The Hilbert Scheme

This article introduces the Hilbert scheme of Grothendieck as a foundational example for studying deformations in local schemes The Hilbert scheme effectively addresses the challenge of characterizing families of closed subschemes within a specified scheme Initially, the author aimed to thoroughly comprehend the proof of a significant theorem, which has evolved into the comprehensive notes and insights presented in this book.

Theorem 1.1 Let Y be a closed subscheme of the projective space X =P n k over a fieldk Then

The Hilbert scheme, denoted as H, is a projective scheme that parametrizes closed subschemes of X sharing the same Hilbert polynomial P as a given subscheme Y It features a universal subscheme W, which is flat over H and contains fibers over closed points h in H that correspond to closed subschemes of X with the same Hilbert polynomial P Additionally, H is considered universal because for any other scheme T, if there exists a closed subscheme W within X×T that is flat over T, all its fibers will also be subschemes of X with the same Hilbert polynomial P.

R Hartshorne, Deformation Theory, Graduate Texts in Mathematics 257, 5DOI 10.1007/978-1-4419-1596-2 2, c Robin Hartshorne 2010

6 1 First-Order Deformations polynomial P, then there exists a unique morphismϕ:T →H such that

(b)The Zariski tangent space to H at the pointy∈H corresponding to Y is given by H 0 (Y,N), whereN is the normal sheaf ofY inX.

If \( Y \) is a locally complete intersection and \( H^1(Y, N) = 0 \), then \( H \) is non-singular at the point \( y \), with a dimension equal to \( h^0(Y, N) = \dim_k H^0(Y, N) \) Additionally, for any locally complete intersection \( Y \), the dimension of \( H \) at \( y \) is at least \( h^0(Y, N) - h^1(Y, N) \).

Parts (a), (b), (c) of this theorem are due to Grothendieck [45] For part (d) there are recent proofs due to Laudal [92] and Mori [109] I do not know whether there is an earlier reference.

This book focuses on studying local theory and does not include a proof of the existence of the Hilbert scheme, as this proof involves different techniques that are not essential for understanding the material presented For readers interested in exploring the proof, numerous sources are available for reference.

[152] Parts (b), (c), (d) of the theorem will be proved in §2, §9, and §11, respectively.

Parts (b), (c), (d) of this theorem illustrate the benefit derived from Grothendieck’s insistence on the systematic use of nilpotent elements Let

Let D = k[t]/t² represent the ring of dual numbers, and consider D as our parameter scheme According to the universal property, flat families Y ⊆ X × D with closed fibers Y correspond to morphisms of schemes Spec D → H that map the unique point to y This correspondence can be interpreted as the Zariski tangent space at H for the point y, denoted as Hom_y(D, H) Consequently, to establish part (b) of the theorem, it suffices to classify the schemes Y ⊆ X × D.

flat overD, whose closed fiber isY, which we will do in§2.

Part (c) of the theorem addresses obstruction theory, which examines the challenges in extending an infinitesimal deformation defined over an Artin ring A to a larger Artin ring Typically, there exists an obstruction that must vanish for an extended deformation to exist For closed subschemes without local obstructions, such as locally complete intersection subschemes, these obstructions are found in H¹(Y, N) When this group is zero, it indicates the absence of obstructions, resulting in a nonsingular moduli space Additionally, the dimension estimate (d) is derived from obstruction theory.

In this section, we will explore the existence of the Hilbert scheme for curves in the projective space P² For an algebraically closed field k, a curve in P²ₖ is defined as a closed subscheme represented by a homogeneous polynomial f(x, y, z) of degree d within the coordinate ring S = k[x, y, z] This polynomial can be expressed in the form f = a₀x^d + a₁x^(d-1)y + + a_nz^d, where a_i ∈ k and n equals d + 2.

−1 sincef has that many terms Consider (a 0 , , a n ) as a point inP n k

(a) Show that curves of degree d in P 2 are in a one-to-one correspondence with points ofP n by this correspondence.

The set \( C \) is defined as a subset of \( P^2 \) based on the equation \( f = a_0 x^d + \ldots + a_n z^d \), where \( x, y, z \) are coordinates in \( P^2 \) and \( a_0, \ldots, a_n \) are coordinates in \( P^n \) This establishes a correspondence where each point \( a \) in \( P^n \) maps to the fiber \( C_a \) in \( P^2 \) above that point Consequently, this relationship is referred to as a tautological family.

For any finitely generated k-algebra A, a family of curves of degree d in P² over A is defined as a closed subscheme X ⊆ P²_A that is flat over A, with fibers above closed points of Spec A being curves in P² It can be demonstrated that the ideal I_X ⊆ A[x, y, z] is generated by a single homogeneous polynomial f of degree d in A[x, y, z].

(d) Conversely, iff ∈ A[x, y, z] is homogeneous of degree d, what is the condition onf for the zero-schemeX defined byf to be flat overA? (Do not assumeA reduced.)

(e) Show that the familyC isuniversalin the sense that for any familyX ⊆P 2 A as in c), there is a unique morphism SpecA→P n such thatX=C × P n SpecA.

(f) For any curveX ⊆ P 2 k of degree d, show that h 0 (N X ) = n and h 1 (N X ) = 0. (Do not assumeX nonsingular.)

1.2 Curves on quadric surfaces inPPP 3 3 3 Consider the familyCof all nonsingular curvesC that lie on some nonsingular quadric surface Q inP 3 and have bidegree (a, b) witha, b >0.

(a) By considering the linear system of curvesCon a fixedQ, and then varyingQ, show that if the total degreedis equal toa+b≥5, then the dimension of the familyCisab+a+b+ 9.

