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NOTES ON HILBERT SCHEMES
DIANE MACLAGAN
Introduction These notes are for lectures on Hilbert schemes in the Summer School on Introduction to explicit methods in algebraic geometry, run September 3–7, 2007 at the University of Warwick as part of the Warwick EPSRC Sympo- sium on Algebraic Geometry. They likely contain many errors, both mathe- matical and typographical; please send any you notice to me at the address D.Maclagan@warwick.ac.uk. Many thanks to those who have already done so. I make no claim for comprehensiveness, and many important areas of this subject are not covered here. Some other references on Hilbert schemes include: The Geometry of Schemes, by Eisenbud and Harris [EH00], Ra- tional Curves on Algebraic Varieties, by J´anos Koll´ar [Kol96], the appen- dix by Iarrobino and Kleiman [IK99] to Power Sums, Gorenstein Alge- bras, and Determinantal Loci by Iarrobino and Kanev, the section on the Hilbert scheme of points in Combinatorial Commutative Algebra by Miller and Sturmfels [MS05], the paper t, q-Catalan numbers and the Hilbert scheme by Haiman [Hai98], and the paper Multigraded Hilbert schemes by Haiman and Sturmfels [HS04]. My treatment has been heavily influenced by these works.
- Lecture 1: Introduction to the Hilbert scheme The Hilbert scheme HilbP (Pn) is a parameter space whose closed points correspond to subschemes of Pn^ with Hilbert polynomial P. The topology on HilbP (Pn) gives a notion of when two subschemes are “close”. Many other moduli spaces are constructed by realizing them as subschemes of the Hilbert scheme. In this lecture we first review the basics of subschemes of Pn^ and Hilbert polynomials, then give the functorial definition of the Hilbert scheme.
1.1. Algebraic preliminaries. Let S = k[x 0 , ,... , xn], where k is a com- mutative ring, and let its irrelevant ideal be m = 〈x 0 ,... , xn〉. A homoge- neous ideal I ⊆ S not containing m determines a closed subscheme of Pn k from the surjection S → S/I (see [Har77, Exercises II.2.14, II.3.12]). In the opposite direction, given a subscheme X ⊂ Pn k , the correspond- ing ideal sheaf IX is the kernel of the map OPn k → OX. The direct sum I = ⊕l≥ 0 H^0 (Pn k , IX (l)) is then a homogeneous ideal of S, because S ∼= ⊕l≥ 0 H^0 (Pn k , O(l)). 1
2 DIANE MACLAGAN
It is important to note that this correspondence between subschemes of Pn k and ideals of S is not a bijection. Essentially this is because the irrelevant ideal m does not correspond to a subscheme of Pn k. More specifically, two ideals I and J correspond to the same subscheme of Pn k if and only if the saturations of I and J with respect to m coincide, so (I : m∞) = (J : m∞), where (I : m∞) := 〈f ∈ S : f mk^ ⊆ I for some k > 0 〉. Note that if (I : m∞) = (J : m∞) then Ik = Jk for k 0. Note also that I ⊆ (I : m∞). An ideal I is called saturated if I = (I : m∞). The saturation of I is the largest ideal corresponding to the same subscheme as I, and there is a one-to-one correspondence between homogeneous saturated ideals of S and subschemes of Pn k. Thus to parameterize subschemes of Pn k , it suffices to parameterize homogeneous saturated ideals of S. If k is an algebraically closed field, and the ideal I is prime, then the subscheme of Pn k determined by I is the variety V (I) determined by I, which has closed points {x ∈ Pn k : f (x) = 0 for all f ∈ I}.
Example 1.1. Let n = 3, and let I = 〈x 0 x 3 − x 1 x 2 , x 0 x 2 − x^21 , x 1 x 3 − x^22 〉. Then I is a prime ideal whose variety is the twisted cubic in P^3. Let I′^ = 〈x^22 x 3 − x 1 x^23 , x 1 x 2 x 3 − x 0 x^23 , x^21 x 3 − x 0 x 2 x 3 , x^32 − x 0 x^23 , x 1 x^22 − x 0 x 2 x 3 , x 0 x^22 − x 0 x 1 x 3 , x^21 x 2 − x 0 x 1 x 3 , x 0 x 1 x 2 − x^20 x 3 , x^31 − x^20 x 3 , x 0 x^21 − x^20 x 2 〉. Then (I′^ : m∞) = (I′^ : m^3 ) = I, as I′^ = I ∩ m^3. Thus I and I′^ determine the same subscheme of P^3.
