g3. By surjectivity of νϕ1 there is an element f2 ∈ F (A2) which mapsto g2 and f1. By surjectivity of νϕ2 there is an element f3 ∈ F (A3) which mapsto g3 and f2. This proves surjectivity of νϕ1ϕ2 .

In light of the lemma above, we take a principal small extension ϕ : B → Ato test surjectivity of νϕ. Fix b ∈ F (B) with image a ∈ F (A). If α ∈ hR(A)maps to a, that is νA(α) = F (α)(u) = a, then we need to check if (α, b) lifts tohR(B). In other words, we need to find a map β : R→ B such that F (β)(u) = band α factors through β. First we will show that we need not worry about b andthat lifting the map α is enough.

Suppose we find some β lifting α, that is α = ϕ β. Then F (β)(u) = b′

where b′ maps to a. Then we can modify β so as to hit b. Indeed, F (B)→ F (A)has fibers which admit transitive tF ⊗ kerϕ action on them. Note also that sincethe extension ϕ is principal tF ⊗ kerϕ ' tF . So there is a x ∈ tF whose actionon F (B) carries b′ to b. Recall in constructing this action we used the followingsequence of maps

B ×k k[I] ' B ×A Bπ2→ B.

Pick y ∈ tR = hR(k[I]) which maps to x. Then the map R(β,y)→ B ×k k[I]

composed with the sequence of maps above yields a map to B which takes u tob.

Now we focus on finding a lift of α. Choose the smallest n such that α factorsthrough Rn → A. Since the extension B → A is small any lift β of α will factorthrough as Rn+1 → B. So we are looking for a map v to complete the followingdiagram:

Rn+1 B

Rn A

v

In order to use the minimality of the defining ideal of Rn+1 we need to bring Sinto the picture. So observe that S being a power series ring, we can define a liftw completing the diagram:

B

S Rn A

w

4

Therefore, the existence of v is equivalent to completing the following diagram

S Rn ×A B

Rn+1 Rn

w

π1v

If π1 has a section then obviously v exists. Otherwise we use the following lemmato conclude that w is surjective:

Lemma 1.3. If a principal small extension ϕ : B → A doesn’t have a section itis essential. That is, if a map w : S → B surjects onto A after composing withϕ then w itself is surjective.

The proof of this lemma is left for the following week. Let J = kerw. It is clearthat ηJn ⊆ J ⊆ Jn and if we further show that un lifts to S/J = Rn ×A B thenwe may conclude J ∈ In+1 from which it follows by minimality of Jn+1 ∈ In+1

that v exists.To lift un observe that H1 implies the map below is surjective

F (S/J) = F (Rn ×A B)→ F (Rn)×F (A) F (B).

The element (un, b)a belongs to the right hand side. Therefore, there is anelement u′ ∈ F (S/J) mapping to un. This concludes the proof that hR → F issmooth.

5

Deformation Theory

Michael Kemeny∗

Lecture 8

1 Theorem on pro-representabilityWe are now going to prove the final piece of Schlessinger’s theorem.

Theorem 1.1. Let F ∈ Fun∗(Artk, (Sets)) then F is pro-representable iff Fsatisfies H1, H2, H3, H4.

Proof.( =⇒ ) Suppose F is pro-representable, i.e. there is ν : hR

∼−→ F . Clearly thismap is also a hull of F therefore the conditions H1, H2, H3 are satisfied. On theother hand, condition H4 is immediate for any hom functor (by the universalproperty of fiber products).( ⇐= ) Conditions H1, H2, H3 imply that there is a hull ν : hR → F . We aregoing to prove that condition H4 implies that ν is an isomorphism. What mustbe checked is that for any A ∈ Artk the map νA : hR(A)→ F (A) is a bijection.We are going to prove this by induction on the length of A.

Definition 1.2. Let A ∈ Artk. By length of A, denoted lenA, we will meanthe largest integer n ≥ 1 such that there is a sequence of ideals

0 ( I1 ( · · · ( In = A.

If lenA = 1 then A = k and obviously νk is a bijection (between two sets ofsize one). If lenA′ = n ≥ 2 then take a principal small extension

0→ I → A′ → A→ 0.

Necessarily lenA = n− 1 and by induction hypothesis we may assume that νAis bijective. Combining this result with smoothness we get

hR(A′) hR(A)×F (A) F (A

′) ' F (A′).

