Identifying limits of ideals of points in the case of projective space

We study the closure of the locus of radical ideals in the multigraded Hilbert scheme associated with a standard graded polynomial ring and the Hilbert function of a homogeneous coordinate ring of points in general position in projective space. In the case of projective plane, we give a sufficient condition for an ideal to be in the closure of the locus of radical ideals. For projective space of arbitrary dimension we present a necessary condition. The paper is motivated by the border apolarity lemma which connects such multigraded Hilbert schemes with the theory of ranks of polynomials.


Introduction
Let S * = C[x 0 , . . . , x n ] be a polynomial ring and F ∈ S * d for some positive integer d. The (Waring) rank of F is min{r ∈ Z|F = L d 1 + . . . + L d r for some L 1 , . . . , L r ∈ S * 1 }. Finding the rank of a general homogeneous polynomial of S * of a given degree is a classical problem studied, among others, by Sylvester. For a given polynomial F ∈ S * d and integer r, the condition that F has Waring rank at most r can be expressed algebraically by means of the classical apolarity lemma (see [28,Thm. 5.3]). We recall this here. We may consider the polynomial ring S = C[α 0 , . . . , α n ] acting on S * by partial differentiation. We denote this action by . Given F ∈ S * one may consider the annihilator ideal Ann(F ) in S consisting of all those polynomials θ for which θ F = 0. Then the condition that F has rank at most r is equivalent to the existence of a homogeneous, radical ideal I ⊆ Ann(F ) such that S/I has Hilbert polynomial r.
Since, for a fixed integer r, the locus of points in P(S * d ) corresponding to polynomials of rank at most r need not be closed, in the context of algebraic geometry, it is more natural to consider the Zariski closure of this locus. In that way, one obtains the secant variety σ r (ν d (P(S * 1 ))). The polynomials in S * d corresponding to points of σ r (ν d (P(S * 1 ))) are said to have border rank at most r. In a recent paper [6], a version of the apolarity lemma for the case of border rank was established. Its statement uses the notion of multigraded Hilbert scheme introduced in [24]. For a fixed number of variables n + 1 and positive integer r, one considers the multigraded Hilbert scheme Hilb hr,n S[P n ] where h r,n is the generic Hilbert function of the homogeneous coordinate ring of r points in P n (see below for a precise definition of h r,n ). This scheme has a distinguished irreducible component called Slip r,n which is the closure of the locus of points corresponding to radical ideals. Then F ∈ S * d has border rank at most r if and only if there is a point [I] ∈ Slip r,n such that I ⊆ Ann(F ). The border apolarity lemma was shown to be an effective tool in the study of ranks of polynomials. See for instance [12], [20] and [26] as a demonstration of its applicability.
As a consequence of the border apolarity lemma, the understanding of the closure involved in the definition of the secant varieties can be shifted to the problem of understanding the closure of the locus of radical ideals in the corresponding multigraded Hilbert scheme. This paper is concerned with finding conditions (necessary or sufficient) for a point [I] ∈ Hilb hr,n S[P n ] to be in Slip r,n . In fact, the theory works in greater generality. One may replace C by an algebraically closed field of arbitrary characteristic and replace P n by any smooth, projective toric variety X. Then the homogeneous coordinate ring S of P n is replaced by the more general notion of a Cox ring of X (see [13]). In this paper we restrict our attention to the case of projective space, but we intend to investigate analogous problems in greater generality in the future. Proposition 1.2 suggests that pathologies of the multigraded Hilbert schemes Hilb hr,n S[P n ] occur for much simpler examples than for the corresponding Hilbert schemes Hilb r (P n ). Therefore, it is expected that these pathologies might be studied more easily in the case of multigraded Hilbert schemes. Theorem 2.7 is then applied in Section 3 to the study of polynomials with small border rank. It seems that the criterion of Theorem 2.7 can be substantially generalized. This is a work in progress by the author.
In Section 4, for a fixed positive integer r, we consider functions f : Z → Z satisfying the condition: We also comment in Remarks 4.18 and 4.19 that natural generalizations of this result to the case of arbitrary dimension n are not true.