The Hilbert polynomial of a homogeneous ideal of S, or a subscheme of Pn k , is an invariant of an ideal/subscheme that will determine the connected components of the Hilbert scheme. For simplicity, we assume that k is a field from now on. The Hilbert polynomial is determined from the Hilbert function of the ideal. This is the function HS/I : N → N given by
HS/I (t) = dimk(S/I)t,
where (S/I)t is the tth graded piece of the S-module (and thus vector space over k) S/I. The key fact is that the function HS/I agrees with a poly- nomial PS/I for large t, so HS/I (t) = PS/I (t) for t 0. The polynomial PS/I (t) is called the Hilbert polynomial of S/I. If X is the subscheme of Pn corresponding to I, then PS/I (t) = χ(OX (t)). There are many different proofs of the fact that PS/I exists. The one we give here, while not the shortest, contains some important ideas that we will return to often. The key idea is to first reduce to the case where I is a monomial ideal, and then give a combinatorial proof in the monomial case. This will be a repeated theme. The reduction to the monomial case uses the theory of Gr¨obner bases, and we now summarize the facts we need. Anyone unfamiliar with Gr¨obner bases is urged to spend an evening or two with the first few chapters of the classic [CLO07]. More geometric details are found in [Eis95, Chapter 15]. Fix w ∈ Nn+1. For f =
u∈Nn+1^ cux
u, set the initial term of f to be
inw(f ) =
cuxu, where the sum is over those u ∈ Nn+1^ with w · u maximal among those u with cu 6 = 0. For example, if n = 3, w = (1, 0 , 0 , 1), and
4 DIANE MACLAGAN
inw(g) ∈ inw(I). From this contradiction we see that the standard monomials are linearly independent in S/I, so form a basis. Thus when I is homogeneous we have HS/I = HS/ inw (I). This shows that it suffices to prove the existence of the Hilbert polynomial for monomial ideals. To do this, we first note that the Hilbert function of the polynomial ring S is HS (t) =
(t+n n
for t ≥ 0, which is a polynomial of degree n. This shows the existence of the Hilbert polynomial for polynomial rings. We next show that the standard monomials of a monomial ideal can be partitioned into translates of the monomials in smaller polynomial rings.
Definition 1.2. Let I be a monomial ideal. A Stanley decomposition for S/I is a finite decomposition of the standard monomials of I into disjoint sets of the form (xu, σ) = {xu+v^ : supp(xv) ⊆ σ}, where σ ⊆ { 0 , 1 ,... , n}, and supp(xv) = {i : vi > 0 }.
Example 1.3. Let I = 〈x^20 x 1 〉 ⊂ k[x 0 , x 1 ]. Then one Stanley decomposition for S/I is {(1, { 0 }), (x 1 , { 1 }), (x 0 x 1 , { 1 })}. Another Stanley decomposition is {(1, { 1 }), (x 0 , { 0 }), (x 0 x 1 , ∅), (x 0 x^21 , { 1 })}.
Recall that if f ∈ S and I is an ideal in S then (I : f ) = {g ∈ S : gf ∈ I}, and (I : f ∞) = {g ∈ S : gf k^ ∈ I for some k ≥ 0 }.
Lemma 1.4. Let I be a monomial ideal. Then a Stanley decomposition for S/I exists.
Proof. Let ki = min{k : (I : xki ) = (I : x∞ i )}, and let k =
∑n i=0 ki.^ The proof is by induction on n and k. When n = 0 we must have I = 〈xl 0 〉 for some l. Then ∪l j−=0^1 (xj 0 , ∅) is a Stanley decomposition for S/I. If k = 0 then I = Pσ = 〈xi : i 6 ∈ σ〉 is a monomial prime ideal, and {(1, σ)} is a Stanley decomposition for S/I. Consider the short exact sequence 0 → S/(I : xi) → S/I → S/(I, xi) → 0.
Note that S/(I, xi) is isomorphic to the quotient of a smaller polynomial ring, missing xi, by a monomial ideal, so by induction a Stanley decomposition {(xuj^ , σj ) : 1 ≤ j ≤ s} for S/(I, xi) exists. Also, the invariant ki for (I : xi) is smaller than that for I, while all other kj are no larger, so again by induction a Stanley decomposition {(xvj^ , τj ) : 1 ≤ j ≤ t} for S/(I : xi) exists. Then {(xuj^ , σj ) : 1 ≤ j ≤ s} ∪ {(xixvj^ , τj ) : 1 ≤ j ≤ t} is a Stanley decomposition for S/I.
If {(xuj^ , σj ) : 1 ≤ j ≤ s} is a Stanley decomposition for S/I, then
HS/I (t) =
∑^ s
j=
HSσj (t − |uj |),
where Sσj is the polynomial ring k[xi : i ∈ σj ], and |uj | =
∑n i=0(uj^ )i. The fact that the Hilbert function of S/I eventually agrees with a polynomial thus follows from the fact that the Hilbert function of a polynomial ring agrees with a polynomial for nonnegative values.
NOTES ON HILBERT SCHEMES 5
Example 1.5. Let S = k[x 0 , x 1 , x 2 , x 3 ], and let I = 〈x 0 x 3 − x 1 x 2 , x 0 x 2 − x^21 , x 1 x 3 −x^22 〉. To compute the Hilbert polynomial of S/I, we first compute a Gr¨obner basis for I and thus compute an initial ideal. For w = (1, 0 , 0 , 1) the given generating set is a Gr¨obner basis, so J = inw(I) = 〈x 0 x 3 , x 0 x 2 , x 1 x 3 〉. A Stanley decomposition for S/J is {(1, { 2 , 3 }), (x 1 , { 1 , 2 }), (x 0 , { 0 , 1 })}, so for t ≥ 1 we have HS/I (t) = Hkx 2 ,x 3 + Hk[x 1 ,x 2 ](t − 1) + Hk[x 0 ,x 1 ](t − 1) = t + 1 + t + t = 3t + 1. Thus PS/I (t) = 3t + 1.