That is, νA′ is surjective. We now would like to show that νA′ is injective.Consider the following diagram:

hR(A′) F (A′)

hR(A) F (A)∼

∗TEX by Emre Sertöz. Please email suggestions or corrections to emresertoz@gmail.com

1

Recall that on hR(A′) we have a I ⊗ tR action that is transitive on the fibers ofhR(A

′ → A). Similarly we have a I ⊗ tF action on F (A′) that is transitive onthe fibers of F (A′ → A).

Bonus exercise. Identifying the additive groups I ⊗ tR and I ⊗ tF with theisomorphism νk[ε] show that the diagram above is equivariant under these actions.

Exercise. Condition H4 implies that the action of I ⊗ tF on the fibers ofF (A′ → A) is in fact free.

Now suppose x, y ∈ hR(A′) map to the same element z ∈ F (A′). Due tothe bijectivity of the bottom horizontal arrow, x and y lie in the same fiber ofhR(A

′ → A). Hence there is an element v ∈ I⊗ tF such that v ·x = y. Using theequivariance of the map we conclude v · z = z. The group action on each fiber ofF (A′ → A) is free by the exercise above hence v = 0. Therefore, x = y.

2 Tangent-obstruction theories

2.1 DefinitionsA functor F ∈ Fun∗(Artk, (Sets)) is said to admit a tangent-obstruction theory ifthere exists finite dimensional k-vector spaces T1 and T2 satisfying the followingthree conditions.

Condition 1

For every small extension 0→M → Bπ→ A→ 0 there exists an “exact sequence”

of sets:T1 ⊗kM F (B) F (A) T2 ⊗kM

F (π) ob

Squiggly arrow denotes a T1 ⊗kM action on the set F (B) and exactness meansthe following are satisfies:

1. ob−1(0) = Im(F (π)) (Exactness at F (A)).

2. The map F (π) is invariant under the action of T1 ⊗M . Furthermore, theaction of T1 ⊗M on each fiber is transitive. (Exactness at F (B)).

Condition 2

If A = k then the action of T1 ⊗M on F (B) is free. Note that, A = k impliesB ' k[M ]. Using the proof of Proposition 2.3 we get a canonical isomorphismof vector spaces:

F (k[M ]) ' T1 ⊗M.

Condition 3

The association of an exact sequence of sets to a small extension given inCondition 1 is functorial. To describe what this means we need to definewhat morphisms are between these objects. A morphism of small extensions

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ϕ : e1 → e2 is simply a triplet (ϕM , ϕB , ϕA) making the following diagramcommutative:

e1 : 0 M B A 0

e2 : 0 M ′ B′ A′ 0

ϕM ϕB ϕA

In general, an exact sequence of sets will consist of the following dataG S1 → S2 → H where G and H are groups, S1 and S2 are sets. Theconditions of exactness are exactly as before. Hence a morphism of two exactsequences of sets ψ : f1 → f2 is given by four morphisms (ψG, ψ1, ψ2, ψH) suchthat the following diagram “commutes”:

f1 : G S1 S2 H

f2 : G′ S′1 S′2 H ′

ψG ψ1 ψ2 ψH

Here, by commutativity we mean that the map ψ1 is equivariant with respect toψG and for the rest of the diagram the notion of commutativity applies as usual.

Condition 1 assigns to a morphism ϕ : e1 → e2 the following morphism(Id⊗ϕM , F (ϕB), F (ϕA), Id⊗ϕM ). Here we require this assignment to respectthe composition of morphisms.

2.2 Main theorem we are aiming forOur goal for the next few weeks is to prove the following theorem.

Theorem 2.1. Let F ∈ Fun∗(Artk, (Sets)) be pro-represented by R ∈ Ârtk.Then the following hold.

• F admits a tangent-obstruction theory with T1 = tF = tR.

• Let d = dim tF . Recall that R ' k[[x1, . . . , xd]]/J for some J ⊆ η2 whereη = (x1, . . . , xn). We may take T2 = J/ηJ .

First note that if J does not lie in η2 then dim tR < d which can not happen.Observe also that finding a tangent-obstruction theory for a pro-representablefunctor F reveals quite a bit about R ∈ Ârtk representing F . Nevertheless, wesaid that “we may take” T2 = J/νJ because, in general, the obstruction spaceT2 is not unique for F .

Exercise. Suppose F admits a tangent-obstruction theory (T1, T2). Show thatfor any k-vector space T ′2 containing T2 we may define a tangent-obstructiontheory for F using the pair (T1, T ′2).

