On the Canonical Ideal of a Set of Points

by Martin Kreuzer (Regensburg)

Sunto: Dato un insieme 𝕏 di s punti nello spazio proiettivo, si costruisce un esplicito ideale canonico ℑ nel suo anello di coordinate R. Si descrivono le componenti omogenee di ℑ e la struttura della mappa di moltiplicazione R_{σ} ⊗ ℑ_{σ+1} → ℑ_{2σ+1}, dove σ = max{i | H_{𝕏}(i) < s}. Tra le applicazioni ci sono varie caratterizzazioni di insiemi di punti coomologicamente uniformi, disuguaglianze nelle loro funzioni di Hilbert, il calcolo del primo modulo delle sizigie di ℑ in casi particolari, una generalizzazione della “trasformata di Gale” a trasformate canoniche di grado superiore e infine alcune osservazioni sui codici MDS.

AMS Classification: primary 13 C 13, secondary 14 C 99, 14 M 05, 13 D 40

Abstract: For a set 𝕏 of s points in projective space, we construct an explicit canonical ideal ℑ in its homogeneous coordinate ring R. We describe the homogeneous components of ℑ and the structure of the multiplication map R_{σ}⊗ℑ_{σ+1} ⟶ ℑ_{2σ+1}, where σ = max{i | H_{𝕏}(i) < s}. Applications include various characterizations of cohomologically uniform sets of points, inequalities for their Hilbert functions, the computation of the first syzygy module of ℑ in special cases, a generalization of the Gale transform to higher canonical transforms, and some remarks about MDS codes.

Introduction

Given a set of points 𝕏 = {P_{1}, …, P_{s}} in projective space ℙ^{d} over an algebraically closed field K, we are interested in studying relations between the geometry of the configuration of the points and the algebraic structure of certain ideals and modules over the homogeneous coordinate ring R = K[X_{0}, …, X_{d}]/I_{𝕏} of 𝕏. In three previous papers [6], [10], and [11], we saw that many geometric properties of 𝕏 are encoded in the R-module structure of the canonical module

ω_{R} ≅ \underline{Hom}_{K[x_{0}]}(R, K[x_{0}])(−1)

of R, where we assume 𝕏 ∩ Z(X_{0}) = ∅ and let x_{0} = X_{0} + I_{𝕏} ∈ R.

The purpose of this paper is to refine those methods by noting that, for reduced 0-dimensional subschemes 𝕏 ⊆ ℙ^{d} as above, there exists an ideal ℑ_{R/K[x_{0}]} ⊆ R which is – up to a shift in degrees – isomorphic to the canonical module. Such an ideal is called a canonical ideal of 𝕏. In section 1 the ideal ℑ_{R/K[x_{0}]} is constructed in an analogous manner to the local case (cf. [9]), but with additional care taken to keep all maps and modules homogeneous and to make them completely explicit.

The canonical ideal of 𝕏 constructed in this way depends on the choice of the element x_{0} ∈ R, but in a very manageable fashion (cf. Prop. 1.8). It has the advantage of being contained in R_{≥σ_{𝕏}+1} = ⊕_{i≥σ_{𝕏}+1}R_{i}, where σ_{𝕏} = max{i ∈ ℤ | dim_{K}(R_{i}) < s}. By [6], 1.13, this part of the ring R can be described precisely using the separators f_{1}, …, f_{s} ∈ R_{σ_{𝕏}+1} of 𝕏: we have a K-basis for each homogeneous component R_{i} with i ≥ σ_{𝕏} +1, and an explicit description of the multiplication of R in terms of those bases. This in turn allows us to completely describe the homogeneous components ℑ_{σ_{𝕏}+1} and ℑ_{2σ_{𝕏}+1} of ℑ = ℑ_{R/K[x_{0}]} by constructing K-bases for them (cf. Cor. 1.10 and Cor. 1.11), and to compute the matrix of the multiplication map R_{σ_{𝕏}} ⊗ ℑ_{σ_{𝕏}+1} ⟶ ℑ_{2σ_{𝕏}+1} (cf. Cor. 1.14).

In the remaining part of the paper we show several applications of the canonical ideal. The main application is the solution of a question about the Hilbert function of cohomologically uniform sets of points posed in [11]. This uniformity condition was introduced in [11] as an intermediate condition between 1-uniformity and Δ_{𝕏}-uniformity. Here Δ_{𝕏} = H_{𝕏}(σ_{𝕏} + 1) − H_{𝕏}(σ_{𝕏}) is the last increase of the Hilbert function H_{𝕏}(i) = dim_{K}(R_{i}) of 𝕏, and 𝕏 is called n-uniform, if every subset 𝕐 ⊆ 𝕏 with s − n ≤ #𝕐 ≤ s satisfies H_{𝕐} = min{H_{𝕏}, #𝕐}. Cohomological uniformity is defined by the non-existence of a splitting X = 𝕐∪· 𝕐′ such that ∑_{P_{i}∈𝕐} K · Lf_{i} ∩ ∑_{P_{j}∈𝕐′} K · Lf_{j} = (0), where Lf_{i} is the image of f_{i} in R/(x_{0}). We characterize cohomological uniformity in terms of the structure of the canonical ideal in the following way.

0.1. Theorem. For a set of points 𝕏 ⊆ ℙ^{d}, the following conditions are equivalent. a) 𝕏 is cohomologically uniform.

b) The multiplication map R_{σ_{𝕏}} ⊗ℑ_{σ_{𝕏}+1} ⟶ ℑ_{2σ_{𝕏}+1} is nondegenerate and surjective.

It is then an immediate consequence of [10], 2.6 and 3.1 that Δ_{𝕏}-uniform points are cohomologically uniform and cohomologically uniform sets of points are 1-uniform.

At the end of section 2 we also relate cohomological uniformity in the case of d + 2 ≤ s ≤ \binom{d+2}{2} points to the condition that 𝕏 does not split linearly, i.e. that there are no two linear subspaces L_{1}, L_{2} ⊆ ℙ^{d} such that 𝕏 ⊆ L_{1}∪L_{2} and L_{1}∩L_{2} ≠ ∅ (cf. Prop. 2.7). We obtain another proof of the result of [2] which says in our situation that 𝕏 does not split linearly if and only if the canonical ideal is generated by its elemets of degree two. An example shows that cohomological uniformity generalizes linear splitting in a suitable way to point sets with σ_{𝕏} ≥ 2 (cf. Ex. 2.8).

The main result of section 3 is the following affirmative soultion of [11], Question, p. 248.

0.2. Theorem. If 𝕏 ⊆ ℙ^{d} is a cohomologically uniform set of points, then

H_{𝕏}(n) + H_{𝕏}(σ_{𝕏} − n) ≤ s − Δ_{𝕏} + 1

for all n ∈ {0, …, σ_{𝕏}}.

In order to prove this theorem, we characterize cohomological uniformity of 𝕏 by the existence of elements r ∈ R_{σ_{𝕏}} and φ ∈ ℑ_{σ_{𝕏}+1} such that ℑ_{2σ_{𝕏}+1} is of the form

ℑ_{2σ_{𝕏}+1} = R_{σ_{𝕏}}φ ⊕ Krφ_{2} ⊕ ⋯ ⊕ Krφ_{Δ_{𝕏}}

for any K-basis {φ, φ_{2}, …, φ_{Δ_{𝕏}}} of ℑ_{σ_{𝕏}+1} (cf. Prop. 3.9). The core part of the proof is to show that one can in fact choose r = ℓ^{σ_{𝕏}} for a generic element ℓ ∈ R_{1} in this characterization (cf. Prop. 3.12). Then the desired inequalities follow easily (cf. Cor. 3.3). For a discussion of the meaning of Theorem 0.2 and for examples we refer the reader to [11].

In the fourth section we apply our knowledge of the canonical ideal in the case of d+1 < s < \binom{d+2}{2} cohomologically uniform points in order to give a description of its first syzygy module. Both for the syzygy module of ℑ, considered as a P = K[X_{0}, …, X_{d}]- module, and of ℑ, considered as an R-module, we provide explicit homogeneous systems of generators (cf. Lemma 4.2.b and Prop. 4.3), and we show how one can compute those elements effectively (cf. Rem. 4.7). In the case of s = d+3 ≥ 6 cohomologically uniform points in ℙ^{d}, we prove that the R-syzygy module of ℑ is generated by its homogeneous elements of lowest degree (cf. Prop. 4.4), and we relate this to the analogous result of [3] for the P -syzygy module of ℑ (cf. Rem. 4.5).

Finally, in section 5, we use the homogeneous components of the canonical ideal to generalize the Gale transform of a set of s points 𝕏 (see for instance [5]). We show that the Gale transform can be defined explicitly by representing a K-basis of ℑ_{2σ} in terms of the separators, and that it consists of s distinct points, if and only if every subset of s − 2 points of 𝕏 spans ℙ^{d} (cf. Prop. 5.2). More generally, we can use the homogeneous components ℑ_{2σ+1−i} for i = 1, …, σ to define higher “canonical transforms” κ_{i}(𝕏) of 𝕏. After describing those components explicitly (cf. Prop. 5.1), we formulate again the precise uniformity condition on 𝕏 to ensure that κ_{i}(𝕏) consists of s distinct points. We end the paper by connecting the uniformity of 𝕏 to the uniformity of κ_{σ}(𝕏) (cf. Prop. 5.10), and by interpreting this result in the language of Coding Theory (cf. Cor. 5.11).

Acknowledgements. Mayor parts of this paper were done while the author was enjoying the hospitality of the Departments of Mathematics of Queen’s University (Kingston, Canada) and Universitá di Genova (Italy). He wishes to express his deep gratitude to those institutions, and in particular to A.V. Geramita (Kingston) and L. Robbiano (Genova). He is also thankful to E. Kunz (Regensburg) for his continuing interest in the subject and for providing the author with the repeated opportunity to lecture about it in the Oberseminar.

1. Construction of the Canonical Ideal

After fixing notations, we describe how to construct the canonical ideal of the projective coordinate ring of a set of points. We follow the procedure outlined in [9],

2. Vortrag, with two major differences: we are dealing with a graded situation, and we want explicit descriptions of the homogeneous components of the canonical ideal.

Throughout this paper we work over an algebraically closed field K of arbitrary characteristic. Our central object of interest is a given set of points 𝕏 = {P_{1}, …, P_{s}} in ℙ^{d}, the d-dimensional projective space over K. The coordinate functions {X_{0}, …, X_{d}} of ℙ^{d} are always chosen in such a manner that 𝕏 ⊆ D_{+}(X_{0}), i.e. such that no point of 𝕏 lies on the hyperplane Z(X_{0}). The projective coordinate ring of 𝕏 in ℙ^{d} is R = K[X_{0}, …, X_{d}]/I_{𝕏}, where I_{𝕏} denotes the homogeneous saturated ideal of 𝕏.

Let us collect a few elementary observations. The ring R = ⊕_{n≥0}R_{n} is a 1-dimen-sional reduced Cohen-Macaulay K-algebra, and x_{0} := X_{0} + I_{𝕏} ∈ R_{1} is not a zerodivisor of R. The Hilbert function H_{𝕏} : ℤ ⟶ ℕ (n ↦ dim_{K} R_{n}) of 𝕏 satisfies

… = 0 = H_{𝕏}(−1) < 1 = H_{𝕏}(0) < … < H_{𝕏}(σ_{𝕏}) < s = H_{𝕏}(σ_{𝕏} + 1) = H_{𝕏}(σ_{𝕏} + 2) = …

for some number σ_{𝕏} ≥ −1. Its first difference function ΔH_{𝕏}(n) := H_{𝕏}(n) − H_{𝕏}(n − 1) therefore has a nonzero value ΔH_{𝕏}(n) ≠ 0 if and only if 0 ≤ n ≤ σ_{𝕏} + 1. We let Δ_{𝕏} := ΔH_{𝕏}(σ_{𝕏} + 1) and write simply σ = σ_{𝕏} and Δ = Δ_{𝕏}, if no confusion can arise.

As in [6] and [10], for i ∈ {1, …, s} we denote by f_{i} ∈ R_{σ+1} a separator of P_{i} in 𝕏, i.e. a function such that f_{i}(P_{j}) = 0 for j ≠ i and f_{i}(P_{i}) ≠ 0. Here we let f(P) := f(1, p_{1}, …, p_{d}), if f ∈ R is a homogeneous element and P = (1 : p_{1} : …: p_{d}) ∈ D_{+}(X_{0}) a closed point. We normalize {f_{1}, …, f_{s}} by requiring f_{i}(P_{i}) = 1 for i = 1, …, s. Now [6], 1.13, says that {x_{0}^{n}f_{1}, …, x_{0}^{n}f_{s}} is a K-basis of R_{σ+1+n} for each n ≥ 0, and [6], 3.2.a, shows that the multiplication in R satisfies rf_{i} = r(P_{i})x_{0}^{n}f_{i} for i = 1, …, s and r ∈ R_{n}.

1.1. Lemma. A homogeneous element r ∈ R_{n} is not a zerodivisor of R if and only if r(P_{i}) ≠ 0 for i = 1, …, s.

Proof: If n ≥ σ_{𝕏} + 1, we have r = r(P_{1})x_{0}^{n−σ−1}f_{1} + ⋯ + r(P_{s})x_{0}^{n−σ−1}f_{s}, and if n ≤ σ_{𝕏}, then rx_{0}^{σ+1−n} = r(P_{1})f_{1} + ⋯ + r(P_{s})f_{s}. In any case, r(P_{i}) = 0 implies rf_{i} = 0, i.e. r is a zerodivisor of R. The converse follows from rf_{i} = r(P_{i})x_{0}^{n}f_{i}. □

The image of f_{i} in \overline{R} := R/(x_{0}) is denoted by Lf_{i} and is called the leading form of f_{i} (i = 1, …, s). By [6], 2.13, we can renumber {P_{1}, …, P_{s}} in such a way that \{Lf_{1}, …, Lf_{Δ}\} is a K-basis of \overline{R}_{σ+1}. Then we write

Lf_{Δ+j} = β_{j1}Lf_{1} + ⋯ + β_{jΔ}Lf_{Δ}

for j = 1, …, s − Δ, and we form the matrix 𝔅 := (β_{ji})^{transp.}.

1.2. Lemma. a) For i = 1, …, Δ we have β_{1i} + ⋯ + β_{s−Δ i} = −1.

b) The elements g_{1}, …, g_{s−Δ} ∈ R_{σ} which satisfy x_{0}g_{i} = f_{Δ+i} − β_{i1}f_{1} − ⋯ − β_{iΔ}f_{Δ} form a K-basis of R_{σ}.

Proof: Part a) follows, if we consider the equation f_{1} + ⋯ + f_{s} = x_{0}^{σ+1} and pass to \overline{R}. Part b) follows from the observation that the elements f_{Δ+i} − β_{i1}f_{1} − ⋯ − β_{iΔ}f_{Δ} with i = 1, …, s − Δ are linearly independent elements of x_{0}R_{σ}. □

Notice that 𝔅 depends on the numbering of {P_{1}, …, P_{s}} in an obvious way. The dependency of 𝔅 on the choice of x_{0} is governed by the following rule.

1.3. Lemma. Let ℓ ∈ R_{1} be a nonzerodivisor, and let β_{ji}^{ℓ} be constructed as above, but using ℓ instead of x_{0}. Then

β_{ji}^{ℓ} = \frac{ℓ(P_{Δ+j})^{σ}}{ℓ(P_{i})^{σ}} · β_{ji}

for i = 1, …, Δ and j = 1, …, s − Δ.

Proof: If f_{1}^{ℓ}, …, f_{s}^{ℓ} denote the separators w.r.t. ℓ, then f_{1}^{ℓ} + ⋯ + f_{s}^{ℓ} = ℓ^{σ+1} yields f_{i}^{ℓ} = ℓ(P_{i})^{σ+1}f_{i} for i = 1, …, s. Now notice that f_{Δ+j}^{ℓ} − \frac{ℓ(P_{Δ+j})^{σ}}{ℓ(P_{1})^{σ}} β_{j1}f_{1}^{ℓ} − ⋯ − \frac{ℓ(P_{Δ+j})^{σ}}{ℓ(P_{Δ})^{σ}} β_{jΔ}f_{Δ}^{ℓ} = ℓ(P_{Δ+j})^{σ+1}f_{Δ+j} − ℓ(P_{Δ+j})^{σ}ℓ(P_{1})β_{j1}f_{1} − ⋯ − ℓ(P_{Δ+j})^{σ}ℓ(P_{Δ})β_{jΔ}f_{Δ} = ℓ(P_{Δ+j})^{σ}ℓg_{j} is an element of ℓR_{σ}. □

At this point we are ready to start with our construction of the canonical ideal. Recall that the integral closure R̃ of R in its total ring of quotients can be described as follows (cf. [13] or [6]):

R̃ ≅ R/𝔭_{1} × ⋯ × R/𝔭_{s} ≅ K[T_{1}] × ⋯ × K[T_{s}]

where 𝔭_{i} is the homogeneous ideal of P_{i} in 𝕏, and T_{i} is the image of x_{0} in R/𝔭_{i}. Using the above notations, the canonical map ı : R ⟶ R̃ satisfies ı(r) = (r(P_{1})T_{1}^{n}, …, r(P_{s})T_{s}^{n}) for r ∈ R_{n}. In particular, ı(x_{0}) = (T_{1}, …, T_{s}) and ı(f_{i}) = (0, …, 0, T_{i}^{σ+1}, 0, …, 0). Since R̃ is graded by deg T_{i} = 1, the map ı is homogeneous of degree zero.

From this description we find that the full ring of quotients Q(R) of R can be identified with Q(R) ≅ K(T_{1}) × ⋯ × K(T_{s}). In the next step we compute Q^{h}(R), the homogeneous ring of quotients of R, i.e. the localization of R w.r.t. the set of all homogeneous nonzerodivisors.

1.4. Proposition. a) Using the above identifications, we have

Q^{h}(R) ≅ K[T_{1}, T_{1}^{−1}] × ⋯ × K[T_{s}, T_{s}^{−1}].

Here an element \frac{f}{g} ∈ Q^{h}(R) with f ∈ R_{m} and a nonzerodivisor g ∈ R_{n} is identified with the tuple

(\frac{f(P_{1})}{g(P_{1})} T_{1}^{m−n}, ⋯, \frac{f(P_{s})}{g(P_{s})} T_{s}^{m−n}) .

b) We have Q^{h}(R) ≅ R_{x_{0}}.

Proof: Let \frac{f}{g} ∈ Q^{h}(R) be as in a). The image of f in Q(R) is (f(P_{1})T_{1}^{m}, …, f(P_{s})T_{s}^{m}), and the image of g in Q(R) is (g(P_{1})T_{1}^{n}, …, g(P_{s})T_{s}^{n}). Therefore \frac{f}{g} is identified in Q(R) as claimed and is contained in K[T_{1}, T_{1}^{−1}] × ⋯ × K[T_{s}, T_{s}^{−1}].

Conversely, for any element (g_{1}, …, g_{s}) ∈ K[T_{1}, T_{1}^{−1}] × ⋯ × K[T_{s}, T_{s}^{−1}] and any n >> 0, we have (T_{1}^{n}g_{1}, …, T_{s}^{n}g_{s}) = ı(f) for some f ∈ R, because dim_{K} R_{m} = s = dim_{K}(K[T_{1}, T_{1}^{−1}] × ⋯ × K[T_{s}, T_{s}^{−1}])_{m} for m >> 0. Thus (g_{1}, …, g_{s}) is the image of \frac{f}{x_{0}^{n}} ∈ Q^{h}(R), proving a) and b). □

In what follows, we let L_{0} := K[x_{0}, x_{0}^{−1}] and L := Q^{h}(R). From 1.4 b) we conclude that L = L_{0} ⊗_{K[x_{0}]} R. Since R is a Cohen-Macaulay ring, it is a free K[x_{0}]-module of rank s. Therefore L is a free L_{0}-module of rank s, and there is an L_{0}-basis {e_{1}, …, e_{s}} of L which is identified with the standard basis of K[T_{1}, T_{1}^{−1}] × ⋯ × K[T_{s}, T_{s}^{−1}] under the isomorphism of 1.4 a). Let σ_{L/L_{0}} : L ⟶ L_{0} (e_{i} ↦ 1) be the canonical trace map.

