Multivariate Polynomial Interpolation for Cubical Zero-Dimensional Schemes

Abstract

We compute the multigraded Hilbert function of general unions of certain degree-8 zero-dimensional schemes, called 2cubes, for the Segre 3-folds and the three-dimensional smooth quadric hypersurface. On the Segre 3-folds, we handle the evaluation at the zero-dimensional scheme of all multigraded addenda of their ring, not just the one associated to the Segre embedding. We discuss the Hilbert function and the index of regularity of the unions of several other low-degree zero-dimensional schemes.

1. Introduction

Another possible title for this paper would have been “On the Hilbert function of cubical zero-dimensional subschemes of a threefold”, which capture the tools we use (algebraic geometry and parts of commutative algebra related to multivariate polynomials).

We use multigraded polynomial rings. We describe the three multigraded polynomial rings and their multigraded addenda for three of the projective threefolds used in this paper. For the case of the projective space (Theorem 1), we use the polynomial ring $\mathbb{K}[x_0,x_1,x_2,x_3]$, where $\mathbb{K}$ is an algebraically closed field of characteristic 0, with each variable having a weight of 1. Hence, for all $d\in \mathbb{N}$, we use the $\left({\displaystyle \genfrac{}{}{0pt}{}{d+3}{3}}\right)$-dimensional vector space $\mathbb{K}{[x_0,x_1,x_2,x_3]}_d$ of all degree d forms in four variables.

For ${\mathbb{P}}^1\times {\mathbb{P}}^1\times {\mathbb{P}}^1$, we use the multigraded polynomial ring $\mathbb{K}[x_0,x_1,y_0,y_1,z_0,z_1]$ in which $x_0$ and $x_1$ have multigrade $(1,0,0)$; $y_0$ and $y_1$ have multigrade $(0,1,0)$; and $z_0$ and $z_1$ have multigrade $(0,0,1)$. See Theorem 2 for this case. For each $(d_1,d_2,d_3)\in {\mathbb{N}}^3$, the vector space $\mathbb{K}{[x_0,x_1,y_0,y_1,z_0,z_1]}_{(d_1,d_2,d_3)}$ of all forms of multidegree $(d_1,d_2,d_3)$ is a vector space of dimension $(d_1+1)(d_2+1)(d_3+1)$.

For ${\mathbb{P}}^2\times {\mathbb{P}}^1$ (Theorem 3), we use the vector space $\mathbb{K}[x_0,x_1,x_2,y_0,y_1]$ with $x_0$, $x_1$, and $x_2$ of bidegree $(1,0)$, and $y_0$ and $y_1$ of bidegree $(0,1)$. For all $(a,b)\in {\mathbb{N}}^2$, the set $\mathbb{K}{[x_0,x_1,x_2,y_0,y_1]}_{(a,b)}$ is a vector space of dimension $\left({\displaystyle \genfrac{}{}{0pt}{}{a+2}{2}}\right)(b+1)$.

These vector spaces are also vector spaces of sections of a line bundle, respectively, $H^0({\mathbb{P}}^3,{\mathcal{O}}_{{\mathbb{P}}^3}\left(d\right))$, $H^0({\mathbb{P}}^1\times {\mathbb{P}}^1\times {\mathbb{P}}^1,{\mathcal{O}}_{{\mathbb{P}}^1\times {\mathbb{P}}^1\times {\mathbb{P}}^1}(d_1,d_2,d_3))$, $H^0({\mathbb{P}}^2\times {\mathbb{P}}^1,{\mathcal{O}}_{{\mathbb{P}}^2\times {\mathbb{P}}^1}(a,b))$. These multigraded rings and their higher-dimensional versions are the ones used for the study of partially symmetric tensors ([1]).

If X is any of these varieties and $Z\subset X$ is a zero-dimensional scheme, evaluating the vector space, say $H^0\left({\mathcal{O}}_X(a,b)\right)$, at Z means to identify the kernel $H^0\left({\mathcal{I}}_Z(a,b)\right)$ of the evaluation map at Z or at least its dimension $h^0\left({\mathcal{I}}_Z(a,b)\right)$. In these cases, the cohomology of line bundles on X gives that $h^0\left({\mathcal{I}}_Z(a,b)\right)-h^1\left({\mathcal{I}}_Z(a,b)\right)=h^0\left({\mathcal{O}}_X(a,b)\right)-deg\left(Z\right)$.

Now we describe the zero-dimensional schemes which are used for the evaluation of the multivariate polynomials. We describe connected zero-dimensional schemes. In all such cases, say on the variety X, we describe an isomorphism type, say $\mathcal{Z}(X,1)$, of certain connected zero-dimensional schemes, and then for all positive integer t, we evaluate a general union $A\left(t\right)\subset X$ of a general union of t elements of $\mathcal{Z}(X,1)$. Hence, we only consider the “generic” Hilbert function of unions of t elements of $\mathcal{Z}(X,1)$.

Definition 1.

Take integers $n>m>0$. Let $\mathcal{A}\left(1\right)$ be an irreducible family of zero-dimensional schemes associated to an irreducible family of ideals of finite colength $x>0$ of $\mathbb{K}[x_1,\cdots ,x_m]$ and let $\mathcal{B}\left(1\right)$ be an irreducible family of zero-dimensional schemes associated to an irreducible family of ideals of finite colength $y>0$ of $\mathbb{K}[x_{m+1},\cdots ,x_n]$. Let $\mathcal{Z}\left(1\right)$ be the irreducible family of degree $xy$ schemes of the form $A\times B$, $A\in \mathcal{A}\left(1\right)$ and $B\in \mathcal{B}\left(1\right)$. Let X be an integral n-dimensional projective variety. Fix $P\in X_{\mathrm{reg}}$ and let Z be a zero-dimensional scheme with $Z_{\mathrm{red}}=\left\{P\right\}$. We say that Z is of type $\mathcal{A}\left(1\right)\times \mathcal{B}\left(1\right)$ if there is an isomorphism between $\mathbb{K}\left[[x_1,\cdots ,x_n]\right]$ and the formal completion ${\widehat{\mathcal{O}}}_{X,P}$ of the local ring ${\mathcal{O}}_{X,P}$ such that the ideal of Z is associated to the ideal of and element of $\mathcal{Z}\left(1\right)$.

Take $n=2$ and $m=1$ with $\mathcal{A}\left(1\right)$ of degree-two and $\mathcal{B}\left(1\right)$ of degree three. We say that any $B\in \mathcal{A}\left(1\right)\times \mathcal{B}\left(1\right)$ is a brick. For any irreducible projective surface X and any positive integer t let $\mathcal{B}(X,t)$, and denote the set of all union of t bricks with their reduction t and different smooth points of X. The set $\mathcal{B}(X,t)$ is parametrized by an irreducible quasi-projective variety. Section 8 describes the Hilbert function of the general element of $\mathcal{B}({\mathbb{P}}^2,t)$ (Theorem 5). There are many other interesting connected zero-dimensional schemes on a germ of smooth surfaces, and they help to compute the dimensions of the secant varieties of tangential and osculating varieties of embedded integral surfaces ([2,3,4]). Take $n=2$, $m=1$ with $\mathcal{A}\left(1\right)=\mathcal{B}\left(1\right)$ associated to the family of degree-two connected schemes. These schemes are studied in [5,6].

For Theorems 1–4, we use the following connected degree-eight zero-dimensional schemes.

Definition 2.

Let X be a three-dimensional variety. Fix a smooth point p of X. Let μ be the maximal ideal of the local ring ${\mathcal{O}}_{X,p}$. A degree-eight zero-dimensional scheme is called a 2cube if its colength 8 ideal $I\subset {\mathcal{O}}_{X,p}$ is of the form $(x^2,y^2,z^2)$ for some regular generators x, y, and z of μ.

A 2cube fits in Definition 1 with $n=3$, $m=2$, $x=4$, $y=2$ taking as $\mathcal{A}\left(1\right)$ the family of all 2-squares and as $\mathcal{B}\left(1\right)$ the family of all degree-two connected schemes. Hence, a 2cube is isomorphic to the product of three connected degree-two schemes.

Several other types of zero-dimensional schemes are introduced in [7].

In [6], Prop. 3.10 and 3.11, the authors studied another problem. They described the “worst” Hilbert function for a particular case of zero-dimensional schemes, called tiles, and proved that they have many different indices of regularity. They proved that the set of their indices of regularity contains a connected set of integers, all the integers between the lowest one (with the “worst” Hilbert function in all degrees) and a certain value.

Question 1: Is this the case also for the zero-dimensional schemes, 2cubes and bricks, discussed in the paper you are reading?

Explicitly, identify the lowest index of regularity, classify the schemes with that index of regularity and show that a subset of them gives the worst Hilbert function in all degrees. This is performed for bricks in Section 8 (Propositions 8 and 9).

J. Briançon described all connected zero-dimensional schemes with an embedding dimension of two and degree of, at most, six ([8], Ch. IV). In the last part of Section 8, we discuss this topic for higher-dimensional varieties X. The interested reader may look at all cases. A key step to prove the results for a fixed type of connected zero-dimensional scheme is to determine enough residual sequences to be able to use the so-called Horace Method ([1,9]).

Open Project 2: Adapt these notions to the multigraded case.

In the following four theorems, we prove that the Hilbert function of general unions of 2cubes is as optimized as possible.

Theorem 1.

Take $X:={\mathbb{P}}^3$. Fix positive integers d and t. Let $A\left(t\right)$ be a general union of t 2cubes. Then, either $h^0\left({\mathcal{I}}_{A\left(t\right)}\left(d\right)\right)=0$ or $h^1\left({\mathcal{I}}_{A\left(t\right)}\left(d\right)\right)=0$.

Theorem 2.

Take $X:={\mathbb{P}}^1\times {\mathbb{P}}^1\times {\mathbb{P}}^1$. For all positive integers $d_1$, $d_2$, $d_3$ and t either $h^0\left({\mathcal{I}}_{A\left(t\right)}(d_1,d_2,d_3)\right)=0$ or $h^1\left({\mathcal{I}}_{A\left(t\right)}(d_1,d_2,d_3)\right)=0$, where $A\left(t\right)$ is a general union of t 2cubes.

Theorem 3.

Take $X:={\mathbb{P}}^2\times {\mathbb{P}}^1$. For all positive integers a, b, and t, either $h^0\left({\mathcal{I}}_{A\left(t\right)}(a,b)\right)=0$ or $h^1\left({\mathcal{I}}_{A\left(t\right)}(a,b)\right)=0$, where $A\left(t\right)$ is a general union of t 2cubes.

Theorem 4.

Let $X:=Q_3\subset {\mathbb{P}}^4$ be the smooth quadric 3-fold. For all integers $t>0$ and $d\ge 3$ either $h^0\left({\mathcal{I}}_{A\left(t\right)}\left(d\right)\right)=0$ or $h^1\left({\mathcal{I}}_{A\left(t\right)}\left(d\right)\right)=0$, where $A\left(t\right)$ is a general union of t 2cubes.

Theorems 1–3 cover all Segre 3-folds. Theorem 4 gives the corresponding result for another homogeneous projective 3-fold.

Open project 3: Extend the result to the 3-dimensional complete flag $F(1,2)$.

Open project 4: Up to isomorphisms, there are five (respectively, nine) zero-dimensional schemes of degree five (respectively, six) connected zero-dimensional schemes of a smooth quasi-projective surface ([8], pp. 74–80). In [8], Ch. IV, there is enough information to obtain all possible residual sequence in the sense of Notation 3. From these sequences, the interested reader may compute the Hilbert function of their general unions, say, for $X={\mathbb{P}}^2$ or $X={\mathbb{P}}^1\times {\mathbb{P}}^1$.

Our result are based on an algebraically closed field $\mathbb{K}$ with characteristic zero. In Section 9, we discuss the case of other fields, e.g., the rational or the reals.