For our next result we need the following lemma.

Lemma 2.2. If F admits a tangent-obstruction theory then for any k-vectorspace M the set F (k[M ]) admits a canonical k-vector space structure. In partic-ular tF is a vector space.

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Proof. We are going to use Lemma 3.2 of Lecture 6 which states that our claimholds if for any two k-vector spaces M1 and M2 the canonical map below is abijection:

F (k[M1]×k k[M2])→ F (k[M1])× F (k[M2]).

Let U = F (k[M1]×k k[M2]) and Vi = F (k[Mi]) for i = 1, 2. By abuse of notationdenote F (πi) : U → Vi also by πi. Choose a tangent-obstruction theory (T1, T2)and use the following morphism of the exact sequences:

0 M1 ⊕M2 k[M1]×k k[M2] k 0

0 Mi k[Mi] k 0

πi

Using Conditions 1 and 3 we get a functorial assignment to these morphisms.Moreover, Condition 2 guarantees that the resulting actions are free. Fix theimage of F (k) in U and in Vi’s to get the following diagram where the horizontalmaps are a result of acting on these distinguished points and are bijective:

(M1 ⊕M2)⊗ T1 U

Mi ⊗ T1 Vi

Since M1 ⊕M2 ' M1 ×M2 we also get U ' V1 × V2 via the canonical maps,which was to be proven.

Proposition 2.3. Let F ∈ Fun∗(Artk, (Sets)) be a functor admitting a tangent-obstruction theory (T1, T2). Then there is a canonical isomorphism T1 ' tF .

Proof. In fact we can prove a little more with no extra work. For any k-vectorspace M there is a canonical isomorphism M ⊗ T1 ' F (k[M ]), where the latterhas a vector space structure by the previous lemma.

We fix the point x0 ∈ F (k[M ]) which is the image of F (k). This establishes abijection f :M ⊗ T1 → F (M) as in the proof of the previous lemma. The pointhere is to show that this bijection is in fact an isomorphism of vector spaces.We will show that the map is additive leaving the (easier) proof that it respectsscaling by k to the reader.

Let V = F (M) and W = M ⊗ T1. We wish to show that the followingdiagram commutes, where the vertical arrows denote addition:

W ×W V × V

W V

f×f

f

To do this we break the diagram into two parts. Recall that the addition mapV ×V → V is defined by V ×V ' F (k[M ]×k F [M ])→ V where the latter mapis the functorial image of (something akin to) addition of the two components.First we want to prove that the composition of the natural maps

W ×W ' (M ⊕M)⊗ T1 → F (k[M ]×k F [M ])→ V × V

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is still f × f . This can be checked by applying functoriality on the following twoexact sequences corresponding to the two projections:

0 M ⊕M k[M ]×k k[M ] k 0

0 M k[M ] k 0

πi

Therefore, we need only show that the following diagram commutes:

W ×W F (k[M ]×k k[M ])

W V

Which follows by functoriality applied to the following (note that it is differentfrom the ones above):

0 M ⊕M k[M ]×k k[M ] k 0

0 M k[M ] k 0

add

This proves that f :W → V is linear.

In practice the main reason we want a tangent-obstruction theory for apro-representable functor F is because we get dimension estimates on the pro-representing ring R. These dimension estimates are extremely useful. With thatin mind, next week we will prove that if F = hR then

dimT1 − dimT2 ≤ dimR ≤ dimT1

for any tangent-obstruction theory (T1, T2) of F .

5

Deformation Theory

Michael Kemeny∗

Lecture 9

1 Deformations of a schemeConventions. By a ring we will mean a noetherian commutative ring withunity. By a scheme we will mean a locally noetherian, seperated scheme.

Definition 1.1. Let X be a scheme over an algebraically closed field k. LetA ∈ Artk. An infinitesemal deformation of X over A is a cartesian diagram

X Y

Spec k SpecA

ιY

where Y → SpecA is flat.

Definition 1.2. An isomorphism of two infinitesimal deformations Y, Y ′ →SpecA of X is an isomorphism ϕ : Y → Y ′ over A such that ϕ ιY = ιY ′ . Hence,there is a commutative diagram as follows

X

Y Y ′

SpecA

ιY ιY ′

ϕ

∼

Definition 1.3. Let f : A → B ∈ Artk. The pullback of a deformationY → SpecA of X via f is defined to be f∗Y → SpecB together with the mapf∗ιY : X → f∗Y induced by the universal property of the pullback.