A necessary condition in the case of projective space
We fix a positive integer n and write S for S[P n ] = k[α 0 , . . . , α n ]. Recall from Section 1 that we denote the irrelevant ideal (α 0 , . . . , α n ) ⊆ S by m. Moreover, given a homogeneous ideal a ⊆ S we denote by √ a its radical and by a its saturation with respect to m.

Algebraic results
In this subsection we present some algebraic results concerning homogeneous ideals that will be used in the proof of Theorem 1.1.
Given a homogeneous ideal a in S we will consider the condition: there exists a positive integer d such that S d ⊆ a.
( * ) The following lemma gives some simple properties of condition ( * ).
(iv) If a 1 , . . . , a m are homogeneous ideals such that a i + a j satisfies ( * ) for all 1 i < j m, then The opposite inclusion follows from the fact that a ∩ b is contained in both a and b and thus, its saturation is contained in both a and b.
Let m 3 and assume that (iv) is established for all integers smaller than m. By part (i) we have Applying the inductive hypothesis for m − 1 this can be rewritten as It follows from part (iii) that a 1 · . . . · a m−1 + a m satisfies ( * ). Therefore, from part (i) and inductive hypothesis for m = 2, we obtain as claimed. Lemma 2.2. Let I, J be homogeneous ideals of S and k be a positive integer. Then: Proof. (i) It is enough to consider a minimal set of homogeneous generators of I and take d 0 to be the maximum of degrees of elements of this set, i.e. d 0 = max{j | β 1,j (S/I) = 0}, where β i,j are graded Betti numbers.
After the above algebraic preparation we are able to compute the Hilbert polynomial of a power of a homogeneous radical ideal defining a closed, zero dimensional subscheme of projective space. Lemma 2.3. Let I ⊆ S be a homogeneous radical ideal such that the Hilbert polynomial of the quotient algebra S/I is constant, equal to r for some positive integer r. Then for a positive integer k, the Hilbert polynomial of S/I k is constant equal to r · dim k S k−1 .
Proof. Let Proj S/I ⊆ P n be the set of (distinct) points P 1 , . . . , P r and define p i to be the prime ideal defining P i . Then I = p 1 ∩ . . . ∩ p r . Define J = p 1 · . . . · p r . Since for all 1 i < j m we have p i + p j = m, it follows from Lemma 2.1(iv) that I = J. Hence I k = J k by Lemma 2.2(ii). It is enough to show that the Hilbert polynomial of S/J k is r · dim k S k−1 . Let K = p k 1 ∩ . . . ∩ p k r . Observe that p k i + p k j satisfies ( * ) for every 1 i < j m by Lemma 2.1(ii). Therefore, K = J k by Lemma 2.1(iv). Thus, it is enough to consider the Hilbert polynomial of S/K. Since, as a set, Proj S/K is the disjoint union of r-points P 1 , . . . , P r it is enough to show that the degree of Proj S/p k i is dim k S k−1 for every i = 1, . . . , r. This is clear, since up to a linear change of variables p i = (α 1 , . . . , α n ). Remark 2.4. In Lemma 2.3 the assumption that I is a radical ideal cannot be weakened to the assumption that I is saturated. Indeed, I = (α 2 0 , α 0 α 1 , α 2 1 ) ⊆ S = k[α 0 , α 1 , α 2 ] is a saturated ideal and S/I has Hilbert polynomial 3, but S/I 2 has Hilbert polynomial 10.
In Lemma 2.3 we have calculated the Hilbert polynomial of S/I k for a homogeneous radical ideal I defining a zero dimensional closed subscheme of a projective space and a positive integer k. The following proposition provides a bound on the least degree, from which the Hilbert function of S/I k agrees with the Hilbert polynomial of S/I k . Recall that for a finitely generated Z-graded S-module M , its regularity reg M is defined to be reg M = max{j − i | β i,j (M ) = 0}.
Proof. By Lemma 2.3, the Hilbert polynomial of S/I k is r · dim k S k−1 . Hence, by [16,Thm. 4.2], it is enough to show that ke + k − 1 reg S/I k . Since reg S/I = e by [16,Thm. 4.2], it follows from the definition of regularity that reg I = e + 1. Thus, reg I k ke + k by [11,Thm. 6]. Therefore, reg S/I k ke + k − 1.  In what follows, we will consider k with A-module structure given by A → A n /nA n ∼ = k. By the definition of universal ideal sheaf we have (A[α 0 , . . . , α n ]/J)⊗ A k ∼ = S/I. Therefore, from the universal property of kernel, there is an induced map J ⊗ A k → I and it is surjective by snake lemma applied to the diagram