Our main theorem requires several lemmas which handle the low-degree cases needed to start the inductive procedure (the Horace Lemma). We easily complete it with full proofs. Our aim for performing this task is to refine the mathematical tools used for interpolations over zero-dimensional schemes. However, for future projects, it seems far better to adapt the existing computer algebra systems to handle exactly these zero-dimensional schemes and the space of multivariate polynomials. It should be faster and the only way when, unless the potential cases are in the dozens or thousands.

In Section 2, we introduce the tools and notation which are often used in these topics when the tools come from algebraic geometry, e.g., residual exact sequences. In Section 3, we give the general lemmas that allow us to compute the Hilbert functions of the unions of several types of zero-dimensional schemes. A key tool is the notion of the residual sequence of a connected zero-dimensional scheme Z. The residual sequence of 2cubes are computed in Remarks 5 and 6 and Proposition 3. The residual sequence of bricks is described in Remark 17. Remark 18 considers this zero-dimensional scheme when they are embedded in manifolds of a higher dimension. In Section 4, Section 5, Section 6 and Section 7, we prove Theorems 1–4, respectively. Bricks and other planar zero-dimensional schemes are studied in Section 8. In Section 9, we briefly consider the case of real zero-dimensional schemes. In the last section, we discuss our results and their possible extensions.

We thank the referees for the suggestions that improved the exposition.

2. Preliminaries

Unless otherwise stated, we work over an algebraically closed base field $\mathbb{K}$ with characteristic zero.

For any projective scheme E, any coherent sheaf F on E and any integer $i\ge 0$, let $H^i(E,F)$ denote the i-th sheaf-theoretic cohomology group of F. Since E is projective, $H^i(E,F)$ is a finite-dimensional $\mathbb{K}$-vector space. Set $h^i(E,F):={dim}_{\mathbb{K}}{H}^i(E,F)$. If E is the 3-fold X, we often write $H^i\left(F\right)$ and $h^i\left(F\right)$ instead of $H^i(X,F)$ and $h^i(X,F)$, respectively. Among the coherent sheaves on E, there are the line bundles.

For any quasi-projective variety T, let $T_{\mathrm{reg}}$ denote the set of its smooth points. For any zero-dimensional scheme $Z\subset T$ let $Z_{\mathrm{red}}$ denote its reduction. The set $Z_{\mathrm{red}}$ is finite, and $P\in Z_{\mathrm{red}}$ if and only if there is a connected component A of Z such that $A_{\mathrm{red}}=\left\{P\right\}$. Now assume that T is projective. For any linear subspace $V\subseteq H^0(T,\mathcal{L})$, let $V(-Z)$ denote the set of all $f\in V$ such that $f_{|Z}=0$, i.e., set $V(-Z):=V\cap H^0(T,{\mathcal{I}}_Z\otimes \mathcal{L})$.

Remark 1.

Let T be an integral projective variety of dimension $n\ge 1$. Let $\mathcal{L}$ be a line bundle on T and $V\subseteq H^0(T,\mathcal{L})$ a linear subspace. Fix zero-dimensional schemes $Z\subset W\subset T$. If $V(-Z)=0$, then $V(-W)=0$. If $dimV(-W)=dimV-deg\left(W\right)$, then $dimV(-Z)=dimV-deg\left(Z\right)$.

Let M be an integral projective variety, $Z\subset M$ a zero-dimensional scheme, and $D\subset M$ an effective Cartier divisor of M. The residual scheme ${\mathrm{Res}}_D\left(Z\right)$ of Z with respect to D is the closed subscheme of M with ${\mathcal{I}}_Z:{\mathcal{I}}_D$ as its ideal sheaf. We have ${\mathrm{Res}}_D\left(Z\right)\subseteq Z$ and $deg\left(Z\right)=deg(Z\cap D)+deg\left({\mathrm{Res}}_D\left(Z\right)\right)$. For all line bundles $\mathcal{L}$ on M, we have an exact sequence, often called the residual exact sequence of D:

$$0\to {\mathcal{I}}_{{\mathrm{Res}}_D\left(Z\right)}\otimes \mathcal{L}(-D)\to {\mathcal{I}}_Z\otimes \mathcal{L}\to {\mathcal{I}}_{Z\cap D,D}\otimes {\mathcal{L}}_{|D}\to 0$$

(1)

Let X be an integral quasi-projective variety. Set $n:=dimX$. Fix $P\in X_{\mathrm{reg}}$ and take a positive integer m. Let $Z\subset X$ be a zero-dimensional scheme. As in [7], Th. 3.3, we say that Z is an m-hypercube of X with P as its reduction if there is a system of local coordinates $x_1,\cdots ,x_n$ of the regular ring ${\mathcal{O}}_{X,x}$ such that Z has $x_1^m,\cdots ,x_n^m$ as its local equations. If Z is an m-hypercube, then $deg\left(Z\right)=m^n$. Let $Z(X_{\mathrm{reg}},m,P)$ denote the set of all m-hypercubes $Z\subset X$ such that $Z_{\mathrm{red}}=\left\{P\right\}$. Set $Z(X_{\mathrm{reg}},m):={\cup }_{P\in X_{\mathrm{reg}}}Z(X_{\mathrm{reg}},m,P)$. The sets $Z(X_{\mathrm{reg}},m,P)$ and $Z(X_{\mathrm{reg}},m)$ are irreducible quasi-projective varieties, and hence it makes sense to study the properties of their general members.

If $P\in X_{\mathrm{reg}}$ and m is a positive integer, the m-point $(mP,X)$ of X is the closed subscheme of X with ${\left({\mathcal{I}}_P\right)}^m$ as its ideal sheaf. We have $(mP,X)=\left\{P\right\}$ and $deg\left((mP,X)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{m+n-1}{n}}\right)$. We often write $mP$ instead of $(mP,X)$ but always write $(mP,Y)$ if Y is a hypersurface of X with $P\in Y_{\mathrm{reg}}$.

Remark 2.

Let T be an integral projective variety of dimension $n\ge 1$. For any $p\in T_{\mathrm{reg}}$ and any positive integer t, the set $\mathcal{C}(T,p,t)$ of all degree t curvilinear subschemes of T with p as their reduction. The set $\mathcal{C}(T,p,t)$ is non-empty, and it is parametrized by an irreducible quasi-projective variety. For all positive integers s and $t_i$, $1\le i\le s$, let $\mathcal{C}(T,t_1,\cdots ,t_s)$ denote the set of all zero-dimensional schemes $A\subset T_{\mathrm{reg}}$ with s connected components, say $A=A_1\cup \cdots \cup A_s$, with $A_i$ curvilinear and $deg\left(A_i\right)=t_i$ for all i. Since $T_{\mathrm{reg}}$ is irreducible and each $\mathcal{C}(T,p,x)$ is irreducible, $\mathcal{C}(T,t_1,\cdots ,t_s)$ is non-empty and irreducible. The short note [10] may be stated in the following way. Let $\mathcal{L}$ be a line bundle on T and $V\subseteq H^0(T,\mathcal{L})$ a linear subspace. Then, $dimV(-Z)=max\{0,dimV-t_1-\cdots -t_s\}$ for a general $Z\in \mathcal{C}(T,t_1,\cdots ,t_s)$.

Remark 3.

Fix a linear space $V\subseteq H^0\left(\mathcal{L}\right)$ such that $dimV\le 2$. Since a 2-point contains a connected degree-two zero-dimensional scheme, Remark 2 gives $V(-2P)=0$ for a general $P\in X_{\mathrm{reg}}$. Hence, $V(-A)=0$ for a general $A\in \mathcal{Z}(X,1)$.

Remark 4.

Fix an integral projective 3-fold X and a line bundle $\mathcal{L}$ on X such that $h^1\left(\mathcal{L}\right)=0$. By Remark 1, to prove that for each $t>0$ either $h^0({\mathcal{I}}_{A\left(t\right)}\otimes \mathcal{L})=0$ or $h^0({\mathcal{I}}_{A\left(t\right)}\otimes \mathcal{L})=0$, it is sufficient to test $t\in \{\lfloor h^0\left(\mathcal{L}\right)/8\rfloor ,\lceil h^0\left(\mathcal{L}\right)/8\rceil \}$. For bricks, i.e., for degree 6 schemes, on any projective variety X, it is sufficient to test the integers $t\in \lfloor h^0\left(\mathcal{L}\right)/6\rfloor ,\lceil h^0\left(\mathcal{L}\right)/6\rceil \}$. For schemes whose connected components have degree z, it is sufficient to test the integers $t\in \{\lfloor h^0\left(\mathcal{L}\right)/z\rfloor ,\lceil h^0\left(\mathcal{L}\right)/z\rceil \}$.

We use the following result ([7], Th. 3.1).

Proposition 1.

For all integers d and t, let $U\left(t\right)\subset {\mathbb{P}}^2$ be a general union of t 2-squares of ${\mathbb{P}}^2$. Then, either $h^0({\mathbb{P}}^2,{\mathcal{I}}_{U\left(t\right)}\left(d\right))=0$ or $h^1({\mathbb{P}}^2,{\mathcal{I}}_{U\left(t\right)}\left(d\right))=0$.

We need the following result ([6], Th. 4.5).

Proposition 2.

Fix $(x,y)\in {\mathbb{N}}^2\setminus \left\{(0,0)\right\}$ and integers $d_1\ge d_2>0$. Let $A\subset {\mathbb{P}}^1\times {\mathbb{P}}^1$ be a general union of x 2-points and y 2-squares. Exclude the case $y=0$, $d_2=2$, $d_1$ even and $x=d_1+1$. Then, either $h^0({\mathbb{P}}^1\times {\mathbb{P}}^1,{\mathcal{I}}_A(d_1,d_2))=0$ or $h^1({\mathbb{P}}^1\times {\mathbb{P}}^1,{\mathcal{I}}_A(d_1,d_2))=0$.

The exceptional case with $y=0$ of Proposition 2 is well known (case $n=0$ of [11], case ${\mathbb{P}}^1\times {\mathbb{P}}^1$ of [12], Th. 3.1).

Notation 1.

For any integral projective 3-dimensional variety X, let $A\left(s\right)$ denote a general union of s 2cubes with the convention $A\left(0\right)=\varnothing$. Note that $A\left(s\right)\subset X_{\mathrm{reg}}$ and $deg\left(A\right(s\left)\right)=8s$. For any integral projective surface Y and any $s\in \mathbb{N}$ let $\left(A\right(s),Y)$ denote the general union of s 2-squares of $Y_{\mathrm{reg}}$.

Notation 2.

Let X be a quasi-projective variety. $Z\subset X$ be a connected a zero-dimensional scheme such that its reduction, P, is a smooth points of X. Let $D\subset X$ be an effective divisor such that $P\in D_{\mathrm{reg}}$. The residual sequence of Z with respect to D is the sequence $(a_1,a_2,\cdots$ with $a_1:=deg(D\cap Z)$ and $a_i:=deg(Z\cap iD)-deg(Z\cap (i-1)D)$ for all $i\ge 1$. We have $a_i\ge a_{i+1}$ for all $i\ge 1$, $a_i=0$ for all $i\gg 0$ and $deg\left(Z\right)={\sum }_{i\ge 1}{a}_i$. Hence, $a_i=0$ for all $i\ge deg\left(Z\right)$ and if $a_i=0$, then $a_j=0$ for all $j>i$. We often write $(a_1,\cdots ,a_i,0)$ if $a_{i+1}=0$.

Let $Z\subset {\mathbb{P}}^n$ be a zero-dimensional scheme, $Z\ne \varnothing$. The index of regularity of Z is the minimal integer $t\ge -1$ such that $h^1\left({\mathcal{I}}_Z\left(t\right)\right)=0$. If Z has index of regularity a and $t>a$, then $h^1\left({\mathcal{I}}_Z\left(t\right)\right)=0$. We discuss the index of regularity of bricks and other zero-dimensional schemes in the last part of Section 8.