Definition 1.4. We define the functor of infinitesimal deformations of X asDefX ∈ Fun∗(Artk, (Sets)) such that

DefX(A) = Infinitesimal deformations of X over A / ∼

where the equivalence is that of isomorphism.∗TEX by Emre Sertöz. Please email suggestions or corrections to emresertoz@gmail.com

1

Definition 1.5. A deformation (Y → SpecA, ιY ) of X is called trivial if itis isomorphic to the pullback of the “deformation” (X → Spec k, IdX) via thestructure morphism sA : SpecA→ Spec k. A deformation Y → SpecA of X iscalled locally trivial if there exists an open cover Ui of Y such that Ui is atrivial deformation of Ui ×SpecA Spec k.

There is another way to phrase local triviality. Observe that as a morphismof topological spaces ιY : |X| → |Y | is a homeomorphism for any infinitesimaldeformation (Y, ιY ) (see the proof of the following lemma). Therefore, if Ui is anopen subscheme in X then we can restrict the structure sheaf of Y to the openset |Ui| to get an open subscheme U ′i of Y . Hence, an infinitesimal deformation(Y, ιY ) is locally trivial iff there exists an open cover Ui of X such that theinduced open subscheme U ′i ⊂ Y is a trivial deformation of Ui for each i.

Notation. We define the subfunctor Def∗X of DefX which gives the locallytrivial deformations of X.

Technically speaking, we can skirt around the following lemma to prove thefinal theorem of this lecture as well as its corollary. However, we provide it herefor completeness. Those of you who want to take the exam should keep in mindthat the following lemma is not part of the curriculum.

Lemma 1.6 ([Ser06] pg. 23, Lemma 1.2.3). An infinitesimal deformation of anoetherian affine scheme is noetherian affine.

Proof. Let Z0 = SpecR0 be a noetherian affine scheme over k and A ∈ Artk.Consider an infinitesimal deformation of Z0:

Z0 Z

Spec k SpecA

j

Step 1

We will first show that j is a closed immersion with coherent nilpotent idealof vanishing. This is local on the codomain so we will assume Z = SpecBwith A → B flat. Then Z0 = Spec(B ⊗A k). Tensoring the exact sequence0→ mA → A→ k → 0 with the flat A-algebra B we get:

0 mA ⊗A B B B ⊗A k 0

Since mA is nilpotent in A so is mA ⊗A B.

Step 2

We will show that j is a homeomorphism. This is local on the codomain sowe may assume Z affine. A closed immersion of an affine scheme cut out by anilpotent ideal is clearly a homeomorphism.

A quicker way to see this is to note Z0 ' V (N) as a subscheme of Z andthat |V (N)| = |Z| by definition of V (N).

Step 3

2

If r is the smallest positive integer such that Nr = 0 then we have

Z = V (Nr) ⊃ V (Nr−1) ⊃ · · · ⊃ V (N) = Z0.

Therefore, our main result is true by induction on r if it is true when r = 2.Assuming r = 2 we note that N obtains the structure of an OZ0

-module. Weuse the homeomorphism j and the fact that cohomology of modules can insteadbe computed by treating them as Z-modules to conclude that

H1(Z,N) = H1(Z0, N) = 0

where the last equality follows because Z0 is affine.Consequently we have the following exact sequence, where R := H0(Z,OZ):

0 N(Z) R R0 0

Let Z ′ = SpecR, the exact sequence above gives us a map Z0 → Z ′ which is ahomeomorphism. Moreover the map R =→ H0(Z,OZ) gives us a map of schemesθ : Z → Z ′ which fits into a commutative diagram as follows:

Z Z ′

Z

θ

j

Since the two arrows from Z are homeomorphisms, so must θ. It remains toshow that OZ′ → OZ is an isomorphism.

Step 4

Observe that Z is quasi-compact because θ identifies the underlying topologicalspace with that of Z ′, which is affine hence quasi-compact. Moreover, becauseZ ′ is seperated, intersection of any two affine subschemes of Z ′ are affine andin particular quasi-compact. For any f ∈ R note that the open sets Zf ⊂ Zand D(f) ⊂ Z ′ coincide. In particular we have shown that for any f, g ∈ Rthe intersection Zf ∩ Zg is quasi-compact. Now we may apply Ex. II.2.16d of[Har77] which states that Γ(Zf ,OZ) ' Rf . This implies that OZ′ → OZ is anisomorphism.