Proof of Theorem 1.1
Hence also the map J k ⊗ A k → I k is surjective. The snake lemma applied to the diagram implies that the dotted arrow induced by the universal property of cokernel is injective. Since it is clearly surjective, it is an isomorphism. Thus and the claim of the theorem follows from Proposition 2.5.

Ideals of subschemes contained in a line
In this subsection we prove Theorem 2.7 which is a consequence of Theorem 1.1. It provides a necessary condition for an ideal [I] ∈ Hilb hr,n S defining a subscheme of P n contained in a line to be in Slip r,n . We start with the following scheme-theoretic result. Lemma 2.6. Let X be a scheme locally of finite type over k. Let Z 1 , Z 2 be irreducible closed subsets of X of dimensions d 1 , d 2 , respectively. Let P ∈ Z 1 ∩ Z 2 be a closed point of the intersection and let d = dim k T P X.
Recall from Section 1, that h r,n is the Hilbert function of S/I where I is a saturated ideal of r points of P n in general position. Let e = min{a ∈ Z | h r,n (a) = r}. We claim that under the assumptions n 2, r 4 we have Since n 2 we get 2+e−1 2 n+e−1 n . Therefore, since n+e−1 n < r, we obtain 2r > e(e + 1) and thus, 2r e(e + 1) + 2. It follows that 2r − 2 e 2 + e. If e 2 then e 2 + e 2e + 2 which gives the claimed relation 2r − 2 2e + 2. If e = 1 then 2r − 2 2e + 2 since r 4 by assumption.
Let n 2 and r 4 be integers. Consider the following condition on homogeneous ideals I of S with the Hilbert function of the quotient algebra S/I equal to h r,n : Observe that condition ( †) is non-trivial only in degree r − 2 since I r−1 = I r−1 . Proof. Up to a linear change of variables, I = (α 0 , . . . , α n−2 , F r (α n−1 , α n )) where F r is non-zero and homogeneous of degree r. Suppose that [I] ∈ Slip r,n but it does not satisfy ( †). Consider the monomial order > w,lex on S given by the weight vector w = (1, 1, . . . , 1, 2, 2) ∈ Z n+1 and a lexicographic monomial order lex (see [14,Ex. 2.4.11]). Let I 0 be the initial ideal of I with respect to > w,lex . Then We shall show that H S/I 2 0 (2r−2) = r(n+1)−1 which contradicts Theorem 1.1 and Equation (1)  Now we will prove the second part of the theorem. Let g : Z → Z be defined by Consider the multigraded Hilbert scheme Hilb g S . We have morphisms Let V ⊆ Hilb r (P n ) be the closed subset defined by the set of its closed points H S/IZ (a) = r for a r − 1. Therefore, we need to choose a codimension one subspace of (I Z ) r−2 . It follows that W is irreducible of dimension n+r−2 n + 2n − 2. Let X be the closed subset of W , whose closed points satisfy ( †). This locus is irreducible of dimension (2(n − 1) + r) + (n − 1)(r − 2) − 1 = nr − 1 since now the codimension one subspace of (I Z ) r−2 as above must contain the subspace ((I Z ) 1 ) 2 · m r−4 . We claim that W ∩ f 1 (Slip r,n ) = X set-theoretically. Note that this will finish the proof. By the first part of the proof, we know that It follows from the assumption on the characteristic of k and [15,Thm. 15.23] that I 0 = (α 0 , . . . , α n−2 , α r n−1 ) and where the generators of degree r − 2 are all monomials divisible by one of the forms α 0 , . . . , α n−2 except for α n−2 α r−3 n . Therefore, to conclude the proof, it is enough to show that dim k Hom S (I 0 , S/I 0 ) 0 n+r−2 n + 2n − 1. This is computed in Lemma 2.8.
Proof. Let n = (α 0 , . . . , α n−2 ) and pick φ ∈ Hom S (I 0 , S/I 0 ) 0 . For every positive integer a we identify the k-vector space (S/I 0 ) a with the k-vector space spanned by monomials of degree a in S that are not in Similarly, considering the generators . Together, these estimates show that dim k Hom S (I 0 , S/I 0 ) 0 n+r−2 n − r + 1 + r + 2(n − 1) where: (1) n+r−2 n − r corresponds to the choices of coefficients at α n−2 α r−3 n for every monomial of degree r − 2 in I 0 ; (2) 1 + r corresponds to the choice of coefficient at α r−1 n−1 for φ(α n−2 α r−2 n ) and the choice of φ(α r n−1 ); We end this section with the proof of Proposition