3. General Lemmas

Remark 5.

Set $R:=\mathbb{K}\left[\right[x,y,z\left]\right]$. R is a three-dimensional regular local ring. Set $I:=(x^2,y^2,z^2)$ and ${\mathcal{O}}_Z:=R/I$. The degree-eight vector space ${\mathcal{O}}_Z$ is the structural sheaf of the degree-eight zero-dimensional scheme Z, which is a 2cube. The vector space ${\mathcal{O}}_Z$ is spanned by the images of the monomials 1, x, y, z, $xy$, $xz$, $yz$ and $xyz$. For any non-zero linear form $\lambda \in R$, set $H_{\lambda }:=\{\lambda =0\}$.

Write $H_1:=H_z$, $H_2:=H_{y+z}$ and $H_3:=H_{x+y+z}$. Note that $2H_1\supset Z$, $y+z$ has the property that there is another non-zero linear form $y-z$ such that $(y+z)(y-z)\in I$, while no such linear form exists for $H_3$. We have ${Res}_{H_1}\left(Z\right)\subset H_1$ and ${Res}_{H_1}\left(Z\right)$ is a 2-square of $H_1$. We have ${Res}_{H_2}\left(Z\right)\subset H_{y-z}$. Since ${Res}_{H_2}\left(Z\right)$ has degree four and no germ of a smooth curve intersects it with a degree of at least three, ${Res}_{H_2}\left(Z\right)$ is a 2-square of $H_{y-z}$. Since $deg(H_2\cap H_{y-z})=2$, we obtain $deg({Res}_{H_2}\left(Z\right)\cap H_2)=2$ and that ${Res}_{H_2}\left(Z\right)\cap H_2$ is the degree-two connected zero-dimensional scheme, with zero as its reduction and contained in the line $H_2\cap H_{y-z}$. Hence, $deg\left({Res}_{2H_2}\left(Z\right)\right)=2$. Since ${Res}_{H_2}\left(Z\right)\subset H_{y-z}$, we obtain ${Res}_{H_2}\left(Z\right)\cap H_2={Res}_{2H_2}\left(Z\right)$. Note that ${(x+y+z)}^3\equiv xyz$ modulo I, i.e., $deg\left({Res}_{3H_2}\left(Z\right)\right)=1$ and that ${(x+y+z)}^4\in I$. We obtain $deg({Res}_{H_3}\left(Z\right)\cap H_3))=2$ and $deg({Res}_{2H_2}\left(Z\right)\cap H_3)=1$. Hence, the residual sequence of Z with respect to $H_3$ is $(4,2,1,1,0)$.

(a) Take a general Z and set $H_2^{\prime }:=H_{\lambda }$ with λ as the sum of two of the variables whose square defines Z. We claim that the degree-two connected zero-dimensional scheme ${Res}_{H_2^{\prime }}\left(Z\right)\cap H_2^{\prime }$ is the general degree-two connected scheme contained in $H_2^{\prime }$ with P as its reduction. For $H_{y+z}$, the residual scheme is $(2P,H_{y-z}\cap H_2)$. Instead of Z, we take $Z_{\mu }$ with ideal $(x^2,y^2,{(y+\mu z)}^2)$. We have $(y+z)(cy+dz)=a{y}^2+b{(y+\mu z)}^2$ if and only if $c=a+b$, $d={\mu }^2$ and $d+c=2b\mu$. Varying μ, we obtain a moving $H_2\cap H_{cy+dz}$.

(b) Step (a) and the semicontinuity theorem for cohomology give the following observation. For a general Z and a general sum $H_3^{\prime }$ of three of the variables whose square defines Z, the degree-two connected zero-dimensional scheme ${Res}_{H_3^{\prime }}\left(Z\right)\cap H_3^{\prime }$ is the general degree-two connected zero-dimensional scheme contained in $H_2^{\prime }$ with P as its reduction.

(c) Take Z and $H_3^{\prime }$ as in step (b). In this step, we prove that the scheme ${Res}_{2H_3^{\prime }}\left(Z\right)$ is a general degree-two connected zero-dimensional with P as its reduction. Indeed, in the proof of step (a), it is sufficient to take $Z_{\mu +\alpha +\beta }$ defined by the ideal $({(x+\alpha y+\beta z)}^2,y^2,{(y+\mu z)}^2)$.

Notation 3.

Take Z and $H_{\lambda }$ as in Remark 5. If Z has residual sequence $(4,4,0)$ (respectively, $(4,2,2,0)$ and $(4,2,1,1,0)$), then we say that Z is of type I (respectively, type II and type III) with respect to $H_{\lambda }$.

Proposition 3.

Fix a threefold X and $P\in X_{\mathrm{reg}}$. Let $Y\subset X$ be a surface such that $p\in Y_{\mathrm{reg}}$. Let Z be the general 2cube such that $Z_{\mathrm{red}}=\left\{P\right\}$. Then, $deg(Y\cap Z)=3$ and $Z\cap Y=(2P,Y)$.

Proof. 

Since $deg(C\cap Y\cap Z)=2$ for each germ of curve $C\subset Y$ which is smooth at p ([7], Prop. 2.15), $deg(Z\cap Y)\le 4$ ([7], Cor. 2.11). Take the setup of Remark 5 with $P=(0,0,0)$. Let $f\in \mathbb{K}\left[\right[x,y,z\left]\right]$ be an equation of the germ of Y at P. Let $I\subset \mathbb{K}\left[\right[x,y,z\left]\right]$ be the colength eight ideal of Z. Set $J:=(I,f)$. Since $2P\subset Z$, $Y\cap Z$ contains $(2P,Y)$. Hence, to prove the proposition, it is sufficient to prove that J has colength three in $\mathbb{K}\left[\right[x,y,z\left]\right]$. Since Y is smooth at P, the ideal $J_1$ of $Z\cap Y$ in ${\mathcal{O}}_{Y,P}$ has the property that ${\mathcal{O}}_{Y,p}/J_1\cong \mathbb{K}\left[[x,y,z]\right]/J$ contains 1 and exactly 2 linearly independent linear forms. Take $f=x+y+z+xy+xz+yz$. Set $\alpha :=x(y+z)$, $\beta :=y(x+z)$ and $\gamma :=z(x+y)$. Since $xyf\in J$, $xyz\in J$. Since $xf\in J$, $\alpha +xyz\in J$. Since $xyz\in J$, $\alpha \in J$. In the same way, from $yf\in J$ we obtain $\beta \in J$, and from $zf\in J$ we obtain $\gamma \in J$. Since $\alpha -\beta +\gamma =2zx$ and $\alpha -\beta -\gamma =2zy$, we obtain $zx\in J$ and $zy\in J$. In the same way, we see that $xy\in J$. Hence, $x+y+z\in J$ and the vector space $\mathbb{K}\left[\right[x,y,z\left]\right]/J$ is spanned by 1, x, and y. □

Remark 6.

Take Z, Y, and $f=x+y+z+xy+xz+yz$ as in the proof of Proposition 3. In this remark, we compute the residual exact sequence of Z with respect to Y. Note that $f^4\in I$. Set $K:=(f^2,I)$ and $K_1:=(f^3,I)$. We have $K_1=(xyz,I)$ and hence $deg(\mathbb{K}\left[[x,y,z]\right]/K_1)=7$. The ideal K has four linearly independent degree-two forms, and all higher-degree forms are contained in K. Hence, $deg\left(\mathbb{K}\right[[x,y,z]]/K)=6$. Hence, the residual exact sequence of Z with respect to Y is $(3,3,1,1,0)$. The three-dimensional vector space $K/J$ is generated by f, $xf$, and $yf$. Thus, ${Res}_Y\left(Z\right)\cap Z=(2P,Y)$. Note that ${Res}_{2Y}\left(Z\right)$ is a connected degree-two zero-dimensional scheme with P as its reduction.

Remark 7.

Take the setup of the Open project 4. The case $Z_1$ with residual sequence $(z,0)$ is the curvilinear case, and it is covered by Remark 2 for all embedded surfaces. The triple point ([8], case (9) at p. 80) is known for ${\mathbb{P}}^2$ ([9]) and all Hirzebruch surfaces $F_n$ ([11]), the case $n=0$ being the case ${\mathbb{P}}^1\times {\mathbb{P}}^1$).

Proposition 4.

Let X be an integral projective 3-fold, $\mathcal{L}$ a line bundle on X, and $V\subseteq H^0\left(\mathcal{L}\right)$ a linear subspace. Fix a general $P\in X_{\mathrm{reg}}$ and assume $dimV(-3P)\le dimV(-2P)-2$. Then, $V(-A)⊊V(-2P)$ for a general 2cube A of X with P as its reduction.

Proof. 

Let $E\subseteq V(-3P)$ denote the closure of the union of all schemes $A\cap 3P$ with A a 2cube of X with P as its reduction. To prove that $dimV(-3P)\le dimV(-2P)-2$, it is sufficient to prove that E contains a colength 1 subscheme of $3P$. Since $2P$ is contained in any 2cube with P as its reduction, E is contained between $2P$ and $3P$. Take a surface $Y\subset X$ which is smooth at P. Taking $Y\in |{\mathcal{O}}_X\left(c\right)|$ with $c\gg 0$, we may find Y such that the restriction map $V\to V_{|Y}$ is injective. Every 2-square of Y with P as its reduction is the intersection with Y of a 2cube of X. By the generality of P, we may assume that P is general in Y. By Proposition 3 and Remark 5, we have $Y\cap E=Y\cap 3P$. Thus, it is sufficient to prove that by varying T, we cover a colength 1 subscheme of $3P$. This is true because this subscheme of $3P$ is the vanishing of an Hessian matrix (here, we use that the Hessian of a cubic form in three variables identically vanishes if and only if it only depends on two variables, up to a linear change of coordinates [13,14]). □

The following result is characteristic free.

Lemma 1.

Let M be an integral projective variety of dimension $>1$, $\mathcal{L}$ a line bundle, $D\subset M$ an integral Cartier divisor, $Z\subset M$ a zero-dimensional scheme, and x a positive integer. Take a general $S\subset D$ such that $\#S=x$.

(a) Assume $h^0(M,{\mathcal{I}}_{{Res}_D\left(Z\right)}\otimes \mathcal{L}(-D))\le h^0(M,{\mathcal{I}}_Z\otimes \mathcal{L})-x$. Then, we have $h^0(M,{\mathcal{I}}_{Z\cup S}\otimes \mathcal{L})=h^0(M,{\mathcal{I}}_Z\otimes \mathcal{L})-x$.

(b) Assume $h^0(M,{\mathcal{I}}_{{Res}_D\left(Z\right)}\otimes \mathcal{L}(-D))=0$. Then, we have $h^0(M,{\mathcal{I}}_{Z\cup S}\otimes \mathcal{L})=max\{0,h^0(M,{\mathcal{I}}_Z\otimes \mathcal{L})-x\}.$

Proof. 

Since $dimM>1$, $dimD=dimM-1\ge 1$. Since we fixed the general $S\subset D$ after fixing D, $S\cap Z=\varnothing$. Obviously, $h^0(M,{\mathcal{I}}_{Z\cup S}\otimes \mathcal{L})\ge max\{0,h^0(M,{\mathcal{I}}_Z\otimes \mathcal{L})-x\}$.

First assume $x=1$. With this assumption, parts (a) and (b) are true because $h^0(M,{\mathcal{I}}_{Z\cup S}\otimes \mathcal{L})=h^0(M,{\mathcal{I}}_Z\otimes \mathcal{L})$ if and only if D is contained in the base locus of ${\mathcal{I}}_Z\otimes \mathcal{L}$.