1.2. Functorial definition. It is natural to worry that there might be many ways of constructing a scheme (or variety) whose closed points correspond to subschemes of Pn. This is taken care of by requiring that the Hilbert scheme be a fine moduli space, which thus carries a universal bundle. To define a fine moduli space, we need the notions of a representable functor and of the functor of points of a scheme. The key idea is that a scheme is determined by its morphisms to other schemes.
Definition 1.6. Let X be a scheme. The functor hX from the opposite of the category of schemes to the category of sets is given by
hX (Y ) = Mor(Y, X),
and if f : Y → Z is a morphism of schemes, then
hX (f ) : Mor(Z, X) → Mor(Y, X)
is the induced map of sets. The functor hX is the functor of points of the scheme X.
Note that if Y = Spec(k) for k a field then hX (Y ) is the set of k-valued points of X.
Definition 1.7. A functor F : (schemes)◦^ → sets is representable if F ∼= hX for some scheme X. The scheme X is unique if it exists. This follows from the categorical result known as Yoneda’s lemma.
Lemma 1.8 (Yoneda’s Lemma). Let C be a category and let X, X′^ be objects of C.
(1) If F is any contravariant functor from C to the category of sets, the natural transformations from Mor(−, X) to F are in natural corre- spondence with the elements of F (X). (2) If the functors Mor(−, X) and Mor(−, X′) from C to the category of sets are isomorphic, then X ∼= X′. More generally, the maps of functors from Mor(−, X) to Mor(−, X′) are the same as maps from X to X′; that is the functor h : C → Fun(C◦, (sets)) sending X to hX is an equivalence of C with a full subcategory of the category of functors. The second part of this lemma immediately proves that if F is repre- sentable then the scheme representing it is unique. The functor F is in fact determined by its values on affine schemes.
NOTES ON HILBERT SCHEMES 7
induced map Π : Mor(−, Y ) → Mor(−, Y ′) satisfies ΨY ′^ = Π ◦ ΨY. Such a Y is then unique up to canonical isomorphism. The disadvantage, though, is that we do not get a nice universal family as for a fine moduli space. Another possibility, beyond the reach of these notes, is use a stack description.
We now describe the moduli problem defining the Hilbert scheme.
Definition 1.12. Fix a base scheme S. The Hilbert functor is the functor hP : (schemes)◦^ → (sets) that associates to any scheme B over S the set of subschemes Y ⊆ PnB flat over B whose fibers over points of B have Hilbert polynomial P.
We will assume here for simplicity that S = Spec(k) for k a field. Taking S = Spec(Z) allows great generality.
Theorem 1.13. There is a scheme HilbP (Pn) that represents hP.
We sketch the proof in the next lecture.
Remark 1.14. One can also consider the Hilbert scheme Hilb(X) where X is a projective scheme. Loosely, one embeds X into some projective space, and constructs Hilb(X) as a subscheme of Hilb(PN^ ). One then shows that this construction also represents some functor, thus showing that it is in- dependent of the choice of embedding into projective space. In Lecture 4 we will consider the Hilbert scheme of points in affine space, which is an analogous construction.
1.3. Exercises 1. The following are more exercises than any of you would want to do this week. So before beginning, look at your notes and decide which aspect you would like to understand better. Then read through all the exercises, before choosing which one to start with. Hopefully there is something for everyone!
(1) Let S = k[x 0 , x 1 , x 2 , x 3 ], where k is a field. Compute the saturation of I = 〈x^23 , x 2 x 3 , x 1 x 3 , x 0 x 3 , x 1 x 2 , x^30 〉. (2) (a) Show that (I : m∞) = (I : mk) for some k > 0, and that if (I : mk) = (I : mk+1) for some k > 0, then (I : mk) = (I : m∞). Here (I : mk) = {f ∈ S : f g ∈ I for all g ∈ mk}. (b) Show that (I : m∞) = ∩ni=0(I : x∞ i ). (c) (For those who know more about Gr¨obner bases) Show that a generating set for (I : x∞ i ) is obtained by computing a Gr¨obner basis for I with respect to the reverse lexicographic order with xi smallest, and then dividing out any power of xi dividing an element. This explains how saturation can be computed in a computer algebra system. (3) Show that the Hilbert polynomial of Pn^ is P (t) =
(n+t n
(4) Let S = k[x 0 , x 1 , x 2 ]. Compute the Hilbert polynomial of S/I for the following ideals. (a) I = 〈x^20 , x 1 x 2 〉, (b) I = 〈x^30 , x^20 x^31 , x 0 x 1 x 2 , x 0 x^42 , x^41 x^32 , x^61 , x^72 〉,
8 DIANE MACLAGAN
(c) I = 〈x^21 − x 0 x 2 〉, (d) I = 〈 3 x 0 x 1 − 2 x^21 − 3 x 0 x 2 + x 1 x 2 + x^22 , 9 x^20 − 4 x^21 − 18 x 0 x 2 + 8 x 1 x 2 + 5x^22 , x^31 − 3 x 1 x^22 + 2x^32 〉. (Hint: You’ll probably want to use a computer algebra package). (5) Show that I and (I : m∞) have the same Hilbert polynomial. (6) In this exercise we sketch a more straight-forward proof of the exis- tence of the Hilbert polynomial, assuming the existence of finite free resolutions. We can extend the definition of the Hilbert function to arbitrary S-modules by setting HM (t) = dimk Mt. (a) Show that the Hilbert function is additive on short exact se- quences, in the sense that if
0 → M ′^ → M → M ′′^ → 0
is a short exact sequence of S-modules, then HM = HM ′^ + HM ′′^. (b) The S-module S[a] is the polynomial ring S with the grading shifted, so S[a]t = Sa+t. Thus the 1 of S[a] has degree −a. Show that the Hilbert function of S[a] agrees with a polynomial for t 0. (c) Deduce the existence of the Hilbert polynomial for any finitely generated S-module from the existence of a finite free resolution
0 → ⊕j S[−bnj ]βnj^ → · · · → S[−b 1 j ]β^1 j^ → S[−b 0 j ]β^0 j^ → M → 0.