(b) Ifa, b≥3, show thatH 0 (C,N C ) has the same dimensionab+a+b+ 9, using the exact sequence of normal bundles

Show thatN C/Q ∼=O C (C 2 ) is nonspecial, i.e., itsH 1 is zero, so you can compute itsH 0 by Riemann–Roch Then note thatN Q | C ∼=O C (2), and compute itsH 0 using the exact sequence

The vanishing theorems for H1 of line bundles on Q, as outlined in [57, III, Ex 5.6], lead to the conclusion that for a, b ≥ 3, the family C represents an open subset of an irreducible component of the Hilbert scheme This component has a dimension of ab + a + b + 9 and is smooth at every point.

(d) What goes wrong with this argument ifa= 2 andb≥4? Cf (Ex 6.4).

A complete intersection curve C in projective space P³ over a field k is defined by a homogeneous ideal I in k[x, y, z, w] generated by two homogeneous polynomials Specifically, the curve is characterized by polynomials of degrees a and b, where both a and b are greater than or equal to 1.

The complete intersection curve C is characterized by a degree of d = ab and an arithmetic genus of g = 1/2 ab(a + b - 4) + 1 Its dualizing sheaf, ω_C, is isomorphic to O_C(a + b - 4) For any integers a and b greater than or equal to 1, a typical complete intersection curve of this type is nonsingular The family of such curves is irreducible and has a dimension of 2 when a equals b, specifically (a + 3 choose 3) - 2, or (a + 3 choose 3) + b + 3 when a does not equal b.

(b) The normal sheaf isN C ∼=O C (a)⊕ O C (b) Using the resolution

The dimension of \( H^0(N_C) \) is equal to the dimension of the family of complete intersection curves defined by polynomials of degrees \( a \) and \( b \) This indicates that the family forms a nonsingular open subset of an irreducible component within the Hilbert scheme.

The limit of a flat family of complete intersection curves in projective space P³ does not necessarily result in a complete intersection curve, indicating that the open set of the Hilbert scheme comprising these curves may lack closure For instance, by selecting a value λ from the field k, specifically λ = 0 or λ = 1, one can examine a family of complete intersection curves defined over the ring k[t, t⁻¹] through the equations tyz - wx = 0 and yw - t(x - z)(x - λz) = 0.

(a) Show that for anyt= 0, these equations define a nonsingular cureC t of degree

(b) Now extend this family to a flat family over all ofk[t], and show that the special

The fiber C₀ overt=0 represents the combination of a nonsingular plane cubic curve and a line that intersects the cubic curve at a single point, while not lying in the same plane Additionally, it can be demonstrated that C₀ does not qualify as a complete intersection As C₀ is a singular curve within a flat family where the general member is nonsingular, it is classified as a smoothable singular curve.

Structures over the Dual Numbers

The twisted cubic curve in projective space P3 is defined parametrically by the coordinates (x0, x1, x2, x3) = (u^3, tu^2, t^2u, t^3) for (t, u) in P1 More broadly, any curve derived from this curve through a linear change of coordinates in P3 is also referred to as a twisted cubic curve.

Any nonsingular curve of degree 3 and genus 0 in P3 is identified as a twisted cubic curve, which represents a family of dimension 12 For these curves, the condition H0(C, NC) = 12 holds true, indicating that twisted cubic curves constitute a nonsingular open subset within an irreducible component of the Hilbert scheme for curves characterized by the Hilbert polynomial 3z + 1.

The subscheme \( Y \subset P^3 \), which consists of a disjoint union of a plane cubic curve and a point, demonstrates that these schemes represent a nonsingular open subset of the Hilbert scheme of curves characterized by the Hilbert polynomial \( 3z + 1 \) Notably, this component exhibits a dimension of 15.

The flat family of twisted cubic curves converges to a curve \( Y_0 \), which is supported on a nodal cubic curve and features an embedded point at the node This curve \( Y_0 \) lies within the closure of both irreducible components, indicating that it corresponds to a singular point on the Hilbert scheme Furthermore, it can be demonstrated that \( h^0(N_{Y_0}/\mathbb{P}^3) = 16 \), thereby confirming that \( Y_0 \) is indeed a singular point of the Hilbert scheme To support this, the homogeneous ideal of \( Y_0 \), given by \( I(z^2, yz, xz, y^2 w - x^2 (x + w)) \), has a resolution within the polynomial ring \( R_k[x, y, z, w] \).

Tensor withB=R/I, then dualize and sheafify to get a resolution

Compute explicitly with the sections ofO Y 0 (2) andO Y 0 (3), which all come from polynomials inR, to show thath 0 (N Y / P 3 ) = 16.

Note.The structure of this Hilbert scheme is studied in detail in the paper [134].

1.7 Let Cbe a nonsingular curve in P n that isnonspecial, i.e.,H 1 (O C (1)) = 0. Show that the Hilbert scheme is nonsingular at the point corresponding toC.Hint:

Utilize the Euler sequence for the tangent bundle on projective space \( P^n \) restricted to the complex numbers \( \mathbb{C} \) Additionally, apply the exact sequence that connects the tangent bundle of the curve \( C \), the tangent bundle of \( P^n \), and the normal bundle of \( C \).

2 Structures over the Dual Numbers

The initial focus in studying deformation questions involves structures over the dual numbers, denoted as D = k[t]/t² This entails establishing a structure—such as a scheme, a scheme with a subscheme, or a scheme accompanied by a sheaf—over a field k, and subsequently classifying the extensions of this structure over the dual numbers These extensions are referred to as first-order deformations.

To maintain an even distribution of our structure across the base, we will consistently consider the extended structure to be flat over D This concept of flatness aligns with the technical definition that reflects the intuitive understanding of deformation.

In this section we will apply this study to Situations A, B, and C. Recall that a moduleM isflatover a ringAif the functorN →N⊗ A M is exact on the category ofA-modules A morphism of schemesf :X →Y is

flatif for every pointx∈X, the local ringO x,X is flat over the ringO f(x),Y

A sheaf ofO X -modulesF isflatoverY if for everyx∈X, its stalkF x is flat overO f(x),Y

Lemma 2.1 A moduleM over a noetherian ring A is flat if and only if for every prime idealp⊆A,Tor A 1 (M, A/p) = 0.