Now assume $x>1$ and that the result is true for lower values of x. Take a general $S^{\prime }\subset D$ such that $\#S^{\prime }=x-1$. Since ${\mathrm{Res}}_D(Z\cup S^{\prime })={\mathrm{Res}}_D\left(Z\right)$, the inductive assumption gives $h^0(M,{\mathcal{I}}_{Z\cup S^{\prime }}\otimes \mathcal{L})=max\{0,h^0(M,{\mathcal{I}}_Z\otimes \mathcal{L})-x+1\}$. Take as S the union of $S^{\prime }$ and a general point of D. Apply the case $x=1$ to the zero-dimensional scheme $Z\cup S^{\prime }$. □

Lemma 2.

Fix a line bundle $\mathcal{L}$ on X, a zero-dimensional scheme $A\subset X$, an irreducible Cartier divisor $D\subset X$, and an integer $c>0$. Assume $h^0({\mathcal{I}}_{{Res}_D\left(A\right)}\otimes \mathcal{L}(-D))\le h^0({\mathcal{I}}_A\otimes \mathcal{L})-c$ and $h^0({\mathcal{I}}_{{Res}_{2D}\left(A\right)}\otimes \mathcal{L}(-2D))\le h^0({\mathcal{I}}_A\otimes \mathcal{L})-2c$. Take a general $(p_1,\cdots ,p_c)\in D^c$. For $i=1,\cdots ,c$ let $v_i$ be a general connected degree-two subscheme of X with $p_i$ as its reduction. Then, $h^0({\mathcal{I}}_{A\cup v_1\cup \cdots \cup v_c}\otimes \mathcal{L})=h^0({\mathcal{I}}_A\otimes \mathcal{L})-2c$.

Proof. 

For all integers $1\le j\le c$, set $S\left[j\right]:=\{p_1,\cdots ,p_j\}$ and $V\left[j\right]:=v_1\cup \cdots \cup v_j$. Since each $p_i$ is general in D, $p_i$ is a smooth point of D and hence of X. Since $v_i$ is general with the only restriction that its reduction is $p_i$ and $p_i\in D_{\mathrm{reg}}$, ${\mathrm{Res}}_D\left(v_i\right)=\left\{p_i\right\}$. Lemma 1 for $S_c$ gives $h^0({\mathcal{I}}_{A\cup S\left[c\right]}\otimes \mathcal{L})=h^0({\mathcal{I}}_A\otimes \mathcal{L})-c$. Set $S\left[0\right]=v\left[0\right]:=\varnothing$. We assume that the lemma fails and that c is the first integer for which it fails, i.e., we assume $h^0({\mathcal{I}}_{A\cup V[c-1]}\otimes \mathcal{L})=h^0({\mathcal{I}}_A\otimes \mathcal{L})-2c+2$ and $h^0({\mathcal{I}}_{A\cup V\left[c\right]}\otimes \mathcal{L})\ge h^0({\mathcal{I}}_A\otimes \mathcal{L})-2c+1$.

(a) In this step, we prove that $h^0({\mathcal{I}}_{A\cup V[c-1]\cup p_c}\otimes \mathcal{L})=h^0({\mathcal{I}}_A\otimes \mathcal{L})-2c+1$. Since $p_c$ is general in the irreducible divisor D, to conclude the proof of this step, it is sufficient to prove that D is not contained in the base locus of ${\mathcal{I}}_{A\cup V[c-1]}\otimes \mathcal{L}$. Assume that D is contained in the base locus, i.e., assume $h^0({\mathcal{I}}_A\otimes \mathcal{L})-2c+2=h^0({\mathcal{I}}_{{\mathrm{Res}}_D\left(A\right)\cup S[c-1]}\otimes \mathcal{L}(-D))$. Since $\#S[c-1]=c-1$, we have $h^0({\mathcal{I}}_{{\mathrm{Res}}_D\left(A\right)\cup S[c-1]}\otimes \mathcal{L}(-D))\ge h^0({\mathcal{I}}_{{\mathrm{Res}}_D\left(A\right)}\otimes \mathcal{L}(-D))-c+1$. We obtain $h^0({\mathcal{I}}_{{\mathrm{Res}}_D\left(A\right)}\otimes \mathcal{L})\ge h^0({\mathcal{I}}_A\otimes \mathcal{L})-c+1$, contradicting one of our assumptions.

(b) Step (a) and the assumption $h^0({\mathcal{I}}_{A\cup V\left[c\right]}\otimes \mathcal{L})\ge h^0({\mathcal{I}}_A\otimes \mathcal{L})-2c+1$ give $H^0({\mathcal{I}}_{A\cup V[c-1]\cup p_c}\otimes \mathcal{L})=H^0({\mathcal{I}}_{A\cup V[c-1]\cup v_c}\otimes \mathcal{L})$. Since this is true for all $v_c$ with $p_c$ as its reduction, we obtain that each element of $|{\mathcal{I}}_{A\cup V[c-1]\cup p_c}\otimes \mathcal{L}|$ is singular at $p_c$. Since (for a fixed A and a fixed $V[c-1]$) we may take $p_c$ general in D, we see that each $M\in |{\mathcal{I}}_{A\cup V[c-1]}\otimes \mathcal{L}\left)\right|$)| is singular at all points of $D\cap M$. Consider the restriction $\Delta :=|{\mathcal{I}}_{A\cup V[c-1]}\otimes \mathcal{L}$)${|}_{|D}$ of $|{\mathcal{I}}_{A\cup V[c-1]}\otimes \mathcal{L}$)| to D. We obtain that all members of the linear system $\Delta$ are everywhere singular divisors of D. By the theorem of Bertini, this is only possible if $\Delta =\varnothing$, i.e., if D is contained in the base locus of $|{\mathcal{I}}_{A\cup V[c-1]}\otimes \mathcal{L}\left)\right|$), or if $dim\Delta =0$. Let $\rho :H^0({\mathcal{I}}_{A\cup V[c-1]}\otimes \mathcal{L})\to H^0(D,{\mathcal{I}}_{(A\cap \cup V[c-1\left]\right)\cap D,D}\otimes {\mathcal{L}}_{|D})$ denote the restriction map. We have $\ker \left(\rho \right)=H^0({\mathcal{I}}_{{\mathrm{Res}}_D\left(A\right)\cup S[c-1]}\otimes \mathcal{L}(-D))$. Hence, we obtain the inequality

$$h^0({\mathcal{I}}_{{\mathrm{Res}}_D\left(A\right)\cup S[c-1]}\otimes \mathcal{L}(-D))+1\ge h^0({\mathcal{I}}_{A\cup V[c-1]}\otimes \mathcal{L})=h^0({\mathcal{I}}_A\otimes \mathcal{L})-2c+2.$$

If $c=1$, we obtain $h^0({\mathcal{I}}_{{\mathrm{Res}}_D\left(A\right)}\otimes \mathcal{L}(-D))\ge h^0(A\otimes \mathcal{L})-1$, which is a contradiction. Now assume $c\ge 2$. Let x be the maximal integer in $\{0,\cdots ,c-1\}$ such that

$$h^0({\mathcal{I}}_{{\mathrm{Res}}_D\left(A\right)\cup S\left[x\right]}\otimes \mathcal{L}(-D))=h^0({\mathcal{I}}_{{\mathrm{Res}}_D\left(A\right)}\otimes \mathcal{L}(-D))-x.$$

If $x=c-1$, we obtain $h^0({\mathcal{I}}_{{\mathrm{Res}}_D\left(A\right)\cup S[c-1]}\otimes \mathcal{L}(-D))=h^0({\mathcal{I}}_{{\mathrm{Res}}_D\left(A\right)}\otimes \mathcal{L}(-D))-c+1$, and hence $h^0({\mathcal{I}}_{{\mathrm{Res}}_D\left(A\right)}\otimes \mathcal{L}(-D))\ge h^0({\mathcal{I}}_A\otimes \mathcal{L})-c$, which is a contradiction. Now assume $x\le c-2$. Since $S[c-1]$ is general in D, we obtain

$$\begin{array}{cc}\hfill & h^0({\mathcal{I}}_{{\mathrm{Res}}_D\left(A\right)}\otimes \mathcal{L}(-D))-x=h^0({\mathcal{I}}_{{\mathrm{Res}}_{2D}\left(A\right)}\otimes \mathcal{L}(-2D))\hfill \\ \hfill & h^0({\mathcal{I}}_{{\mathrm{Res}}_D\left(A\right)}\otimes \mathcal{L}(-D))-x+1\ge h^0({\mathcal{I}}_A\otimes \mathcal{L})-2c+2.\hfill \end{array}$$

Hence, we obtain the inequality $h^0({\mathcal{I}}_{{\mathrm{Res}}_{2D}\left(A\right)}\otimes \mathcal{L}(-2D))\ge h^0({\mathcal{I}}_A\otimes \mathcal{L})-2c+1$, which is a contradiction. □

Remark 8.

Lemma 2 is applied to the double residue of a general union of c 2cubes of type III with respect to D. If we require the stronger condition $h^0({\mathcal{I}}_A\otimes \mathcal{L})-h^0({\mathcal{I}}_{{Res}_D\left(A\right)}\otimes \mathcal{L}(-D))\ge 2c$, we may use c type 2cubes and do not involve $\mathcal{L}(-2D)$. We may apply it to 2cubes of residual sequence $(3,3,1,1,0)$ with respect to D.

Remark 9.

Take the setup of Lemma 2 without the assumptions

$$\begin{array}{cc}\hfill & h^0({\mathcal{I}}_{{\mathrm{Res}}_D\left(A\right)}\otimes \mathcal{L}(-D))\le h^0({\mathcal{I}}_A\otimes \mathcal{L})-c\hfill \\ \hfill & h^0({\mathcal{I}}_{{\mathrm{Res}}_{2D}\left(A\right)}\otimes \mathcal{L}(-2D))\le h^0({\mathcal{I}}_A\otimes \mathcal{L})-2c.\hfill \end{array}$$

Set $x:=h^0({\mathcal{I}}_A\otimes \mathcal{L})-h^0({\mathcal{I}}_{{Res}_{2D}\left(A\right)}\otimes \mathcal{L}(-2D))$ and assume $x>2c$. First assume x even. Using $x/2$ instead of c in Lemma 2, we see that $h^0({\mathcal{I}}_{v_1\cup \cdots \cup v_{x/2}\cup A}\otimes \mathcal{L})=h^0({\mathcal{I}}_A\otimes \mathcal{L})-x$ if $h^0({\mathcal{I}}_{{Res}_D\left(A\right)}\otimes \mathcal{L}(-D))\le h^0({\mathcal{I}}_A\otimes \mathcal{L})-x/2$. Now assume x is odd and write $x=2y+1$. To prove that $h^0({\mathcal{I}}_{v_1\cup \cdots \cup v_y\cup p_{y+1}\cup A}\otimes \mathcal{L})=h^0({\mathcal{I}}_A\otimes \mathcal{L})-x$, it is sufficient to assume the inequalities

$$\begin{array}{cc}\hfill & h^0({\mathcal{I}}_{{\mathrm{Res}}_{2D}\left(A\right)}\otimes \mathcal{L}(-2D))\le h^0({\mathcal{I}}_A\otimes \mathcal{L})-2y\hfill \\ \hfill & h^0({\mathcal{I}}_{{\mathrm{Res}}_D\left(A\right)}\otimes \mathcal{L}(-D))\le h^0({\mathcal{I}}_A\otimes \mathcal{L})-y-1.\hfill \end{array}$$

In Section 8, we use the case in which some of the connected components of the zero-dimensional scheme A have residual sequence $(2,2,1,1,0)$ and $(3,1,1,1,0)$ with respect to D. In those cases, we need ${Res}_{xD}\left(A\right)$ for $x=1,2,3,4,5$. In Section 8, we have $dimX=2$, and hence D is an integral curve. By Remark 2 for each $x=0,1,2,3,4,5$, the scheme $D\cap {Res}_{xD}\left(A\right)$ has the expected postulation for the corresponding linear system on D.