Applications to border ranks of polynomials
In this section we assume that the base field k is the field of complex numbers C since we will cite papers in which this is assumed. Let n be a positive integer and S = C[α 0 , . . . , α n ] be the polynomial ring with standard Z-grading. We will consider the dual polynomial ring S * = C[x 0 , . . . , x n ] with the structure of an S-module on S * given by partial differentiation. We will denote this action by . Given a homogeneous polynomial F ∈ S * we denote by Ann(F ) the ideal {θ ∈ S|θ F = 0}. We recall various definitions of ranks of polynomials and state some versions of apolarity lemmas. See [31] for a discussion of related results and [6] for the case of border rank.
Let F ∈ S * d . We define the rank of F to be r(F ) = min{r ∈ Z|F = L d 1 + . . . + L d r for some L 1 , . . . , L r ∈ S * 1 }.
This can be generalized in various ways. We define the smoothable rank of F to be sr(F ) = min{r ∈ Z|[F ] ∈ ν d (R) for a smoothable zero-dimensional closed subscheme and ν d (R) is the projective linear span of the scheme ν d (R). We define the cactus rank of F to be It follows from these definitions, that cr(F ) sr(F ) r(F ). Recall that the r-th secant variety of the d-th Finally, we define the border rank of F to be br(F ) = min{r ∈ Z|[F ] ∈ σ r (ν d (P(S * 1 )))}.
We will use the following two apolarity lemmas. We always have br(F ) sr(F ) and we say that F is wild, if the inequality is strict. Wild polynomials are more difficult to control using standard, existing methods. Therefore, new methods need to be developed in order to study them effectively. For example, see [2,Prop. 11] and its applications, [8,Rmk. 1.5] and [20].
We will study wild homogeneous polynomials F such that br(F ) deg(F ) + 2. If br(F ) deg(F ) + 1 then F is not wild. This is established in [4, Prop. 2.5] based on a result in [2]. Therefore, we will assume that br(F ) = deg(F ) + 2.

Polynomials in three variables of small border rank
In this subsection we will prove the following proposition which shows that there are no wild homogeneous polynomials F in three variables of border rank at most deg(F ) + 2. Proof. By [4, Prop. 2.5] we may assume that br(F ) = d + 2. From Proposition 3.2 we obtain that there is an ideal I ⊆ Ann(F ) such that [I] ∈ Slip d+2,2 . If S/I has a Hilbert function different from h d+2,1 then I d = I d and therefore, cr(F ) d + 2 by Corollary 3.3. The Hilbert scheme Hilb d+2 (P 2 ) is irreducible, hence cr(F ) = sr(F ). Moreover, the smoothable rank is bounded from below by the border rank. As br(F ) = d + 2 by assumption, we get that cr(F ) = sr(F ) = d + 2.

Remark 3.5.
In [2, page 37] in the paragraph above Remark 13, there is an example that suggests that there could exist a wild polynomial in σ 8 (ν 6 (P 2 )). Proposition 3.4 shows that there is no such polynomial.
It would be interesting to address the following questions related to Proposition 3.4.