Theorem 1.7. Any infinitesimal deformation of a smooth affine scheme istrivial.

Proof. Using Lemma 1.6 we reduce to deformations of algebras. Let B0 be asmooth algebra over k and let A ∈ Artk. Consider a deformation of algebras

B0 B

k A

We wish to show that B is isomorphic to B0 ⊗k A in a way that respects themap B ⊗A k ' B0.

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We will do this by induction on dimA. The base case dimA = 1 impliesA = k and there is nothing to prove. If dimA ≥ 2, choose 0 6= t ∈ mA andconsider the following principal small extension

0 (t) A A′ 0

We have the following diagram

B B ⊗A A′ B0

A A′ k

By induction hypothesis B⊗AA′ ' B0⊗kA′. There is a map B0⊗kA→ B0⊗kA′and we wish to lift the map B → B0 ⊗k A′ to B0 ⊗k A.

Since A→ B is flat and the only geometric fiber of the map is smooth weconclude that B is smooth over A (see [Har77] Thm III.10.2). Moreover we havethe following exact sequence

0 B0 ⊗k (t) B0 ⊗k A B0 ⊗k A′ 0

where the kernel is a square zero ideal because (t) ⊂ A is. Therefore we may usethe smoothness criterion ([Ser06] Theorem C.9) to conclude that

homA(B,B0 ⊗k A)→ homA(B,B0 ⊗k A′)

is surjective. In particular, the lift we are after exists.

Exercise. Compare the definition of smoothness for functors with this liftingproperty of smooth algebras.

It remains to show that the map B → B0 ⊗k A is an isomorphism. ClearlyB0 ⊗k A is free and thus flat over A, the map is an isomorphism on the specialfiber from which it follows that the map itself is an isomorphism by Lemma 3.1of Lecture 5 (or [Ser06] Lemma A.4).

We will now study locally trivial deformations. Something that is locallytrivial is formed by gluing trivial objects by isomorphisms. Therefore thefollowing lemma will be crucial for our understanding of the functor Def∗X .Note however that this is a simplified version of [Ser06] Lemma 1.2.6. We willeventually need the stronger version but this will do for now.

Lemma 1.8. Let A be a k-algebra and A[ε] = A×k k[ε]. Denote by Aut∗k[ε](A[ε])the group of k[ε]-algebra automorphisms of A[ε] that induce the identity onA[ε]/(ε) ' A. Denote by Derk(A) the k-derivations of A into itself.

There is a natural isomorphism of groups

Aut∗k[ε](A[ε])∼−→ Derk(A).

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Proof. Since there is a natural splitting A[ε] = A⊕εA as a k-vector space, we maydefine two projections from A[ε] to A which we will denote by π1 and π2, suchthat the first projection is a morhpism of algebras whereas the second projectionis only k-linear. Letting ψ ∈ Aut∗k[ε](A[ε]), we may write ψ = ψ1 + ε · ψ2 whereψi = πi ψ.

By k[ε]-linearity we have ψ(a+ εb) = ψ(a) + εψ(b). Expanding this out givesthe following identities:

ψ1(a+ εb) = ψ1(a)

ψ2(a+ εb) = ψ2(a) + ψ1(b).

Finally we combine this with our assumption that ψ descends to the identity onA to conclude ψ1 = π1. Therefore, given our restrictions on ψ only d := ψ2|A isnot completely determined.

We now prove that d is a k-derivation of A. Indeed, for c ∈ k we haveψ(c) = cψ(1) and thus d(c) = 0. If a, b ∈ A then expanding out ψ(ab) = ψ(a)ψ(b)gives d(ab) = a d(b) + bd(a).

Conversely, begin with a k-derivation d : A→ A and define

ψ := π1 + ε(d +π2) : A[ε]→ A[ε].

It is left as an exercise to show that ψ is a k[ε]-algebra automorphism.We have thus established a bijection between automorphisms of A[ε] and

k-derivations of A. It is clear that this bijection respects the group structureand is thus a group isomorphism.

Let X be a scheme, then we denote HomOX(Ω1

X ,OX) by TX and call it thetangent bundle of X.

Theorem 1.9. Let X be a variety over k (ie. integral, seperated scheme of finitetype over k). Then the tangent space of the functor of locally trivial deformationsof X is canonically isomorphic to H1(X, TX).