The functor N → N ⊗ A M is precisely defined when Tor 1 (M, N) equals zero for all A-modules N Given that Tor commutes with direct limits, it is sufficient to establish that Tor 1 (M, N) equals zero for all finitely generated A-modules.

In a Noetherian ring A, a finitely generated module N has a filtration with quotients structured as A/p_i for various prime ideals p_i within A By applying the exact sequence of Tor, we can deduce that if Tor 1(M, A/p) equals zero for all prime ideals p, then Tor 1(M, N) also equals zero for all modules N This leads to the conclusion that M is a flat module.

In the sequel, we will often make use of the following result, which is a special case of the “local criterion of flatness.”

Proposition 2.2 Let A → A be a surjective homomorphism of noetherian rings whose kernelJ has square zero Then anA -moduleM is flat over A if and only if

(2)the natural mapM ⊗ A J →M is injective.

Proof Note that since J has square zero, it is an A-module and we can identifyM ⊗ A J withM ⊗ A J.

IfM is flat overA , then (1) follows by base extension, and (2) follows by tensoringM with the exact sequence

If the module M satisfies the specified conditions, it is essential to demonstrate that Tor A 1 (M, A/p) equals zero for every prime ideal p in A Given that J is nilpotent, it is included within the prime ideal p By considering p as the quotient prime ideal p/J, we can proceed with the proof.

A, we can write a diagram of exact sequences

2 Structures over the Dual Numbers 11 Tensoring withM we obtain

According to hypothesis (2), the second horizontal sequence is exact on the left, which implies that the Tors at the top are isomorphic due to the snake lemma Since the second Tors is zero as stated in hypothesis (1), it follows that the first Tors is also zero, as required.

In this article, we explore the deformation problem, specifically Situation A, where X is a scheme over a field k and Y is a closed subscheme of X We define a deformation of Y over a disc D in X as a closed subscheme Y within the product space X × D that is flat over D, with the condition that the fiber Y × D k equals Y Our goal is to classify all possible deformations of Y over D.

In the affine case, we analyze the relationship between an ak-algebra B and an ideal I ⊆ B, which defines Y Our goal is to identify ideals I ⊆ B = B[t]/t² that ensure B/I is flat over D, while also maintaining that the image of I in B = B/tB remains as I.

Note that (B /I )⊗ D k=B/I SinceB is automatically flat overk, by (2.2) the flatness ofB /I overDis equivalent to the exactness of the sequence

SupposeI is such an ideal, and consider the diagram

0 0 0 where the exactness of the bottom row implies the exactness of the top row.

Proposition 2.3 states that to provide an ideal I ⊆ B, ensuring that B/I is flat over D, is equivalent to defining a homomorphism ϕ ∈ Hom_B(I, B/I) Notably, when ϕ equals zero, it corresponds to the trivial deformation where I is expressed as I = I ⊕ tI within the context of B being isomorphic to B ⊕ tB.

Proof We will make use of the splitting B = B⊕tB as B-modules, or, equivalently, of the sectionσ:B→B given byσ(b) =b+ 0ãt, which makes

For any element \( x \) in \( I \), we can lift it to an element of \( I \) expressed as \( x + ty \) for some \( y \) in \( B \), utilizing the splitting of \( B \) Although different liftings may vary by a term of the form \( tz \) where \( z \) is in \( I \), the image \( \bar{y} \) in \( B/I \) remains uniquely defined Consequently, this process establishes a mapping \( \phi: I \to B/I \) by sending \( x \) to \( \bar{y} \).

It is clear from the construction that it is aB-module homomorphism.

Conversely, supposeϕ∈Hom B (I, B/I) is given Define

I ={x+ty|x∈I, y∈B, and the image ofy inB/I is equal toϕ(x)}.

Then one checks easily thatI is an ideal ofB , that the image ofI in B is

I, and that there is an exact sequence

Therefore there is a diagram as before, where this time the exactness of the top row implies the exactness of the bottom row, and hence thatB /I is flat overD.

The T i Functors

In this section we will present the construction and main properties of the T i functors introduced by Lichtenbaum and Schlessinger [96] For any ring homomorphism A → B and any B-module M they define functors

T i (B/A, M), for i= 0,1,2 WithA andB fixed these form a cohomological functor inM, giving a nine-term exact sequence associated to a short exact

3 TheT i Functors 19 sequence of modules 0 → M → M → M → 0 On the other hand, if

In the context of three rings and homomorphisms A→B→C, when M is a C-module, there exists a nine-term exact sequence of T_i functors linked to the homomorphisms A → B, A → C, and B → C These functors are primarily utilized to explore the deformations of rings and schemes, as outlined in Situation D Specifically, the classification of ring deformations is associated with a certain T_1 group, while obstructions are represented in a T_2 group Additionally, the vanishing of the T_1 functor is crucial for characterizing smooth morphisms, whereas the vanishing of the T_2 functor identifies locally complete intersection morphisms.

Construction 3.1.LetA→B be a homomorphism of rings and letM be a

B-module Here we will construct the groupsT i (B/A, M) fori= 0,1,2 The rings are assumed to be commutative with identity, but we do not impose any finiteness conditions yet.

First choose a polynomial ringR=A[x] in a set of variablesx={x i }(pos- sibly infinite) such thatB can be written as a quotient ofR as anA-algebra.

LetIbe the ideal defining B, so that there is an exact sequence

Second choose a free R-moduleF and a surjectionj:F→I→0 and let

Having chosenRandFas above, the construction proceeds with no further choices LetF 0 be the submodule ofF generated by all “Koszul relations” of the formj(a)b−j(b)afora, b∈F Note thatj(F 0 ) = 0 soF 0 ⊆Q.