Quartics in four variables of small border rank
In this subsection we will prove the following proposition which shows that there are no wild homogeneous quartics F in four variables of border rank at most 6.
There is a polynomial θ ∈ C[α 2 , α 3 ] 3 such that θ C 1 = θ C 2 = 0. By a linear change of variables in C[α 2 , α 3 ] we may assume that θ is one of the following: We study this case by case. We will further simplify F by a linear change of variables and in each case we will find a homogeneous ideal J ⊆ Ann(F ) whose initial ideal with respect to the lex order with α 2 > α 3 > α 1 > α 0 will be saturated and the Hilbert polynomial of the corresponding quotient algebra will be 6. Thus, cr(F ) 6 by Lemma 4.12 and Proposition 3.1. In each case the given set of generators of J will be a Gröbner basis. We may assume that Ann(F ) 1 = 0 by Proposition 3.4 and [5, §3.1].
We start with case 1. Up to a linear change of variables in S * we have one of the cases: Corresponding to the above cases, the following ideals contained in Ann(F ) show that the cactus rank of F is at most 6: . Now we consider case 2, namely we assume that α 2 2 α 3 C 1 = α 2 2 α 3 C 2 = 0. Then, up to a linear change of variables in S * and excluding possibilities already considered in case 1, we have one of the cases: Corresponding to the above cases, the following ideals contained in Ann(F ) show that the cactus rank of G is at most 6: Finally we consider case 3, that is we assume that α 2 α 3 (α 2 − α 3 ) C 1 = α 2 α 3 (α 2 − α 3 ) C 2 = 0. Then, up to a linear change of variables in S * and excluding possibilities considered in case 1, we have one of the cases: Corresponding to the above cases, the following ideals show that the cactus rank of G is at most 6: B If a = 0 then we are in case 2.B. Therefore, assume that a = 0. Take

Wild quintic in four variables of border rank 7
In Proposition 3.6 we showed that there are no wild quartics in four variables of border rank 6. In this subsection, we give an example of a wild quintic in four variables of border rank 7.
Remark 3.8. The annihilator ideal Ann(C) of a concise cubic C has a minimal generator of degree 3 (see [6,Thm. 5.4] for a vast generalization). Therefore, the form given in Equation (3) should be compared to the form given in [7,Thm. 4.5].

A sufficient condition in the case of projective plane
In this section we are concerned with P 2 , so in order to simplify notation, we will write S instead of S[P 2 ]. Let T = k[α 1 , α 2 ] ⊆ k[α 0 , α 1 , α 2 ] = S. Fix a positive integer r and recall from Section 1 that the irrelevant ideal (α 0 , α 1 , α 2 ) ⊆ S is denoted by m.
The main result is Theorem 1.3 which we establish in Subsection 4.2. Proof of the theorem is based on Proposition 4.3 which is presented in Subsection 4.1.