1.5. Proposition. The homomorphism Σ : L ⟶ \underline{Hom}_{L_{0}}(L, L_{0}) (1 ↦ σ_{L/L_{0}}) is an isomorphism of graded L-modules.

Proof: Let φ : L ⟶ L_{0} be an L_{0}-linear map and g_{i}(x_{0}, x_{0}^{−1}) := φ(e_{i}) for i = 1, …, s. Then φ = (g_{1}(T_{1}, T_{1}^{−1}), …, g_{s}(T_{s}, T_{s}^{−1})) · σ_{L/L_{0}} shows that Σ is surjective. The map Σ is also injective, since from ψ = (g_{1}(T_{1}, T_{1}^{−1}), …, g_{s}(T_{s}, T_{s}^{−1})) · σ_{L/L_{0}} = 0 we obtain 0 = ψ(e_{i}) = g_{i}(x_{0}, x_{0}^{−1}) for i = 1, …, s, and hence (g_{1}(T_{1}, T_{1}^{−1}), …, g_{s}(T_{s}, T_{s}^{−1})) = 0.

□

Definition. Now we consider the following homomorphism of graded R-modules

Φ : \underline{Hom}_{K[x_{0}]}(R, K[x_{0}]) ⟶ \underline{Hom}_{L_{0}}(L, L_{0}) ⟶^{Σ^{−1}} L
φ ⟼ φ ⊗ id_{L_{0}}

Its image is a homogeneous fractional R-ideal ℭ_{R/K[x_{0}]} of L which is called the Dedekind complementary module of R with respect to x_{0}.

It is not difficult to see (cf. [6], 3.1) that the Hilbert function of ℭ := ℭ_{R/K[x_{0}]} ≅ \underline{Hom}_{K[x_{0}]}(R, K[x_{0}]) is given by H_{ℭ}(n) = s − H_{𝕏}(−n − 1) for all n ∈ ℤ.

1.6. Proposition. (Explicit description of ℭ_{R/K[x_{0}]})

Let n ≥ 0, and let φ ∈ \underline{Hom}_{K[x_{0}]}(R, K[x_{0}])_{−σ−1+n}. For i = 1, …, s we write φ(f_{i}) = c_{i}x_{0}^{n} with c_{i} ∈ K. Then

Φ(φ) = (c_{1}T_{1}^{−σ−1+n}, …, c_{s}T_{s}^{−σ−1+n}) ∈ L.

Proof: Since ı : R ⟶ L satisfies ı(f_{i}) = (0, …, 0, T_{i}^{σ+1}, 0, …, 0), the map φ ⊗ id_{L_{0}} : L ≅ R ⊗_{K[x_{0}]} L_{0} ⟶ L_{0} is given by (φ ⊗ id_{L_{0}})(x_{0}^{σ+1}e_{i}) = (φ ⊗ id_{L_{0}})((0, …, 0, T_{i}^{σ+1}, 0, …, 0)) = φ(f_{i}) = c_{i}x_{0}^{n}. Therefore we have φ ⊗ id_{L_{0}} = x_{0}^{−σ−1+n}(c_{1}, …, c_{s})σ_{L/L_{0}} in \underline{Hom}_{L_{0}}(L, L_{0}). Thus the claim follows from the description of Σ^{−1} given in the proof of 1.5. □

1.7. Corollary. The set x_{0}^{2σ+2} · ℭ_{R/K[x_{0}]} ⊆ L is an ideal of R.

Proof: Clearly this set is an R-submodule of L. We shall show that it is contained in the image of R in L. Let φ ∈ \underline{Hom}_{K[x_{0}]}(R, K[x_{0}])_{−σ−1+n} with n ≥ 0, and let φ(f_{i}) = c_{i}x_{0}^{n} with c_{i} ∈ K for i = 1, …, s. Then we have

\begin{aligned}x_{0}^{2σ+2}Φ(φ) & = x_{0}^{2σ+2} · (c_{1}T_{1}^{−σ−1+n}, …, c_{s}T_{s}^{−σ−1+n}) = \\ & = (c_{1}T_{1}^{σ+1+n}, …, c_{s}T_{s}^{σ+1+n}), \end{aligned}

and this element is the image of x_{0}^{n}(c_{1}f_{1} + … + c_{s}f_{s}) ∈ R in L. □

Definition. The ideal ℑ_{R/K[x_{0}]} := x_{0}^{2σ+2} · ℭ_{R/K[x_{0}]} of R is called the canonical ideal of 𝕏 with respect to x_{0}. Since the choice of x_{0} will remain fixed, we shall also denote this ideal by ℑ := ℑ_{R/K[x_{0}]}.

Later we shall show that, in general, σ_{𝕏} + 2 is the smallest number n such that x_{0}^{n} · ℭ_{R/K[x_{0}]} is contained in R, but in special cases also smaller numbers n can suffice for this to be true. It is also clear that ℑ_{R/K[x_{0}]} depends on the choice of the linear nonzerodivisor x_{0} of R. The next proposition makes this dependency explicit.

1.8. Proposition. If ℓ ∈ R_{1} is a nonzerodivisor of R, then x_{0} · ℭ_{R/K[x_{0}]} = ℓ · ℭ_{R/K[ℓ]} and ℓ^{2σ+1} · ℑ_{R/K[x_{0}]} = x_{0}^{2σ+1} · ℑ_{R/K[ℓ]} ⊆ R.

In other words, for an element f = x_{0}^{n}(c_{1}f_{1}+…+c_{s}f_{s}) of R_{σ+1+n}, with c_{i} ∈ K and n ≥ 0, we have f ∈ ℑ_{R/K[ℓ]} if and only if x_{0}^{n}(ℓ(P_{1})^{−2σ−1}c_{1}f_{1} + … + ℓ(P_{s})^{−2σ−1}c_{s}f_{s}) ∈ ℑ_{R/K[x_{0}]}.

Proof: In view of the definition, it suffices to prove the first claim. Because of symmetry reasons, we only show x_{0} ·ℭ_{R/K[x_{0}]} ⊆ ℓ·ℭ_{R/K[ℓ]}. Let φ : R ⟶ K[x_{0}] be a homogeneous K[x_{0}]-linear map of degree −σ_{𝕏} −1+n with n ≥ 0, and let φ(f_{i}) = c_{i} x_{0}^{n} with c_{i} ∈ K for i = 1, …, s. Since φ(R_{σ−n}) ⊆ K[x_{0}]_{−1} = (0), we have λ_{1}c_{1} + … + λ_{s}c_{s} = 0 whenever λ_{1}, …, λ_{s} ∈ K and λ_{1}f_{1} + … + λ_{s}f_{s} = x_{0}^{n+1}g for some g ∈ R_{σ−n}. By 1.6, we have Φ(x_{0}φ) = (c_{1}T_{1}^{−σ+n}, …, c_{s}T_{s}^{−σ+n}) in L.

Now let U_{i} be the image of ℓ in R/𝔭_{i}, and consider the representation L = K[U_{1}, U_{1}^{−1}] × ⋯ × K[U_{s}, U_{s}^{−1}]. Since U_{i} = ℓ(P_{i})T_{i} for i = 1, …, s, the element Φ(x_{0}φ) is given by (c_{1}ℓ(P_{1})^{σ−n}U_{1}^{−σ+n}, …, c_{s}ℓ(P_{s})^{σ−n}U_{s}^{−σ+n}) in this representation.

Next we let ψ : R ⟶ K[ℓ, ℓ^{−1}] be the homogeneous K[ℓ]-linear map of degree −σ_{𝕏} −1+n such that ψ(f_{i}^{(ℓ)}) = c_{i}ℓ(P_{i})^{σ−n}ℓ^{n} for i = 1, …, s. Here f_{1}^{(ℓ)}, …, f_{s}^{(ℓ)} are the normalized separators with respect to ℓ. Using 1.6 again, we see that (c_{1}ℓ(P_{1})^{σ−n}U_{1}^{−σ+n}, …, c_{s}ℓ(P_{s})^{σ−n}U_{s}^{−σ+n}) is also the image of ℓψ in L.

Therefore it suffices to show im ψ ⊆ K[ℓ]. For this we need to prove ψ(R_{σ−n}) = 0. Let g ∈ R_{σ−n}, and let x_{0}^{n+1}g = λ_{1}f_{1} + … + λ_{s}f_{s} with λ_{1}, …, λ_{s} ∈ K. By the proof of 1.3 we know f_{i}^{(ℓ)} = ℓ(P_{i})^{σ+1}f_{i} for i = 1, …, s. Therefore we have ℓ^{n+1}g = λ_{1}ℓ(P_{1})^{n+1}f_{1} + … + λ_{s}ℓ(P_{s})^{n+1}f_{s} = λ_{1}ℓ(P_{1})^{−σ+n}f_{1}^{(ℓ)} + … + λ_{s}ℓ(P_{s})^{−σ+n}f_{s}^{(ℓ)}. Thus we obtain ℓ^{n+1}ψ(g) = ψ(ℓ^{n+1}g) = λ_{1}ℓ(P_{1})^{−σ+n}ψ(f_{1}^{(ℓ)}) + … + λ_{s}ℓ(P_{s})^{−σ+n}ψ(f_{s}^{(ℓ)}) = ℓ^{n}(λ_{1}c_{1} + … + λ_{s}c_{s}) = 0. Altogether we get ψ(g) = 0, as desired. □

The explanation why ℑ_{R/K[x_{0}]} is important for the study of 𝕏 and why it is called the canonical ideal of 𝕏 is provided by the following proposition and its corollaries. Recall that the graded R-module ω_{R} = \underline{Hom}_{K}(H_{𝔪}^{1}(R), K) ≅ \underline{Hom}_{K[x_{0}]}(R, K[x_{0}])(−1) is the canonical module of R and can be used to characterize many geometric properties of 𝕏 (cf. [10]).

1.9. Proposition. There are isomorphisms of graded R-modules

ω_{R} ≅ ℭ_{R/K[x_{0}]}(−1) ≅ ℑ_{R/K[x_{0}]}(2σ_{𝕏} + 1).

Here an element φ ∈ (ω_{R})_{−σ+n} = \underline{Hom}_{K[x_{0}]}(R, K[x_{0}])_{−σ−1+n} with n ≥ 0 is identified with the element φ(f_{1})f_{1} + … + φ(f_{s})f_{s} of ℑ_{σ+1+n}.

Proof: Since Φ is a monomorphism of graded R-modules and ℑ = x_{0}^{2σ+2}ℭ_{R/K[x_{0}]} ≅ ℭ_{R/K[x_{0}]}(−2σ_{𝕏} − 2), the first claim is clear. For the second claim, let φ ∈ (ω_{R})_{−σ+n} be given by φ(f_{i}) = c_{i} x_{0}^{n}, with n ≥ 0 and c_{i} ∈ K for i = 1, …, s. Then multiplication by x_{0}^{2σ+2} provides an isomorphism ℭ_{R/K[x_{0}]}(−1) ≅ ℭ_{R/K[x_{0}]}(2σ_{𝕏} + 1) which identifies Φ(φ) = (c_{1}T_{1}^{−σ−1+n}, …, c_{s}T_{s}^{−σ−1+n}) with x_{0}^{n}(c_{1}f_{1}+…+c_{s}f_{s}) = φ(f_{1})f_{1}+…+φ(f_{s})f_{s}.

□

As a first application, we can use the isomorphism of this proposition to give explicit descriptions of ℑ_{σ+1} ≅ (ω_{R})_{−σ} and ℑ_{2σ+1} ≅ (ω_{R})_{0}. From what we mentioned earlier, we know the Hilbert funtion of ℑ, namely

H_{ℑ}(n) = s − H_{𝕏}(2σ_{𝕏} + 1 − n)

for all n ∈ ℤ. Thus ℑ_{σ+1} and ℑ_{2σ+1} are the first and last nontrivial homogeneous components of ℑ. In section 5 we shall also give a (somewhat less explicit) description of the remaining nontrivial homogeneous components of ℑ.

1.10. Corollary. The elements π_{i} := f_{i} +β_{1i}f_{Δ+1} +…+β_{s−Δ i}f_{s} such that 1 ≤ i ≤ Δ form a K-basis of ℑ_{σ+1}.

Proof: Since \{Lf_{1}, …, Lf_{Δ}\} is a K-basis of \overline{R}_{σ+1}, the projections π_{1}, …, π_{Δ} defined by π_{i}(f_{j}) := δ_{ij} for j = 1, …, Δ form a K-basis of (ω_{R})_{−σ} = \underline{Hom}_{K[x_{0}]}(R, K[x_{0}])_{−σ−1} (cf. [10], 1.5). For j = 1, …, s−Δ this yields π_{i}(f_{Δ+j}) = π_{i}(β_{j1}f_{1} + … + β_{jΔ}f_{Δ}) = β_{ij}. Now an application of 1.9 finishes the proof. □

1.11. Corollary. The elements h_{i} := x_{0}^{σ}(f_{i+1} −f_{i}) with i = 1, …, s−1 form a K-basis of ℑ_{2σ+1}. An element x_{0}^{σ}(c_{1}f_{1} + … + c_{s}f_{s}) of R_{2σ+1} is contained in ℑ_{2σ+1} if and only if c_{1} + … + c_{s} = 0.

Proof: An element φ of (ω_{R})_{0} ≅ \underline{Hom}_{K[x_{0}]}(R, K[x_{0}])_{−1} yields a K-linear map φ |_{R_{σ+1}}: R_{σ+1} ⟶ K[x_{0}]_{σ} with the property (φ |_{R_{σ+1}})(x_{0}^{σ+1}) = x_{0}^{σ+1}φ(1) ∈ x_{0}^{σ+1}K[x_{0}]_{−1} = (0). Conversely, a K-linear map ψ : R_{σ+1} ⟶ K[x_{0}]_{σ} extends to a K[x_{0}]-linear homogeneous map \overline{ψ} : R ⟶ K[x_{0}] of degree −1 if and only if ψ(x_{0}^{σ+1}) = 0, because dim_{K}(ω_{R})_{0} = s−1 and \overline{ψ} is uniquely determined by \overline{ψ} |_{R_{σ+1}} (cf. [10], 1.4). Now x_{0}^{σ+1} = f_{1} + … + f_{s} implies φ(f_{1}) + … + φ(f_{s}) = 0 for φ ∈ (ω_{R})_{0}, and the claims of the corollary follow from Proposition 1.9. □

Using Corollary 1.10, we can now give the promised examples which show that, in general, one has to multiply ℑ_{R/K[x_{0}]} by x_{0}^{2σ+2} in order to get an ideal of R, but in special cases a lower power of x_{0} may suffice.

1.12. Example. Let char K ≠ 2, 3, and let 𝕏 = {(1 : 0 : 0), (1 : 1 : 0), (1 : 2 : 0)} ⊆ ℙ^{2}. Then 𝕏 is a complete intersection of type (3, 1), i.e. I_{𝕏} = (X_{2}, X_{1}(X_{1} −X_{0})(X_{1} −2X_{0})), its Hilbert function is given by ΔH_{𝕏} : 1 1 1 0 …, and σ_{𝕏} = 1. It is easy to compute f_{1} = \frac{1}{2}(x_{1} − x_{0})(x_{1} − 2x_{0}), f_{2} = −x_{1}(x_{1} − 2x_{0}), and f_{3} = \frac{1}{2}x_{1}(x_{1} − x_{0}) (cf. [6], 1.15). Thus Lf_{2} = β_{11}Lf_{1} with β_{11} = −2, and Lf_{3} = β_{21}Lf_{1} with β_{21} = 1. By Corollary 1.10, the element π_{1} = f_{1} − 2f_{2} + f_{3} is a K-basis of ℑ_{σ+1}. Since 𝕏 is a complete intersection, {π_{1}} is even an R-basis of ℑ. Clearly, π_{1} ∉ x_{0}R_{σ}, so that in this case x_{0}^{2σ+1}ℭ_{R/K[x_{0}]} ⊈ R.

1.13. Example. Let K := ℂ, and let 𝕏 := \{(1 : 0 : 0), (1 : 2 : 0), (1 : 1 + \sqrt{3}i : 0)} ⊆ ℙ^{2}. Again 𝕏 is a complete intersection of type (3, 1), its Hilbert function is ΔH_{𝕏} : 1 1 1 0 …, and σ_{𝕏} = 1. This time we find f_{1} = \frac{1}{2+2\sqrt{3}i}(x_{1} −2x_{0})(x_{1} −x_{0} −\sqrt{3}i x_{0}), f_{2} = \frac{1}{2−2\sqrt{3}i}x_{1}(x_{1} − x_{0} − \sqrt{3}i x_{0}), and f_{3} = −\frac{1}{4}x_{1}(x_{1} − 2x_{0}). Thus Lf_{2} = β_{11}Lf_{1} with β_{11} = \frac{1}{2}(\sqrt{3}i − 1) and Lf_{3} = β_{21}Lf_{1} with β_{21} = −\frac{1}{2}(\sqrt{3}i + 1). Therefore the R-basis π_{1} = f_{1} + β_{11}f_{2} + β_{21}f_{3} of ℑ_{R/K[x_{0}]} satisfies Lπ_{1} = Lf_{1} + β_{11}Lf_{2} + β_{21}Lf_{3} = 0, i.e. we have π_{1} ∈ x_{0}R_{σ}. This means that we could have taken x_{0}^{2σ+1}ℭ_{R/K[x_{0}]} = x_{0}^{3}ℭ_{R/K[x_{0}]} and obtained an ideal of R.

Let us also show that x_{0}^{2}ℭ_{R/K[x_{0}]} ⊈ R here. This fractional ideal starts in degree zero, so that we want to show π_{1} ∉ (x_{0}^{2}). But π_{1} = λx_{0}^{2} = λf_{1} + λf_{2} + λf_{3} for some λ ∈ K implies β_{11} = β_{21} = 1, which is not the case.

In [10], 2.6 and 3.1, the multiplication map R_{σ} ⊗ ℑ_{σ+1} ⟶ ℑ_{2σ+1} has been used to characterize geometrical properties of 𝕏. By applying 1.10 and 1.11, we shall now give an explicit matrix for this map.

1.14. Corollary. Let {g_{1}, …, g_{s−Δ}}, {π_{1}, …, π_{Δ}}, and {h_{1}, …, h_{s−1}} be as in 1.2.b, 1.10, and 1.11, resp. Then the multiplication map R_{σ} ⊗ ℑ_{σ+1} ⟶ ℑ_{2σ+1} is given by

g_{j} · π_{i} = \{ \begin{matrix}β_{j1}h_{Δ+j−1} & for i = 1 and j = 1, …, s − Δ, \\ β_{ji}h_{Δ+j−1} − β_{ji}h_{i−1} & for i = 2, …, Δ and j = 1, …, s − Δ. \end{matrix}

Proof: For i ∈ {1, …, Δ} and j ∈ {1, …, s − Δ} we have x_{0}g_{j}π_{i} = (f_{Δ+j} − β_{j1}f_{1} − ⋯ − β_{jΔ}f_{Δ}) · (f_{1} + β_{Δ+1 i}f_{Δ+1} + ⋯ + β_{s−Δ i}f_{s−Δ}) = x_{0}^{σ}(β_{ji}f_{Δ+j} − β_{ji}f_{i}). From this the claim follows immediately. □

Finally, we remind the reader that in complete analogy with [9], 6.13, one can show the following proposition in our situation.