Proof.

Step 1: From locally trivial families to cohomology classes

Let X → Spec k[ε] be a locally trivial infinitesimal deformation of X. Choosean affine open cover Ui → X such that X|Ui

is a trivial deformation ofUi. Choose isomorphisms θi : Ui × Spec k[ε] → X|Ui

and by pulling backthese isomorphisms we define automorphisms θij := θ−1i θj : Uij × Spec k[ε]→Uij × Spec k[ε] where Uij := Ui ∩ Uj = Ui ×X Uj .

Each Uij is affine because X is seperated and thus Lemma 1.8 correspondsto each automorphisms θij a derivations of Γ(Uij ,OX), equivalently an elementdij ∈ Γ(Uij , TX). Since θijθjkθ−1ik = Id we conclude dij + djk − dik = 0. Thusdij is a Čheck 1-cocycle and therefore gives a class in H1(X, TX).

Step 2: Showing invariance on the isomorphism class

We made three choices in constructing the cohomology class from the isomor-phism class of a locally trivial family. First we chose a representing family, secondwe chose a trivializing cover Ui and then we chose trivializing isomorphisms θi.With one stroke we can show the invariance of the resulting cohomology classon the first and third choices whilst the second one is an easy exercise:

5

Exercise. Show the cohomology class does not depend on the trivializing coverUi.

Let X ′ → Spec k[ε] be another deformation and let Ψ : X ′ → X be anisomorphism of deformations. The cover Ui → X also trivializes X ′ andtherefore we choose trivializing isomorphisms θ′i : Ui × Spec k[ε] → X ′|Ui

. Letd′ij ∈ Γ(Uij , TX) be constructed as above, this time using θ′i’s. Let µi : Ui ×Spec k[ε] → Ui × Spec k[ε] be the isomorphism given by θ−1i Ψθ′i. Observe thefollowing:

θ′i−1θ′j = θ′i

−1ΨΨ−1θ′j

= (θiµi)−1(θjµj)

= µ−1i θ−1i θjµj

Therefore, letting ui ∈ Γ(Ui, TX) be the derivation associated to µi we getd′ij = dij + uj − ui. In other words, the cohomology classes associated to d′ijand dij are equal.

If we take X ′ = X and Ψ = Id then this also proves the independence of theclass on the choice of trivializing isomorphisms θi.

Step 3: From cohomology classes to deformations

Now we would like to reverse the operation and get a locally trivial defor-mation of X from a cohomology class ξ ∈ H1(X, TX). Choose an affine opencover Ui → X. By [Har77] Theorem III.4.5 we may calculate the sheaf co-homology using Čheck cohomology with the cover Ui. Thus ξ = dij withdij ∈ Γ(Uij , TX) and Uij = Ui ∩ Uj as before.

By Lemma 1.8 each dij corresponds to an isomorphism θij : Uij×Spec k[ε]→Uij × Spec k[ε]. We glue the schemes Ui × Spec k[ε] along the open sets Uij ×Spec k[ε] via these isomorphisms θij to obtain a scheme X (see [Har77] Ex.II.2.12).

Since the automorphisms θij don’t interfere with the projection onto Spec k[ε]we can glue these projections to get a map X → Spec k[ε]. Similarly, themaps Ui → X glue to a map X → X because each automorphism θij isan automorphism of deformations, therefore it fixes the special fiber Uij →Uij × Spec k[ε]. This gives us the following diagram

X X

Spec k Spec k[ε]

The map X → Spec k[ε] is flat because flatness is local on the domain andlocally this map is just the projection from a trivial family. Moreover, the mapX → Spec k[ε] is proper because X → Spec k is proper (since any morphismfrom k[ε] to a DVR kills ε the implication is immediate).

Step 4: Invariance on the representative of the cohomology class

We would like to show now that the isomorphism class of the deformationconstructed from ξ ∈ H1(X, TX) is invariant with respect to the choice of a

6

representative in ξ. This requires first that we show invariance under refinementsof the cover we use for Čheck cohomology, which is easy, and then invariance fortwo different representatives dij, d′ij ∈ ξ. Let X correspond to dij and X ′correspond to d′ij. We will show that these two deformations are isomorphic.To do this, use the fact that dij − d′ij = aj − ai where ai ∈ Γ(Ui, TX). Theai’s correspond to isomorphisms αi : Ui × Spec k[ε]→ Ui × Spec k[ε] and theseαi’s glue to an isomorphism X → X ′.