Computing dimension of tangent space
Recall that if M, N are Z-graded T -modules (respectively, S-modules) and M is finitely generated then the Ext groups Ext i T (M, N ) (respectively, Ext i S (M, N )) are Z-graded T -modules (respectively, S-modules) in a natural way (see Consider a monomial ideal I of T such that dim k T /I = r for some r ∈ Z >0 . The goal of this subsection is to compute dim k Hom S (I ex , S/I ex ) 0 where I ex = S · I ⊆ S is the extended ideal. The associations ϕ → ϕ and ψ → ψ are inverse to each other.
We will use Lemma 4.1 in the case that H S/I ex = h r,2 . Then dim k Hom S (I ex , S/I ex ) 0 is the dimension of tangent space to Hilb hr,2 S at the point [I ex ] (see [24,Prop. 1.6]). Moreover, dim k Hom T (I, T /I) = 2r since this is the tangent space to the Hilbert scheme of 0-dimensional length r closed subschemes of A 2 and that scheme is smooth of dimension 2r (see [18]). Therefore, dim k Hom S (I ex , S/I ex ) 0 = 2r − d>0 dim k Hom T (I, T /I) d and we may focus on calculating the latter sum.
Given a monomial ideal I as above we have the associated staircase diagram (see [30]) obtained as follows. For each pair of non-negative integers (s, t) such that α s 1 α t 2 / ∈ I we put a 1 × 1 box with sides parallel to coordinate axis and (s, t) as lower left corner of the box. We will denote the diagram corresponding to I by D I . The set of boxes of D I (identified with the set of monomials outside I) will be denoted by Λ I . We will use the canonical minimal free resolution of T /I as given in [30,Prop. 3.1]. The set of minimal monomial generators of I will be denoted by M I and the generating set of relations (or more precisely the set of their degrees when T is considered with natural Z 2 -grading) used in that resolution will be denoted by R I . Note that #Λ I = dim k T /I = r and #R I = #M I − 1. We will identify monomials of T with lattice points in Z 2 . Given a point u = (s, t) in Z 2 we will write |u| for s + t. We define three functions from integers to integers: λ I (a) =#{u ∈ Λ I | |u| = a}, µ I (a) =β 1,a (T /I) = #{u ∈ M I | |u| = a}, ρ I (a) =β 2,a (T /I) = #{u ∈ R I | |u| = a}. * * Figure 1: Staircase diagram of the ideal I = (α 3 1 , α 2 1 α 2 , α 2 2 ).
Example 4.2. Figure 1 presents the staircase diagram of I = (α 3 1 , α 2 1 α 2 , α 2 2 ). Filled boxes correspond to monomials outside I (i.e. elements of Λ I ), dots correspond to elements of M I (i.e. minimal monomial generators of I) and asterisks correspond to elements of R I (i.e. minimal relations between those generators).
The goal of this subsection is the proof of the following proposition, which will play a key role in the proof of Theorem 1.3.
The proof of Proposition 4.3 is based on the following observation.
From Equation (5) for i = s we deduce that α as−1 1 divides f s−1 . It follows by induction that α ai 1 divides f i for all i. Thus, f 0 = α a0 1 f for some f ∈ T . From Equation (5) we conclude that f i = α ai 1 α bi 2 f for each i.
(ii) We start with showing that Ext 1 T (I, T ) >0 = 0. Consider the canonical minimal graded free resolution of I (see [30,Prop. 3.1]). Applying the functor Hom T (−, T ) to the above resolution, we obtain for every integer c a k-linear map We claim that ψ c is surjective for every c > 0. Observe that ker ψ c ∼ = Hom T (I, T ) c ∼ = T c by part (i). Therefore, the claim is a consequence of the calculation This is equivalent to Equation (4).