1.15. Proposition. The ring R/ℑ is a 0-dimensional Gorenstein ring.

2. Canonical Ideals of Cohomologically Uniform Schemes

In this section we shall apply our knowledge of the canonical ideal in order to study the property of cohomological uniformity introduced in [11]. Recall that a 0-dimensional scheme 𝕏 = {P_{1}, …, P_{s}} ⊆ ℙ^{d} is called n-uniform for some n ≥ 0, if every subset 𝕐 ⊆ 𝕏 consisting of s − n points has the same Hilbert function, namely H_{𝕐} = min{H_{𝕏}, s − n}. In [6], 1-uniform schemes were also called Cayley-Bacharach schemes.

Our main goals are to characterize cohomological uniformity in terms of the structure of the canonical ideal of 𝕏 and to show that this property is intermediate between 1-uniformity and Δ_{𝕏}-uniformity. Towards the end of the section we shall also give a concrete geometrical interpretation of cohomological uniformity in the case of d + 2 ≤ s ≤ \binom{d+2}{2} points with generic Hilbert function.

Let us start with the definition. We continue to use all notations and conventions of section 1.

Definition. We say that 𝕏 splits cohomologically, if we can decompose 𝕏 = 𝕐∪𝕐′ such that 𝕐 ≠ ∅, 𝕐′ ≠ ∅, 𝕐 ∩ 𝕐′ = ∅ and ∑_{P_{i}∈𝕐} K · Lf_{i} ∩ ∑_{P_{i}∈𝕐′} K · Lf_{i} = (0) in \overline{R}_{σ+1}.

If 𝕏 does not split cohomologically, we say that 𝕏 is cohomologically uniform.

The choice of this name is explained in [11], sec. 3. From our next proposition and Lemma 1.3 it follows that the definition is in fact independent of the choice of the linear nonzerodivisor x_{0} ∈ R_{1} (for another proof see [11]).

2.1. Proposition. The following conditions are equivalent.

a) 𝕏 splits cohomologically.

b) We can renumber the points of {P_{1}, …, P_{Δ}} and of {P_{Δ+1}, …, P_{s}} such that the matrix 𝔅 is of the form 𝔅 = \binom{∗ 0}{0 ∗}, where each block may have zero rows or columns and is strictly smaller than 𝔅.

Proof: Suppose condition a) holds. If Lf_{i} = 0 for some i ∈ {1, …, s}, then one of the columns of 𝔅 is zero, i.e. after renumbering {P_{Δ+1}, …, P_{s}} the matrix 𝔅 is of the form 𝔅 = (0 ∗). Therefore we shall assume now that all elements Lf_{1}, …, Lf_{s} are different from zero. Let 𝕏 = 𝕐∪· 𝕐′ be the decomposition according to the above definition. Renumber {P_{1}, …, P_{Δ}} and {P_{Δ+1}, …, P_{s}} such that we have 𝕐 = {P_{1}, …, P_{δ}, P_{Δ+1}, …, P_{Δ+ϵ}} for some 0 ≤ δ ≤ Δ and 0 ≤ ϵ ≤ s − Δ.

Here we cannot have δ = 0, since in that case \overline{R}_{σ+1} agrees with ∑_{P_{i}∈𝕐′} K · Lf_{i}, and for all points P_{j} ∈ 𝕐 we have Lf_{j} = 0. In an analogous way we see that δ < Δ holds. In case ϵ = 0 we have Lf_{j} ∈ ∑_{P_{i}∈𝕐′} K · Lf_{i} for Δ + 1, …, s, so that 𝔅 is of the form 𝔅 = \binom{0}{∗}. Similarly, if ϵ = s − Δ, the matrix 𝔅 is of the form 𝔅 = \binom{∗}{0}. Hence we can assume that 1 ≤ ϵ < s − Δ.

Now we observe that for j = 1, …, ϵ it follows from β_{j δ+1}Lf_{δ+1} + … + β_{jΔ}Lf_{Δ} = Lf_{Δ+j} − β_{j1}Lf_{1} − … − β_{jδ}Lf_{δ} ∈ ∑_{P_{i}∈𝕐} K · Lf_{i} ∩ ∑_{P_{i}∈𝕐′} K · Lf_{i} that β_{j δ+1} = … = β_{jΔ} = 0. Also, for j = ϵ + 1, …, s − Δ it follows from β_{j1}Lf_{1} + … + β_{jδ}Lf_{δ} = Lf_{Δ+j} − β_{j δ+1}Lf_{δ+1} −…−β_{jΔ}Lf_{Δ} ∈ ∑_{P_{i}∈𝕐} K ·Lf_{i} ∩ ∑_{P_{i}∈𝕐′} K ·Lf_{i} that β_{j1} = … = β_{jδ} = 0. Thus the matrix 𝔅 has the required shape.

Conversely, suppose that β_{j1} = … = β_{jΔ} = 0 for some j ∈ {1, …, s − Δ}. Then we can choose 𝕐 = {P_{j}} and 𝕐′ = 𝕏 \ {P_{j}}, and we get the desired decomposition of \overline{R}_{σ+1}. On the other hand, if the matrix 𝔅 is of the form 𝔅 = \binom{∗ 0}{0 ∗} or 𝔅 = \binom{∗}{0} or 𝔅 = \binom{0}{∗}, we let δ ∈ \{1, …, Δ − 1\} be the number of rows of the upper blocks and ϵ ∈ {1, …, s − Δ − 1} the number of columns of the left-hand blocks. Then we can choose 𝕐 = {P_{1}, …, P_{δ}, P_{Δ+1}, …, P_{Δ+ϵ}} and 𝕐′ = {P_{δ+1}, …, P_{Δ}, P_{Δ+ϵ+1}, …, P_{s}}, and we get the desired decomposition of \overline{R}_{σ+1} again. □

This proposition implies that cohomologically uniform schemes have Lf_{i} ≠ 0 for i = 1, …, s, i.e. that they are 1-uniform (cf. [6], 2.6). Let us illustrate the phenomenon of cohomological splitting with an example. We note that condition 2.1.b) yields a computational way to check for this property.

2.2. Example. Consider two skew lines L_{1}, L_{2} ⊆ ℙ^{3}, e.g. L_{1} = Z(X_{2}, X_{3}) and L_{2} = Z(X_{3} − X_{0}, X_{1}). Let 𝕏 = {P_{1}, …, P_{6}} ⊆ ℙ^{3} with {P_{1}, P_{3}, P_{4}} ⊆ L_{1} and {P_{2}, P_{5}, P_{6}} ⊆ L_{2}, e.g. P_{1} = (1 : 0 : 0 : 0), P_{2} = (1 : 0 : 0 : 1), P_{3} = (1 : 1 : 0 : 0), P_{4} = (1 : −1 : 0 : 0), P_{5} = (1 : 0 : 1 : 1), and P_{6} = (1 : 0 : −1 : 1). Here we assume char K ≠ 2. Then 𝕏 has Hilbert function ΔH_{𝕏} : 1 3 2 0 … and σ_{𝕏} = 1. We compute the separators of 𝕏 and find f_{1} = x_{0}^{2} − x_{0}x_{3} − x_{1}^{2}, f_{2} = x_{0}x_{3} − x_{2}^{2}, f_{3} = \frac{1}{2}x_{0}x_{1} + \frac{1}{2}x_{1}^{2}, f_{4} = −\frac{1}{2}x_{0}x_{1} + \frac{1}{2}x_{1}^{2}, f_{5} = \frac{1}{2}x_{0}x_{2} + \frac{1}{2}x_{2}^{2}, and f_{6} = −\frac{1}{2}x_{0}x_{2} + \frac{1}{2}x_{2}^{2}.

Thus \{Lf_{1}, Lf_{2}\} is a K-basis of \overline{R}_{σ+1}, and we see that the matrix 𝔅 of 𝕏 is given by \binom{−1/2 −1/2 0 0}{0 0 −1/2 −1/2}. Using 2.1 we conclude that 𝕏 splits cohomologically in the form 𝕏 = {P_{1}, P_{3}, P_{4}} ∪ {P_{2}, P_{5}, P_{6}}.

The following lemma gives us a different coherence property of the matrix 𝔅 of a cohomologically uniform scheme.

2.3. Lemma. Suppose 𝕏 = {P_{1}, …, P_{s}} ⊆ ℙ^{d} is a cohomologically uniform set of points, and ∼ is an equivalence relation on the set {1, …, s} with the property that i ∼ j whenever i ∈ {1, …, Δ}, j ∈ {Δ + 1, …, s}, and β_{j−Δ i} ≠ 0. Then 1 ∼ 2 ∼ … ∼ s.

Proof: W.l.o.g. let {Δ + 1, …, Δ + t_{1}} with t_{1} ≥ 1 be the set of those numbers j among {Δ + 1, …, s} for which β_{j−Δ 1} ≠ 0. Here we have t_{1} ≥ 1 because of 1.2.a. By assumption we then get 1 ∼ Δ + 1 ∼ … ∼ Δ + t_{1}, and for j ∈ {Δ + t_{1} + 1, …, s} we have β_{j−Δ 1} = 0. Since 𝕏 is cohomologically uniform, Proposition 2.1 yields a number i ∈ {2, …, Δ} such that not all elements of {β_{1i}, …, β_{t_{1}i}} are zero. W.l.o.g. let i = 2 and β_{ν_{1}2} ≠ 0 with ν_{1} ∈ {1, …, t_{1}}. Then we have 2 ∼ Δ + ν_{1} ∼ 1.

W.l.o.g. let {Δ+t_{1}+1, …, Δ+t_{2}} with t_{2} ≥ t_{1} be the set of those numbers j among {Δ + t_{1} + 1, …, s} for which β_{j−Δ 2} ≠ 0. (Notice that we allow t_{2} = t_{1} and an empty set.) Thus we have 2 ∼ Δ+t_{1}+1 ∼ … ∼ Δ+t_{2} and β_{j−Δ 2} = 0 for j = Δ+t_{1}+2, …, s. Since 𝕏 is cohomologically uniform, Proposition 2.1 yields a number i ∈ {3, …, Δ} such that not all elements of the set {β_{1i}, …, β_{t_{2}i}} are zero. W.l.o.g. let i = 3 and β_{ν_{2}3} ≠ 0 with ν_{2} ∈ {1, …, t_{2}}. Then we have 3 ∼ Δ + ν_{2} ∼ 2 ∼ 1.

Continuing in this manner we finally obtain t_{Δ} = s − Δ, since not all elements of {β_{s−Δ 1}, …, β_{s−Δ Δ}} are zero, and we get Δ ∼ Δ − 1 ∼ … ∼ 1. By 2.1, each column of 𝔅 has a nonzero entry, so 1 ∼ … ∼ Δ ∼ Δ + 1 ∼ … ∼ s. □

Now we are ready to prove the main theorem of this section.

2.4. Theorem. Let 𝕏 = {P_{1}, …, P_{s}} ⊆ ℙ^{d} be a set of points. The following conditions are equivalent.

a) 𝕏 is cohomologically uniform.

b) The multiplication map μ : R_{σ} ⊗ℑ_{σ+1} ⟶ ℑ_{2σ+1} is nondegenerate and surjective.

c) The multiplication map μ̃ : R_{σ} ⊗ (ω_{R})_{−σ} ⟶ (ω_{R})_{0} is nondegenerate and surjective.

Proof: “a) ⇒ b)” From Proposition 2.1 it follows that 𝕏 is a Cayley-Bacharach scheme. In [10], 2.6 it was shown that the multiplication map μ̃ is nondegenerate for Cayley-Bacharach schemes. By 1.9, also μ is nondegenerate.

Because of 1.11, we still need to show that the elements h_{i} = x_{0}^{σ}(f_{i+1} − f_{i}) such that 1 ≤ i ≤ s − 1 are in the image of μ. Define a relation ∼ on the set {1, …, s} by i ∼ j ⇔ x_{0}^{σ}(f_{i} − f_{j}) ∈ im μ. Obviously ∼ is an equivalence relation. If i ∈ {1, …, Δ} and j ∈ {Δ + 1, …, s}, then g_{j−Δ}π_{i} = β_{j−Δ i}x_{0}^{σ}(f_{j} − f_{i}) ∈ im μ by 1.14. Hence if β_{j−Δ i} ≠ 0, then i ∼ j. Thus we can apply the lemma and obtain 1 ∼ … ∼ s, i.e. x_{0}^{σ}(f_{j} − f_{i}) ∈ im μ for i, j ∈ {1, …, s}.

“b)⇒a)” Suppose that 𝕏 splits cohomologically in the form 𝕏 = 𝕐 ∪ 𝕐′. Then Δ_{𝕏} ≥ 2, and we can renumber {P_{1}, …, P_{s}} such that 𝕐 = {P_{1}, …, P_{δ}, P_{Δ+1}, …, P_{Δ+ϵ}} with 1 ≤ δ ≤ Δ − 1 and 0 ≤ ϵ ≤ s − Δ. Since the image of μ is generated by the elements g_{j−Δ}π_{i} = β_{j−Δ i} x_{0}^{σ}(f_{j} − f_{i}) with i ∈ {1, …, Δ} and j ∈ {Δ + 1, …, s}, it is already generated by the subset of those elements for which β_{j−Δ i} ≠ 0. As 𝕏 splits cohomologically, that subset is contained in the set

M := {x_{0}^{σ}(f_{j} − f_{i}) | i = 1, …, δ and j = Δ + 1, …, Δ + ϵ}
∪ {x_{0}^{σ}(f_{j} − f_{i}) | i = δ + 1, …, Δ and j = Δ + ϵ + 1, …, s}.

Now let x = x_{0}^{σ}(c_{1}f_{1} + … + c_{s}f_{s}) with c_{1}, …, c_{s} ∈ K be an arbitrary element of im μ. We express x with the generators from M and get

x = ∑_{i=1}^{δ} ∑_{j=Δ+1}^{Δ+ϵ} ζ_{ij} x_{0}^{σ}(f_{j} − f_{i}) + ∑_{i=δ+1}^{Δ} ∑_{j=Δ+ϵ+1}^{s} ζ_{ij} x_{0}^{σ}(f_{j} − f_{i})

with ζ_{ij} ∈ K. By comparing the two representations of x, we find

c_{i} = \{ \begin{matrix}−ζ_{iΔ+1} − … − ζ_{iΔ+ϵ} & for i = 1, …, δ, \\ −ζ_{iΔ+ϵ+1} − … − ζ_{is} & for i = δ + 1, …, Δ, \\ ζ_{1i} + … + ζ_{δi} & for i = Δ + 1, …, Δ + ϵ, \\ ζ_{δ+1i} + … + ζ_{Δi} & for i = Δ + ϵ + 1, …, s. \end{matrix}

Thus we conclude ∑_{P_{i}∈𝕐} c_{i} = 0 and ∑_{P_{i}∈𝕐′} c_{i} = 0. In particular, the element x_{0}^{σ}(f_{Δ} − f_{1}) is not in the image of μ. This contradicts the hypothesis.

“b)⇔ c)” is a consequence of 1.9. □

Using [10], 3.2, this theorem implies that Δ_{𝕏}-uniform schemes are cohomologically uniform. In view of [6], example 3.11, it was asked in [11] whether cohomological uniformity is the right condition to show the inequalities H_{𝕏}(n)+H_{𝕏}(σ_{𝕏}−n) ≤ s−Δ_{𝕏}+1 for n = 1, …, σ_{𝕏} for the Hilbert function of 𝕏. This is the topic of our next section.

But before we want to specialize the situation for a moment and consider schemes 𝕏 consisting of “few” points in ℙ^{d}. More precisely, let 𝕏 = {P_{1}, …, P_{s}} ⊆ ℙ^{d} consist of d + 2 ≤ s ≤ \binom{d+2}{2} points, and suppose that 𝕏 is nondegenerate, i.e. it is not contained in a hyperplane, and that σ_{𝕏} = 1. Then 𝕏 has generic Hilbert function, i.e. the Hilbert function of 𝕏 is H_{𝕏} : 1 d+1 s s ⋯, and we have Δ_{𝕏} = s − d − 1. In this situation [10], 3.2 and 4.2 yield that some conditions of uniformity coincide.

2.5. Remark. The following conditions are equivalent.

a) 𝕏 is in linearly general position, i.e. any subscheme 𝕐 ⊆ 𝕏 consisting of d + 1 points has Hilbert function H_{𝕐} : 1 d+1 d+1 · · ·.

b) 𝕏 is Δ_{𝕏}-uniform.

c) The multiplication map μ : R_{1} ⊗ℑ_{2} ⟶ ℑ_{3} is biinjective, i.e. μ(r ⊗φ) = 0 implies r = 0 or φ = 0.

For i = 1, …, s, let us write P_{i} = (1 : p_{i1} : …: p_{id}) with p_{ij} ∈ K. Because of 1.14 we know how the matrix of the above multiplication map μ depends on the coefficients of the matrix 𝔅 = (β_{ij}). In our situation, this matrix 𝔅 has a particularly simple description.

2.6. Lemma. If 𝕏 is in linearly general position, and if we choose the coordinate system suitably, the matrix 𝔅 = (β_{ij}) is given by

𝔅 = ( \begin{matrix}−1 + p_{11} + ⋯ + p_{1d} & −p_{11} & ⋯ & −p_{1d} \\ ⋮ & ⋮ & & ⋮ \\ −1 + p_{Δ1} + ⋯ + p_{Δd} & −p_{Δ1} & ⋯ & −p_{Δd} \end{matrix} )

Proof: Recall that we have numbered P_{1}, …, P_{s} such that {Lf_{1}, …, Lf_{Δ}} is a K-basis of \overline{R}_{σ+1}. Since 𝕏 is in linearly general position, P_{Δ+1}, …, P_{s} span ℙ^{d}. Thus we may change the coordinate system such that P_{Δ+1} = P_{s−d} = (1 : 0 : …: 0), P_{s−d+1} = (1 : 1 : 0 : …: 0), ..., P_{s} = (1 : 0 : …: 0 : 1). Both {x_{0} − x_{1} − ⋯ − x_{d}, x_{1}, …, x_{d}} and {g_{1}, …, g_{d+1}} are K-bases of R_{1} (cf. 1.2.b). They attain the same values (x_{0} − x_{1} − ⋯ − x_{d})(P_{s−d+i}) = δ_{i0} = g_{1}(P_{s−d+i}) and x_{j}(P_{s−d+i}) = δ_{ij} = g_{j+1}(P_{s−d+i}) for i = 0, …, d and j = 1, …, d at the points P_{s−d}, …, P_{s}. Since those points span ℙ^{d}, we have g_{1} = x_{0} − x_{1} − ⋯ − x_{d}, g_{2} = x_{1}, ..., g_{d+1} = x_{d}, and the claim follows from β_{ij} = −g_{i}(P_{j}) for i = 1, …, d + 1 and j = 1, …, Δ. □

The following notion will be used to explain the geometrical meaning of cohomological splitting in the present situation.

Definition. We say that a reduced 0-dimensional subscheme 𝕏 ⊆ ℙ^{d} splits linearly, if there exist linear subspaces L_{1}, L_{2} ⊆ ℙ^{d} such that 𝕏 ⊆ L_{1} ∪ L_{2} and L_{1} ∩ L_{2} = ∅.