4. ${\mathbb{P}}^{\mathbf{3}}$

In this section, we take $X={\mathbb{P}}^3$ and prove Theorem 1. Let $H\subset {\mathbb{P}}^3$ be a plane and let $Q\subset {\mathbb{P}}^3$ be a smooth quadric surface. Let $\mathcal{Z}\left(t\right):=\mathcal{Z}({\mathbb{P}}^3,t)$ denote the set of all unions of t 2cubes of X with, as their reduction, t distinct points of X. Let $A\left(t\right)$ be a general element of $\mathcal{Z}\left(t\right)$.

Remark 10.

We have $h^0\left({\mathcal{O}}_{{\mathbb{P}}^3}\left(d\right)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{d+3}{3}}\right)$. To prove that for all $t>0$ either $h^0\left({\mathcal{I}}_Z\left(t\right)\right)=0$ or $h^1\left({\mathcal{I}}_Z\left(t\right)\right)$ for a general $Z\in \mathcal{Z}\left(t\right)$, it is sufficient to prove it for $t\in \{\lfloor \left({\displaystyle \genfrac{}{}{0pt}{}{d+3}{3}}\right)/8\rfloor ,\lceil \left({\displaystyle \genfrac{}{}{0pt}{}{d+3}{3}}\right)/8\rceil \}$ (Remark 4).

Remark 11.

Obviously, $h^0\left({\mathcal{I}}_Z\left(1\right)\right)=0$ for all $Z\in \mathcal{Z}\left(t\right)$. Since every $Z\in \mathcal{Z}\left(1\right)$ is the complete intersection of two quadric surfaces, $h^0\left({\mathcal{I}}_Z\left(2\right)\right)=2$ for all $Z\in \mathcal{Z}\left(1\right)$, and hence $h^1\left({\mathcal{I}}_Z\left(2\right)\right)=0$ for all $Z\in \mathcal{Z}\left(1\right)$. Obviously, $h^0\left({\mathcal{I}}_Z\left(2\right)\right)=0$ for all $Z\in \mathcal{Z}\left(2\right)$.

Lemma 3.

For all integers $t>0$ either $h^0\left({\mathcal{I}}_{A\left(t\right)}\left(3\right)\right)=0$ or $h^1\left({\mathcal{I}}_{A\left(t\right)}\left(3\right)\right)=0$.

Proof. 

By Remark 10 and the semicontinuity theorem for cohomology, it is sufficient to prove that $h^1\left({\mathcal{I}}_A\left(3\right)\right)=0$ for a general $A\in \mathcal{Z}\left(2\right)$ with $A_{\mathrm{red}}\subset H$ and $h^0\left({\mathcal{I}}_B\left(3\right)\right)=0$ for a general $B\in \mathcal{Z}\left(3\right)$ with $B_{\mathrm{red}}\subset H$, in which all the connected components of A and B are of type III with respect to H. We have $h^0(H,{\mathcal{O}}_H\left(3\right))=10$. By Proposition 1, we have $h^1(H,{\mathcal{I}}_{H\cap A,H}\left(3\right))=0$ and $h^0(H,{\mathcal{I}}_{H\cap B,H}\left(3\right))=0$. Thus, it is sufficient to prove that $h^1\left({\mathcal{I}}_{{\mathrm{Res}}_H\left(A\right)}\left(2\right)\right)=0$ and $h^0\left({\mathcal{I}}_{{\mathrm{Res}}_H\left(B\right)}\left(2\right)\right)=0$. The scheme ${\mathrm{Res}}_H\left(A\right)\cap H$ (respectively, ${\mathrm{Res}}_H\left(B\right)\cap H$) is a general union of two (respectively, three) connected degree-two schemes. Hence, $h^1(H,{\mathcal{I}}_{{\mathrm{Res}}_H\left(A\right)\cap H,H}\left(2\right))=0$ and $h^0(H,{\mathcal{I}}_{{\mathrm{Res}}_H\left(B\right)\cap H,H}\left(2\right))=0$. By the residual exact value of H, it is sufficient to prove that $h^1\left({\mathcal{I}}_{{\mathrm{Res}}_{2H}\left(A\right)}\left(1\right)\right)=0$ and $h^0\left({\mathcal{I}}_{{\mathrm{Res}}_{2H}\left(B\right)}\left(1\right)\right)=0$. These vanishings are obvious accoridng to Remark 2 (or [10]) because any two points of ${\mathbb{P}}^3$ are contained in a plane and hence, up to a projective transformation, A and B are general unions of degree-two connected zero-dimensional schemes. Alternatively, this follows from Lemma 2. □

Lemma 4.

Take a general $A\in \mathcal{Z}\left(4\right)$ and a general $B\in \mathcal{Z}\left(5\right)$. Then, $h^1\left({\mathcal{I}}_A\left(4\right)\right)=0$ and $h^0\left({\mathcal{I}}_B\left(4\right)\right)=0$.

Proof. 

We specialize A to $A_1=A^{\prime }\cup A^{″}$ with $A^{″}\in \mathcal{Z}\left(1\right)$, $A^{″}\cap H=\varnothing$, $A_{\mathrm{red}}^{\prime }\subset H$, three of the connected components of $A^{\prime }$ being of type I with respect to H and four being of type III. Hence, ${\mathrm{Res}}_H\left({Res}_H\left(A^{\prime }\right)\right)={\mathrm{Res}}_{2H}\left(A^{\prime }\right)$ is a general union of degree-four schemes which are double the residue of type III with respect to H. Using twice the residual exact sequence of H, we reduce to prove that $h^1\left({\mathcal{I}}_{A^{″}\cup v}\left(2\right)\right)=0$, where v is a general degree-two connected zero-dimensional scheme with its reduction being a general point of H. Since $\mathrm{Aut}\left({\mathbb{P}}^3\right)$ acts transitively on the partial flag of points and planes, this is a consequence of Remark 2. Alternatively, we may use Lemma 2.

We specialize B to $B_1=A_1\cup B_2$ with $B_2$ a general 2cube with reduction in H and of type I with respect to H. In the two residual exact sequences of H, we obtain the $h^0$-vanishing. In the last step, we use that $h^0\left({\mathcal{I}}_{A^{″}\cup v}\left(2\right)\right)=0$, because $h^1\left({\mathcal{I}}_{A^{″}\cup v}\left(2\right)\right)=0$ and $deg(A^{″}\cup v)=\left({\displaystyle \genfrac{}{}{0pt}{}{5}{2}}\right)$. □

Lemma 5.

We have $h^i\left({\mathcal{I}}_{A\left(7\right)}\left(5\right)\right)=0$, $i=0,1$.

Proof. 

We have $h^0(H,{\mathcal{I}}_H\left(5\right))=21$ and $h^0(H,{\mathcal{I}}_H\left(4\right))=15$. We degenerate $A\left(5\right)$ to $W:=A\cup B\cup A\left(1\right)$ with A general union of three 2cubes of type III with respect to H and three schemes with residual sequence $(3,3,1,1,0)$ with respect to H. Since $W\cap H$ is a general union of three 2-squares and three 2-points, $h^i(H,{\mathcal{I}}_{W\cap H,H}\left(5\right))=0$. Hence, the residual exact sequence of H shows that it is sufficient to prove $h^i\left({\mathcal{I}}_{{\mathrm{Res}}_H\left(W\right)}\left(4\right)\right)=0$. The scheme $H\cap {\mathrm{Res}}_H\left(W\right)$ is a general union of three degree-two connected schemes and three 2-points. Hence, $h^i(H,{\mathcal{I}}_{H\cap {\mathrm{Res}}_H\left(W\right),H}\left(4\right))=0$, $i=0,1$. Hence, it is sufficient to prove that $h^i\left({\mathcal{I}}_{{\mathrm{Res}}_{2H}\left(W\right)}\left(3\right)\right)=0$, $i=0,1$. The scheme ${\mathrm{Res}}_{2H}\left(W\right)$ is a general union of $A\left(1\right)$ and six degree-two connected degree-two schemes with the only restriction being that their reduction is contained in H. We have $h^1\left({\mathcal{I}}_{A\left(1\right)}\left(2\right)\right)=0$. Use Lemma 2. □

Lemma 6.

Take a general $A\in \mathcal{Z}\left(10\right)$ and a general $B\in \mathcal{Z}\left(11\right)$. Then, $h^1\left({\mathcal{I}}_A\left(6\right)\right)=0$ and $h^0\left({\mathcal{I}}_B\left(6\right)\right)=0$

Proof. 

We specialize A to $A_1=A^{\prime }\cup A^{″}$ with $A^{″}$ general in $\mathcal{Z}\left(3\right)$ and $A_{\mathrm{red}}^{\prime }$ general in H. The generality of $A^{″}$ gives $A^{″}\cap H=\varnothing$. Since $h^0(H,{\mathcal{O}}_H\left(5\right))=21$, we impose that three of the connected components of $A^{\prime }$ are type I with respect to H, one of them being of type III. Using twice the residual exact sequence of H, we reduce to prove that $h^1\left({\mathcal{I}}_{A^{″}\cup w}\left(2\right)\right)=0$, where w is a general union of four degree-two connected zero-dimensional schemes with the only restriction being that its reduction is contained in H. This is performed as in the proof of Lemma 4 using Remark 2 and H.

We specialize B to $B_1=B^{\prime }\cup B^{″}$ with $B^{″}$ general in $\mathcal{Z}\left(4\right)$$B_{\mathrm{red}}^{\prime }\subset H$, with four of the connected components of $B^{\prime }$ being of type I with respect to H and the other ones being of type III with respect to H. It is sufficient to use twice the residual exact sequence of H. □

Lemma 7.

We have $h^i\left({\mathcal{I}}_{A\left(15\right)}\left(7\right)\right)=0$, $i=0,1$.

Proof. 

We have $h^0(H,{\mathcal{O}}_H\left(7\right))=36$. We specialize $A\left(15\right)$ to $A_1=A\cup A\left(6\right)$ with A being a general union of one 2cube of type I and eight 2cubes of type III with respect to H. Since $A\left(6\right)\cap H=\varnothing$, $h^i(H,{\mathcal{I}}_{A\cap H,H}\left(7\right))=0$, $i=0,1$ (Proposition 1). Hence, it is sufficient to prove that $h^i\left({\mathcal{I}}_{{\mathrm{Res}}_H\left(A\right)\cup A\left(6\right)}\left(6\right)\right)=0$. We have $h^0(H,{\mathcal{O}}_H\left(6\right))=28$. We specialize $A\left(6\right)$ to $A_1:=B\cup A\left(4\right)$ with B being two general 2cubes of type III with respect to H. Since $h^i(H,{\mathcal{I}}_{H\cap {\mathrm{Res}}_H\left(A\right)\cup A_1}\left(6\right))=0$, $i=0,1$, it is sufficient to prove that $h^i\left({\mathcal{I}}_{{\mathrm{Res}}_{2H}\left(A\right)\cup {\mathrm{Res}}_H\left(B\right)\cup A\left(4\right)}\left(5\right)\right)=0$, $i=0,1$. We have $h^1\left({\mathcal{I}}_{A\left(4\right)}\left(x\right)\right)=0$ for $x=4,5$ (Lemma 4). Set $W:={\mathrm{Res}}_{2H}\left(A\right)\cup {\mathrm{Res}}_H\left(B\right)$. The scheme $W\cap H$ is a general union of two 2-squares and eight points of H. Hence, $h^1(H,{\mathcal{I}}_{W\cap H}\left(5\right))=0$. The residual exact sequence of H gives $h^1\left({\mathcal{I}}_{A\left(4\right)\cup (W\cap H)}\left(5\right)\right)=0$. Since ${\mathrm{Res}}_{2H}\left(B\right)$ and ${\mathrm{Res}}_{2H}\left(B\right)$ are a general union of two degree-two connected zero-dimensional schemes with their reduction contained in H, it is sufficient to use Lemma 2. □

Lemma 8.