(7) Prove Yoneda’s lemma (8) Give a direct argument that hSpec(Z) is not isomorphic to any functor of points of a different scheme. (9) Flesh out the details of the construction of the Grassmannian. What are the Pl¨ucker relations? Why does the given scheme represent the Grassmann functor?
- Lecture 2: Construction In this lecture we outline the construction of the Hilbert scheme. The proof comes in two parts. First, one constructs a scheme X whose closed points correspond to subschemes of Pn. This is essentially combinatorial commutative algebra. One then shows that X satisfies the desired universal property, and thus represents the functor hP. We will only give details on the first of these steps. The scheme X is constructed as a subscheme of a Grassmannian. The key to the construction of the Hilbert scheme is the fact that there is a uniform degree D = D(P ) for which all ideals I ⊆ S of Hilbert polynomial P are generated in degree at most D. This follows from Gotzmann’s regularity theorem, which uses the notion of Castelnuovo-Mumford regularity. A good reference for the commutative algebra from this lecture is [BH93]. Recall that for a homogeneous ideal the degrees of the minimal generators, and also the degrees of minimal generators for higher syzygy modules, is well-defined. See [Eis95, Chapter 20] for details.
10 DIANE MACLAGAN
if I is generated in degree at most D then I≥D is generated in degree D. Thus we can consider ideals generated in degree D. Let GD be the Grass- mannian Gr(
(n+D n
− P (D), SD). Saturated ideals I with Hilbert polynomial P correspond to closed points in G, where an ideal I corresponds to the k- subspace ID of SD. We now show that the closed points in Gr(N − P (D), N ) corresponding to such ideals are the closed points of a subscheme H of G. This relies on Gotzmann’s persistence theorem, which gives a criterion for the Hilbert function of an ideal to agree with its Hilbert polynomial. This relies on a curious numerical function from N to N depending on a parameter d ∈ N.
Definition 2.6. Given n, d ∈ N, we can write n uniquely as
n =
∑^ t
j=
kj d − j
where kj > kj+1 ≥ 0. The Macaulay upper boundary of n with respect to d is then
n〈d〉^ =
∑^ t
j=
kj + 1 d − j + 1
We note that this is very closely related to the description of the Hilbert polynomial in Theorem 2.5.
Theorem 2.7. Let k ∈ N be such that all minimal generators of I are in degrees less than k. If HS/I (k +1) = HS/I (k)〈k〉, then HS/I (t+1) = HS/I (t)〈t〉 for all t ≥ k.
Note that in particular that the Hilbert polynomial satisfies PS/I (t + 1) = PS/I (t)〈t〉^ for all t 0, so a particular corollary of Theorem 2.7 is that if an ideal J is generated in degrees at most D, and if HS/J (D) = P (D) and HS/J (D + 1) = P (D + 1) then PS/J = P. Let D be the Gotzmann number of P , and let HP be the scheme {(L, M ) ∈ GD × GD+1 : xiL ⊆ M for all i} with the natural induced closed subscheme structure on GD × GD+1.
Theorem 2.8. The scheme HP represents the functor hP.
Remark 2.9. One can also describe HilbP (Pn) directly as a subscheme of GD, by writing S 1 L ⊂ SD+1 in terms of the coordinates on GD, and demanding that dimk(S 1 L) ≤
(n+D+ n
− P (D + 1), which gives determinantal equations. This works because Macaulay’s theorem guarantees the reverse inequality, so such L must actually have dimk(S 1 L) =
(n+D n
− P (D + 1). See also [HS04] and [DB82].
2.1. Exercises 2.
(1) Compute the Gotzmann number for the Hilbert polynomial of I = 〈x^20 , x 1 x 2 〉 ⊂ k[x 0 , x 1 , x 2 ]. Repeat for the Hilbert polynomial of I = 〈x 0 x 3 , x 0 x 2 , x 1 x 2 〉 ⊂ k[x 0 , x 1 , x 2 , x 3 ].