For example, the scheme of 2.2 splits linearly. Like 2.5, the next proposition shows that several notions which are distinct in general, coincide in the case of “few” points. The equivalence of conditions a) and d) follows also from [2], 1.2 and 1.5.

2.7. Proposition. Let 𝕏 ⊆ ℙ^{d} be a nondegenerate, reduced, 0-dimensional subscheme consisting of d + 2 ≤ s ≤ \binom{d+2}{2} points with generic Hilbert function. Then the following conditions are equivalent.

a) 𝕏 does not split linearly.

b) 𝕏 is cohomologically uniform.

c) The multiplication map μ : R_{1} ⊗ ℑ_{2} ⟶ ℑ_{3} is nondegenerate and surjective.

d) 𝕏 is pure, i.e. its canonical ideal ℑ is generated by the elements of ℑ_{2} (cf. [11], section 4).

Proof: “a)⇒b)” The scheme 𝕏 has the Cayley-Bacharach property w.r.t hypersurfaces of degree σ_{𝕏} = 1, because a hyperplane L_{1} containing exactly s − 1 points of 𝕏 and the 0-dimensional linear space L_{2} consisting of the remaining point would constitute a linear splitting. Thus 𝕏 is 1-uniform.

Suppose 𝕏 splits cohomologically, and we have renumbered {P_{1}, …, P_{s}} such that the second part of 2.1.b holds. Then we have g_{j}(P_{i}) = 0 for j ∈ {1, …, ϵ} and i ∈ {δ+1, …, Δ}∪{Δ+ϵ+1, …, s}. Hence the linear subspace L_{1} := Z(g_{1}, …, g_{ϵ}) of ℙ^{d} contains {P_{δ+1}, …, P_{Δ}} ∪ {P_{Δ+ϵ+1}, …, P_{s}}. Analogously it follows that the linear subspace L_{2} = Z(g_{ϵ+1}, …, g_{s−Δ}) of ℙ^{d} contains {P_{1}, …, P_{δ}} ∪ {P_{Δ+1}, …, P_{Δ+ϵ}}. Altogether, L_{1} ∪ L_{2} contains 𝕏 and L_{1} ∩ L_{2} = Z(g_{1}, …, g_{s−Δ}) = Z(⊕_{n≥1}R_{n}) = ∅, contradicting a).

The equivalence “b)⇔c)” is a special case of 2.4. For “c)⇒d)” we note that ℑ_{3} does not contain a minimal generator of ℑ, that we have dim_{K} ℑ_{3} = s − 1, and that dim_{K} ℑ_{4} = s. In view of this and 1.11, it suffices to find one element r = x_{0}^{2}(c_{1}f_{1} + ⋯ + c_{s}f_{s}) in ℑ_{4} such that r ∈ R_{2}ℑ_{2} and c_{1} + ⋯ + c_{s} ≠ 0. For this we can take any nonzero element r′ = c′_{1}f_{1} + ⋯ + c′_{s}f_{s} ∈ ℑ_{2} such that c′_{i} ≠ 0 for some i ∈ {1, …, s} and multiply it by f_{i} ∈ R_{2}.

As “d)⇒c)” is trivially true, we are left with “b)⇒a)”. Suppose there are linear subspaces L_{1}, L_{2} ∈ ℙ^{d} such that 𝕏 ⊆ L_{1} ∪ L_{2} and L_{1} ∩ L_{2} = ∅. W.l.o.g. we may assume that dim L_{1} + dim L_{2} = d − 1. We let ϵ := d − dim L_{1} = dim L_{2} + 1 ∈ {1, …, d}. By a linear change of coordinates, we may also assume L_{1} = Z(X_{0}, …, X_{ϵ−1}) and L_{2} = Z(X_{ϵ}, …, X_{d}). Notice that here we may loose the property that x_{0} ∈ R_{1} is not a zerodivisor. If we denote the number of points of 𝕏 ∩ L_{1} by r ∈ {1, …, s − 1}, we can renumber {P_{1}, …, P_{s}} such that 𝕏 ∩ L_{1} = {P_{1}, …, P_{r}} and 𝕏 ∩ L_{2} = {P_{r+1}, …, P_{s}}. Furthermore, we choose a nonzerodivisor ℓ ∈ R_{1} of R, we construct the normalized separators f_{1}^{ℓ}, …, f_{s}^{ℓ} w.r.t ℓ, and for i=1,…,d we let x_{i} be the image of X_{i} in R_{1}.

Because of 𝕏 ⊆ L_{1} ∪ L_{2}, we have x_{i}x_{j} = 0 for i ∈ {0, …, ϵ − 1} and j ∈ {ϵ, …, d}. Thus f_{i}^{ℓ} has a decomposition f_{i}^{ℓ} = f′_{i} + f′′_{i} such that f′_{i} ∈ K[x_{0}, …, x_{ϵ−1}] ⊆ R and f′′_{i} ∈ K[x_{ϵ}, …, x_{d}] ⊆ R. For every i ∈ {1, …, s}, the definitions of L_{1} and L_{2} imply f′_{i}(P_{j}) = 0 for j ∈ {1, …, r} and f′′_{i}(P_{j}) = 0 for j ∈ {r + 1, …, s}. Since also f′_{i}(P_{j}) = f_{i}^{ℓ}(P_{j}) = 0 for i = 1, …, r and j = r + 1, …, s, we obtain f′_{i} = 0 for i = 1, …, r. Analogously, we have f′′_{i} = 0 for i = r+1, …, s. Altogether we have shown that f_{i}^{ℓ} ∈ K[x_{ϵ}, …, x_{d}] for i = 1, …, r and f_{i}^{ℓ} ∈ K[x_{0}, …, x_{ϵ−1}] for i = r + 1, …, s.

Now we can conclude that ℓ·(Kx_{ϵ}+⋯+Kx_{d}) ⊆ Kf_{1}^{ℓ}⊕⋯⊕Kf_{r}^{ℓ} and ℓ·(Kx_{0}+⋯+ Kx_{ϵ−1}) ⊆ Kf_{r+1}^{ℓ} ⊕ ⋯ ⊕ Kf_{s}^{ℓ}. Hence c_{1}f_{1}^{ℓ} + ⋯ + c_{s}f_{s}^{ℓ} ∈ ℓ · R_{1} for some c_{1}, …, c_{s} ∈ K implies c_{1}f_{1}^{ℓ} + ⋯ + c_{r}f_{r}^{ℓ} ∈ ℓR_{1} and c_{r+1}f_{r+1}^{ℓ} + ⋯ + c_{s}f_{s}^{ℓ} ∈ ℓR_{1}. Thus we find the splitting (R/(ℓ))_{2} = ∑_{i=1}^{r} K · Lf_{i}^{ℓ} ⊕ ∑_{i=r+1}^{s} K · Lf_{i}^{ℓ}, i.e. 𝕏 splits cohomologically.

□

We end this section with an example which shows that not every cohomologically split scheme splits linearly. The second part of this example also demonstrates that cohomological splitting or uniformity may depend on quite subtle geometrical properties of the configuration of the points of 𝕏. This example suggests us to ask if for every cohomologically split scheme 𝕏 = 𝕐 ∪ 𝕐′ there exist disjoint varieties V_{1}, V_{2} ⊆ ℙ^{d} of “small” degree such that 𝕐 ⊆ V_{1} and 𝕐′ ⊆ V_{2}.

2.8. Example. a) Let C_{1}, C_{2} ⊆ ℙ_{ℂ}^{3} be two skew twisted cubics, and let 𝕏 consist of 20 points on C_{1} and 20 points on C_{2}. For instance, we can take C_{1} = {(u^{3} : u^{2}v : uv^{2} : v^{3}) | (u : v) ∈ ℙ_{ℂ}^{1}} and C_{2} = {(2uv^{2} : u^{3} : v^{3} : u^{2}v) | (u : v) ∈ ℙ_{ℂ}^{1}}, and then choose the points corresponding to (u : v) = (1 : i) for i = 1, …, 20 on both C_{1} and C_{2}. The Hilbert function of 𝕏 is H_{𝕏} : 1 4 10 20 26 32 38 40 40 ⋯, so that σ_{𝕏} = 6 and Δ_{𝕏} = 2. The matrix 𝔅 is of the form \binom{∗···∗ 0···0}{0···0 ∗···∗}, and therefore 𝕏 splits in the form 𝕏 = (𝕏 ∩ C_{1}) ∪ (𝕏 ∩ C_{2}). Clearly 𝕏 does not split linearly.

We note the inequality H_{𝕏}(3)+H_{𝕏}(σ_{𝕏} −3) = 40 > 39 = s−Δ_{𝕏} +1, a phenomenon which will be explored more deeply in the next section.

b) If we replace C_{2} by C_{3} = {(u^{3} : uv^{2} : v^{3} : 2u^{2}v) | (u : v) ∈ ℙ_{ℂ}^{1}}, then C_{1} ∩ C_{3} = {(1 : 0 : 0 : 0)}, and the analogously defined scheme 𝕏 has H_{𝕏} : 1 4 10 19 25 31 37 40 · · ·. Since its matrix 𝔅 does not split and has nonzero columns, 𝕏 is cohomologically uniform. We note that the inequalities H_{𝕏}(n) + H_{𝕏}(σ_{𝕏} − n) ≤ s − Δ_{𝕏} + 1 hold in this case for n = 0, …, σ_{𝕏}.

The calculations for this example were done using the program COP for computations with zerodimensional schemes (cf. [1]).

3. Hilbert Functions of Cohomologically Uniform Schemes

As in the previous sections, we let 𝕏 = {P_{1}, …, P_{s}} ⊆ ℙ^{d} be a set of points with canonical ideal ℑ ⊆ R. We have seen that cohomological uniformity of 𝕏 is an intermediate property between 1-unformity and Δ_{𝕏}-uniformity. In [11] it was shown that Δ_{𝕏}-uniformity implies certain inequalities for the Hilbert function of 𝕏 which do not hold in general for 1-uniform schemes (cf. [6], 3.10 and 3.11). Our goal in this section is to confirm the view expressed in [11], section 3, that cohomological uniformity is the correct condition which implies those inequalities.

3.1. Theorem. Let 𝕏 = {P_{1}, …, P_{s}} ⊆ ℙ^{d} be a cohomologically uniform set of points, and suppose that char (K) ∉ {2, …, σ_{𝕏}}. Then we have

H_{𝕏}(n) + H_{𝕏}(σ_{𝕏} − n) ≤ s − Δ_{𝕏} + 1

for all n ∈ {0, …, σ_{𝕏}}.

Notice that if 𝕏 consists of only one point, then σ_{𝕏} = −1 and the theorem does not claim anything. Similarly, the theorem is trivial in case σ_{𝕏} = 0. Therefore we shall assume s ≥ 3 and σ_{𝕏} ≥ 1 for the rest of this section. For proof of Theorem 3.1 we can assume that 𝕏 ⊆ ℙ^{d} is nondegenerate, since the Hilbert function of 𝕏 does not change, if we replace ℙ^{d} by the linear span of 𝕏. This proof relies on a detailed understanding of how those inequalities result from structural properties of the canonical ideal of 𝕏. The next proposition and its corollary provide us with the basic link.

3.2. Proposition. Let r ∈ R_{σ} and φ ∈ ℑ_{σ+1}. The following conditions are equivalent. a) There exist elements φ_{2}, …, φ_{Δ} ∈ ℑ_{σ+1} such that

ℑ_{2σ+1} = R_{σ}φ ⊕ Krφ_{2} ⊕ ⋯ ⊕ Krφ_{Δ}.

b) If φ_{2}, …, φ_{Δ} ∈ ℑ_{σ+1} are such that {φ, φ_{2}, …, φ_{Δ}} is a K-basis of ℑ_{σ+1}, then

ℑ_{2σ+1} = R_{σ}φ ⊕ Krφ_{2} ⊕ ⋯ ⊕ Krφ_{Δ}.

c) We have φ ≠ 0, and if r′ ∈ R_{σ}, φ′ ∈ ℑ_{σ+1} are such that r′φ = rφ′, then r′ = λr and φ′ = λφ for some λ ∈ K.

Proof: First we show that a) implies c). From s − 1 = dim_{K} ℑ_{2σ+1} > Δ_{𝕏} − 1 ≥ dim_{K}(Krφ_{2} ⊕ ⋯ ⊕ Krφ_{Δ}) we conclude that φ ≠ 0. Now suppose that rφ′ = r′φ for some r′ ∈ R_{σ}, φ′ ∈ ℑ_{σ+1}. From the hypothesis it is clear that {φ, φ_{2}, …, φ_{Δ}} are K-linearly independent, and hence form a K-basis of ℑ_{σ+1}. Thus we can write φ′ = λ_{1}φ+λ_{2}φ_{2}+⋯+λ_{Δ}φ_{Δ} with λ_{1}, …, λ_{Δ} ∈ K, and we obtain r(λ_{2}φ_{2}+⋯+λ_{Δ}φ_{Δ}) = (r′ − λ_{1}r)φ ∈ R_{σ}φ ∩ (Krφ_{2} ⊕ ⋯ ⊕ Krφ_{Δ}) = (0). Therefore λ_{2} = ⋯ = λ_{Δ} = 0 and φ′ = λ_{1}φ. Since the hypothesis also implies dim_{K}(R_{σ}φ) = s − Δ, we get r′ − λ_{1}r = 0, i.e. r′ = λ_{1}r.

As “b)⇒a)” is trivially true, we are left with proving “c)⇒b)”. We claim that dim_{K}(R_{σ}φ) = s − Δ, i.e. that no element r′ ∈ R_{σ} \ {0} annihilates φ. Otherwise we could choose φ′ = 0, and r′φ = 0 = rφ′ would imply φ′ = λφ, r′ = λr for λ = 0, contradicting r′ ≠ 0. We also claim that we may assume r ≠ 0. Otherwise r′φ = 0 ⇒ r′ = 0 and [10], 2.6, imply that 𝕏 is 1-uniform with Δ_{𝕏} = 1, so that the claim is obviously true.

Suppose now that φ′ = λ_{2}φ_{2} + ⋯ + λ_{Δ}φ_{Δ} with λ_{2}, …, λ_{Δ} ∈ K satisfies rφ′ = 0. Then rφ′ = r′φ for r′ = 0, and thus r′ = λr, φ′ = λφ with λ = 0, i.e. we have φ′ = 0. This shows dim_{K}(Krφ_{2} ⊕ ⋯ ⊕ Krφ_{Δ}) = Δ_{𝕏} − 1. Finally, it is clear from the hypothesis that (R_{σ}φ) ∩ (Krφ_{2} ⊕ ⋯ ⊕ Krφ_{Δ}) = (0), and the claim follows by adding up dimensions. □

Note that if 𝕏 is 1-uniform and Δ_{𝕏} = 1, then condition c) holds for r = 0 and every φ ∈ ℑ_{σ+1} \ {0}. The existance of r ∈ R_{σ} and φ ∈ ℑ_{σ+1} such that 3.2.a)-c) hold has strong consequences.

3.3. Corollary. a) If there exists an element ℓ ∈ R_{1} such that the equivalent conditions of 3.2 are satisfied for r = ℓ^{σ} and some φ ∈ ℑ_{σ+1}, then

H_{𝕏}(n) + H_{𝕏}(σ_{𝕏} − n) ≤ s − Δ_{𝕏} + 1

for all n ∈ {0, …, σ_{𝕏}}.

b) If there exist r ∈ R_{σ} and φ ∈ ℑ_{σ+1} such that the equivalent conditions of 3.2 are satisfied, then 𝕏 is cohomologically uniform.

Proof: To see a), we choose φ_{2}, …, φ_{Δ} ∈ ℑ_{σ+1} as in 3.2.b. Like in the proof of 3.2, we find Ann_{R}(φ) ∩ R_{σ} = (0). Then we observe that ℓ^{n}(λ_{2}φ_{2} + ⋯ + λ_{Δ}φ_{Δ}) ∈ R_{n}φ for n ∈ {0, …, σ} and λ_{2}, …, λ_{Δ} ∈ K implies ℓ^{σ}(λ_{2}φ_{2} + ⋯ + λ_{Δ}φ_{Δ}) ∈ R_{σ}φ, and therefore λ_{2} = ⋯ = λ_{Δ} = 0. Thus s − H_{𝕏}(σ_{𝕏} − n) = dim_{K} ℑ_{σ+1+n} ≥ dim_{K}(R_{n}φ) + Δ_{𝕏} − 1 = H_{𝕏}(n) + Δ_{𝕏} − 1.

Now we show claim b). From Ann_{R}(φ) ∩ R_{σ} = (0) and [10], 2.6, it follows that 𝕏 is 1-uniform. A look at Theorem 2.4 finishes the proof. □

This corollary reduces the proof of Theorem 3.1 to showing that for every cohomologically uniform scheme 𝕏 there exist ℓ ∈ R_{1} and φ ∈ ℑ_{σ+1} such that 3.2.a)-c) hold for r = ℓ^{σ} and φ. This question will now be reduced to a linear algebra problem.

Recall that we have explicit bases {g_{1}, …, g_{s−Δ}} of R_{σ} (cf. 1.2.b) and {π_{1}, …, π_{Δ}} of ℑ_{σ+1} (cf. 1.10).

3.4. Proposition. Let r ∈ R_{σ} and φ ∈ ℑ_{σ+1}. Write r = a_{1}g_{1} + ⋯ + a_{s−Δ}g_{s−Δ} with a_{1}, …, a_{s−Δ} ∈ K and φ = c_{1}π_{1} + ⋯ + c_{Δ}π_{Δ} with c_{1}, …, c_{Δ} ∈ K. Let b_{i} = a_{1}β_{1i}+⋯+a_{s−Δ}β_{s−Δ i} for i = 1, …, Δ and d_{j} = c_{1}β_{j1}+⋯+c_{Δ}β_{jΔ} for j = 1, …, s−Δ.

Define the matrix

M(r, φ) = ( \begin{matrix}−b_{1} & & 0 & c_{1}β_{11} & ⋯ & c_{1}β_{s−Δ 1} \\ & ... & & ⋮ & & ⋮ \\ 0 & & −b_{Δ} & c_{Δ}β_{1Δ} & ⋯ & c_{Δ}β_{s−Δ Δ} \\ a_{1}β_{11} & ⋯ & a_{1}β_{1Δ} & −d_{1} & & 0 \\ ⋮ & & ⋮ & & ... \\ a_{s−Δ}β_{s−Δ 1} & ⋯ & a_{s−Δ}β_{s−Δ Δ} & 0 & & −d_{s−Δ} \end{matrix} )

Then conditions 3.2.a)-c) are equivalent with rk M(r, φ) = s − 1.