We define a complex ofB-modules, called thecotangent complex,

The relationship between the modules L 2, L 1, and L 0 is established as L 2 = Q/F 0 Initially, L 2 is considered an R-module; however, it is demonstrated to be a B-module When x is an element of the ideal I and a is an element of Q, we can express x as j(x) for some x in F, leading to the equation xa = j(x)a ≡ j(a)x (mod F 0) Since j(a) equals 0 for a in Q, it follows that xa must also equal 0 This confirms that L 2 is indeed a B-module.

Take L 1 =F⊗ R B =F/IF, and let d 2 : L 2 → L 1 be the map induced from the inclusionQ→F.

TakeL 0 =Ω R/A ⊗ R B, whereΩ R/A is the module of relative differentials.

To defined 1 just map L 1 to I/I 2 , then apply the derivation d: R→Ω R/A , which induces aB-module homomorphism I/I 2 →L 0

In the defined complex of B-modules, it is evident that d1 d2 = 0 Additionally, both L1 and L0 are recognized as free B-modules L1 is classified as free due to its derivation from the free R-module F, while L0 is considered free because R is a polynomial ring over A, which makes Ω R/A a free R-module.

For anyB-moduleM we now define the modules

T i (B/A, M) =h i (Hom B (L • , M)) as the cohomology modules of the complex of homomorphisms of the complex

To show that these modules are well-defined (up to isomorphism), we must verify that they are independent of the choices made in the construction.

Lemma 3.2 The modulesT i (B/A, M)constructed above are independent of the choice ofF (keeping R fixed).

Proof IfF andF are two choices of freeR-modules mapping ontoI, then

F⊕F is a third choice, so by symmetry it is sufficient to compareF with

In the context of the free module F⊕F, the mapping j: F → I can be expressed through another map p: F → F, allowing for a transformation of bases in F⊕F By replacing each generator e of F with e - p(e), we can simplify the mapping such that it becomes j on the first factor and zero on the second factor This results in a clear diagrammatic representation of the relationships between these components.

The kernel of the map (j,0): F ⊕ F → I is identified as Q ⊕ F, leading to the conclusion that (F ⊕ F)₀ equals F₀ + IF By denoting the new construction as L•, we find that L₂ = L₂ ⊕ F/IF, L₁ = L₁ ⊕ (F ⊗ R B), and L₀ remains unchanged Given that F ⊗ R B = F/IF is a free B-module, the complex L• is formed by taking the direct sum of L• with the free acyclic complex F ⊗ R B → F ⊗ R B.

Hence when we take Hom of these complexes intoM and then cohomology, the result is the same.

Lemma 3.3 The modulesT i (B/A, M)are independent of the choice ofR.

To demonstrate the proof, let R = A[x] and R = A[y] represent two polynomial rings that have surjective mappings to B As established in the earlier proof, it suffices to compare R with R = A[x, y] Additionally, the mapping from A[y] to B can be factored through another function.

A[x] by a homomorphismp:A[y]→A[x] Then, changing variables inA[x, y], replacing each y i byy i −p(y i ), we may assume that all the y i go to zero in the ring homomorphismA[x, y]→B Then we have the diagram

0→ I →R →B →0 showing that the kernel ofR →B is generated byI and all they-variables.

Since we have already shown that the construction is independent of the choice ofF, we may use anyF’s we like in the present proof Take any free

R-moduleF mapping surjectively toI TakeF a freeR -module on the same number of generators asF, and takeG a freeR -module on the index set of theyvariables Then we have

Observe that since the y i are independent variables, and G has a basis e i going toy i , the kernelQ in the upper row must be generated by (1) things in

Q, (2) things of the formy i a−j(a)e i witha∈F, and (3) things of the form y i e j −y j e i Clearly the elements of types (2) and (3) are in (F ⊕G ) 0 Therefore

Q /(F ⊕G ) 0 is aB-module generated by the image ofQ, soL 2 =L 2

On the other hand,L 1 =L 1 ⊕(G ⊗ R B), andL 0 =L 0 ⊕(Ω A[y]/A ⊗B).

ThusL 1 has an extra freeB-module generated by thee i ,L 0 has an extra free

The B-module generated by dy i is transformed by the map d 1, which takes e i to dy i Similar to the earlier proof, we observe that L • is derived from L • by incorporating a free acyclic complex, resulting in the equivalence of the modules T i (B/A, M).

Remark 3.3.1.Even though the complexL • is not unique, the proofs of (3.2) and (3.3) show that it gives a well-defined element of the derived category of the category ofB-modules.

Theorem 3.4 LetA→B be a homomorphism of rings Then fori= 0,1,2,

T i (B/A,ã)is a covariant, additive functor from the category ofB-modules to itself If

0→M →M →M →0 is a short exact sequence ofB-modules, then there is a long exact sequence

In the language of[57, III, §1], the T i ’s form a truncated δ-functor.

The T i (B/A, M) are established as well-defined covariant additive functors Given a short exact sequence of modules, and noting that the terms L 1 and L 0 of the complex L • are free, we can derive a sequence of complexes.

0→Hom B (L • , M )→Hom B (L • , M)→Hom B (L • , M )→0 that is exact except possibly for the map

The sequence of complexes associated with the 22 1 First-Order Deformations may lack surjectivity, leading to a long exact sequence of cohomology It is important to note that the complex L • is unique, except for the addition of free acyclic complexes, which ensures that the coboundary maps in the long exact sequence are functorial.

Theorem 3.5 Let A→B →C be rings and homomorphisms, and let M be aC-module Then there is an exact sequence ofC-modules

Proof To prove this theorem, we will show that for suitable choices in the construction (3.1), the resulting complexes form a sequence

0→L • (B/A)⊗ B C→L • (C/A)→L • (C/B)→0 that is split exact on the degree 0 and 1 terms, and right exact on the degree

In the context of homological algebra, applying Hom(ã, M) results in a sequence of complexes that is exact at degrees 0 and 1, while remaining left exact at degree 2 This process leads to the derivation of a nine-term exact sequence in cohomology.

First choose a surjection A[x] →B → 0 with kernel I, and a surjection

F → I → 0 with kernel Q, where F is a free A[x]-module, to calculate the functorsT i (B/A, M).