Take a general $A\in \mathcal{Z}\left(20\right)$ and a general $B\in \mathcal{Z}\left(21\right)$. Then, $h^1\left({\mathcal{I}}_A\left(8\right)\right)=0$ and $h^0\left({\mathcal{I}}_B\left(8\right)\right)=0$.

Proof. 

Recall that $\left({\displaystyle \genfrac{}{}{0pt}{}{11}{3}}\right)=165$, $h^0(Q,{\mathcal{O}}_Q\left(8\right))=81$ and $h^0(Q,{\mathcal{O}}_Q\left(6\right))=49$. We specialize A to $A^{\prime }$ with $A_{\mathrm{red}}^{\prime }$ general in Q, 4 (respectively, 16) of the connected components of $A^{\prime }$ being of type I with respect to Q. Using twice the residual exact sequence of Q, we reduce to prove that $h^1\left({\mathcal{I}}_w\left(4\right)\right)=0$, where w is a general union of 16 connected degree-two schemes, with the only restriction being that $w_{\mathrm{red}}\subset Q$. Using twice Q as in the proof of Lemma 4, we obtain $h^0\left({\mathcal{I}}_w\left(4\right)\right)=3$. Hence, $h^1\left({\mathcal{I}}_A\left(8\right)\right)=0$.

We specialize B to $E\cup F$ with E being a general union of 2 2cubes of type I and 16 2cubes of type III with respect to Q, and F being a general union 3 2cubes with residual sequence $(3,3,1,1,0)$ with respect to Q.

We have $h^0(Q,{\mathcal{O}}_Q\left(6\right))=49$. The scheme $Q\cap ({\mathrm{Res}}_Q\left(E\right)\cup {\mathrm{Res}}_Q\left(F\right))$ is a general union of 2 2-squares, 16 connected degree-two schemes and 3 2-points. Hence, $h^i(Q,{\mathcal{I}}_{\cap \left({\mathrm{Res}}_Q(E\cup F)\right),Q}\left(6\right))=0$, $i=0,1$. The residual exact sequence of Q shows that it is sufficient to prove that $h^0\left({\mathcal{I}}_{{\mathrm{Res}}_{2Q}(E\cup F)}\left(4\right)\right)=0$. We have $h^0\left({\mathcal{O}}_X\left(4\right)\right)=35$. The scheme ${\mathrm{Res}}_{2Q}(E\cup F$ is a general union of 19 degree-two connected schemes with the only restriction being that their reduction is contained in Q. Since $h^0(Q,{\mathcal{O}}_Q\left(4\right))=25\ge 19$, it is sufficient to apply Remark 2 and/or Lemma 2 and Remark 9. □

Lemma 9.

Fix an odd integer $d\ge 9$. If Theorem 1 is true for the degrees $d-2$, $d-4$, and $d-6$, then it is true for the integer d.

Proof. 

Set $e:={(d+1)}^2/4$. Since d is odd, e is a positive integer. Since $d\ge 7$, we have $2e\le {(d-1)}^2$. By Remark 10, it is sufficient to test a general $Z\in \mathcal{Z}\left(t\right)$ with $t\in \{\lfloor \left({\displaystyle \genfrac{}{}{0pt}{}{d+3}{3}}\right)/8\rfloor ,\lceil \left({\displaystyle \genfrac{}{}{0pt}{}{d+3}{3}}\right)/8\rceil \}$. Hence, we may assume $t\ge e$.

Let $Q\subset {\mathbb{P}}^3$ be a smooth quadric. Fix a general $S\subset Q$ such that $\#S=e$. Set $f:=({(d-1)}^2/2-2e)/2$. Since d is even, f is an integer. We have $2e+2f={(d-1)}^2$. Since $4e={(d+1)}^2$, $0<f<e$. We degenerate Z to $W:=A\cup F$, where F is a general element of $\mathcal{Z}(t-e)$ (and, in particular, $F\cap Q=\varnothing$) and A is a general union of 2cubes with $A_{\mathrm{red}}=S$, with the restriction that f of the connected components of A are of type I with respect Q. By the semicontinuity theorem for cohomology, it is sufficient to prove that either $h^0\left({\mathcal{I}}_W\left(d\right)\right)=0$ or $h^1\left({\mathcal{I}}_W\left(d\right)\right)=0$. By the generality of A (with the given restriction for f of its connected components), the other $e-f$ connected components are of type III, and $Q\cap A$ is a general union of each connected component of A of type III with respect to Q in the sense of Remark 5. Since S is general in Q and A is general, with the only restriction that $A_{\mathrm{red}}=S$ and we prescribe the type of its connected components, $A\cap Q$ is a general union of e 2-squares of Q. Since $4e=h^0(Q,{\mathcal{O}}_Q\left(d\right))$, Proposition 2 gives $h^i(Q,{\mathcal{I}}_{A\cap Q,Q}\left(d\right))=0$, $i=0,1$. Since $F\cap Q=\varnothing$, the residual exact sequence of Q gives $h^i\left({\mathcal{I}}_W\left(d\right)\right)=h^i\left({\mathcal{I}}_{F\cup {\mathrm{Res}}_Q\left(A\right)}(d-2)\right)$, $i=0,1$. By the generality of S and A with the prescribed types, $Q\cap {\mathrm{Res}}_Q\left(A\right)$ is a general union of f 2-squares and $e-f$ connected degree-two zero-dimensional schemes. Proposition 2 and Remark 2 give $h^i(Q,{\mathcal{I}}_{{\mathrm{Res}}_Q\left(A\right)\cap Q}(d-2))=0$, $i=0,1$. Hence, $h^i\left({\mathcal{I}}_W\left(d\right)\right)=h^i\left({\mathcal{I}}_{F\cup {\mathrm{Res}}_{2Q}\left(A\right)}(d-4)\right)$, $i=0,1$. By assumption for $x\in \{4,6\}$, either $h^0\left({\mathcal{I}}_F(d-x)\right)=0$ or $h^1\left({\mathcal{I}}_F(d-x)\right)=0$. If $h^0\left({\mathcal{I}}_D(d-4)\right)=0$, then $h^0\left({\mathcal{I}}_{F\cup {\mathrm{Res}}_{2Q}\left(A\right)}(d-4)\right)=0$, concluding the proof in this case. Thus, we may assume $h^1\left({\mathcal{I}}_F(d-4)\right)=0$. Hence, $h^0\left({\mathcal{I}}_F(d-4)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{d-1}{3}}\right)-8(t-e)$. The scheme ${\mathrm{Res}}_{2Q}\left(A\right)$ is a general union of $e-f$ connected degree-two zero-dimensional schemes of Q (Remark 5). By Remark 8, it is sufficient to prove that either $h^0\left({\mathcal{I}}_F(d-6)\right)=0$ or $\left({\displaystyle \genfrac{}{}{0pt}{}{d-1}{3}}\right)-8(t-e)-\left({\displaystyle \genfrac{}{}{0pt}{}{d-3}{t}}\right)-8(t-e)\ge 2(e-f)$, i.e., ${(d-3)}^2\ge 2(e-f)$. Since $f\ge 0$ and $e={(d+1)}^2/4$, for $d\ge 13$, it is sufficient to use that $2{(d-3)}^2\ge {(d+1)}^2$. Now assume $d=11$ and hence $e=36$. We have $f=14$, and hence $2(e-f)=44\le 8^2$. Now assume $d=9$, and hence $e=25$ and $f=7$. We have $2(e-f)=36\le 36$. □

Lemma 10.

Fix an even integer $d\ge 10$ and assume that for all positive integers t and x such that $x\in \{d-8,d-6,d-4\}$ either $h^0\left({\mathcal{I}}_{A\left(t\right)}\left(x\right)\right)=0$ or $h^1\left({\mathcal{I}}_{A\left(t\right)}\left(x\right)\right)=0$. Then, for each positive integer t, either $h^0\left({\mathcal{I}}_{A\left(t\right)}\left(d\right)\right)=0$ or $h^1\left({\mathcal{I}}_{A\left(t\right)}\left(d\right)\right)=0$.

Proof. 

By Remark 10 we may assume $t\in \{\lfloor \left({\displaystyle \genfrac{}{}{0pt}{}{d+3}{3}}\right)/8\rfloor ,\lceil \left({\displaystyle \genfrac{}{}{0pt}{}{d+3}{3}}\right)/8\rceil \}$. Recall that for each even integer $2y$, we have $h^0(Q,{\mathcal{O}}_Q\left(2y\right))={(2y+1)}^2=4y^2+4y+1\equiv 1(mod8)$. Hence, $u:=(h^0(Q,{\mathcal{O}}_Q\left(d\right))-9)/4$ is a positive integer.

(a) In this step, we prove that $t\ge u+3$. Since $t\ge \lfloor \left({\displaystyle \genfrac{}{}{0pt}{}{d+3}{3}}\right)/8\rfloor$, it is sufficient to check that $8u+24\le \left({\displaystyle \genfrac{}{}{0pt}{}{d+3}{3}}\right)$. We have $8u+24=6+{(d+1)}^2$. Use that $d\ge 4$.

(b) In this step, we prove that $u\ge 2d$. Indeed, it is sufficient to observe that $8d+9\le {(d+1)}^2$ for all $d\ge 10$.

(c) Take a general $(S,S^{\prime })\subset Q\times Q$ such that $\#S=u$ and $\#S^{\prime }=v$. We degenerate $A\left(t\right)$ to $A(t-u-3)\cup A\cup B$ with A being a general union of $u-2d$ 2cubes of type I with respect to the hypersurface Q and $2d$ 2cubes of type III with respect to Q, and B being a general union of three 2cubes with residual sequence $(3,3,1,1,0)$ with respect to the hypersurface Q. Note that we have $A(t-u-3)\cap Q=\varnothing$ and $h^i(Q,{\mathcal{I}}_{Q\cap (A\cup B),Q}\left(d\right))=0$, $i=0,1$ (Proposition 2). Thus, the residual exact sequence of Q shows that it is sufficient to prove that either $h^0\left({\mathcal{I}}_{A(t-u-3)\cup {\mathrm{Res}}_Q(A\cup B)}(d-2)\right)=0$ or $h^1\left({\mathcal{I}}_{A(t-u-3)\cup {\mathrm{Res}}_Q(A\cup B)}(d-2)\right)=0$. The scheme $Q\cap {\mathrm{Res}}_Q(A\cup B)$ is a general union of $u-2d$ 2-squares, $2d$ degree-two connected zero-dimensional schemes and two 2-points. Hence, $h^i(Q,{\mathcal{I}}_{{\mathrm{Res}}_Q(A\cup B),Q}(d-2))=0$ (Proposition 2). Hence, by the residual exact sequence of Q, it is sufficient to prove that either $h^0\left({\mathcal{I}}_{A(t-u-3)\cup {\mathrm{Res}}_{2Q}(A\cup B)}(d-4)\right)=0$ or $h^1\left({\mathcal{I}}_{A(t-u-3)\cup {\mathrm{Res}}_{2Q}(A\cup B)}(d-4)\right)=0$. Since ${\mathrm{Res}}_{2Q}(A\cup B)$ is a general union of $2d+3$ connected degree-two schemes with the only restriction being that their reduction is contained in Q, it is sufficient to use Lemma 2 and Remark 9 and that for each $x\in \{d-8,d-6,d-4\}$ either $h^0\left({\mathcal{I}}_{A(t-u-3)}\left(x\right)\right)=0$ or $h^1\left({\mathcal{I}}_{A(t-u-3)}\left(x\right)\right)=0$. □

Proof of Theorem 1: 

The case $d=1,2$ is Remark 11. The cases $d=3,4,5,6,7,8$ are Lemmas 3–8. Lemma 9 is the case $d\ge 9$ and odd. Lemma 10 is the case $d\ge 10$ and even. □

5. $\mathbf{X}={\mathbb{P}}^{\mathbf{1}}\times {\mathbb{P}}^{\mathbf{1}}\times {\mathbb{P}}^{\mathbf{1}}$

In this section, we prove Theorem 2. Hence, in this section, we take $X={\mathbb{P}}^1\times {\mathbb{P}}^1\times {\mathbb{P}}^1$ with $\mathcal{L}={\mathcal{O}}_X(d_1,d_2,d_3)$, $d_i>0$, $i=1,2,3$, and $H^0({\mathcal{O}}_X(d_1,d_2,d_3)$ as the linear space for which we perform interpolation. We have $h^0\left(\mathcal{L}\right)=(d_1+1)(d_2+1)(d_3+1)$. Take homogeneous coordinates $x_0,x_1$ (respectively, $y_0,y_1$ and $z_0,z_1$) of the first (respectively, second and third) factor of X. We have $H^0\left({\mathcal{O}}_X(d_1,d_2,d_3)\right)=\mathbb{K}{[x_0,x_1,y_0,y_1,z_0,z_1]}_{(d_1,d_2,d_3)}$. Recall that $h^0\left({\mathcal{O}}_X(d_1,d_2,d_3)\right)=(d_1+1)(d_2+1)(d_3+1)$ for all $(d_1,d_2,d_3)\in {\mathbb{N}}^3$. Let $\mathcal{Z}\left(t\right)$ be the variety of all unions of t disjoint 2cubes with the convention $\mathcal{Z}\left(0\right)=\varnothing$. Let $A\left(t\right)$ denote the general element of $\mathcal{Z}\left(t\right)$.