NOTES ON HILBERT SCHEMES 11
(2) Let I = 〈x 0 x 3 − x 1 x 2 , x 0 x 2 − x^21 , x 1 x 3 − x^22 〉. Compute saturated lexicographic ideal with the same Hilbert polynomial as I. Verify that it is generated in degrees at most the Gotzmann number of PS/I. (3) Let S = k[x 0 , x 1 , x 2 ], and let P (t) = 2. Write down equations de- scribing HilbP (P^2 ). (4) Describe the equations for the Hilbert scheme Hilb 3 t+1(P^3 ). (5) Assuming Macaulay’s theorem (Proposition 2.4), show that every Hilbert polynomial can be written in the form of Theorem 2.5. Hint: What do Stanley decompositions of lexicographic ideals look like? Show that such a decomposition is unique. Hint: This is a purely numerical property. Fix a large t, and look at the corresponding decomposition of P (t). Can you identify a 1 in this case? (6) Show that there is no bound on the regularity of S/I with Hilbert polynomial P if I is not assumed to be saturated. (7) This question outlines a proof of Theorem 2.5. (a) Note first that it suffices to show that the bound given in Theo- rem 2.5 bounds the regularity of saturated monomial ideals with Hilbert polynomial P. (Hint: Gr¨obner degeneration and upper semicontinuity - skip this part if you don’t know these words). (b) A Stanley filtration for S/I, where I is a monomial ideal, is a Stanley decomposition {(xui^ , σi) : 1 ≤ i ≤ s} with the extra property that {(xui^ , σi) : 1 ≤ i ≤ j} is a Stanley decomposition for S/(I, xuj+1^ ,... , xus^ ) for 1 ≤ j ≤ s. Show that if I is a monomial ideal then a Stanley filtration for S/I exists. (c) What can you say about regularity in short exact sequences? Deduce that if {(xui^ , σi) : 1 ≤ i ≤ s} is a Stanley filtration for S/I, where I is a saturated ideal, then reg(S/I) ≤ maxi |ui|, where the sum is over all i with |σi| > 0. (d) Show that there is always a Stanley decomposition {(xui^ , σi) : 1 ≤ i ≤ s} for S/I with |ui| ≤ i − 1, and max{i : σi 6 = ∅} ≤ D. Conclude Gotzmann’s theorem. (e) It is an open question whether the upper bound on the regularity of S/I given in (7c) is ever not sharp for some Stanley filtration, and (weaker) whether there always a Stanley decomposition for which the maximum |ui| is at most the regularity.
- Lecture 3: Connectedness and Pathologies
3.1. Connectedness. Little is known about the global structure of the Hilbert scheme. The one uniform fact that is known is Hartshorne’s the- orem that the Hilbert scheme is always connected. We now outline the proof of this result. The key idea of the proof is to first reduce to showing that all monomial ideals in HilbP (Pn) live on the same connected component. This can be done, for example, by using a Gr¨obner degeneration. This re- duces connectedness to a more combinatorial problem, as monomial ideals
NOTES ON HILBERT SCHEMES 13
lexicographic term order ≺ and let g ∈ GL(n + 1, k) lie in the Zariski open set U for computing the revlex gin. Pick a map ψ from A^1 to GL(n + 1, k) that has g and the identity in its image. This shows that I and gI lie in the same connected component. Then in≺(gI) lies in the same irreducible component as gI by the Gr¨obner degeneration, so I lies in the same connected component as the Borel-fixed ideal gin≺(I).
This proof can be refined to show that I actually lives in the same irre- ducible component as a Borel-fixed ideal. Proposition 3.5 reduces the prob- lem of showing connectedness to showing that all Borel-fixed ideals live on the same connected component, by showing we can “walk” from each one to the lex-segment ideal. We now sketch a proof of the fact that the Hilbert scheme is connected. We follow the paper of Alyson Reeves [Ree95], who analyzed Hartshorne’s approach to prove the stronger result that the radius of the component- graph of the Hilbert schemes is at most d + 1, where d is the degree of the Hilbert polynomial (and thus the dimension of the subschemes of Pn k being parameterized). We restrict here the base field k being algebraically closed of characteristic zero.
Definition 3.6. The component graph of the Hilbert scheme HilbP (Pn) has vertices the irreducible components of HilbP (Pn), and an edge connecting two vertices if and only if the corresponding components intersect. The radius of the Hilbert scheme is the minimum over all vertices v of the component graph of the maximum distance from v to any other vertex.
By a result of Reeves and Stillman [RS97] the saturated lexicographic ideal is a smooth point of HilbP (Pn), so lies on exactly one irreducible component, which we call the lexicographic component.
Theorem 3.7. [Ree95] Let d be the degree of the Hilbert polynomial P. Then the distance from any vertex to the vertex of the lexicographic component is at most d + 1, so the radius of the Hilbert scheme HilbP (Pn) is at most d + 1.
The key idea of the proof is to do a construction known as the distraction which takes a Borel-fixed ideal to another Borel-fixed ideal that is closer to the lexicographic ideal.
Definition 3.8. Let I ⊂ S be a monomial ideal. The polarization of I is the following monomial ideal p(I) in the polynomial ring k[zij : 0 ≤ i ≤ n, j ≥ 0] in infinitely many variables:
p(I) = 〈
∏^ n
i=
∏^ ui
j=
zij : xu^ is a minimal generator of I〉.
Note that p(I) is a squarefree monomial ideal, and thus radical. Define σ : k[zij ] → S by σ(zij ) = xi −αij xn, where αij ∈ k. The distraction of I is then σ(p(I)).