Next choose a surjection B[y]→ C → 0 with kernel J, and a surjection

G→J →0 of a freeB[y]-moduleGwith kernelP, to calculateT i (C/B, M).

To calculate the functorsT i forC/A, take a polynomial ringA[x, y] in the x-variables and the y-variables Then A[x, y]→B[y]→C gives a surjection ofA[x, y]→C If Kis its kernel then there is an exact sequence

In the context of polynomials in y with coefficients in I, we consider free A[x, y]-modules F and G defined on the same index sets By selecting a lifting of the map from G to J that extends to a map from G to K, we can combine this with the natural map from F to K This process results in a surjection from the direct sum of F and G to K, with S representing the kernel of this surjection.

Now we are ready to calculate Out of the choices thus made there are induced maps of complexes

On the degree 0 level we have Ω A[x]/A ⊗C→Ω A[x,y]/A ⊗C→Ω B[y]/B ⊗C.

These are freeC-modules with bases{dx i }on the left,{dy i }on the right, and

{dx i , dy i } in the middle So this sequence is clearly split exact.

On the degree 1 level we have

F⊗C→(F ⊕G )⊗C→G⊗C, which is split exact by construction.

On the degree 2 level we have

The right-hand map is surjective due to the surjectivity of the map from S to P The composition of these two maps results in zero While we do not assert injectivity for the left-hand map, our proof is completed by demonstrating the exactness in the middle.

Let \( s = f + g \) be an element of \( S \), and assume its image in \( P \) is contained in \( G_0 \) We need to demonstrate that \( s \) can be expressed as a sum of an element in \( (F \oplus G)_0 \) and an element from the image of \( Q[y] \) In the mapping from \( S \) to \( P \), the element \( f \) maps to 0, while \( g \), the image of \( g \), belongs to \( G_0 \) Consequently, \( g \) can be represented as a linear combination of expressions \( j(a)b - j(b)a \) for \( a, b \in G \) By lifting \( a \) and \( b \) to elements \( a' \) and \( b' \) in \( G \), the expressions \( j(a')b' - j(b')a' \) will also reside in \( S \).

Letg be g minus a linear combination of these expressionsj(a )b −j(b )a

We get a new element s = f +g in S, differing from s by something in (F ⊕G ) 0 , and where now g is in the kernel of the map G →G, which is

In the context of group theory, we can express g as a sum of elements xh, where x belongs to set I and h is an element of group G By mapping x from set F to itself via the function j, we establish that xh can be rewritten as j(x)h, which is equivalent to j(h)x modulo F0 Consequently, we find that s is congruent to f plus the summation of j(h)x modulo (F ⊕ G)0 This expression resides within the intersection of sets F and S, thus confirming its membership in Q[y].

Now we will give some special cases and remarks concerning these functors.

Proposition 3.6 For any A → B and any M, T 0 (B/A, M) Hom B (Ω B/A , M) = Der A (B, M) In particular, T 0 (B/A, B) Hom B (Ω B/A , B)is thetangent module T B/A of B overA.

Proof WriteB as a quotient of a polynomial ring R, with kernel I Then there is an exact sequence [57, II, 8.4A]

SinceF →I is surjective, there is an induced surjective mapL 1 →I/I 2 →0. Thus the sequence

L 1 →L 0 →Ω B/A →0 is exact Taking Hom(ã, M), which is left exact, we see that T 0 (B/A, M) Hom B (Ω B/A , M).

Proposition 3.7 If B is a polynomial ring over A, then T i (B/A, M) = 0 fori= 1,2 and for allM.

Proof In this case we can take R = B in the construction Then I = 0,

F = 0, so L 2 = L 1 = 0, and the complex L • is reduced to the L 0 term. ThereforeT i = 0 fori= 1,2 and anyM.

Remark 3.7.1.We will see later that the vanishing of theT 1 functor char- acterizes smooth morphisms (4.11).

Proposition 3.8 If A →B is a surjective ring homomorphism with kernel

I, then T 0 (B/A, M) = 0 for all M, and T 1 (B/A, M) = Hom B (I/I 2 , M).

In particular,T 1 (B/A, B) = Hom B (I/I 2 , B)is the normal module N B/A of

Proof In this case we can takeR=A, so that L 0 = 0 ThusT 0 = 0 for any

0→Q→F →I→0, tensored withB, gives an exact sequence

There is also a surjective mapQ⊗ A B→Q/F 0 , since the latter is aB-module, so we have an exact sequence

Taking Hom(ã, M) shows thatT 1 (B/A, M) = Hom B (I/I 2 , M).

A useful special case is the following.

Corollary 3.9 IfAis a local ring andB is a quotientA/I, whereI is gene- rated by a regular sequencea 1 , , a r , thenT 2 (B/A, M) = 0 for allM.

Proof Indeed, in this case, since the Koszul complex of a regular sequence is exact [104, 16.5], we findQ=F 0 in the construction of theT i -functors Thus

Remark 3.9.1.We will see later that the vanishing of theT 2 functor char- acterizes relative local complete intersection morphisms (4.13).

Another useful special case is given by the following proposition.

Proposition 3.10 SupposeA =k[x 1 , , x n ] and B =A/I Then for any

M there is an exact sequence

0→T 0 (B/k, M)→Hom(Ω A/k , M)→Hom(I/I 2 , M)→T 1 (B/k, M)→0 and an isomorphism

Proof Write the long exact sequence ofT i -functors for the compositionk→

A→B and use (3.6), (3.7), and (3.8) The same works for any base ring k, not necessarily a field.

In this section, we do not impose any finiteness assumptions on rings and modules However, it is evident that if A is a Noetherian ring, B is a finitely generated A-algebra, and M is a finitely generated B-module, then the B-modules T i (B/A, M) are also finitely generated Specifically, we can consider R as a polynomial ring in a finite number of variables over A.