Remark 12.

Fix positive integers $d_1$, $d_2$ and $d_3$. To prove that for all positive integer t, either $h^0\left({\mathcal{I}}_Z(d_1,d_2,d_3)\right)=0$ or $h^1\left({\mathcal{I}}_Z(d_1,d_2,d_3)\right)=0$ for a general $Z\in \mathcal{Z}\left(t\right)$, it is sufficient to prove it for the integers $t=\lfloor (d_1+1)(d_2+1)(d_3+1)\rfloor$ and $t=\lceil (d_1+1)(d_2+1)(d_3+1)\rceil$ (Remark 4).

Lemma 11.

We have $h^i\left({\mathcal{I}}_{A\left(1\right)}(1,1,1)\right)=0$, $i=0,1$.

Proof. 

Since no element of $H^0\left({\mathcal{O}}_X(1,1,1)\right)$ is the square of a non-zero linear form, $h^0\left({\mathcal{I}}_{A\left(1\right)}(1,1,1)\right)=0$. Since $h^0\left({\mathcal{O}}_X(1,1,1)\right)=8=deg\left(A\left(1\right)\right)$, $h^1\left({\mathcal{I}}_{A\left(1\right)}(1,1,1)\right)=0$. □

Lemma 12.

We have $h^1\left({\mathcal{I}}_{A\left(1\right)}(2,1,1)\right)=0$ and $h^0\left({\mathcal{I}}_{A\left(2\right)}(2,1,1)\right)=0$.

Proof. 

We have $h^0\left({\mathcal{O}}_X(2,1,1)\right)=12$. Take $T\in |{\mathcal{O}}_X(1,0,0)|$. We have $T\cong {\mathbb{P}}^1\times {\mathbb{P}}^1$, $h^0(T,{\mathcal{O}}_T(2,1,1))=h^0\left({\mathcal{O}}_X(2,1,1)\right)-h^0\left({\mathcal{O}}_X(1,1,1)\right)=4$ and the isomorphism between T and ${\mathbb{P}}^1\times {\mathbb{P}}^1$ maps ${\mathcal{O}}_{{\mathbb{P}}^1\times {\mathbb{P}}^1}(1,1)$ to ${\mathcal{O}}_T(1,1,1)\cong {\mathcal{O}}_T(2,1)$. We specialize $A\left(1\right)$ to a 2cube of type I with respect to T. Since $h^0\left({\mathcal{O}}_X(0,1,1)\right)=4$, applying twice the residual exact sequence of T, we obtain $h^1\left({\mathcal{I}}_{A\left(1\right)}(2,1,1)\right)=0$. Take $T_1\in \left|{\mathcal{O}}_X(1,0,0)\right|$ such that $T_1\ne T$. We specialize $A\left(2\right)$ to $A\cup A_1$ with A of type III with respect to T and $A_1$ of type III with respect to $T_1$. Using first the residual exact sequence of T and then the one of $T_1$, we reduce to prove that $h^0\left({\mathcal{I}}_{{\mathrm{Res}}_T\left(A\right)\cup {\mathrm{Res}}_{T_1}\left(A_1\right)}(0,1,1)\right)=0$. We have $h^i\left({\mathcal{I}}_{{\mathrm{Res}}_{2T}\left(A\right)\cup {\mathrm{Res}}_{2T_1}\left(A_1\right)}(0,1,1)\right)$, $i=0,1$ by Lemma 2. □

Lemma 13.

We have $h^1\left({\mathcal{I}}_{A\left(2\right)}(2,2,1)\right)=0$ and $h^0\left({\mathcal{I}}_{A\left(3\right)}(2,2,1)\right)=0$.

Proof. 

We have $h^0\left({\mathcal{O}}_X(2,2,1)\right)=18$. By Remark 2, it is sufficient to prove that $h^1\left({\mathcal{I}}_{A\left(2\right)}(2,2,1)\right)=0$. We degenerate $A\left(2\right)$ to a general union A of 2 cubes of type I with respect to D. We obtain $h^1(D,{\mathcal{I}}_{D\cap A}(2,2,1))=h^1(D,{\mathcal{I}}_{{\mathrm{Res}}_D\left(A\right)}(2,2,0))=0$. We conclude, because $h^i\left({\mathcal{O}}_X(2,2,-1)\right)=0$, $i=0,1$. □

Remark 13.

Fix $(d_1,d_2,d_3)\in {\mathbb{N}}^2$. Take $D\in |{\mathcal{O}}_X(0,0,1)|$. We have $D\cong {\mathbb{P}}^1\times {\mathbb{P}}^1$ with an isomorphism identifying ${\mathcal{O}}_D(d_1,d_2,d_3)$ with ${\mathcal{O}}_{{\mathbb{P}}^1\times {\mathbb{P}}^1}(d_1,d_2))$. The Künneth formula gives $h^1\left({\mathcal{O}}_X(d_1,d_2,d_3-1)\right)=0$. Hence, the long cohomology exact sequence of the exact sequence

$$0\to {\mathcal{O}}_X(d_1,d_2,d_3-1)\to {\mathcal{O}}_X(d_1,d_2,d_3)\to {\mathcal{O}}_D(d_1,d_2,d_3)\to 0$$

gives

$$h^0(D,{\mathcal{O}}_D(d_1,d_2,d_3))=h^0\left({\mathcal{O}}_X(d_1,d_2,d_3)\right)-h^0\left({\mathcal{O}}_X(d_1,d_2,d_3-1)\right)=(d_1+1)(d_2+1).$$

Lemma 14.

We have $h^1\left({\mathcal{I}}_{A\left(3\right)}(2,2,2)\right)=0$ and $h^0\left({\mathcal{I}}_{A\left(4\right)}(2,2,2)\right)=0$.

Proof. 

We have $h^0\left({\mathcal{O}}_X(2,2,2)\right)=27$. Take $D\in |{\mathcal{O}}_X(0,0,1)|$. Remark 13 gives $h^0(D,{\mathcal{O}}_D(2,2,2))=27-18=9$ and $h^0(D,{\mathcal{O}}_D(2,2,1))=9$.

We specialize $A\left(3\right)$ to $A_1:=A\cup A\left(1\right)$ with A being a general union of two 2cubes of type III with respect to D. By Proposition 2, we have $h^1(D,{\mathcal{I}}_{A_1\cap D,D}(2,2,2))=0$ and $h^0(D,{\mathcal{I}}_{A_1\cap D,D}(2,2,2))=1$. Then, we specialize ${\mathrm{Res}}_D\left(A_1\right)$ to $A_2:={\mathrm{Res}}_D\left(A\right)\cup A^{\prime }$ with $A^{\prime }$ being a general 2cube of type I with respect to D. By Proposition 2 and Remark 2, we have $h^1(D,{\mathcal{I}}_{A_2\cap D,D}(2,2,1))=0$ and $h^1(D,{\mathcal{I}}_{A_2\cap D,D}(2,2,1))=0$. Hence, to conclude the proof of the vanishing of $h^1\left({\mathcal{I}}_{A\left(3\right)}(2,2,2)\right)$, it is sufficient to prove that $h^1\left({\mathcal{I}}_{{\mathrm{Res}}_{2D}\left(A\right)\cup {\mathrm{Res}}_D\left(A^{\prime }\right)}(2,2,0)\right)=0$. Since $h^i\left({\mathcal{O}}_X(2,2,-1)\right)=0$, $i=0,1$, the long cohomology exact sequence of the residual sequence of D shows that it sufficient to prove that $h^1(D,{\mathcal{I}}_{({\mathrm{Res}}_{2D}\left(A\right)\cup {\mathrm{Res}}_D\left(A^{\prime }\right)})\cap D,(2,2,0))=0$. Since ${\mathrm{Res}}_{2D}\left(A\right)$ are two general points of D, it is sufficient to use that ${\mathrm{Res}}_D\left(A\right)$ is a 2-square of D.

To prove that $h^0\left({\mathcal{I}}_{A\left(4\right)}(2,2,2)\right)=0$, we modify the case for $A\left(3\right)$, adding at the beginning another 2cube $A^{\prime \prime }$ of type III with respect to D but taking $A^{\prime }$ of type III with respect to D. At the third step for $H^0\left({\mathcal{O}}_X(2,2,0)\right)$, we have two general points of D, a general degree-two scheme with, as its reduction, a point of D. □

Proposition 5.

Assume that at least two of the integers $d_i$ are odd. Then, for each t either $h^1\left({\mathcal{I}}_{A\left(t\right)}(d_1,d_2,d_3)\right)=0$ or $h^0\left({\mathcal{I}}_{A\left(t\right)}(d_1,d_2,d_3)\right)=0$.

Proof. 

With no loss of generality, we may assume that $d_1$ and $d_2$ are odd. Since $d_1$ and $d_2$ are odd, $(d_1+1)(d_2+1)\equiv 0(mod4)$. Set $e:=(d+1)(d_2+1)/4$. By Remark 12, it is sufficient to test the cases $t\in \{\lfloor (d_1+1)(d_2+1)(d_3+1)/8\rfloor ,\lceil d_1+1)(d_2+1)(d_3+1)/8\rceil \}$. Note that $t\ge e$ with equality if and only if $d_3=1$. Take $D\in |{\mathcal{O}}_X(0,0,1)|$. We have $D\cong {\mathbb{P}}^1\times {\mathbb{P}}^1$.

(a) Assume $d_3$ to be odd. We degenerate $A\left(t\right)$ to $A\cup A(t-e)$ with A a general union of e 2cubes of type I with respect to D. First assume $d_3=1$. Since $h^i\left({\mathcal{O}}_X(d_1,d_2,-1)\right)=0$, $i=0,1$, we obtain $h^i\left({\mathcal{I}}_A(d_1,d_2,1)\right)=0$, $i=0,1$, concluding this case. Now assume $d_3\ge 3$ and that the theorem is true for ${\mathcal{O}}_X(d_1,d_2,d_3-2)$. Using twice the residual exact sequence of D, we complete the proof of this case with the inductive assumption $h^i\left({\mathcal{I}}_{A(t-e)}(d_1,d_2,d_3-2)\right)=0$, $i=0,1$.