14 DIANE MACLAGAN
Lemma 3.9. For sufficiently generic choice of αij the distraction σ(p(I)) has the same Hilbert function as I. In fact, there is a Gr¨obner degeneration from σ(p(I)) to I.
We note that the second sentence of the lemma follows from the first, as it is straightforward to observe that the lexicographic initial ideal of σ(p(I)) contains I, so if they have the same Hilbert function they must be equal. The plan to show connectedness is then to start with an arbitrary ideal, take the gin, take the distraction of the gin with a sufficiently general choice of αij , take its revlex gin, and then repeat, taking distractions and then revlex gins. All ideals obtained in this fashion are clearly in the same connected component of the Hilbert scheme, so it suffices to check that after a finite number of steps we obtain the lexicographic ideal. This finite number will be bounded by d + 1, proving Theorem 3.7. This is accomplished by analyzing the irreducible components of σ(p(I). For sufficiently generic αij these are linear subspaces of Pn.
Definition 3.10. An irreducible component of σ(p(I)) is in lexicographic position if it is an irreducible component of σ(p(L)), where L is the lexico- graphic ideal with the same Hilbert polynomial as I.
To prove the d + 1 bound, the key of Reeves’ argument is:
Proposition 3.11. Let I be a saturated Borel-fixed ideal such that all irre- ducible components of σ(p(I)) of dimension at least i + 1 are in lexicographic position. Let J be the saturation of ginrevlex(σ(p(I))). Then σ(p(J)) has all irreducible components of dimension at least i in lexicographic position.
This proves the radius bound, since any saturated Borel-fixed ideal with Hilbert polynomial P has all components of σ(p(I)) of dimension at most d, so trivially satisfies the hypotheses of the proposition for i = d. Thus after d + 1 iterations of the (ginrevlex(σ(p(I))) : m∞) procedure we have an ideal whose distraction has all components in lexicographic position. This means its distraction equals the distraction of the lexicographic ideal (as containment would imply a smaller Hilbert polynomial), and thus that the ideal equals the lexicographic ideal.
Remark 3.12. We note that there are many other proof of this result. Reeves’ proof outlined above assumes that the base scheme is a field of characteristic zero. The characteristic assumption was removed in the Keith Pardue’s thesis [KP94]. A substantially different proof with the characteristic assumption, appears in the work of Peeva and Stillman [PS05], which is also related to the work of Daniel Mall [Mal00]. See [Fum05] for extensions of Mall’s work.
3.2. Pathologies. It is somewhat expected that connectedness is the only positive geometric property to be shared by all Hilbert schemes. This belief is expressed in the book [HM98] by the following law.
16 DIANE MACLAGAN
(a) Let P (t) = t + 1 be the Hilbert polynomial of a line. What is HilbP (P^3 )? (b) Let P (t) = 2t + 1 be the Hilbert polynomial of a conic. What is HilbP (P^2 )? (c) Explain geometrically why you expect these answers. (2) Let P (t) = d be a constant polynomial. List the saturated monomial ideals in HilbP (P^1 ). Which of these are Borel-fixed? What does this tell you about HilbP (P^1 )? What is HilbP (P^1 )? (3) Show that an ideal I is fixed by the T ∼= (k∗)n^ action on HilbP (Pn) if and only if I is monomial. Conclude that Borel-fixed ideals are mono- mial. Finish the proof of the “only if” direction of Proposition 3. by showing that the conditions of the second part of that proposition are satisfied by a Borel-fixed ideal. (4) Show that the lexicographic ideal is Borel-fixed. (5) Check that the saturation of a Borel-fixed ideal is Borel-fixed. (6) Let ≺ be the reverse-lexicographic term order. Compute by hand the generic initial ideal of I = 〈x 0 x 1 〉 ⊂ k[x 0 , x 1 ] with respect to ≺. What is the corresponding set U ⊂ GL(2, k)? (7) It is essentially never possible to compute the generic initial ideal exactly by adding extra variables for the elements of GL(n + 1, k), as the complexity of Gr¨obner basis calculations increases with the number of variables. In practice one chooses a “random” element g ∈ GL(n + 1, k), and computes inw(g · I). Do this for I = 〈x 0 x 3 − x 1 x 2 , x 0 x 2 −x^21 , x 1 x 3 −x^22 〉 ⊂ k[x 0 , x 1 , x 2 , x 3 ] using a computer algebra system, and verify that the result is Borel fixed. Warning: Random g chosen with small entries may well not be sufficiently generic. One implemented algorithm to compute gins is to try 50 different g, and take the most common answer. (8) Let P (t) = 3. List all Borel-fixed ideals in HilbP^ (P^2 ). Compute the distraction of each. What does Reeves’ walk do on this Hilbert scheme?