A, which is therefore noetherian Then I is finitely generated and we can takeF to be a finitely generatedR-module Thus the complexL • consists of finitely generatedB-modules, whence the result.

Notation In the sequel we will often denote the modules T i (B/A, B) and

T i (B/k, B) byT B/A i and T B/k i , or evenT B i , if there is no confusion as to the base FurthermoreT B/A 0 will be writtenT B/A , the tangent module ofB over

A Similarly for the sheaves T i (X/Y,O X ) and T i (X/k,O X ) (see (Ex 3.5)), we will writeT X/Y i andT X/k i , or even T X i The sheafT X 0 will be writtenT X , the tangent sheaf ofX.

The Infinitesimal Lifting Property

The definition of nonsingularity can be confusing, as it may not be clear that this property remains consistent regardless of the affine embedding applied Therefore, establishing an intrinsic criterion for nonsingularity is essential for clarity and understanding.

Proposition 4.1 A schemeX of finite type over an algebraically closed field k is nonsingular if and only if the local ring O P,X is a regular local ring for every pointP ∈X [57, I, 5.1; II, 8.14A].

Using differentials we have another characterization of nonsingular vari- eties.

Proposition 4.2 Let X be a scheme overk algebraically closed ThenX is nonsingular if and only if the sheaf of differentialsΩ X/k 1 is locally free of rank n= dimX at every point ofX [57, II, 8.15].

This result is closely related to the original definition using the Jacobian criterion The generalization of the Jacobian criterion describes when a closed subschemeY of a nonsingular scheme X overk is nonsingular.

Proposition 4.3 Let Y be an irreducible closed subscheme of a nonsingular schemeX overk algebraically closed, defined by a sheaf of idealsI ThenY is nonsingular if and only if

(2)the sequence of differentials [57, II, 8.12]

0→ I/I 2 →Ω 1 X/k ⊗ O Y →Ω Y /k 1 →0 is exact on the left.

Furthermore, in this caseI is locally generated by n−r= dimX−dimY elements, andI/I 2 is locally free onY of rankn−r [57, II, 8.17].

Nonsingular schemes possess a unique feature in deformation theory known as the infinitesimal lifting property This concept arises when considering a morphism f: Y → X of schemes alongside an infinitesimal thickening Y ⊆ Y, where Y is a closed subscheme defined by a nilpotent ideal I The central inquiry is whether a lifting g: Y → X exists, such that g restricted to Y equals f While this lifting may not generally be guaranteed, it holds true when both Y and X are affine, and X is nonsingular This characteristic of nonsingular schemes effectively distinguishes them for all morphisms f: Y → X.

Proposition 4.4, known as the Infinitesimal Lifting Property, states that for a nonsingular affine scheme \(X\) of finite type over a field \(k\), and a morphism \(f: Y \to X\) from an affine scheme \(Y\) over \(k\), if \(Y\) is an infinitesimal thickening of itself, then there exists a lifting morphism \(g: Y \to X\) that coincides with \(f\) on \(Y\).

Proof (cf [57, II, Ex 8.6]).First we note that Y is also affine [57, III,

In algebraic terms, we can rephrase the problem by defining X as Spec A and Y as Spec B Consequently, the function f is represented as a ring homomorphism, which we denote as f: A → B.

B is a quotient ofB by an idealIwithI n = 0 for somen The problem is to

find a homomorphismg :A →B liftingf, i.e., such thatg followed by the projectionB →B is f.

If we filter I by its powers and consider the sequence B = B /I n →

B /I n −1 → ã ã ã → B /I 2 → B /I, it will be sufficient to lift one step at a time Thus (changing notation) we reduce to the caseI 2 = 0.

Since X is of finite type over k, we can express A as a quotient of the polynomial ring P = k[x₁, , xₙ] by an ideal J By composing the projection from P to A with f, we obtain a homomorphism from P to B This homomorphism can be lifted to h: P → B, as the variables xᵢ can be mapped to any liftings of their images in B, reflecting the fact that the polynomial ring is a free object in the category of k-algebras.

Now h induces a map h : J → I, and since I 2 = 0, this gives a map ¯h :

Next we note that the homomorphism P →A gives an embedding ofX in an affinen-spaceA n k By (4.3), we obtain an exact sequence

0→J/J 2 →Ω P /k 1 ⊗ P A→Ω 1 A/k →0, and note that these modules correspond to locally free sheaves onX, hence are projectiveA-modules Via the mapsh, f, we get aP-module structure on

B , andA-module structures onB, I Applying the functor Hom A (ã, I) to the above sequence gives another exact sequence

Let θ ∈ Hom P (Ω P /k 1 , I) be an element whose image is ¯h∈ Hom A (J/J 2 , I).

We define a new map \( h: P \to B \) as \( h = h - \theta \), considering \( \theta \) as a k-derivation of \( P \) to the module \( I \) It can be claimed that \( h \) serves as a ring homomorphism that lifts \( f \) and satisfies \( h(J) = 0 \) This conclusion follows from lemma (4.5) To demonstrate that \( h(J) = 0 \), let \( y \in J \); then \( h(y) = h(y) - \theta(y) \) Focusing on \( y \) modulo \( J^2 \), we find that \( h(y) = \theta(y) \) by the selection of \( \theta \), which leads to \( h(y) = 0 \) Consequently, since \( h(J) = 0 \), \( h \) can be descended to yield the desired homomorphism \( g: A \to B \) that lifts \( f \).

Lemma 4.5 Let B →B be a surjective homomorphism of k-algebras with kernelI of square zero Let R→B be a homomorphism of k-algebras.

(a)If f, g:R→B are two liftings of the map R→B to B , thenθ=g−f is ak-derivation ofR toI.

(b)Conversely, if f : R → B is one lifting, and θ : R → I a derivation, theng=f+θis another homomorphism ofR toB lifting the given map

If the set of liftings from R to B as k-algebra homomorphisms is nonempty, it forms a principal homogeneous space under the addition action of the group Der k (R, I), which is equivalent to Hom R (Ω R/k , I) Additionally, I has a natural structure as both a B-module and an R-module since I² is not equal to zero.