Using Theorem 1.7 together with Theorem 1.9 we have the following.

Corollary 1.10. If X is a smooth variety over k then its first order deformationsare in bijection with H1(X, TX).

References[Har77] R. Hartshorne, Algebraic geometry. Springer-Verlag, New York-Heidelberg,

1977, pp. xvi+496, Graduate Texts in Mathematics, No. 52.

[Ser06] E. Sernesi, Deformations of algebraic schemes, ser. Grundlehren derMathematischen Wissenschaften [Fundamental Principles of Mathe-matical Sciences]. Springer-Verlag, Berlin, 2006, vol. 334, pp. xii+339.

7

Deformation Theory

Michael Kemeny∗

Lecture 10

If X is a smooth variety over k then we know that DefX(k[ε]) ' H1(X, TX).We will give a similar expression for the tangent space of the deformation functorfor any (integral) variety.

1 Background on extensionsThe main reference for this section is Section III.6 of [Har77].

Let X be a scheme over k and Mod(X) be the category of OX -modules.An object I of Mod(X) is called injective if the functor HomX(−, I) is exact.Recall that Mod(X) has enough injectives. This means that for any objectF ∈ Mod(X) we can find an exact sequence (called an injective resolution of F):

0 F I0 I1 I2 . . .

where In ∈ Mod(X) is injective for all n ≥ 0.Fix G ∈ Mod(X). We will define ExtiX(G,−) to be the ith right derived

functor of HomX(G,−).Using the resolution above ExtiX(G,F) is, by definition, the ith cohomology

of the following complex:

0 HomX(G, I0) HomX(G, I1) HomX(G, I2) . . .

where HomX(G, In) is considered to be the nth term of the complex. Observethat Ext0X(G,F) = HomX(G,F).

IfX is a scheme over k then ExtiX(G,F) has a natural k-vector space structure.Indeed, in general the set of homomorphisms between two OX -modules has thenatural structure of a Γ(X,OX)-module. Thus, so will their Ext.

1.1 Useful exerciseExercise III.6.1 in [Har77] gives a concrete description of Ext1X(G,F). An exactsequence

ξ : 0 F H G 0

is called an extension of G by F . Two extensions are considered isomorphic ifthere is an isomorphism of short exact sequences that is an equality on F and G.There is a natural k-vector space structure on the set of isomorphism classes ofextensions. To get an idea on how this structure is defined see p.12-13 of [Ser06].

∗TEX by Emre Sertöz. Please email suggestions or corrections to emresertoz@gmail.com

1

Using the extension ξ we get a long exact sequence:

0 HomX(G,F) HomX(G,H) HomX(G,G) Ext1X(G,F) . . .δ

Define [ξ] := δ(IdG) ∈ Ext1X(G,F). This gives a bijection (and even a k-vectorspace isomorphism) between isomorphism classes of extensions of G by F andelements of Ext1X(G,F).

1.2 Ext sheavesLet Hom(G,F) denote the hom-sheaf which is a OX -module. Do not confusethis with the Γ(X,OX)-module HomX(G,F).

The functor Hom(G,−) gives a left-exact functor from Mod(X) to itself andits ith-right derived functor is denoted by ExtiX(G,−).

If X is a variety then there is a relation between the cohomology groupsof the sheaves ExtiX with the modules ExtiX . This relationship is given by aspectral sequence called local-to-global Ext-sequence. In particular, it gives thefollowing exact sequence:

0 H1(X,Hom(G,F)) Ext1X(G,F) H0(X,Ext1X(G,F))

H2(X,Hom(G,F)) Ext2X(G,F)

See Huybrecths “Fourier-Mukari transforms in algebraic geometry” section 2.3 orthe Wikipedia page on the “Grothendieck spectral sequence” and the referencestherein (also see the page on “Five term exact sequence”).

2 Main theoremWe are now in a position to state the main theorem we are aiming for.

Theorem 2.1. Let X be a variety over k (in particular X is integral). Then

DefX(k[ε]) ' ExtX(Ω1X ,OX).

Remember that TX = Hom(Ω1X ,OX), therefore the first part of the local-to-

global Ext-sequence gives us

0 H1(X, TX) Ext1X(Ω1X ,OX)

We proved that the first term parametrizes the first order locally trivial de-formations of X. This injection corresponds to the inclusion of locally trivialdeformation functor into the deformation functor.