- Lecture 4: Hilbert schemes of points on surfaces We saw in the previous lecture that we should generally expect the Hilbert scheme to be as nasty as we can imagine. This result does not, however, cover absolutely all Hilbert schemes, and there are still some that are nice (meaning smooth and irreducible). In particular, Hilbert schemes of points on smooth surfaces are always smooth and irreducible, by a result of Fogarty [Fog68]. By a Hilbert scheme of points we mean one where the Hilbert polynomial is that of a finitely collection of points, so P (t) = d for all t, where d is a constant. We will consider the related local picture of the Hilbert scheme of d points in the affine plane Hilbd(A^2 ). This parameterizes all artinian ideals I in the polynomial ring S = k[x, y] with dimk(S/I) = d. Note that these ideals need no longer be homogeneous. In this lecture we will show that Hilbd(A^2 ) is
NOTES ON HILBERT SCHEMES 17
smooth and irreducible, following the proof of Haiman [Hai98]. It is impor- tant to emphasize that this is only scratching the surface of what is known about Hilbert schemes of points on smooth surfaces, and we could spend more than the entire week on this topic alone. In particular, we will not touch on the description of the Betti numbers of Hilbd(A^2 ) for all d, and the Heisenberg algebra action on the homology of Hilbd(A^2 ). See [Nak99] for an introduction to these topics. The proof that Hilbd(A^2 ) is smooth and irreducible has four steps: (1) Show that Hilbd(A^2 ) is connected by showing that every ideal lives in the same irreducible component as a monomial ideal, and all mono- mial ideals live in the same irreducible component (the “good com- ponent” of the Hilbert scheme of points). (2) Reduce to showing that all monomial ideals on Hilbd(A^2 ) are smooth points. (3) Show that the dimension of the good component of Hilbd(A^2 ) is at least 2d. (4) Give a combinatorial description of the tangent space to a monomial ideal in Hilbd(A^2 ), and show that the dimension of this space is at most 2d. Together these steps show that Hilbd(A^2 ) is smooth and connected, and thus smooth and irreducible. The last of these steps has the most content. Most of this plan generalizes to Hilbd(An), with the exception of the last step showing an upper bound on the dimension of the tangent space. We can thus use this outline to give examples of smooth and singular Hilbd(An) for n > 2.
4.1. Step 1: Connectedness. Let I be an ideal in Hilbd(A^2 ), so dimk(S/I) = d. For any term order ≺, the ideal J = in≺(I) satisfies dimk(S/J) = dimk(S/I) = d, and lives on the same irreducible component of Hilbd(A^2 ) as I, since the Gr¨obner degeneration from I to J is a flat family, so gives rise to a map from A^1 to Hilbd(A^2 ), the image of which must lie in one irreducible component. Thus every ideal lives on the same irreducible component of Hilbd(A^2 ) as a monomial ideal. To show that all monomial ideals live in the same connected component, we will show that they all live in the same connected component as the ideal J = 〈x, yd〉. Let I = 〈yv^1 ,... , xui^ yvi^ ,... , xul^ 〉 be a monomial ideal in Hilbd(A^2 ), where we set u 1 = vl = 0. Let xayb^ be a socle element for S/I with b 6 = 0, so xa+1yb^ and xayb+1^ both live in I. If no such element exists, we must have I = J. Consider the ideal I′^ = 〈xui^ yvi^ : 1 ≤ i ≤ l − 1 〉 + 〈xayb^ − xul^ 〉. One can check (using Buchberger’s S-pair criterion) that I = inw(I′) for any w ∈ N^2 with aw 1 + bw 2 < ulw 1. Let w′^ = N (b, ul − a) − w for N 0, and set I 2 = inw′ (I′). We can repeat this construction for I 3 , I 4 ,.... The exponent of the minimal generator of the form xc^ increases at each step, since xul^6 ∈ I 2 in the formulation above, so this procedure must terminate at some ideal Ij , at which point we must have Ij = J. Since each step is a pair of Gr¨obner
NOTES ON HILBERT SCHEMES 19
a subfunctor of the Hilbert functor that is represented by UM. Since the standard monomials of an initial ideal of I form a k-basis for S/I, the set of all UM cover Hilbd(A^2 ). The scheme UM is affine, and we now describe its defining equations. Let S = {xu^1 ,... , xud^ } be the set of standard monomials of M. Let B be the set of monomials xu^ ∈ M for which xu/xi 6 ∈ M for some i with ui > 0. The set B is the border of M. Set b = |B|. Let I be an ideal in UM. Since S is a basis for S/I, for each xu^ ∈ B, there is a unique polynomial fu ∈ I of the form
fu = xu^ −
v∈S
γvu xv,
where γvu ∈ k. Let R = k[zuv : xu^ ∈ B, xv^ ∈ S] be the coordinate ring of Abd. Each point in UM gives a point in Abd^ by taking the ideal to the vector γvu. Form the d × d multiplication matrix Xi with rows and columns indexed by S with the (xu, xv)th entry equal to zv u+ ei if xixv^ ∈ M , equal to 1 if xu^ = xixv^6 ∈ M , and equal to zero otherwise. Then the fact that x 1 and x 2 commute mean that we must have X 1 X 2 − X 2 X 1 = 0. This is also sufficient to guarantee that the ideal generated by the fu for specific values of the zuv has colength d. So the ideal IM generated by the entries of the commutator X 1 X 2 − X 2 X 1 defines the affine scheme UM ⊂ Abd.