Proof (a) Letf, g:R→B and letθ=g−f As ak-linear map,θfollowed by the projectionB →B is zero, soθsends RtoI Letx, y∈R Then θ(xy) =g(xy)−f(xy)

=xθ(y) +yθ(x), the last step being because g(x) and f(y) act in I just like x, y Thusθ is a k-derivation ofR toI.

(b) Conversely, givenf andθas above, let g=f+θ Then g(xy) =f(xy) +θ(xy)

=g(x)g(y), where we note thatθ(x)θ(y) = 0, sinceI 2 = 0 Thusg is a homomorphism of

For a converse to (4.4), we need only a special case of the infinitesimal lifting property.

Proposition 4.6 LetX be a scheme of finite type overkalgebraically closed. Suppose that for every morphismf :Y →Xof apunctual schemeY (meaning

Y is the Specof a local Artin ring), finite over k, and for every infinitesimal thickeningY ⊆Y with ideal sheaf of square zero, there is a liftingg:Y →X. ThenX is nonsingular.

To demonstrate that the local ring O P,X is a regular local ring for each closed point P in X, it suffices to address an algebraic question Specifically, consider a local k-algebra A, which is essentially of finite type over k with a residue field k It is assumed that for every homomorphism f: A → B, where B is an appropriate ring, certain properties hold that are essential for the regularity of the local ring.

In the context of local artinian algebras, the concept of 30 1 First-Order Deformations indicates that for any thickening 0→I →B →B →0, where I^2 = 0, a lifting g:A→B exists, confirming that A is a regular local ring Furthermore, a minimal set of generators for the maximal ideal m can be denoted as a1, , an.

A Then there is a surjective homomorphismf of the formal power series ring

P =k[[x 1 , , x n ]] to ˆA, the completion ofA, sending x i toa i , and creating an isomorphism ofP/n 2 toA/m 2 , wheren= (x 1 , , x n ) is the maximal ideal ofP.

The surjections \( P/n_{i+1} \to P/n_i \), defined by ideals of square zero, allow us to construct a sequence of maps starting from \( A \to A/m^2 \cong P/n_2 \) By progressively lifting these maps, we obtain mappings \( A \to P/n_i \) for each \( i \), leading to a map \( g \) into the inverse limit \( P \) Transitioning to \( \hat{A} \), we establish maps \( P \to f \hat{A} \to g P \), ensuring that \( g \circ f \) is an isomorphism on \( P/n_2 \) Consequently, \( g \circ f \) emerges as an automorphism.

P (Ex 4.1) Henceg◦f has no kernel, sof is injective Butf was surjective by construction, sof is an isomorphism, and ˆAis regular From this it follows thatAis regular, as required.

Corollary 4.7 states that a local ring A, which is essentially of finite type over an algebraically closed field k with residue field k, is classified as a regular local ring if and only if it possesses the infinitesimal lifting property for local Artin rings B that are finite over k.

The following result shows that infinitesimal deformations of nonsingular affine schemes are trivial.

Corollary 4.8 states that if X is a nonsingular affine scheme over a field k and A is a local Artin ring over k, then a scheme X that is flat over Spec A and satisfies the condition that X × A k is isomorphic to X must also be isomorphic to the trivial deformation of X over A This implies that the structure of X is preserved in its deformation, ensuring that the properties of the original scheme are maintained in the flat extension.

We apply the identity map of X to the infinitesimal thickening defined by the isomorphism X × A k ∼= X, leading to a lifting p: X → X such that p◦i = id X The maps from X to both X and Spec A define a product map: X → X × k A Since both schemes are flat over A and this map restricts to the identity on X, it follows that the map is an isomorphism.

The infinitesimal lifting property for nonsingular varieties over an algebraically closed field can be generalized to the relative case of a morphism of schemes, which provides a characterization of smooth morphisms Grothendieck considers the infinitesimal lifting property as one of the equivalent definitions of smooth morphisms.

Next we investigate the relation between nonsingularity and the T i functors.

Theorem 4.9 Let X = SpecB be an affine scheme over k algebraically closed Then X is nonsingular if and only if T 1 (B/k, M) = 0 for all B-modules M Furthermore, ifX is nonsingular, then also T 2 (B/k, M) = 0 for allM.

To demonstrate the proof, let B be expressed as a quotient of the polynomial ring A = k[x₁, , xₙ] over a field k It follows that the spectrum of A, denoted Spec A, is nonsingular By applying criterion (4.3), we can conclude that the variety X is nonsingular if and only if the conormal sequence is satisfied.

Deformations of Rings

(b) This family is still not trivial over the complete local ring k[[t]] at the origin, because one can look at thej-invariant over the field of fractions of this ring.

The projective completion of this family in P2 is significant, even when considered over the Artin ring C = k[t]/t^n for any n ≥ 2 This is due to the fact that the computation of the j-invariant can be effectively carried out within the ring C.

The affine family over the Artin ring C is trivial for any n, as indicated by equation (4.8) Your challenge is to establish a clear isomorphism between this family over C and the trivial family defined by y² = x(x - 1)(x - λ).

(e) Finda, b, c, d ∈ C such that the transformation x = (ax+b)/(cx+d) sends

(f) Substitute forx in the equation y 2 =x (x −1)(x −(λ+t)) and show that the result can be written as y 2 (cx+d) 3 a(a−c)(a−c(λ+t))=x(x−1)(x−λ).

(g) Now, using the fact thattis nilpotent, show that one can findf(x, t) andg(x, t) inC[x] such that the substitutions x =x+tf(x, t), y =y(1 +tg(x, t)), bring the equation into the formy 2 =x(x−1)(x−λ).

(h) Show that the transformation (x, y)→(x , y ) is an automorphism of the ring

C[x, y], so the two families are isomorphic overC.