2.1 Preliminaries for the main theoremWe need two lemmas before we start proving the theorem next week. The firstgives a criterion for flatness over artinian local rings.

2

Remark. It is well known that a finite module over a local ring is flat iff it isfree. The following is a stronger result, eliminating the finiteness condition, forArtinian rings. Furthermore, we give a simple criterion to check for flatness.

Lemma 2.2. Let A ∈ Artk and let M be an A-module. Then M is flat iff

TorA1 (M,k) = 0.

Moreover, if M is flat then it is free.

Proof.( =⇒ ) If M is flat then for any ideal I ⊂ A we have TorA1 (M,A/I) = 0. In

particular for I = mA.(⇐= ) We are going to show that M is free and hence flat.Consider the following exact sequence:

0 mA A k 0 (1)

Let xi ⊂M be such that the images of this subset forms a basis in M ⊗A k.Using Nakayama’s lemma for nilpotent maximal ideals we conclude that xigenerate M .

Suppose there is a relation amongst xi, i.e. there exists fi ⊂ A such that∑fixi = 0 (note that almost all terms have to be zero for the sum to make

sense). We wish to show that in fact all fi are zero. Since xi restrict to a basis onM ⊗ k, the fi’s must vanish under the map A→ k. Therefore, for any relationfi, each fi belongs to mA.

We are now going to show that if all relations belong to mjA then in fact allrelations belong to mj+1

A . Since mA is a nilpotent ideal, this means that thereare no non-zero relations amongst xi and that M is free with generators xi.

We established the base case j = 1 above. (If we take the obvious j = 0 tobe the base case then the induction step below will not work. Why?) Observethat if all relations belong to mjA then xi restricts to a free basis of M/mjAM .

Take a relation fi ⊂ mjA and consider the natural map

mA ⊗AM −→ mA/mj+1A ⊗A/mj

AM/mjA

If∑i fi ⊗ xi = 0 then its image under this map must also be zero. But xi maps

to a free basis on M/mjA so this implies that the image of each fi in mA/mj+1A

must be zero, hence fi ∈ mj+1A . So we need only show that for any relation∑

i fixi = 0 the tensor∑i fi ⊗ xi is also zero.

To see this, tensor the equation (1) with M over A. Using the fact thatTorA1 (M,k) = 0 we conclude mA ⊗AM → mAM is an isomorphism.

The following lemma may be considered as a flatness criterion for “thickenings”of deformations via principal small extensions.

Lemma 2.3. Suppose A → B ∈ Artk is a principal small extension. Letf : A→ C be a morphism of k-algebras. The morphism f is flat iff the followingtwo conditions are satisfied:

• f ′ : B → B ⊗A C is flat

• ker(C → B ⊗A C) ' k ⊗A C.

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Proof.( =⇒ ) Flatness is preserved under base change. Hence, if f is flat so is

f ′. Tensor the exact sequence 0→ k → A→ B → 0 (treated as a sequence ofA-modules) with the flat A-module C to get

0 k ⊗A C C B ⊗A C 0

This gives us the second condition.( ⇐= ) Now assume the two listed conditions are satisfied. The second of

the conditions imply that TorA1 (B,C) = 0 as the first map in the following exactsequence is injective:

k ⊗A C C B ⊗A C 0

Now tensor the exact sequence 0→ mB → B → k → 0 seen as A-modules withC:

mB ⊗A C B ⊗A C k ⊗A C 0

We could have viewed the exact sequence as B-modules and tensored withB⊗AC instead, to get the same exact sequence. Since B⊗AC is flat over B, themap mB ⊗AC → B⊗AC is injective. Using the Tor-exact sequence we concludethat TorA1 (C,B)→ TorA1 (C, k) is surjective. But TorA1 (C,B) ' TorA1 (B,C) = 0and hence TorA1 (C, k) = 0. Applying the previous lemma finishes the proof.

References[Har77] R. Hartshorne, Algebraic geometry. Springer-Verlag, New York-Heidelberg,

1977, pp. xvi+496, Graduate Texts in Mathematics, No. 52.

[Ser06] E. Sernesi, Deformations of algebraic schemes, ser. Grundlehren derMathematischen Wissenschaften [Fundamental Principles of Mathe-matical Sciences]. Springer-Verlag, Berlin, 2006, vol. 334, pp. xii+339.

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