Proof: Using 1.14, we calculate rπ_{i} = a_{1}g_{1}π_{i} + ⋯ + a_{s−Δ}g_{s−Δ}π_{i} = a_{1}x_{0}^{σ}(β_{1i}f_{Δ+1} − β_{1i}f_{i})+⋯+a_{s−Δ}x_{0}^{σ}(β_{s−Δ i}f_{s} −β_{s−Δ i}f_{i}) = x_{0}^{σ}(−b_{i}f_{i} +a_{1}β_{1i}f_{Δ+1} +⋯+a_{s−Δ}β_{s−Δ i}f_{s}) for i = 1, …, Δ and g_{j}φ = c_{1}g_{j}π_{1} + ⋯ + c_{Δ}g_{j}π_{Δ} = c_{1}x_{0}^{σ}(β_{j1}f_{Δ+j} − β_{j1}f_{1}) + ⋯ + c_{Δ}x_{0}^{σ}(β_{jΔ}f_{Δ+j} −β_{jΔ}f_{Δ}) = x_{0}^{σ}(−c_{1}β_{j1}f_{1} −⋯−c_{Δ}β_{jΔ}f_{Δ} +d_{j}f_{Δ+j}) for j = 1, …, s−Δ. Therefore the columns of the matrix

( \begin{matrix}T_{1}^{2σ+1} & & & & 0 \\ & . & . & . \\ 0 & & & & T_{s}^{2σ+1} \end{matrix} ) · M(r, φ)

represent the images of rπ_{1}, …, rπ_{Δ}, −g_{1}φ, …, −g_{s−Δ}φ under the canonical injection ı : R ⟶ R̃ ≅ K[T_{1}] × ⋯ × K[T_{s}] of section 1. Thus rk M(r, φ) = s − 1 means that rℑ_{σ+1} + R_{σ}φ is (s−1)-dimensional. This is equivalent to dim_{K}((rℑ_{σ+1}) ∩ (R_{σ}φ)) = 1 and hence to conditions 3.2.a)-c). □

3.5. Proposition. Let r = a_{1}g_{1} + ⋯ + a_{s−Δ}g_{s−Δ} ∈ R_{σ} with a_{1}, …, a_{s−Δ} ∈ K. The following conditions are equivalent.

a) There exists an element φ ∈ ℑ_{σ+1} such that conditions 3.2.a)-c) hold.

b) The matrix

𝔐(r) = ( \begin{matrix}−b_{1} & & 0 & a_{1}β_{11} & ⋯ & a_{s−Δ}β_{s−Δ 1} \\ & ... & & ⋮ & & ⋮ \\ 0 & & −b_{Δ} & a_{1}β_{1Δ} & ⋯ & a_{s−Δ}β_{s−Δ Δ} \\ β_{11}Y_{1} & ⋯ & β_{1Δ}Y_{Δ} & −D_{1} & & 0 \\ ⋮ & & ⋮ & & ... \\ β_{s−Δ 1}Y_{1} & ⋯ & β_{s−Δ Δ}Y_{Δ} & 0 & & −D_{s−Δ} \end{matrix} )

has rank s − 1.

Here b_{i} = a_{1}β_{1i} + ⋯ + a_{s−Δ}β_{s−Δ i} for i = 1, …, Δ, the symbols Y_{1}, …, Y_{Δ} denote independent variables over K, and D_{j} = β_{j1}Y_{1} + ⋯ + β_{jΔ}Y_{Δ} for j = 1, …, s − Δ.

Proof: This follows from the preceding proposition, since 𝔐(r) has rank s − 1 if and only if one of its specializations M(r, φ)^{transp.} has rank s − 1. □

Now we can prove Theorem 3.1 in some special cases.

3.6. Corollary. a) If in the matrix 𝔅 of 𝕏 there is a row (β_{1i}, …, β_{s−Δ i}), all of whose entries are nonzero, and if ℓ ∈ R_{1} is a nonzerodivisor, then rk 𝔐(ℓ^{σ}) = s − 1 and the Hilbert function of 𝕏 satisfies the inequalities of 3.1.

b) If 𝕏 is Δ_{𝕏}-uniform, then rk 𝔐(ℓ^{σ}) = s − 1 for all nonzerodivisors ℓ ∈ R_{1}.

Proof: To show a), we may assume that β_{11} ≠ 0, ..., β_{s−Δ 1} ≠ 0. Then we cancel the first row and the first column of 𝔐(ℓ^{σ}). The coefficient of Y_{1}^{s−Δ} in the determinant of the resulting matrix is (−ℓ(P_{Δ+1})^{σ}β_{11}) ⋯ (−ℓ(P_{s})^{σ}β_{s−Δ 1}) ≠ 0. For the proof of claim b) we recall that g_{j}π_{i} = β_{ji}(h_{Δ+j−1} − δ_{i1}h_{i−1}) by 1.14. Thus [10], 3.2 implies that all entries of 𝔅 are nonzero and we can apply a). □

In view of 3.6.a it is natural to ask if rk 𝔐(ℓ^{σ}) = s − 1 holds for all nonzerodivisors ℓ ∈ R_{1}. Our next example demonstrates that this is not the case and that we can only hope to show rk 𝔐(ℓ^{σ}) = s − 1 for generic ℓ ∈ R_{1}.

3.7. Example. Let 𝕏 ⊆ ℙ^{3} consist of the following six points: P_{1} = (1 : 1 : 1 : 0), P_{2} = (1 : 2 : 0 : 1), P_{3} = (1 : 0 : 0 : 0), P_{4} = (1 : 1 : 0 : 0), P_{5} = (1 : 0 : 1 : 0), and P_{6} = (1 : 0 : 0 : 1). We calculate H_{𝕏} : 1 4 6 6 ⋯, so that σ_{𝕏} = 1 and Δ_{𝕏} = 2. We also calculate that the matrix 𝔅 of 𝕏 is given by

𝔅 = ( \begin{matrix}1 & −1 & −1 & 0 \\ 2 & −2 & 0 & −1 \end{matrix} )

Since the first two columns have nonzero entries, and since no column is zero, we know from 2.1 that 𝕏 is cohomologically uniform. But we may check that

rk 𝔐(x_{0}^{σ}) = rk ( \begin{matrix}1 & 0 & 1 & −1 & −1 & 0 \\ 0 & 1 & 2 & −2 & 0 & −1 \\ Y_{1} & 2Y_{2} & −Y_{1} − 2Y_{2} & 0 & 0 & 0 \\ −Y_{1} & −2Y_{2} & 0 & Y_{1} + 2Y_{2} & 0 & 0 \\ −Y_{1} & 0 & 0 & 0 & Y_{1} & 0 \\ 0 & −Y_{2} & 0 & 0 & 0 & Y_{2} \end{matrix} ) = 4 = s − 2.

Next we shall show that the converse of Corollary 3.3.b holds. This will allow us to prove Theorem 3.1 in even more cases.

In the sequel Y_{1}, …, Y_{Δ} and Z_{1}, …, Z_{s−Δ} will denote sets of independent variables over K, and we let B_{i} = β_{1i}Z_{1} + ⋯ + β_{s−Δ i}Z_{s−Δ} for i = 1, …, Δ as well as D_{j} = β_{j1}Y_{1} + ⋯ + β_{jΔ}Y_{Δ} for j = 1, …, s − Δ.

3.8. Lemma. Let 1 ≤ u < Δ_{𝕏}, let 1 ≤ v < s − Δ_{𝕏}, and let M_{u,v} be the matrix

M_{u,v} = ( \begin{matrix}−B_{1} & & 0 & β_{11}Z_{1} & ⋯ & β_{v1}Z_{v} \\ & ... & & ⋮ & & ⋮ \\ 0 & & −B_{u} & β_{1u}Z_{1} & ⋯ & β_{vu}Z_{v} \\ β_{11}Y_{1} & ⋯ & β_{1u}Y_{u} & −D_{1} & & 0 \\ ⋮ & & ⋮ & & ... \\ β_{v1}Y_{1} & ⋯ & β_{vu}Y_{u} & 0 & & −D_{v} \end{matrix} )

Then det M_{u,v} = 0 implies that 𝕏 splits cohomologically.

Proof: We proceed by induction on u + v. In case u + v = 2, we are considering the matrix M_{1,1} = ( \begin{matrix}−B_{1} & β_{11}Z_{1} \\ β_{11}Y_{1} & −D_{1} \end{matrix} ). Since det M_{1,1} = 0, none of the variables Y_{2}, …, Y_{Δ}, Z_{2}, …, Z_{s−Δ} may occur in B_{1} or D_{1}. Hence β_{12} = … = β_{1Δ} = 0 and β_{21} = … = β_{s−Δ 1} = 0, i.e. 𝔅 splits.

In order to prove the induction step, we grade the polynomial ring K[Y_{1}, …, Y_{Δ}, Z_{1}, …, Z_{s−Δ}] by deg Y_{1} = … = deg Y_{Δ} = deg Z_{1} = … = deg Z_{v} = 0 and deg Z_{v+1} = … = deg Z_{s−Δ} = 1. We expand det M_{u,v} and look at its leading form. Note that Z_{v+1}, …, Z_{s−Δ} only occur in B_{1}, …, B_{u}, and renumber P_{1}, …, P_{u} such that {1, …, u′} with 1 ≤ u′ ≤ u are precisely those indices i among {1, …, u} for which β_{v+1 i} = … = β_{s−Δ i} = 0.

Here we must have u′ ≥ 1, since u′ = 0 would imply that the leading form of det M_{u,v} is ±(β_{v+1 1}Z_{v+1} + ⋯ + β_{s−Δ 1}Z_{s−Δ}) ⋯ (β_{v+1 u}Z_{v+1} + ⋯ + β_{s−Δ u}Z_{s−Δ}) · D_{1} ⋯ D_{v} ≠ 0, contradicting det M_{u,v} = 0. If u′ ≥ 1, the leading form of det M_{u,v} is ±(β_{v+1 u′+1}Z_{v+1} + ⋯ + β_{s−Δ u′+1}Z_{s−Δ}) ⋯ (β_{v+1 u}Z_{v+1} + ⋯ + β_{s−Δ u}Z_{s−Δ}) · det M_{u′,v}. Thus if u′ < u, the claim follows from the induction hypothesis.

We are left with the case u′ = u, i.e. with the case that β_{v+1 i} = … = β_{s−Δ i} = 0 for i = 1, …, u. In this case we grade the polynomial ring K[Y_{1}, …, Y_{Δ}, Z_{1}, …, Z_{s−Δ}] by deg Y_{1} = … = deg Y_{u} = deg Z_{1} = … = deg Z_{s−Δ} = 0 and deg Y_{u+1} = … = deg Y_{Δ} = 1. Again we consider the leading form of det M_{u,v} w.r.t. this grading. Note that Y_{u+1}, …, Y_{Δ} only occur in D_{1}, …, D_{v}, and renumber P_{Δ+1}, …, P_{Δ+v} in such a way that {1, …, v′} with 1 ≤ v′ ≤ v are precisely those indices j among {1, …, v} for which β_{j u+1} = … = β_{jΔ} = 0.

Here we have v′ ≥ 1, since v′ = 0 would imply that the leading form of det M_{u,v} is ±(β_{1 u+1}Y_{u+1}+⋯+β_{1Δ}Y_{Δ}) ⋯ (β_{v u+1}Y_{u+1}+⋯+β_{vΔ}Y_{Δ})·B_{1} ⋯ B_{u} ≠ 0, contradicting det M_{u,v} = 0. If v′ ≥ 1, the leading form of det M_{u,v} is ±(β_{v′+1 u+1}Y_{u+1} + ⋯ + β_{v′+1 Δ}Y_{Δ}) ⋯ (β_{v u+1}Y_{u+1} + ⋯ + β_{vΔ}Y_{Δ}) · det M_{u,v′}. Thus if v′ < v, the claim follows from the induction hypothesis.

Finally we are left with the case u′ = u and v′ = v, i.e. with the case that β_{v+1 i} = … = β_{s−Δ i} = 0 for i = 1, …, u and β_{j u+1} = … = β_{jΔ} = 0 for j = 1, …, v. Obviously, in this case the matrix 𝔅 of 𝕏 splits, too. □

3.9. Proposition. Let 𝕏 ⊆ ℙ^{d} be a reduced, 0-dimensional subscheme. The following conditions are equivalent.

a) 𝕏 is cohomologically uniform.

b) There are elements r ∈ R_{σ}, φ ∈ ℑ_{σ+1} such that conditions 3.2.a)-c) hold.

Proof: Because of 3.3.b, it suffices to show “a)⇒b)”. By 3.5, we have to find an element r = a_{1}g_{1} + ⋯ + a_{s−Δ}g_{s−Δ} ∈ R_{σ} such that rk 𝔐(r) = s − 1. Thus we have to show that the matrix

𝔐 = ( \begin{matrix}−B_{1} & & 0 & β_{11}Z_{1} & ⋯ & β_{s−Δ 1}Z_{s−Δ} \\ & ... & & ⋮ & & ⋮ \\ 0 & & −B_{Δ} & β_{1Δ}Z_{1} & ⋯ & β_{s−Δ Δ}Z_{s−Δ} \\ β_{11}Y_{1} & ⋯ & β_{1Δ}Y_{Δ} & −D_{1} & & 0 \\ ⋮ & & ⋮ & & ... \\ β_{s−Δ 1}Y_{1} & ⋯ & β_{s−Δ Δ}Y_{Δ} & 0 & & −D_{s−Δ} \end{matrix} )

has rank s − 1. Let 𝔐′ be the matrix which is obtained from 𝔐 by deleting the last row and the last column. We shall show that 𝕏 splits cohomologically, if det 𝔐′ = 0.

Notice that Z_{s−Δ} appears only in the entries −B_{1}, …, −B_{Δ} of 𝔐′. W.l.o.g. let u ∈ {1, …, Δ} be such that β_{s−Δ 1} = … = β_{s−Δ u} = 0 and β_{s−Δ u+1} ≠ 0, ..., β_{s−Δ Δ} ≠ 0. Here u = Δ is impossible because of Prop. 2.1. The leading coefficient of det 𝔐′ w.r.t. Z_{s−Δ} is (−β_{s−Δ u+1}) ⋯ (−β_{s−Δ Δ}) · det M_{u,s−Δ−1}. Hence det M_{u,s−Δ−1} = 0, and an application of the lemma finishes the proof. □

Thus we have found yet another characterization of cohomological uniformity. It allows us to determine the generic rank of 𝔐(ℓ^{σ}) in case σ_{𝕏} = 1. Here and later we use the phrase “for generic ℓ ∈ R_{1}” to mean that there exists a Zariski-open subset U of the affine space 𝔸(R_{1}) such that the claimed property holds whenever ℓ corresponds to a closed point of U.

3.10. Corollary. Let X ⊆ ℙ^{d} be a reduced 0-dimensional subscheme with σ_{𝕏} = 1. For generic ℓ ∈ R_{1} we have rk 𝔐(ℓ) = s − 1.

Proof: It suffices to apply the proposition and to note that the condition rk 𝔐(ℓ) = s−1 defines an open subset of 𝔸(R_{1}). □

Our next lemma and proposition constitute the heart of the proof of Theorem 3.1.

3.11. Lemma. Let 𝕏 ⊆ ℙ^{d} be a nondegenerate, reduced, 0-dimensional subscheme with σ_{𝕏} ≥ 2, suppose that char (K) ∉ {2, …, σ_{𝕏}}, and suppose there are i ∈ {1, …, Δ}, j ∈ {1, …, s − Δ}, and pairwise distinct elements ν_{1}, …, ν_{j} ∈ {1, …, s − Δ} such that ℓ(P_{Δ+ν_{1}})^{σ}β_{ν_{1} i} + ⋯ + ℓ(P_{Δ+ν_{j}})^{σ}β_{ν_{j} i} = 0 for generic ℓ ∈ R_{1}. Then we have β_{ν_{1} i} = … = β_{ν_{j} i} = 0.

Proof: Let ℙ^{N} = ℙ(A_{σ}) with N = \binom{d+σ}{d} be the projective space associated with A_{σ} = K[X_{0}, …, X_{d}]_{σ}. If we use the lexicographically ordered set of monomials {X_{0}^{σ}, X_{0}^{σ−1}X_{1}, …, X_{d}^{σ}} of degree σ as a K-basis of A_{σ}, and if L = λ_{0}X_{0} + ⋯ + λ_{d}X_{d} ∈ A_{1} with λ_{0}, …, λ_{d} ∈ K is a linear form, then L^{σ} ∈ A_{σ} is given in this basis by the coordinate tuple (λ_{0}^{σ}, λ_{0}^{σ−1}λ_{1}, …, λ_{d}^{σ}), where the coefficient of λ_{0}^{ν_{0}} ⋯ λ_{d}^{ν_{d}} is the multinomial coefficient \binom{σ}{ν_{0},...,ν_{d}}. Since char (K) ∉ \{2, …, σ\}, all of those coefficients are nonzero. In the lexicographically ordered basis \{\binom{σ}{ν_{0},...,ν_{d}}X_{0}^{ν_{0}} ⋯ X_{d}^{ν_{d}} | ν_{0} + ⋯ + ν_{d} = σ\} of A_{σ}, the element L^{σ} is then given by the coordinate tuple (λ_{0}^{σ}, λ_{0}^{σ−1}λ_{1}, …, λ_{d}^{σ}). The set of all elements of the form L^{σ} such that L ∈ A_{1} is therefore precisely the set of closed points of a Veronese variety V, namely of the σ^{th} Veronese embedding of ℙ^{d} in ℙ^{N}.

Now we consider the linear subspace Λ = ℙ((I_{𝕏})_{σ}) of ℙ^{N}. It has codimension codim(Λ, ℙ^{N}) = s − Δ. Since I_{𝕏} is a radical ideal, it follows from L^{σ} ∈ (I_{𝕏})_{σ} for some L ∈ A_{1} that L ∈ (I_{𝕏})_{1}, i.e. that L = 0, since 𝕏 is nondegenerate. Altogether we conclude Λ ∩ V = ∅. Next we let π : ℙ^{N} \ Λ ⟶ ℙ^{s−Δ−1} = ℙ(R_{σ}) be the morphism given by ⟨F ⟩ ⟼ ⟨F + I_{𝕏}⟩ for F ∈ A_{σ} \ (I_{𝕏})_{σ}, i.e. we let π be the projection from ℙ^{N} with center Λ. Since Λ ∩ V = ∅, the image \overline{V} = π(V) ⊆ ℙ^{s−Δ−1} is a closed subscheme (cf. [4], 14.2), and since V is well-known to be irreducible and nondegenerate, the same is true for \overline{V}. Thus the set of all closed points of ℙ(R_{σ}) of the form ℓ^{σ} with ℓ ∈ R_{1} is not contained in a hyperplane of ℙ(R_{σ}).

If we look at the coordinate system of ℙ(R_{σ}) which corresponds to the K-basis {g_{1}, …, g_{s−Δ}} of R_{σ}, we can use the equation ℓ^{σ} = ℓ(P_{Δ+1})^{σ}g_{1} + ⋯ + ℓ(P_{s})^{σ}g_{s−Δ} for ℓ ∈ R_{1} to conclude that the closed point ⟨ℓ^{σ}⟩ ∈ ℙ(R_{σ}) is in that coordinate system given by the tuple (ℓ(P_{Δ+1})^{σ} : …: ℓ(P_{s})^{σ}). By the hypothesis, it is for generically chosen ℓ ∈ R_{1} contained in the variety H = Z(β_{ν_{1} i}X_{ν_{1}} + ⋯ + β_{ν_{j} i}X_{ν_{j}}) ⊆ ℙ(R_{σ}). Since \overline{V} is nondegenerate, the variety H cannot be a hyperplane, i.e. we have to have H = ℙ(R_{σ}) and β_{ν_{1} i} = ⋯ = β_{ν_{j} i} = 0. □

3.12. Proposition. If 𝕏 ⊆ ℙ^{d} is a nondegenerate, reduced, 0-dimensional, cohomologically uniform scheme, and if char (K) ∉ {2, …, σ_{𝕏}}, then rk 𝔐(ℓ^{σ}) = s−1 for generic ℓ ∈ R_{1}.