In this section we use the T i functors to study deformations of arbitrary schemes (Situation D) We will see that the deformations of an affine scheme

X = SpecB over k are given by T 1 (B/k, B), and that the deformations of a nonsingular scheme X are given by H 1 (X,T X ), where T X is the tangent sheaf As an application, we study deformations of cones.

A deformation of a scheme X over an Artin ring A is defined as a scheme that is flat over A, accompanied by a closed immersion i: X → X, where the induced map i × A k: X → X × A k is an isomorphism Two deformations, X1 with immersion i1 and X2 with immersion i2, are considered equivalent if there exists an isomorphism f: X1 → X2 over A that is compatible with the immersions, satisfying the condition i2 = f ◦ i1.

A deformation of a scheme \(X\) over a base \(A\) is defined as a scheme that is flat over \(A\) and for which the product \(X \times_A k\) is isomorphic to \(X\), without specifying the isomorphism However, this definition is less functorially well-behaved compared to the established definition, leading to complications in the set of such deformations.

First-order deformations can be obtained by dividing by the action of the automorphism group of X over k The presence of automorphisms complicates deformation problems compared to Situation A, where isomorphisms of closed subschemes are simply equal and no automorphisms exist For further details on the significance of automorphisms, refer to §18.

We start by considering deformations of affine schemes Let B be a k-algebra A deformation of SpecB over the dual numbers D is then a

D-algebra B , flat over D, together with a homomorphism B → B induc- ing an isomorphism B ⊗ D k → B Because of (2.2) the flatness of B is equivalent to the exactness of the sequence

In this context, we consider B on the right as rings and B on the left as an ideal of square 0, which functions as a B-module Additionally, B serves as both a D-algebra and a k-algebra By disregarding the D-algebra structure of B, we can treat it simply as a k-algebra through the inclusion of k into D As demonstrated in (2.7), the D-algebra structure of B can be uniquely recovered in alignment with the exact sequence (∗) This recovery requires us to define multiplication by t, which involves transitioning from B on the right to B on the left through the inclusion of B.

Equivalence classes of deformations of the k-algebra B over D correspond directly to equivalence classes of exact sequences In this context, B is considered an extension of the k-algebra B by the B-module B.

This article explores a broader context by defining an extension of B by M as A-algebras through an exact sequence, where A represents a ring, B is an A-algebra, and M is a B-module.

0→M →B →B→0, whereB →Bis a homomorphism ofA-algebras, andM is an ideal inB with

M 2 = 0 Two such extensionsB , B areequivalentif there is an isomorphism

B →B compatible in the exact sequences with the identity maps onB and

M Thetrivialextension is given byB =B⊕M made into a ring by the rule (b, m)ã(b 1 , m 1 ) = (bb 1 , bm 1 +b 1 m).

Theorem 5.1 states that for a ring A, an A-algebra B, and a B-module M, there exists a natural one-to-one correspondence between the equivalence classes of extensions of B by M as A-algebras and the elements of the group T 1 (B/A, M) Additionally, the trivial extension is associated with the zero element of this group.

In this article, we explore a surjective map A[x]→B from a polynomial ring over A to B, utilizing a set of generators {e i } for the B-module M We introduce a corresponding set of indeterminates y={y i } with the same index set as {e i } Furthermore, we highlight that for any extension of B by M, it is possible to identify a suitable polynomial ring A[x, y].

5 Deformations of Rings 37 a surjective ring homomorphismf : A[x, y]→B , not unique, that makes a commutative diagram

0 0 0 where the two outer vertical arrows are determined by the construction Here (y) denotes the ideal in A[x, y] generated by the y i , and the map (y)→ M sendsy i toe i

In this article, we outline a two-step process for classifying quotients f: A[x, y] → B that form a specific diagram Initially, we aim to determine the number of distinct ways to express a given extension B as a quotient of A[x, y] By addressing this ambiguity, we can effectively describe the set of extensions B To facilitate this, we enhance the existing diagram by incorporating a top row that includes the kernels of the vertical arrows.

Giving B as a quotient of A[x, y] is equivalent to giving the ideal I in

The inclusion of the ring A[x] into A[x, y] establishes a splitting of the middle row, demonstrating a natural one-to-one correspondence between the set of diagrams and the group Hom A[x](I, M), which is equivalent to Hom B(I/I², M).

In the second step, we utilize (4.5) to demonstrate that the set of potential mappings \( f: A[x, y] \rightarrow B \) constitutes a principal homogeneous space under the action of \( \text{Der}_A(A[x], M) \) It's important to note that the inclusion \( A[x] \rightarrow A[x, y] \) implies that any mapping \( f: A[x, y] \rightarrow B \) inherently defines a corresponding mapping \( g: A[x] \rightarrow B \), and this mapping is uniquely determined by \( f \).

Now write the long exact sequence ofT i functors (3.5) for the three rings

A→A[x]→B and the moduleM The part that interests us is

The first term, as defined in equation (3.6), represents Hom A[x] (Ω A[x]/A , M), which corresponds to the module of derivations Der A (A[x], M) The second term, according to equation (3.8), is Hom B (I/I² , M) The fourth term, as indicated in equation (3.7), equals zero Consequently, T 1 (B/A, M) is identified as the cokernel of a natural map.

Under the interpretations above we see that this cokernel is the set of dia- grams A[x, y] → B as above, modulo the ambiguity of choice of the map

A[x, y] → B , and so T 1 (B/A, M) is in one-to-one correspondence with the set of extensionsB , as required.

Corollary 5.2 Letkbe a field and letBbe ak-algebra Then the set of defor- mations of B over the dual numbers is in natural one-to-one correspondence with the groupT 1 (B/k, B).

The theorem discussed earlier establishes that the deformations are uniquely linked to the k-algebra extensions of B by B.

Next we consider deformations of a nonsingular variety We use ˇCech cohomology of an open covering, knowing that deformations of affine non- singular varieties are trivial.