Proof of Theorem 1.3
This subsection is devoted to proving Theorem 1.3. We will start with a series of lemmas. The first one concerns Hilbert functions of saturated ideals defining zero-dimensional closed subschemes of P 2 . Proof. We may assume that I = (1).
(i) We shall prove that there is an element f ∈ S 1 that is a non-zero divisor on S/I. Let p 1 , . . . , p k be the associated primes of S/I. It is enough to show that  by the proof of (i). We claim that also (S/I) d ·f − → (S/I) d+1 is an isomorphism. It is injective since f is a non-zero divisor on S/I. Let g ∈ (S/I) d+1 . Then g = α 0 h 0 + α 1 h 1 + α 2 h 2 for some h 0 , h 1 , h 2 ∈ (S/I) d . By assumptions, there are k 0 , k 1 , k 2 ∈ (S/I) d−1 such that f k 0 = h 0 , f k 1 = h 1 , and f k 2 = h 2 . It follows that g = f (α 0 k 0 + α 1 k 1 + α 2 k 2 ).
We will use the following observation to show that certain Ext groups vanish. Proof. Apply the functor Hom S (−, k) to a minimal graded free resolution P • of M . The Ext groups Ext i S (M, k) can be computed as cohomology groups of the obtained complex. Since the i-th differential in P • maps P i into mP i−1 , the differentials in the complex Hom S (P • , k) are zero. Therefore, dim k Ext i S (M, k) e = dim k Hom S (P i , k) e = β i,−e (M ). The next lemma enables us to consider only ideals I for which the Hilbert function of S/I satisfies a more restrictive condition (⋆⋆) instead of (⋆). → Hilb r (P 2 ) from Remark 4.7. Therefore, we may assume that r < dim k S e . We claim that it is enough to consider the case that f (e − 1) = h r,2 (e − 1) = dim k S e−1 . Indeed, otherwise f (e − 1) = h r,2 (e − 1) = r and this contradicts Lemma 4.5 since f (e) = d < r = f (e − 1). Using Lemma 4.5(i) we obtain dim k S e−1 d and moreover by Lemma 4.5(ii) this inequality is strict, since r = f (e + 1) > d. The following lemma will enable us to reduce the proof of Theorem Note that A i α 2 = B i but we have written I f in the form as above since it will be more convenient in the proof of the following lemma. We will now find a saturated ideal K f such that the initial ideal of K f with respect to an appropriate monomial order is I f . Let Lemma 4.11. In the above notation, the initial ideal of K f with respect to the lex order > with α 0 > α 2 > α 1 is I f . In particular, S/K f has Hilbert function h r,2 .
Proof. All S-polynomials of the generators displayed in Equation (8) are in the ideal (C e+2 , . . . , C 0 ). It follows that this set of generators satisfies the Büchberger criterion (see [14,Thm. 6 in Ch. 2 §6]) and is thus, a Gröbner basis. In particular, the initial ideal in < (K f ) is I f so S/K f has Hilbert function h r,2 by Lemma 4.10.
In order to prove that K f is saturated we will use the following result. Proof. Let f ∈ a. Then there is an integer l such that (α l 0 f, α l 1 f, α l 2 f ) ⊆ a. Therefore, Thus, in < (f ) ∈ in < (a) which finishes the proof.
Lemma 4.13. In the above notation, K f is a saturated ideal. In particular, [I f ] ∈ Slip r,2 .
Proof. Let > be the lex order with α 0 > α 2 > α 1 . From Lemmas 4.10, 4.11 and 4.12 we obtain Suppose that K f = K f . Then I f = in < (K f ) in < (K f ) ⊆ J f , where the containment in the middle is strict since the two ideals have different Hilbert functions by assumption. Since I f , J f differ only in degree e, it follows that there is an element g ∈ S e ∩ K f such that in < (g) does not belong to the set of monomial generators of I f of degree e. From Equations (6) and (9) we get that g = e i=d−s a i A i for some a i ∈ k. We assumed that in < (g) / ∈ I f . Thus, by Equation (7), we have a i = 0 for some i ∈ {d − s, . . . , r − s − 1}. Furthermore, we may assume that a i = 0 for i = r − s, . . . , e by Equation (8). Multiplying g by α 2 0 and using the generators of K f given in Equation (8)  We claim that it is not possible. By Equation (9), it is enough to show that no monomial of the form B j α 0 for j ∈ {0, . . . , r − d − 1} is in J f . This is clear since monomials of degree e + 2 in J f that are divisible by α 0 are also divisible by α r−d Lemmas 4.14 -4.17 are devoted to this calculation. In what follows, we will consider k as a Z-graded S-module via isomorphism k ∼ = S/m. Then J f /I f ∼ = k(−e) ⊕(r−d) . We have Ext 1 S (J f /I f , S/J f ) 0 ∼ = (Ext 1 S (k, S/J f ) e ) ⊕(r−d) so it is enough to compute the dimension of Ext 1 S (k, S/J f ) e as a k-vector space. Applying the functor Hom S (−, S/J f ) e to the Koszul resolution of k we obtain the following complex: We need to show that the cohomology at (S/J f ) ⊕3 e+1 is an (r − d)-dimensional k-vector space. We will denote the map (S/J f ) e → (S/J f ) ⊕3 e+1 by d 0 and the map (S/J f ) ⊕3 e+1 → (S/J f ) ⊕3 e+2 by d 1 . Let h 1 , h 2 , h 3 ∈ S e+1 be such that d 1 (h 1 , h 2 , h 3 ) = 0, where h i is the class of h i in the quotient ring S/J f . Let h 1 = α 0 h ′ 1 (α 0 , α 1 , α 2 )+ h ′′ 1 (α 1 , α 2 ). Then (−α 1 h ′′ The ideal J f is monomial. Therefore, since (J f : α 0 ) = J f and h ′′ 1 does not depend on α 0 , we get ( Lemma 4.16. In the above notation, dim k Hom S (J f , S/I f ) 0 = 2r.
However, one could expect that existence of more irreducible components of Hilb r (P n ) is the only obstacle. The following example shows that the Hilbert function of S/I may satisfy condition (⋆ ⋆ ⋆) for [I] ∈ Hilb hr,n S[P n ]