Proof: Because of 3.10 it suffices to consider the case σ_{𝕏} ≥ 2. We renumber the points P_{Δ+1}, …, P_{s} such that β_{11} ≠ 0, ..., β_{t_{1}1} ≠ 0, β_{t_{1}+1 1} = … = β_{s−Δ 1} = 0 for some 1 ≤ t_{1} ≤ s − Δ. Note that t_{1} ≥ 1, since β_{11} + ⋯ + β_{s−Δ 1} = −1 (cf. 1.2.a). Next we renumber P_{2}, …, P_{Δ} such that one of β_{12}, …, β_{t_{1}2} is not zero. This is possible, because 𝔅 does not split. Then we renumber P_{Δ+t_{1}+1}, …, P_{s} such that β_{t_{1}+1 2} ≠ 0, ..., β_{t_{2}2} ≠ 0, β_{t_{2}+1 2} = … = β_{s−Δ 2} = 0 for some t_{1} ≤ t_{2} ≤ s − Δ, and we renumber P_{3}, …, P_{Δ} such that one of β_{13}, …, β_{t_{2}3} is not zero. Continuing this way, we find numbers 1 ≤ t_{1} ≤ t_{2} ≤ ⋯ ≤ t_{Δ−1} ≤ s−Δ such that the matrix 𝔅 of 𝕏 looks as follows:

𝔅=( \begin{matrix}β_{11} & ⋯ & β_{t_{1}1} & 0 & ⋯ & 0 & 0 & ⋯ & 0 & ⋯ & 0 & ⋯ & 0 \\ β_{12} & ⋯ & β_{t_{1}2} & β_{t_{1}+1 2} & ⋯ & β_{t_{2}2} & 0 & ⋯ & 0 & ⋯ & 0 & ⋯ & 0 \\ ⋮ & & ⋮ & ⋮ & & ⋮ & β_{t_{2}+1 3} & ⋯ & β_{t_{3}3} & ⋯ & ⋮ & & ⋮ \\ ⋮ & & ⋮ & ⋮ & & ⋮ & ⋮ & & ⋮ & & 0 & ⋯ & 0 \\ β_{1Δ} & ⋯ & β_{t_{1}Δ} & β_{t_{1}+1 Δ} & ⋯ & β_{t_{2}Δ} & β_{t_{2}+1 Δ} & ⋯ & β_{t_{3}Δ} & ⋯ & β_{t_{Δ−1}+1 Δ} & ⋯ & β_{s−Δ Δ} \end{matrix} )

Notice that t_{Δ} = s − Δ, because no column of 𝔅 is zero by 2.1. For j = 1, …, s − Δ we let λ_{j} = ℓ(P_{Δ+j})^{σ}. In addition, we have arranged this matrix such that γ_{2} = λ_{1}β_{12} + ⋯ + λ_{t_{1}}β_{t_{1}2} ≠ 0, ..., γ_{Δ} = λ_{1}β_{1Δ} + ⋯ + λ_{t_{Δ−1}}β_{t_{Δ−1}Δ} ≠ 0 for generic ℓ ∈ R_{1}.

In the matrix 𝔐(ℓ^{σ}) we add columns 2, …, Δ + t_{1} to column 1 and obtain the following matrix 𝔐_{0} =

0  0  ⋯  0  λ_{1}β_{11} ⋯ λ_{t_{1}}β_{t_{1}1}  0  ⋯  0
δ_{2}  −b_{2}  0  λ_{1}β_{12} ⋯ λ_{t_{1}}β_{t_{1}2} λ_{t_{1}+1}β_{t_{1}+1 2} ⋯ λ_{s−Δ}β_{s−Δ 2}
⋮  ...  ⋮  ⋮  ⋮  ⋮
δ_{Δ}  0  −b_{Δ}  λ_{1}β_{1Δ} ⋯ λ_{t_{1}}β_{t_{1}Δ} λ_{t_{1}+1}β_{t_{1}+1 Δ} ⋯ λ_{s−Δ}β_{s−Δ Δ}
0  β_{12}Y_{2} ⋯ β_{1Δ}Y_{Δ} −D_{1}  0  0  ⋯  0
⋮  ⋮  ⋮  ...  ⋮  ⋮
0  β_{t_{1}2}Y_{2} ⋯ β_{t_{1}Δ}Y_{Δ}  0  −D_{t_{1}}  0  ⋯  0
D_{t_{1}+1} β_{t_{1}+1 2}Y_{2} ⋯ β_{t_{1}+1 Δ}Y_{Δ} 0 ⋯ 0  −D_{t_{1}+1}  0
⋮  ⋮  ⋮  ⋮  ⋮  ...
D_{s−Δ} β_{s−Δ 2}Y_{2} ⋯ β_{s−Δ Δ}Y_{Δ}  0 ⋯ 0  0  −D_{s−Δ}

If Δ_{𝕏} = 1, we have t_{1} = s − Δ, and the claim rk 𝔐(ℓ^{σ}) = rk 𝔐_{0} = s − 1 follows from (−D_{1}) ⋯ (−D_{s−Δ}) ≠ 0. Therefore we shall assume Δ_{𝕏} ≥ 2 from now on.

Let the matrix 𝔐_{1} be obtained from 𝔐_{0} by deleting its first row and its first column. We shall show rk 𝔐(ℓ^{σ}) = rk 𝔐_{0} = s − 1 by proving det 𝔐_{1} ≠ 0. Notice that Y_{1} appears only in the entries −D_{1}, …, −D_{t_{1}} of 𝔐_{1}. Let the matrix 𝔐_{2} be obtained form 𝔐_{1} by deleting the columns and rows containing those entries. Then the coefficient of Y_{1}^{t_{1}} in det 𝔐_{1} is (−β_{11}) ⋯ (−β_{t_{1}1}) · det 𝔐_{2}. Hence det 𝔐_{2} ≠ 0 implies det 𝔐_{1} ≠ 0.

More generally, for each i ∈ {2, …, Δ} we define a submatrix 𝔐_{i} of 𝔐_{1} by 𝔐_{i} =

−b_{i}  0  λ_{t_{i−1}+1}β_{t_{i−1}+1 i} ⋯ λ_{t_{i}}β_{t_{i}i}  0  ⋯  0
...  ⋮  ⋮  ⋮  ⋮
0  −b_{Δ} λ_{t_{i−1}+1}β_{t_{i−1}+1 Δ} ⋯ λ_{t_{i}}β_{t_{i}Δ} λ_{t_{i}+1}β_{t_{i}+1 Δ} ⋯ λ_{s−Δ}β_{s−Δ Δ}
β_{t_{i−1}+1 i}Y_{i} ⋯ β_{t_{i−1}+1 Δ}Y_{Δ} −D_{t_{i−1}+1}  0  0  ⋯  0
⋮  ⋮  ...  ⋮  ⋮
β_{t_{i}i}Y_{i} ⋯ β_{t_{i}Δ}Y_{Δ}  0  −D_{t_{i}}  0  ⋯  0
0  ⋯ β_{t_{i}+1 Δ}Y_{Δ}  0  ⋯ 0  −D_{t_{i}+1}  0
⋮  ⋮  ⋮  ⋮  ...
0  ⋯ β_{s−Δ Δ}Y_{Δ}  0  ⋯ 0  0  −D_{s−Δ}

Using downward induction on i ∈ {2, …, Δ}, we shall show that if det 𝔐_{i} = 0, then the matrix 𝔅 splits. For cohomologically uniform schemes 𝕏, this proves det 𝔐_{2} ≠ 0.

We begin the induction by looking at the case i = Δ. If det 𝔐_{Δ} = 0, then t_{Δ−1} < s − Δ, since t_{Δ−1} = s − Δ implies det 𝔐_{Δ} = −b_{Δ} ≠ 0 for generic ℓ ∈ R_{1}. Starting from column 2, we add all columns of 𝔐_{Δ} to column 1, and we obtain the matrix

𝔐′_{Δ} = ( \begin{matrix}−γ_{Δ} & λ_{t_{Δ−1}+1}β_{t_{Δ−1}+1 Δ} & ⋯ & λ_{s−Δ}β_{s−Δ Δ} \\ 0 & −D_{t_{Δ−1}+1} & & 0 \\ ⋮ & & ... \\ 0 & 0 & & −D_{s−Δ} \end{matrix} )

which has det 𝔐′_{Δ} = det 𝔐_{Δ} ≠ 0, since −γ_{Δ} = −λ_{1}β_{1Δ} − ⋯ − λ_{t_{Δ−1}}β_{t_{Δ−1} Δ} ≠ 0 for generic ℓ ∈ R_{1}.

Finally we prove the induction step. We assume i < Δ and det 𝔐_{i} = 0. If t_{i−1} = t_{i}, then the only nonzero entry in the first row of 𝔐_{i} is −b_{i}, and we have det 𝔐_{i} = (−b_{i}) · det 𝔐_{i+1}. In this case the claim follows from the induction hypothesis, because b_{i} ≠ 0 for generic ℓ ∈ R_{1}. Thus we may also assume t_{i−1} < t_{i}. Then we add columns 2, …, Δ − i + 1 + t_{i} − t_{i−1} of 𝔐_{i} to column 1, and we obtain the following matrix 𝔐′_{i} =

−γ_{i}  0  ⋯  0  λ_{t_{i−1}+1}β_{t_{i−1}+1 i} ⋯ λ_{t_{i}}β_{t_{i}i} 0 ⋯ 0
η_{i+1}  −b_{i+1}  0 λ_{t_{i−1}+1}β_{t_{i−1}+1 i+1} ⋯ λ_{t_{i}}β_{t_{i} i+1} ∗ ⋯ ∗
⋮  ...  ⋮  ⋮  ⋮  ⋮
η_{Δ}  0  −b_{Δ} λ_{t_{i−1}+1}β_{t_{i−1}+1 Δ} ⋯ λ_{t_{i}}β_{t_{i}Δ} ∗ ⋯ ∗
0 β_{t_{i−1}+1 i+1}Y_{i+1} ⋯ β_{t_{i} Δ}Y_{Δ}  −D_{t_{i−1}+1}  0  0 ⋯ 0
⋮  ⋮  ⋮  ...  ⋮  ⋮
0  β_{t_{i} i+1}Y_{i+1} ⋯ β_{t_{i} Δ}Y_{Δ}  0  −D_{t_{i}}  0 ⋯ 0
D_{t_{i}+1} β_{t_{i}+1 i+1}Y_{i+1} ⋯ β_{t_{i}+1 Δ}Y_{Δ}  0  ⋯ 0 −D_{t_{i}+1}  0
⋮  ⋮  ⋮  ⋮  ⋮  ...
D_{s−Δ} β_{s−Δ i+1}Y_{i+1} ⋯ β_{s−Δ Δ}Y_{Δ}  0  ⋯ 0  0  −D_{s−Δ}

Here the elements “∗” have to be replaced by the corresponding entries of 𝔐_{i}. Notice that the only entries of 𝔐′_{i} containing Y_{i} are −D_{t_{i−1}+1}, …, −D_{t_{i}}. Let the matrix 𝔐′′_{i} be obtained form 𝔐′_{i} by deleting the rows and columns containing those entries. Then the coefficient of Y_{i}^{t_{i}−t_{i−1}} in det 𝔐′_{i} is (−β_{t_{i−1}+1 i}) ⋯ (−β_{t_{i}i}) · det 𝔐′′_{i}, and det 𝔐_{i} = det 𝔐′_{i} = 0 implies det 𝔐′′_{i} = 0. We observe that −γ_{i} = −λ_{1}β_{1i} − ⋯ − λ_{t_{i−1}}β_{t_{i−1} i} ≠ 0 is the only nonzero entry in the first row of 𝔐′′_{i}, and that det 𝔐′′_{i} = (−γ_{i}) · det 𝔐_{i+1}. Thus the claim follows from the induction hypothesis. □

In view of 3.3.a, 3.5, and 3.12, the proof of Theorem 3.1 is now complete.

4. The Syzygy Module of the Canonical Ideal

In this section we give a description of the first syzygy module of the canonical ideal in the case of d + 1 < s < \binom{d+2}{2} points 𝕏 = \{P_{1}, …, P_{s}\} in ℙ^{d}. We shall assume that 𝕏 is nondegenerate, cohomologically uniform and satisfies σ_{𝕏} = 1. We use the notations introduced in the previous sections. Our first goal is to find an explicit homogeneous system of generators of the first syzygy module of ℑ.

The Hilbert function of ℑ is H_{ℑ} : 0 0 s−d−1 s−1 s s ⋯, and a K-basis of ℑ_{2} is given by {π_{1}, …, π_{Δ}}, where Δ = s − d − 1. Since 𝕏 is cohomologically uniform, Prop. 2.7 implies that {π_{1}, …, π_{Δ}} is a minimal homogeneous system of generators of ℑ. Now we choose a generic element φ = c_{1}f_{1} + ⋯ + c_{s}f_{s} ∈ ℑ_{2} with c_{1}, …, c_{s} ∈ K. By using the representation φ = c_{1}π_{1} + ⋯ + c_{Δ}π_{Δ} and Prop. 2.1, we see that we may assume c_{1}, …, c_{s} ≠ 0. After a generic change of coordinates, we may also assume by Prop. 3.9 that

ℑ_{3} = R_{1}φ ⊕ Kx_{0}π_{2} ⊕ ⋯ ⊕ Kx_{0}π_{Δ}.

Therefore there are unique elements ℓ_{ij} ∈ R_{1} and λ_{ij}^{(2)}, …, λ_{ij}^{(Δ)} ∈ K such that

x_{i}π_{j} = ℓ_{ij}φ + λ_{ij}^{(2)}x_{0}π_{2} + ⋯ + λ_{ij}^{(Δ)}x_{0}π_{Δ}

for i = 1, …, d and j = 2, …, Δ. Since {φ, π_{2}, …, π_{Δ}} is a minimal homogeneous system of generators of ℑ, we can use those relations to describe the R-module

S_{R} := Syz_{R}^{1}(φ, π_{2}, …, π_{Δ}) = {(r_{1}, …, r_{Δ}) ∈ R^{Δ} | r_{1}φ + r_{2}π_{2} + ⋯ + r_{Δ}π_{Δ} = 0}

explicitly.

4.1. Remarks. a) From the exact sequence of graded R-modules

0 ⟶ Syz_{R}^{1}(φ, π_{2}, …, π_{Δ}) ⟶ R(−2)^{Δ} ⟶^{ϵ} ℑ ⟶ 0

with ϵ(e_{1}) = φ and ϵ(e_{i}) = π_{i} for i = 2, …, Δ, we obtain H_{S_{R}} : 0 0 0 (Δ−1)d (Δ− 1)s (Δ−1)s · · ·.

b) The elements

σ_{ij} = (ℓ_{ij}, λ_{ij}^{(2)}x_{0}, …, λ_{ij}^{(j−1)}x_{0}, λ_{ij}^{(j)}x_{0} − x_{i}, λ_{ij}^{(j+1)}x_{0}, …, λ_{ij}^{(Δ)}x_{0}) ∈ R(−2)^{Δ}

such that 1 ≤ i ≤ d and 2 ≤ j ≤ Δ form a K-basis of (S_{R})_{3}, since they are K-linearly independent and H_{S_{R}}(3) = (Δ − 1)d.

c) Let P = K[X_{0}, …, X_{d}] and S_{P} = Syz_{P}^{1} (φ, π_{2}, …, π_{Δ}) = {(F_{1}, …, F_{Δ}) ∈ P^{Δ} | F_{1}φ + F_{2}π_{2} + ⋯ + F_{Δ}π_{Δ} = 0}. From the exact sequence of graded P -modules

0 ⟶ Syz_{P}^{1} (φ, π_{2}, …, π_{Δ}) ⟶ P (−2)^{Δ} ⟶ ℑ ⟶ 0

we obtain an exact sequence of graded R-modules

0 ⟶ Tor_{P}^{1} (ℑ, R) ⟶ Syz_{P}^{1} (φ, π_{2}, …, π_{Δ}) ⊗ R ⟶ Syz_{R}^{1}(φ, π_{2}, …, π_{Δ}) ⟶ 0.

In particular, we get H_{S_{P}} (3) = (Δ − 1)d = H_{S_{R}}(3). Therefore any set of preimages

Σ_{ij} = (L_{ij}, λ_{ij}^{(2)}X_{0}, …, λ_{ij}^{(j)}X_{0} − X_{i}, λ_{ij}^{(Δ)}X_{0}) ∈ P (−2)^{Δ}

of the elements σ_{ij} ∈ R(−2)^{Δ} forms a K-basis of (S_{P})_{3}.

d) From the short exact sequence 0 ⟶ I_{𝕏} ⟶ P ⟶ R ⟶ 0 we get a long exact sequence of graded P -modules

0 ⟶ Tor_{P}^{1} (ℑ, R) ⟶ I_{𝕏} ⊗_{P} ℑ ⟶ ℑ ⟶^{id} ℑ ⟶ 0

from which we conclude that Tor_{P}^{1} (ℑ, R) ≅ I_{𝕏} ⊗_{P} R ⊗_{R} ℑ ≅ (I_{𝕏}/I_{𝕏}^{2}) ⊗_{R} ℑ. Thus the second exact sequence of c) translates in degree four into a short exact sequence of K-vector spaces

0 ⟶ (I_{𝕏})_{2} ⊗ ℑ_{2} ⟶^{ı} Syz_{P}^{1} (φ, π_{2}, …, π_{Δ})_{4} ⟶ Syz_{R}^{1}(φ, π_{2}, …, π_{Δ})_{4} ⟶ 0

where ı is given by ı(F_{1} ⊗ φ + F_{2} ⊗ π_{2} + ⋯ + F_{Δ} ⊗ π_{Δ}) = (F_{1}, …, F_{Δ}) for F_{1}, …, F_{Δ} ∈ (I_{𝕏})_{2}.

Next we determine how big a part of Syz_{P}^{1} (φ, π_{2}, …, π_{Δ})_{4} is generated by the elements Σ_{ij}.

4.2. Lemma. a) The set {Σ_{ij} | 1 ≤ i ≤ d, 2 ≤ j ≤ Δ} generates a subspace of Syz_{P}^{1} (φ, π_{2}, …, π_{Δ})_{4} which is at least (Δ − 1)(\frac{1}{2}d^{2} + \frac{3}{2}d)-dimensional.

b) The P -module Syz_{P}^{1} (φ, π_{2}, …, π_{Δ}) is generated by the elements

{Σ_{ij} | 1 ≤ i ≤ d, 2 ≤ j ≤ Δ} ∪ {(F, 0, …, 0) | F ∈ (I_{𝕏})_{2}} ∪ {Θ_{j} | 2 ≤ j ≤ Δ}

where Θ_{j} = (Q_{j}, 0, …, 0, −X_{0}^{2}, 0, …, 0) and Q_{j} is a preimage in P_{2} of the unique element q_{j} ∈ R_{2} such that x_{0}^{2}π_{j} = q_{j}φ.

Proof: “a)” Let ≤_{σ} be the lexicographical term ordering on P, where X_{0} <_{σ} X_{1} <_{σ} ⋯ <_{σ} X_{d}. On P (−2)^{Δ} we define a module term ordering ≤_{τ} in the following way. If t_{1}, t_{2} ∈ P are power products of variables, and if i, j ∈ {1, …, Δ}, then we let t_{1}e_{i} ≤_{τ} t_{2}e_{j} if and only if one of four conditions is satisfied:

1) i = j = 1 and t_{1} ≤_{σ} t_{2}

2) i = 1 and j > 1

3) i > 1, j > 1, and t_{1} <_{σ} t_{2}

4) 1 < i ≤ j and t_{1} = t_{2}

It is easy to check that ≤_{τ} is in fact a module term ordering (cf. [4], p. 324) and that the elements Σ_{ij} have leading terms Lt_{τ} (Σ_{ij}) = (0, …, 0, −X_{i}, 0, …, 0). Now it is a standard fact that, for U = ⟨{Σ_{ij} | 1 ≤ i ≤ d, 2 ≤ j ≤ Δ}⟩ ⊆ P (−2)^{Δ}, we have

dim_{K}(U_{4}) = dim_{K}(Lt_{τ} (U))_{4} ≥ dim_{K}⟨{Lt_{τ} (Σ_{ij}) | 1 ≤ i ≤ d, 2 ≤ j ≤ Δ}⟩_{4}

(cf. [4], 15.26). The last number is obviously (Δ − 1)(\frac{1}{2}d^{2} + \frac{3}{2}d)).