Example 4.1. For convenience we set S = k[x, y] here, and use the notation
zxy for z(1(0,,0)1). Let d = 3, and let M = 〈x^2 , xy, y^2 〉. Then the matrices X 1 and X 2 are
X 1 =
0 zx 2 1 z
xy 1 1 zx 2 x z xy x 0 zx
2 y z xy y
, X 2 =
0 z 1 xy zy
2 1 0 zxxy zy 2 x 1 zyxy zy 2 y
so X 1 X 2 − X 2 X 1 equals
0 zx
2 1 z
xy x +^ z
xy 1 z xy y −^ z
xy 1 z x^2 x −^ z
y^2 1 z x^2 y z
x^2 1 z
2 x +^ z
xy 1 z y^2 y −^ z
xy 1 z xy x −^ z
y^2 1 z xy y 0 z 1 xy + zxxy zyxy − zx 2 y z y^2 x z
y^2 1 +^ z x^2 x z y^2 x +^ z xy x z y^2 y −^ (z xy x ) (^2) − zy^2 x z xy y 0 zx 2 y z xy x +^ z
xy y^2 −^ z
x^2 1 −^ z xy y z x^2 x −^ z y^2 − yzx^2 y z x^2 y z y^2 x −^ z
xy 1 −^ z xy x z xy y
Thus UM is the subscheme of A^9 defined by the ideal 〈zx
2 1 z
xy x +^ z
xy 1 z xy y − zxy 1 zx 2 x −^ z
y^2 1 z x^2 y , z x^2 1 z 2 x +^ z
xy 1 z y^2 y −^ z
xy 1 z xy x −^ z
y^2 1 z xy y , z
xy 1 +^ z xy x z xy y −^ z x^2 y z y^2 x , z
y^2 1 + zx 2 x z y^2 x +z xy x z y^2 y −(z xy x ) (^2) −zy^2 x z xy y , z x^2 y z xy x +z
xy y^2 −z
x^2 1 −z xy y z x^2 x −z y^2 −yzx^2 y , z x^2 y z y^2 x −
zxy 1 − zxxy zxyy 〉 ⊂ k[zx 2 1 , z
x^2 x , z
x^2 y , z
xy 1 , z
xy x , z
xy y , z
y^2 1 , z
y^2 x , z
y^2 y ]. The generators for IM are of the form
f (^) qp =
xu∈S
zx^1 x
p xu^ z
x 2 xu xq^ −^
xv^ ∈S
zx^2 x
p xv^ z
x 1 xv xq^ ,
where p, q ∈ N^2 with xp^ = xp 11 xp 22 and xq^ = xq 11 xq 22 ∈ S. Note that f (^) qp = 0 when both x 1 xp, x 2 xp^6 ∈ M. The dimension of the cotangent space of UM at
20 DIANE MACLAGAN
Figure 1. Two equivalent arrows
the origin of Abd^ (representing the point M ∈ Hilbd(A^2 )) is given by
dimk m/m^2 ,
where m = 〈zvu : xu^ ∈ B, xv^ ∈ S〉. This is equal to
dimk〈zuv 〉/〈(zuv )^2 + IM 〉 = dimk〈zvu 〉/〈(zvu )^2 + f˜ (^) qp : xu^ ∈ B, xv, xp, xq^ ∈ S〉,
where f˜ (^) qp is the degree one part of f (^) qp. Now f˜ (^) qp = zx
px 1 xq−e^2 +^ z
xpx 1 x 2 xq^ −^ z
xpx 2 xq−e^1 − zx
px 1 x 2 xq^ , where the first term only shows up if^ x^2 divides^ x q (^) and xpx 1 ∈ M ,
the second only shows up if xpx 1 6 ∈ M , the third only shows up if x 1 divides xq^ and xpx 2 ∈ M , and the last term only shows up if xpx 2 6 ∈ M. We thus see that f˜ (^) qp has either one or two variables, and the variables zdc occurring
have c − d constant. If f˜ (^) qp = zcd − zc ′ d′^ then^ c ′ (^) = c ± ei. We draw the variable zvu as an arrow in Z^2 with tail u and head v. The degree one part of the polynomial ring has k-basis the set of all arrows with tail in B and head in S. The relations coming from f˜ (^) qp are that two arrows are equivalent if one can be obtained from the other by moving the first horizontally or vertically keeping the tail in M and the head in S. This is illustrated in Figure 1 There is a distinguished representative for each nonzero equivalence class of arrows consisting of an arrow zuv where either xu/x 1 6 ∈ M and x 2 xv^ ∈ M , or the equivalent statement with 1 and 2 reversed. Given such an arrow, let φ(zvu ) = gcd(xu, xv) ∈ S. Note that the map from distinguished arrows to S is two-to-one, so we conclude that the number of distinguished arrows is at most 2|S| = 2d. This shows that the dimension of the tangent space to a monomial ideal is at most 2d. Combined with the lower bound from the previous subsection, we conclude that the dimension of the tangent space is the dimension of Hilbd(A^2 ), so Hilbd(A^2 ) is smooth.
4.5. Exercises 4.
(1) List all monomial ideals I in k[x, y] with dimk(k[x, y]/I) = 4. Ver- ify that for each of these the tangent space to Hilb^4 (A^2 ) at I is 8- dimensional. Repeat if desired with 4 replaced by 5. (2) Compute the equations for UM for each of the monomial ideals M from your answer to the previous question. (3) Show that every monomial ideal in Hilbd(A^2 ) lives in the same irre- ducible component. (4) When d = 2, the variety (A^2 )^2 /S 2 is the quotient of affine space by an abelian group, and is thus a toric variety. Can you describe it?