“b)” By subtracting suitable multiples of the syzygies Σ_{ij}, every P -syzygy of (φ, π_{2}, …, π_{Δ}) can be reduced to one of the form (F, γ_{2}X_{0}^{2}, …, γ_{Δ}X_{0}^{2}) with F ∈ P_{2} and γ_{2}, …, γ_{Δ} ∈ K. Then one can subtract ∑_{j=2}^{Δ} γ_{j}Θ_{j}. The result is a syzygy of the form (G, 0, …, 0) with G ∈ P_{2}. Since Ann_{R}(φ) = (0), this implies G ∈ (I_{𝕏})_{2}. It is well-known that it follows from σ_{𝕏} = 1 = α_{𝕏} − 1 that Syz_{P}^{1} (φ, π_{2}, …, π_{Δ}) is generated by its homogeneous elements of degrees three and four (cf. e.g. [10], 5.1). □

4.3. Proposition. For j = 2, …, Δ let θ_{j} = (q_{j}, 0, …, 0, −x_{0}^{2}, 0, …, 0) ∈ R(−2)^{Δ} be the image of Θ_{j} (cf. 4.2.b). The graded R-module Syz_{R}^{1}(φ, π_{2}, …, π_{Δ}) is generated by the homogeneous elements {σ_{ij} | 1 ≤ i ≤ d, 2 ≤ j ≤ Δ} of degree three and the homogeneous elements {θ_{j} | 2 ≤ j ≤ Δ} of degree four.

Proof: The claim is a consequence of Remark 4.1.c and Lemma 4.2.b. □

Notice that the elements σ_{ij} are minimal generators of Syz_{R}^{1}(φ, π_{2}, …, π_{Δ}), whereas it is not clear how many of the elements θ_{j} are actually needed. In the case Δ = 2, i.e. for s = d + 3 points in ℙ^{d}, we can in fact do without the elements θ_{j}.

4.4. Proposition. Let d ≥ 3, and let 𝕏 ⊆ ℙ^{d} be a set of s = d + 3 cohomologically uniform points with generic Hilbert function ΔH_{𝕏} : 1 d 2. Then the R-module Syz_{R}^{1}(φ, π_{2}) is generated by its homogeneous elements of degree three.

Proof: We choose the coordinate system and the element φ ∈ ℑ_{2} as at the beginning of this section. For 1 ≤ i ≤ d we write x_{i}π_{2} = ℓ_{i}φ + λ_{i}x_{0}π_{2} with ℓ_{i} ∈ R_{1} and λ_{i} ∈ K, and we let σ_{i} = (ℓ_{i}, λ_{i}x_{0} − x_{i}) ∈ R(−2)^{2}. Furthermore, we let q ∈ R_{2} be the unique element such that x_{0}^{2}π_{2} = qφ. By Prop. 4.3, we need to show that (−q, x_{0}^{2}) ∈ ∑_{i=1}^{d} P_{1}σ_{i}. Since Ann_{R}(φ) = (0), this amounts to showing that x_{0}^{2} is an element of the ideal J = (λ_{1}x_{0} − x_{1}, …, λ_{d}x_{0} − x_{d}) of R.

As a first step, we prove that not all elements ℓ_{1}, …, ℓ_{d} ∈ R_{1} can be zero. Otherwise (λ_{i}x_{o} − x_{i})π_{2} = 0 for i = 1, …, d and π_{2} = f_{2} + β_{12}f_{3} + ⋯ + β_{d+1 2}f_{d+3} imply that (λ_{1}x_{0} − x_{1}, …, λ_{d}x_{0} − x_{d}) equals 𝔭_{2}, the ideal of P_{2} in 𝕏. Since 𝕏 does not split cohomologically, Lemma 2.3 yields β_{j2} ≠ 0 for some j ∈ {1, …, d + 1}. Since not all elements of 𝔭_{2} vanish at P_{j+2}, there exists a k ∈ {1, …, d + 1} such that (λ_{k}x_{0} − x_{k})(P_{j+2}) ≠ 0 Now (λ_{k}x_{0} − x_{k})π_{2}(P_{j+2}) = (λ_{k}x_{0} − x_{k})(P_{j+2})β_{j2} = 0 implies (λ_{k}x_{0} − x_{k})(P_{j+2}) = 0, a contradiction.

Let Q = (1 : λ_{1} : …: λ_{d}) ∈ ℙ^{d}. In the next step, we prove Q ∉ 𝕏. Suppose Q = P_{1}, and let 𝔭_{1} ⊆ R be the ideal of P_{1} in 𝕏. If we use the equation ℓ_{i}φ + (λ_{i}x_{0} − x_{i})π_{2} = 0, as well as λ_{i}x_{0} − x_{i} ∈ J = 𝔭_{1} and π_{2} = f_{2} + β_{12}f_{3} + ⋯ + β_{d+1 2}f_{d+3} ∈ 𝔭_{1}, we obtain ℓ_{i}φ + 𝔭_{1}^{2} = 0 in R/𝔭_{1}^{2}. Since the residue class of φ ∉ 𝔭_{1} is not a zero divisor in that ring, we arrive at ℓ_{i} ∈ 𝔭_{1}^{2}, which means ℓ_{i} = 0, because ℓ_{i} ∈ R_{1} and 𝔭_{1}^{2} is generated in degree two. But the first step shows that ℓ_{i} = 0 cannot hold simultaneously for i = 1, …, d.

Next we suppose Q ∈ {P_{2}, …, P_{s}}. Then the equation ℓ_{i}φ + (λ_{i}x_{0} − x_{i})π_{2} = 0 shows that ℓ_{i} vanishes at P_{1} and Q, i.e. there are elements c_{ij} ∈ K such that ℓ_{i} = ∑_{j=1}^{d} c_{ij}(λ_{j}x_{0} − x_{j}) for i = 1, …, d. Furthermore, by passing to the ring R/J^{2} and arguing as before, we see that π_{2}(Q) ≠ 0, i.e. Q has to be P_{2} or one of the points P_{j} such that 3 ≤ j ≤ d + 3 and β_{j−2 2} ≠ 0. Using the representation of ℓ_{i} ∈ J, we get equations ∑_{j=1}^{d} c_{ij}ℓ_{j}φ + ℓ_{i}π_{2} = 0. Again the conditions ℓ_{i} ∈ 𝔭_{1}, π_{2} ∈ 𝔭_{1}, and φ ∉ 𝔭_{1} can only be satisfied if ∑_{j=1}^{d} c_{ij}ℓ_{j} = 0 and ℓ_{i}π_{2} = 0. Now ℓ_{i}φ + (λ_{i}x_{0} − x_{i})π_{2} = 0 implies that ℓ_{i}(P_{ν}) = 0 for all ν ∈ {1, …, d + 3} such that π_{2}(P_{ν}) = 0, while ℓ_{i}π_{2} = 0 means ℓ_{i}(P_{ν}) = 0 for all ν ∈ {1, …, d + 3} such that π_{2}(P_{ν}) ≠ 0. Thus we get ℓ_{i} = 0 for i = 1, …, d, contradicting the first step.

Altogether we have shown that Q ∉ 𝕏, and we can consider the Hilbert function of 𝕐 = 𝕏 ∪ {Q}. We claim that it is given by ΔH_{𝕐} : 1 d 3. The only other possibility is ΔH_{𝕐} : 1 d 2 1. In that case the canonical Ideal ℑ_{𝕐} of 𝕐 starts in degree three and satisfies H_{ℑ_{𝕐}}(3) = 1, H_{ℑ_{𝕐}}(4) = 3. Thus a nonzero element of (ℑ_{𝕐})_{3} has a nontrivial annihilator, i.e. 𝕐 is not a Cayley-Bacharach scheme. Therefore there exists an i ∈ {1, …, s} such that ΔH_{𝕐\{P_{i}}} : 1 d 1 1 and ΔH_{𝕏\{P_{i}}} : 1 d 1. Now [7], 5.2 shows that Q and three points of 𝕏 lie on a line in ℙ^{d}. Since the other d points of 𝕏 span at most a (d − 1)-dimensional linear space,we get that 𝕏 splits linearly, in contradiction with our hypothesis.

Thus we have H_{𝕐} : 1 d+1 d+4 d+4 ⋯, and the formula H_{R/J}(i) = H_{𝕏}(i) + H_{{Q}}(i) − H_{𝕐}(i) proves H_{R/J}(2) = 0, so that x_{0}^{2} ∈ J, as was to be shown. □

It is instructive to compare the preceding result to a similar one for the P -module Syz_{P}^{1} (φ, π_{2}) which follows from [3].

4.5. Remark. Let d ≥ 3, and let 𝕏 ⊆ ℙ^{d} be a set of s = d + 3 points. Then the following conditions are equivalent.

a) The set 𝕏 is cohomologically uniform, has generic Hilbert function ΔH_{𝕏} : 1 d 2, and Syz_{P}^{1} (φ, π_{2}) is minimally generated by (Δ − 1)d homogeneous elements of degree three and d−1 homogeneous elements of degree four. (The number d−1 is the minimal possible one, here.)

b) No d + 2 points of 𝕏 are on a hyperplane and no d points of 𝕏 are on a linear subspace of ℙ^{d} of dimension d − 2.

The implication “a)⇒b)” follows from 2.7 and [3], 4.3. Conversely, [3], 1.3 shows that 𝕏 has generic Hilbert function, [3], 4.1 and our Prop. 2.7 imply that 𝕏 is cohomologically uniform, and [3], 4.3 yields the remaining claim.

It is an elementary exercise to verify directly that condition 4.5.b implies that 𝕏 does not split linearly. Our next example shows that it is in fact a stronger hypothesis.

4.6. Example. Let 𝕏 ⊆ ℙ^{3} consist of three points on a line and three generically chosen points. Then we have ΔH_{𝕏} : 1 3 2, and 𝕏 is cohomologically uniform. But Syz_{P}^{1} (φ, π_{2}) needs three minimal generators in degree four besides its three minimal generators of degree three. The module Syz_{R}^{1}(φ, π_{2}), however, is minimally generated in degree three by Prop. 4.4.

To conclude the discussion of Syz_{R}^{1}(φ, π_{2}, …, π_{Δ}), we show how the elements σ_{ij} and θ_{j} can be computed effectively in terms of the constants β_{ij}, the separators f_{1}, …, f_{s} ∈ R_{2}, and the coordinates of the points. We note that the numbers β_{ij} and the separators f_{1}, …, f_{s} ∈ R_{2} can be found as a by-product of the Buchberger-Möller Algorithm (cf. [12], sect. 1.2).

4.7. Remark. Let us write x_{0}ℓ_{ij} = ℓ_{ij}^{(1)}f_{1} + ⋯ + ℓ_{ij}^{(s)}f_{s} with ℓ_{ij}^{(1)}, …, ℓ_{ij}^{(s)} ∈ K. Recall that we have P_{i} = (1 : p_{i1} : …: p_{id}) for i = 1, …, s and φ = c_{1}f_{1} + ⋯ + c_{s}f_{s} with c_{Δ+k} = c_{1}β_{k1} + ⋯ + c_{Δ}β_{kΔ} for k = 1, …, s − Δ. For i = 1, …, d and j = 2, …, Δ we express both sides of x_{i}π_{j} = ℓ_{ij}φ+λ_{ij}^{(2)}x_{0}π_{2} +⋯+λ_{ij}^{(Δ)}x_{0}π_{Δ} in the K-basis {x_{0}f_{1}, …, x_{0}f_{s}} of R_{3} (cf. [6], 1.13). We get

\begin{aligned}x_{i}π_{j} & = x_{0}^{−1}(p_{1i}f_{1} + ⋯ + p_{si}f_{s})(f_{j} + β_{1j}f_{Δ+1} + ⋯ + β_{s−Δ j}f_{s}) \\ & = p_{ji}x_{0}f_{j} + p_{Δ+1 i}β_{1j}x_{0}f_{Δ+1} + ⋯ + p_{si}β_{s−Δ j}x_{0}f_{s} \end{aligned}

as well as

ℓ_{ij}φ + λ_{ij}^{(2)}x_{0}π_{2} + ⋯ +λ_{ij}^{(Δ)}x_{0}π_{Δ} = ℓ_{ij}^{(1)}c_{1}x_{0}f_{1} + ⋯ + ℓ_{ij}^{(s)}c_{s}x_{0}f_{s}
+ (λ_{ij}^{(2)}x_{0}f_{2} + λ_{ij}^{(2)}β_{12}x_{0}f_{Δ+1} + ⋯ + λ_{ij}^{(2)}β_{s−Δ 2}x_{0}f_{s}) + ⋯
+ (λ_{ij}^{(Δ)}x_{0}f_{Δ} + λ_{ij}^{(Δ)}β_{1Δ}x_{0}f_{Δ+1} + ⋯ + λ_{ij}^{(Δ)}β_{s−Δ Δ}x_{0}f_{s})
= ℓ_{ij}^{(1)}c_{1}x_{0}f_{1} + (ℓ_{ij}^{(2)}c_{2} + λ_{ij}^{(2)})x_{0}f_{2} + ⋯ + (ℓ_{ij}^{(Δ)}c_{Δ} + λ_{ij}^{(Δ)})x_{0}f_{Δ}
+ (ℓ_{ij}^{(Δ+1)}c_{Δ+1} + λ_{ij}^{(2)}β_{12} + ⋯ + λ_{ij}^{(Δ)}β_{1Δ})x_{0}f_{Δ+1} + ⋯
+ (ℓ_{ij}^{(s)}c_{s} + λ_{ij}^{(2)}β_{s−Δ 2} + ⋯ + λ_{ij}^{(Δ)}β_{s−Δ Δ})x_{0}f_{s}

Comparing coefficients yields ℓ_{ij}^{(1)} = 0, λ_{ij}^{(k)} = −ℓ_{ij}^{(k)}c_{j} for k ∈ {2, …, Δ} \ {j} and λ_{ij}^{(j)} = p_{ji} − ℓ_{ij}^{(j)}c_{j}. Substituting this into the remaining equations yields

(∗)  (p_{Δ+k i} − p_{ji})β_{kj} = ℓ_{ij}^{(Δ+k)}c_{Δ+k} − ℓ_{ij}^{(2)}c_{2}β_{k2} − ⋯ − ℓ_{ij}^{(Δ)}c_{Δ}β_{kΔ}

for k = 1, …, s − Δ. Hence the elements λ_{ij}^{(2)}, …, λ_{ij}^{(Δ)} are uniquely determined by ℓ_{ij}^{(2)}, …, ℓ_{ij}^{(s)}. The condition ℓ_{ij}^{(1)} = 0 means that the hyperplane Z(ℓ_{ij}) contains P_{1}, and that we can find m_{1}, …, m_{d} ∈ K such that ℓ_{ij} = m_{1}(x_{1} −p_{11}x_{0})+⋯+m_{d}(x_{d} −p_{1d}x_{0}). Therefore we have ℓ_{ij}^{(k)} = m_{1}(p_{k1} − p_{11}) + ⋯ + m_{d}(p_{kd} − p_{1d}) for k = 2, …, s. Now equations (∗) yield s − Δ = d + 1 linear equations (∗∗) for m_{1}, …, m_{d}. But those equations are not linearly independent, because adding them up gives ∑_{k=1}^{s−Δ} p_{Δ+k i}β_{kj} + p_{ji} = ∑_{k=1}^{s−Δ} ℓ_{ij}^{(Δ+k)}c_{Δ+k} + ℓ_{ij}^{(2)}c_{2} + ⋯ + ℓ_{ij}^{(Δ)}c_{Δ} = ∑_{k=1}^{s}(x_{i}π_{j})(P_{k}) + ∑_{k=1}^{s} ℓ_{ij}^{(k)}c_{k} = 0 in view of x_{i}π_{j}, ℓ_{ij}φ ∈ ℑ_{3} and 1.11.

Altogether, it follows from 3.9 and the above that the system of linear equations

(∗∗) (p_{Δ+k i} − p_{ji})β_{jk} = [m_{1}(p_{Δ+k 1} − p_{11}) + ⋯ + m_{d}(p_{Δ+k d} − p_{1d})]c_{Δ+k}
− [m_{1}(p_{21} − p_{11}) + ⋯ + m_{d}(p_{2d} − p_{1d})]c_{2}β_{k2} − ⋯
− [m_{1}(p_{Δ1} − p_{11}) + ⋯ + m_{d}(p_{Δd} − p_{1d})]c_{Δ}β_{kΔ}
= (p_{Δ+k 1}c_{Δ+k} − p_{11}c_{1}β_{k1} − ⋯ − p_{Δ1}c_{Δ}β_{kΔ})m_{1} + ⋯
+ (p_{Δ+k d}c_{Δ+k} − p_{1d}c_{1}β_{k1} − ⋯ − p_{Δd}c_{Δ}β_{kΔ})m_{d}

for k = 1, …, s − Δ uniquely determines m_{1}, …, m_{d} ∈ K, and hence ℓ_{ij} ∈ R_{1} as well as λ_{ij}^{(2)}, …, λ_{ij}^{(Δ)} ∈ K. Thus it suffices to solve the linear system of equations (∗∗) in order to compute the syzygies σ_{ij}.

Furthermore, if we want to find for j ∈ {2, …, Δ} the unique element q_{j} ∈ R_{2} such that x_{0}^{2}π_{2} = q_{j}φ, we write q_{j} = γ_{j1}f_{1} + ⋯ γ_{js}f_{s} with γ_{j1}, …, γ_{js} ∈ K, and we compare coefficients in

x_{0}^{2}π_{j} = x_{0}^{2}(f_{j} + β_{1j}f_{Δ+1} + ⋯ + β_{d+1 j}f_{s}) = q_{j}φ = x_{0}^{2}(γ_{j1}c_{1}f_{1} + ⋯ + γ_{js}c_{s}f_{s}).

Thus γ_{jk} = \frac{1}{c_{j}} δ_{jk} for k = 1, …, Δ and γ_{jk} = \frac{1}{c_{k}} β_{k−Δ j} for k = Δ + 1, …, s. This effectively computes the syzygies θ_{j} = (q_{j}, −x_{0}^{2}) in terms of the elements β_{ij} and the separators f_{1}, …, f_{s}.

5. Canonical Transforms

The canonical ideal can also be used to generalize the Gale transform of a set of points. In [5], it was shown that this transform may be defined using the elements of ℑ_{2σ}. In this section we shall describe similar “canonical tansforms” based on each of the homogeneous components ℑ_{σ+1}, …, ℑ_{2σ}. We start by characterizing those components as explicitly as we can. The notations and assumptions are the same as in section 1. In particular, we let 𝕏 = {P_{1}, …, P_{s}} ⊆ ℙ^{d} be a set of points with homogeneous coordinate ring R and canonical ideal ℑ ⊆ R.

5.1. Proposition. For all i ∈ {0, …, σ}, the homogeneous component ℑ_{σ+1+i} of the canonical ideal consists precisely of those elements φ = c_{1}x_{0}^{i}f_{1} + ⋯ + c_{s}x_{0}^{i}f_{s} ∈ R_{σ+1+i} such that c_{1}, …, c_{s} ∈ K and r(P_{1})c_{1} + ⋯ + r(P_{s})c_{s} = 0 for all r ∈ R_{σ−i}.

In other words, if we apply the monomorphisms ı_{j} : R_{j} ⟶ K^{s} given by r ↦ (r(P_{1}), …, r(P_{s})) for j ≥ 0, the homogeneous component ℑ_{σ+1+i} of the canonical ideal corresponds to the orthogonal space of R_{σ−i} with respect to the standard pairing.

Proof: If we consider φ ∈ ℑ_{σ+1+i} ≅ (ω_{R})_{−σ−i} as a homogeneous K[x_{0}]-linear map φ : R ⟶ K[x_{0}] of degree −σ − 1 + i, we have φ(R_{σ−i}) ⊆ K[x_{0}]_{−1} = (0). For every element r ∈ R_{σ−i}, this implies φ(x_{0}^{i+1}r) = x_{0}^{i+1}φ(r) = 0. Since we have x_{0}^{i+1}r = r(P_{1})f_{1} + ⋯ + r(P_{s})f_{s}, we get r(P_{1})φ(f_{1}) + ⋯ + r(P_{s})φ(f_{s}) = 0. Now Proposition 1.9 yields φ(f_{j}) = c_{j} for j = 1, …, s, and therefore r(P_{1})c_{1} + ⋯ + r(P_{s})c_{s} = 0.

The reverse implication follows from the fact that the orthogonal space of ı_{σ−i}(R_{σ−i}) has dimension s − dim_{K}(R_{σ−i}) = dim_{K}(ℑ_{σ+1+i}). □

The next proposition provides the basis for our definition of the Gale transform which differs slightly from the one in [5].

5.2. Proposition. Suppose 𝕏 ⊆ ℙ^{d} has the property that any subset of s − 2 points of 𝕏 spans ℙ^{d}. We choose a K-basis {t_{1}, …, t_{s−d−1}} of ℑ_{2σ} and write t_{i} = γ_{i1}x_{0}^{σ−1}f_{1} + ⋯ + γ_{is}x_{0}^{σ−1}f_{s} with γ_{i1}, …, γ_{is} ∈ K for i = 1, …, s − d − 1. Then any two columns of the matrix Γ = (γ_{ij}) are K-linearly independent.

Proof: We assume that there are indices μ, ν ∈ {1, …, s} and an element λ ∈ K \ {0} such that μ ≠ ν and γ_{iμ} = λγ_{iν} for i = 1, …, s − d − 1. Let us denote the homogeneous coordinate ring of the subscheme 𝕐 = 𝕏 \ {P_{ν}} of 𝕏 by R_{𝕐} = K[X_{0}, …, X_{d}]/I_{𝕐}, and its canonical ideal by ℑ_{𝕐} ⊆ R_{𝕐}. Since 𝕐 spans ℙ^{d}, we have

dim_{K}(ℑ_{𝕐})_{2σ_{𝕐}} = (s − 1) − d − 1.

We note that σ_{𝕏} − 1 ≤ σ_{𝕐} ≤ σ_{𝕏} and ω_{R_{𝕐}} ≅ \{φ ∈ ω_{R} | \overline{I_{𝕐}} · φ = 0\}, where \overline{I_{𝕐}} is the image of I_{𝕐} in R (cf. [10], 1.3.d). Now we distinguish two cases.

Case 1: σ_{𝕐} = σ_{𝕏}. In this case we have (ℑ_{𝕐})_{2σ_{𝕐}} ≅ {c_{1}x_{0}^{σ−1}f_{1} + ⋯ + c_{s}x_{0}^{σ−1}f_{s} ∈ ℑ_{2σ} | c_{ν} = 0\}, because \overline{I_{𝕐}} is given by \overline{I_{𝕐}} = (f_{ν})^{sat.} and (\overline{I_{𝕐}})_{2σ} = Kx_{0}^{σ−1}f_{ν}. Since (ℑ_{𝕐})_{2σ_{𝕐}} ⊊ ℑ_{2σ}, we may assume γ_{1ν} ≠ 0. Then we find

(ℑ_{𝕐})_{2σ_{𝕐}} = K · (γ_{1ν}t_{2} − γ_{2ν}t_{1}) ⊕ ⋯ ⊕ K · (γ_{iν}t_{s−d−1} − γ_{s−d−1 ν}t_{1})

and all elements of (ℑ_{𝕐})_{2σ_{𝕐}} are of the form c̃_{1}x_{0}^{σ−1}f_{1} +⋯+c̃_{s}x_{0}^{σ−1}f_{s} with c̃_{1}, …, c̃_{s} ∈ K and c̃_{μ} = c̃_{ν} = 0.

Case 2: σ_{𝕐} = σ_{𝕏} − 1. Again we have (ℑ_{𝕐})_{2σ_{𝕐}} ≅ (ω_{R_{𝕐}})_{−1} ⊆ (ω_{R})_{−1} ≅ ℑ_{2σ_{𝕏}}, and the same argument as above shows that all elements c̃_{1}x_{0}^{σ−1}f_{1} + ⋯ + c̃_{s}x_{0}^{σ−1}f_{s} ∈ (ℑ_{𝕐})_{2σ_{𝕐}} satisfy c̃_{μ} = c̃_{ν} = 0.

Im both cases all elements of (ℑ_{𝕐})_{2σ_{𝕐}} are also contained in the canonical ideal of 𝕐′ = 𝕐 \ {P_{μ}}. Then dim_{K}(ℑ_{𝕐′})_{2σ} = (s − 2) − (d − 1) − 1 contradicts the assumption that the points of 𝕐′ span ℙ^{d}. □

5.3. Definition. Let 𝕏 = {P_{1}, …, P_{s}} ⊆ ℙ^{d} be a set of points with the property that any subset consisting of s − 2 points of 𝕏 spans ℙ^{d}, and let Γ be the matrix defined in Proposition 5.2. Then the set of points κ(𝕏) ⊆ ℙ^{s−d−2} defined by the columns of the matrix Γ is called the Gale transform (or the associated set of points) of 𝕏.

By Proposition 5.2, the set κ(𝕏) consists of s distinct points of ℙ^{s−d−1}. It is clear that κ(𝕏) does not depend on the choice of the linear nonzerodivisor x_{0} ∈ R (see the proof of 1.3), and that it changes by a coordinate transformation of ℙ^{s−d−1}, if we choose a different basis for ℑ_{2σ} in Proposition 5.2. In [5], Gale transforms were defined assuming only that subsets of s − 1 points of 𝕏 span ℙ^{d}. Out next example shows that in this case κ(𝕏) does not necessarily consist of s distinct points in ℙ^{s−d−1}.

5.4. Example. Let K be a field of characterstic char (K) ≠ 2, and let 𝕏 ⊆ ℙ^{3} consist of the following six points: P_{1} = (1:1:0:0), P_{2} = (1:0:1:0), P_{3} = (1: −1:0:0), P_{4} = (1:0: −1:0), P_{5} = (1:0:0:1), and P_{6} = (1:0:0: −1). Then the Hilbert function of 𝕏 is H_{𝕏} : 1 4 6 6 ⋯, and we have σ_{𝕏} = 1 as well as Δ_{𝕏} = 2. A basis of \overline{R}_{σ+1} is given by Lf_{1} and Lf_{2}, and we have Lf_{3} = Lf_{1}, Lf_{4} = Lf_{2}, Lf_{5} = Lf_{6} = −Lf_{1} − Lf_{2}. Thus the homogeneous component ℑ_{2} = ℑ_{σ+1} = ℑ_{2σ} has the K-basis {π_{1}, π_{2}}, where π_{1} = f_{1} + f_{3} − f_{5} − f_{6} and π_{2} = f_{2} + f_{4} − f_{5} − f_{6}.

It is clear that any five points of 𝕏 span ℙ^{3}. But the Gale transform κ(𝕏) ⊆ ℙ^{1} consists only of the three points (1:0), (0:1), and (1:1).

If σ_{𝕏} = 1, as in the previous example, the description of κ(𝕏) can be simplified considerably.

5.5. Remark. Let 𝕏 ⊆ ℙ^{d} be a set of s points such that σ_{𝕏} = 1 and such that any s − 2 points of 𝕏 span ℙ^{d}. We form the matrix

𝔅̃ = ( \begin{matrix}1 & & & & 0 \\ & . & . & . \\ 0 & & & & 1 \end{matrix} | 𝔅 )

Then, by Corollary 1.10, the Gale transform of 𝕏 is the set of points κ(𝕏) ⊆ ℙ^{Δ−1} defined by the columns of the matrix 𝔅̃.

Next we want to generalize the Gale transform by using other components of the canonical ideal as well. Again we are interested in the correct conditions under which those constructions yield sets of s distinct points. To be able to formulate those conditions, we introduce the following notion of uniformity.

5.6. Definition. Let 1 ≤ i ≤ s − 1 and 1 ≤ j ≤ σ_{𝕏}. We say that 𝕏 is (i, j)-uniform, if every subscheme 𝕐 ⊆ 𝕏 consisting of deg 𝕐 = s − i points satisfies H_{𝕐}(j) = H_{𝕏}(j).

This notion generalizes most uniformity conditions for 0-dimensional schemes considered in [10] and elsewhere. Recall that 𝕏 is called i-uniform, if every subset 𝕐 ⊆ 𝕏 of degree s − i ≤ deg 𝕐 ≤ s satisfies H_{𝕐} = min{H_{𝕏}, deg 𝕐}.

5.7. Remark. a) A set of points 𝕏 ⊆ ℙ^{d} is i-uniform, if and only if 𝕏 is (i, σ)-uniform. In particular, 𝕏 is a Cayley-Bacharach scheme, if and only if 𝕏 is (1, σ)-uniform, and 𝕏 is in uniform position, if and only if 𝕏 is (s − 1, σ)-uniform.

b) If 𝕏 ⊆ ℙ^{d} is nondegenerate, then 𝕏 is in linearly general position (i.e. any d + 1 points of 𝕏 span ℙ^{d}), if and only if 𝕏 is (s − d − 1, 1)-uniform. More generally, 𝕏 is in lineraly general position of i^{th} order (cf. [10]), if and only if 𝕏 is (s − \binom{d+i}{d}, i)-uniform.

5.8. Proposition. Let 𝕏 ⊆ ℙ^{d} be a set of s points, let 1 ≤ i ≤ σ, let r = s − H_{𝕏}(i), and let {t_{1}, …, t_{r}} be a K-basis of ℑ_{2σ+1−i}. For j = 1, …, r we write t_{j} = γ_{j1}x_{0}^{σ−i}f_{1} + ⋯ + γ_{js}x_{0}^{σ−i}f_{s} with γ_{j1}, …, γ_{js} ∈ K, and we form the matrix Γ_{i} = (γ_{jk}). Then the following conditions are equivalent.

a) 𝕏 is (2, i)-uniform.

b) The columns of the matrix Γ_{i} are pairwise K-linearly independent.

Proof: Suppose b) holds. Let 1 ≤ j < k ≤ s. Clearly, condition b) does not depend on the choice of the basis {t_{1}, …, t_{r}} of ℑ_{2σ+1−i}. If we choose it suitably, the j^{th} and k^{th} columns of Γ_{i} are given by (1, 0, …, 0) and (0, 1, 0, …, 0), respectively. Then we get

dim_{K}{c_{1}x_{0}^{σ−i}f_{1} + ⋯ + c_{s}x_{0}^{σ−i}f_{s} ∈ ℑ_{2σ+1−i} | c_{1}, …, c_{s} ∈ K, c_{j} = c_{k} = 0} =

= dim_{K} ℑ_{2σ+1−i} − 2 = r − 2 (∗)

This implies dim_{K}(ω_{R_{𝕐}})_{−i} = dim_{K}(ω_{R})_{−i} − 2 for all subschemes 𝕐 ⊆ 𝕏 with deg 𝕐 = s − 2 and canonical module ω_{R_{𝕐}}. Since we have H_{𝕐}(i) = deg 𝕐 − dim_{K}(ω_{R_{𝕐}})_{−i} = (s − 2) − (dim_{K}(ω_{R})_{−i} − 2) = H_{𝕏}(i), we see that 𝕏 is (2, i)-uniform.

Conversely, given a), the same calculation shows that equation (∗) holds for all 1 ≤ j < k ≤ s, i.e. that columns j and k of Γ_{i} are K-linearly independent. □

5.9. Definition. Let 𝕏 ⊆ ℙ^{d} be a (2, i)-uniform set of s points, let i ∈ {1, …, σ}, let r = s − H_{𝕏}(i), and let Γ_{i} be the matrix defined in the previous proposition. Then the set of s distinct points κ_{i}(𝕏) ⊆ ℙ^{r} is called the i^{th} canonical transform of 𝕏.

It is clear that κ_{1}(𝕏) = κ(𝕏) is the Gale transform of 𝕏, and κ_{i}(𝕏) is determined up to a coordinate transformation in ℙ^{r}. It seems to be a largely unexplored question how properties of the embedding 𝕏 ⊆ ℙ^{s} are related to properties of κ_{i}(𝕏) ⊆ ℙ^{r}. Our next proposition gives a small result in this direction.

5.10. Proposition. Let 𝕏 ⊆ ℙ^{d} be a set of s points. Then 𝕏 is Δ_{𝕏}-uniform (or (Δ_{𝕏}, σ_{𝕏})-uniform in the terminology of Definition 5.6), if and only if κ_{σ}(𝕏) ⊆ ℙ^{Δ−1} is in linearly general position (i.e. if and only if any subset of Δ_{𝕏} points of κ_{σ}(𝕏) spans ℙ^{Δ−1}).

Proof: By Corollary 1.10, the set κ_{σ}(𝕏) is given by the columns of the matrix 𝔅̃ defined in Remark 5.5. Those columns are the coordinate vectors of Lf_{1}, …, Lf_{s} in the K-basis \{Lf_{1}, …, Lf_{Δ}\} of \overline{R}_{σ+1}. Now the claim follows from [10], 3.4. □

We end this section (and this paper) by pointing out some connections between the material presented above and Coding Theory. Let p > 0 be a prime number, e > 0, q = p^{e}, and 𝔽_{q} the field with q elements. In Coding Theory, a linear subspace C ⊆ 𝔽_{q} is called a linear code. The number s is called the length of C, the number d(C) = dim_{𝔽_{q}}(C) is called the dimension of C, and the minimal number m(C) of nonzero components of a nonzero vector of C is called the minimal distance of C. With those notations, C is also called an [s, d(C)]_{q}-code. It satisfies the Singleton bound m(C) ≤ s − d(C) + 1, and if it achieves equality there, it is called an MDS-code (“maximum distance separable”).

In [8], J.P. Hansen described the following way to associate linear codes to a set of 𝔽_{q}-rational points 𝕏 = {P_{1}, …, P_{s}} ⊆ ℙ^{d}(𝔽_{q}) ∩ D_{+}(X_{0}) ⊆ ℙ_{𝔽_{q}}^{d}. If R = 𝔽_{q}[X_{0}, …, X_{d}]/I_{𝕏} is the homogeneous coordinate ring of 𝕏 and 1 ≤ r ≤ σ_{𝕏}, then the image of the map

\begin{aligned}Φ_{r} : & R_{r} & ⟶ & 𝔽_{q}^{s} \\ & f & ⟼ & (f(P_{1}), …, f(P_{s})) \end{aligned}

is called the r^{th} associated Reed-Muller code of 𝕏 and denoted by C_{r}(𝕏) = im Φ_{r}. In case r = 1, the code C(𝕏) = im Φ_{1} is also called the associated linear code of 𝕏. In [8], Prop. 6 and Thm. 8, the uniformity of 𝕏 and invariants of C_{r}(𝕏) were interrelated as follows.

5.11. Proposition For all 1 ≤ r ≤ σ_{𝕏} and i ≥ 1, we have m(C_{r}(𝕏)) ≥ s − i + 1, if and only if 𝕏 is (s − i, r)-uniform. In particular, C_{r}(𝕏) is an MDS-code, if and only if 𝕏 is (s − H_{𝕏}(r))-uniform.

If we specialize to the case r = σ_{𝕏}, we see that C_{σ}(𝕏) is the linear code generated by the vectors (−β_{j1}, …, −β_{jΔ}, 0, …, 0, 1, 0, …, 0) for j = 1, …, s−Δ. Thus C_{σ}(𝕏) is precisely the kernel of the linear map with matrix 𝔅̃ (defined as in 5.5). In Coding Theory, one says that 𝔅̃ is a parity check matrix for C_{σ}(𝕏). From Propositions 5.10 and 5.11 we get the following relation between the linear code C_{σ}(𝕏) and the canonical transform κ_{σ}(𝕏).

5.12. Corollary. For a set of 𝔽_{q}-rational points 𝕏 ⊆ ℙ_{𝔽_{q}}^{d} as above, the following conditions are equivalent.

a) The linear code C_{σ}(𝕏) is an MDS-code.

b) The set of points 𝕏 is Δ_{𝕏}-uniform.

c) The σ^{th} canonical transform κ_{σ}(𝕏) is in linearly general position.

References

[1] S. Beck and M. Kreuzer, How to compute the canonical module of a set of points, in: L. Gonzáles-Vega and T. Recio (eds.), Algorithms in Algebraic Geometry and Applications, Proc. Conf. MEGA ’94, Santander 1994, Progress in Math. 143, Birkhäuser Verlag, Basel 1996, p. 51 – 78

[2] M.P. Cavaliere, M.E. Rossi, and G. Valla, Quadrics through a set of points and their syzygies, Math. Z. 218 (1995), 25 – 42

[3] M.P. Cavaliere, M.E. Rossi, and G. Valla, On Green-Lazarsfeld and minimal resolution conjecture for n + 3 points in ℙ^{n}, J. Pure Appl. Algebra 85 (1993), 105 – 117

[4] D. Eisenbud, Commutative algebra with a view toward algebraic geometry, Graduate Texts in Math. 150, Springer, New York 1995

[5] D. Eisenbud and S. Popescu, Gale duality and free resolutions of ideals of points, preprint 1996

[6] A.V. Geramita, M. Kreuzer, and L. Robbiano, Cayley-Bacharach schemes and their canonical modules, Trans. Amer. Math. Soc. 339 (1993), 163 – 189

[7] A.V. Geramita, P. Maroscia, and L.G. Roberts, The Hilbert function of a reduced k-algebra, J. London Math. Soc. (2) 28 (1983), 443 – 452

[8] J.P. Hansen, Points in uniform position and maximum distance separable codes, in: F. Orecchia and L. Chiantini (eds.), Zero-Dimensional Schemes, Proc. Conf. Ravello 1992, de Gruyter, Berlin 1994, p. 205 – 211

[9] J. Herzog and E. Kunz, Der kanonische Modul eines Cohen-Macaulay Rings, Lect. Notes in Math. 238, Springer, Heidelberg 1971

[10] M. Kreuzer, On the canonical module of a 0-dimensional scheme, Can. J. Math. 46 (1994), 357 – 379

[11] M. Kreuzer, Some applications of the canonical module of a 0-dimensional scheme, in: F. Orecchia and L. Chiantini (eds.), Zero-Dimensional Schemes, Proc. Conf. Ravello 1992, de Gruyter, Berlin 1994, p. 243 – 252

[12] T. Mora and L. Robbiano, Points in affine and projective spaces, in: D. Eisenbud and L. Robbiano (eds.), Computational Algebraic Geometry and Commutative Algebra, Proc. Conf. Cortona 1991, Symposia Math. 34, Cambridge University Press, Cambridge 1993, p. 106 – 150

[13] F. Orecchia, Points in generic position and the conductor of curves with ordinary singularities, J. London Math. Soc. (2) 24 (1981), p. 85 – 96

Address of the author:

Fakultät für Mathematik Universität Regensburg D-93040 Regensburg, Germany

E-Mail:

Martin.Kreuzer@mathematik.uni-regensburg.de