(b) Assume $d_3$ is even. Hence, $d_3\ge 2$. Using the construction of step (a), we reduce to the case $d_3=2$. We have $h^0\left({\mathcal{O}}_X(d_1,d_2,2)\right)=3(d+1)(d_2+1)\equiv (d_1+1)(d_2+1)(mod8)$. We degenerate the scheme $A\left(t\right)$ to $B\cup A(t-e)$, with B being a general union e 2cubes of type II with respect to D. By the residual exact sequence of D, we reduce to prove that either $h^0\left({\mathcal{I}}_{{\mathrm{Res}}_D\left(B\right)\cup A(t-e)}(d_1,d_2,1)\right)=0$ or $h^1\left({\mathcal{I}}_{{\mathrm{Res}}_D\left(B\right)\cup A(t-e)}(d_1,d_2,1)\right)=0$.

(b1) Assume that the integer $e/4$ is even. In this case, we have $t=3e/2$ and $h^0\left({\mathcal{I}}_{{\mathrm{Res}}_D\left(B\right)\cup A(t-e)}(d_1,d_2,1)\right)=h^1\left({\mathcal{I}}_{{\mathrm{Res}}_D\left(B\right)\cup A(t-e)}(d_1,d_2,1)\right)$. We degenerate the scheme ${\mathrm{Res}}_D\left(B\right)\cup A(t-e)$ to ${\mathrm{Res}}_D\left(B\right)\cup E$, with E being a general union of $e/2$ 2cubes of type I with respect to D. Since $({\mathrm{Res}}_D\left(B\right)\cup E)\cap D$ is a general union of $e/2$ 2-squares and e degree-two connected zero-dimensional subschemes of D, Proposition 2 and Remark 2 give $h^i(D,{\mathcal{I}}_{D\cap ({\mathrm{Res}}_D\left(B\right)\cup E),D}(d_1,d_2,1))=0$, $i=0,1$. Hence, the residual exact sequence of D shows that it is sufficient to prove that $h^i\left({\mathcal{I}}_{{\mathrm{Res}}_{2D}\left(B\right)\cup {\mathrm{Res}}_D\left(E\right)}(d_1,d_2,0)\right)=0$, $i=0,1$. We have ${\mathrm{Res}}_{2D}\left(B\right)\cup {\mathrm{Res}}_D\left(E\right)\subset D$ and ${\mathcal{O}}_D(d_1,d_2)\cong {\mathcal{O}}_{{\mathbb{P}}^1\times {\mathbb{P}}^1}(d_1,d_2)$. Since ${\mathrm{Res}}_{2D}\left(B\right)$ is a general union of e degree-two connected zero-dimensional schemes of D and $e/2$ general 2-squares of D, $h^i\left({\mathcal{I}}_{{\mathrm{Res}}_{2D}\left(B\right)\cup {\mathrm{Res}}_D\left(E\right)}(d_1,d_2,0)\right)=0$, $i=0,1$ (Proposition 2).

(b2) Assume $e/4$ is odd. We have $t\in \{e+(e-1)/2,e+(e+1)/2\}$. We use the proof of step (b). We degenerate each connected component of $A(t-e)$ to a 2cube of some type with respect to D, obtaining that either $h^0\left({\mathcal{I}}_{{\mathrm{Res}}_{2D}\left(B\right)\cup {\mathrm{Res}}_D\left(E\right)}(d_1,d_2,0)\right)=0$ (if $t=e+(e+1)/2$) or $h^1\left({\mathcal{I}}_{{\mathrm{Res}}_{2D}\left(B\right)\cup {\mathrm{Res}}_D\left(E\right)}(d_1,d_2,0)\right)=0$ (if $t=e+(e-1)/2$). □

Proposition 6.

Assume that not all positive integers $d_1$, $d_2$ and $d_3$ have the same parity. Then, for each $t\in \mathbb{N}$, either $h^0\left({\mathcal{I}}_{A\left(t\right)}(d_1,d_2,d_3)\right)=0$ or $h^1\left({\mathcal{I}}_{A\left(t\right)}(d_1,d_2,d_3)\right)=0$.

Proof. 

By Proposition 5, we may assume that two of the integers $d_i$’s are even, say $d_2$ and $d_3$. Hence, $d_1$ is odd. Take $D\in |{\mathcal{O}}_X(0,0,1)|$.

(a) Assume $d_1\equiv 3(mod4)$. Hence, $(d_1+1)(d_2+1)\equiv 0(mod4)$. We repeat the proof of Proposition 5 with $e:=(d_1+1)(d_2+1)/4$.

(b) Assume $d_1\equiv 1(mod4)$. Hence, $(d_1+1)(d_2+1)\equiv 2(mod4)$ and $h^0\left({\mathcal{O}}_X(d_1,d_2,d_3)\right)/8\notin \mathcal{Z}$. Set $u:=((d_1+1)(d_2+1)-6)/4$. Take a general $S\subset D$ such that $\#S=u$, and a general $S^{\prime }\subset D$ such that $\#S^{\prime }=2$. Recall that

$$h^0(D,{\mathcal{O}}_D(d_1,d_2,d_3))=h^0\left({\mathcal{O}}_X(d_1,d_2,d_3)\right)-h^0\left({\mathcal{O}}_X(d_1,d_2,d_3-1)\right)=(d_1+1)(d_2+1).$$

(b1) In this step, we prove that $t\ge u+2$. Since $8t\ge (d_1+1)(d_2+1)(d_3+1)-7$ and $d_3\ge 2$, it is sufficient to prove that $8u+16\le 3(d_1+1)(d_2+1)-7$. We have $4u=(d_1+1)(d_2+1)-6$ and hence $8u=2(d_1+1)(d_2+1)-12$, concluding the proof of step (b1).

(b2) By step (b1), we may assume $t\ge u+2$. We degenerate the scheme $A\left(t\right)$ to $W:=A\cup B\cup A(t-u)$ with $A_{\mathrm{red}}=S$, and each connected component of A has some type I, II or III with respect to D, $B_{\mathrm{red}}=S^{\prime }$ and each connected component of B as residual sequence $(3,3,1,1,0)$ (Remark 6). Since $(A\cap B)\cap D$ is a general union of u 2-squares and two 2-points and $D\cong {\mathbb{P}}^1\times {\mathbb{P}}^1$, $h^i(D,{\mathcal{I}}_{D\cap W,D}(d_1,d_2,d_3))=0$, $i=0,1$. Hence, it is sufficient to prove that either $h^0\left({\mathcal{I}}_{{\mathrm{Res}}_D\left(W\right)}(d_1,d_2,d_3-1)\right)=0$ or $h^1\left({\mathcal{I}}_{{\mathrm{Res}}_D\left(W\right)}(d_1,d_2,d_3-1)\right)=0$. The scheme $D\cap {\mathrm{Res}}_D\left(W\right)$ is the union of $2S^{\prime }$ and u connected components with being of either degree four (if the corresponding connected component of A has type I) or two (if type II or III).

(b3) Assume $d_3\ge 4$ and that the proposition is true for all even integers $\le d_3-2$. We assume that all connected components of A have type I with respect to D. Hence, $h^i(D,{\mathcal{I}}_{{\mathrm{Res}}_D\left(W\right),D}(d_1,d_2,d_3-1))=0$, $i=0,1$. Hence, it is sufficient to prove that at least one of the integers $h^0\left({\mathcal{I}}_{{\mathrm{Res}}_{2D}\left(W\right)}(d_1,d_2,d_3-2)\right)$ or $h^1\left({\mathcal{I}}_{{\mathrm{Res}}_{2D}\left(W\right)}(d_1,d_2,d_3-2)\right)$ vanishes. We have ${\mathrm{Res}}_{2D}\left(W\right)={\mathrm{Res}}_{2D}\left(B\right)\cup A(t-u-2)$. By the inductive assumption, either $h^0\left({\mathcal{I}}_{A(t-u-2)}(d_1,d_2,d_3-2)\right)=0$ or $h^1\left({\mathcal{I}}_{A(t-u-2)}(d_1,d_2,d_3-2)\right)=0$. If $h^0\left({\mathcal{I}}_{A(t-u-2)}(d_1,d_2,d_3-2)\right)=0$, then $h^0\left({\mathcal{I}}_{{\mathrm{Res}}_{2D}\left(W\right)}(d_1,d_2,d_3-2)\right)=0$. Thus, we may assume that $h^1\left({\mathcal{I}}_{A(t-u-2)}(d_1,d_2,d_3-2)\right)=0$. The scheme ${\mathrm{Res}}_{2D}\left(B\right)$ is a general union of two connected zero-dimensional schemes with the only restriction being that its reduction is two general points of D. If $d_3\ge 6$, then the inductive assumption gives that either $h^0\left({\mathcal{I}}_{A(t-u-2)}(d_1,d_2,d_3-4)\right)=0$ or $h^1\left({\mathcal{I}}_{A(t-u-2)}(d_1,d_2,d_3-4)\right)=0$. In this case, to prove $h^0\left({\mathcal{I}}_{{\mathrm{Res}}_{2D}\left(W\right)}(d_1,d_2,d_3-2)\right)=0$ or $h^1\left({\mathcal{I}}_{{\mathrm{Res}}_{2D}\left(W\right)}(d_1,d_2,d_3-2)\right)=0$, it is sufficient to use Lemma 2. Now assume $d_3=4$. We have

$$h^0\left({\mathcal{I}}_{A(t-u-2)}(d_1,d_2,0)\right)=max\{0,(d_1+1)(d_2+1)-4(t-u-2)\}.$$

Hence, we may apply Lemma 2.

(b4) Assume $d_3=2$. Using $d_2$ instead of $d_3$ as in step (b1), we reduce to prove the case $d_2=2$. Write $d_1=4x+1$ with $x\in \mathbb{N}$. We have $h^0\left({\mathcal{O}}_X(d_1,2,2)\right)=36x+18$ and $u=3x$. If x is even, then we have $h^0\left({\mathcal{O}}_X(d_1,d_2,d_3)\right)\equiv 2(mod8)$. If x is odd, then we have $h^0({\mathcal{O}}_X(d_1,d_2,d_3)\equiv 6mod8$. Note that in the case x even if we show the proposition for $t=3x/2+2$, then the case with $t=3x/2+3$ follows from the case $t=3x/2+2$ (with $h^0=2$) and Remark 2. If we perform the first step as in (b1), then in the second step, we degenerate $A(t-u-2)$ to the union, U, $t-u-2$ 2cubes of type III with respect to u. The scheme ${\mathrm{Res}}_D(A\cup B)\cap D$ is the union of two 2-points and the u first residue, which are of either degree two or degree four. We have $3(d_1+1)=6+4x$. To compensate for the additional degree $4(t-u-2)$ scheme, we need to prove that $4(t-u-2)+2u\le 4u$, i.e., $4t\le 6u+8=18x+8$, which is true for $t=\lfloor h^0\left({\mathcal{O}}_X(d_1,2,2)\right)/8\rfloor$. Then, for $H^0\left({\mathcal{O}}_X(d_1,2,0)\right)\cong H^0({\mathbb{P}}^1\times {\mathbb{P}}^1,{\mathcal{O}}_{{\mathbb{P}}^1\times {\mathbb{P}}^1}(d_1,2))$, we use Remark 2. We see that this is sufficient to conclude the case x to be even. Now assume $t=\lceil h^0\left({\mathcal{O}}_X(d_1,2,2)\right)/8\rceil$ and x is odd. Hence, $8t=h^0\left({\mathcal{O}}_X(d_1,2,2)\right)+6$. Although we have two degrees inside D, more than the one needed to prove the $h^0$-vanishing, we still have ${\mathrm{Res}}_{2D}(A\cup B)\cup {\mathrm{Res}}_D\left(U\right)$ big enough to kill the vector space $H^0\left({\mathcal{O}}_X(d_1,2,0)\right)$, which is isomorphic to $H^0({\mathbb{P}}^1\times {\mathbb{P}}^1,{\mathcal{O}}_{{\mathbb{P}}^1\times {\mathbb{P}}^1}(d_1,2))$, and hence it has dimension $3(d_1+1)$. □

Proposition 7.

Assume that all $d_1$, $d_2$, and $d_3$ are even. Then, for each $t\in \mathbb{N}$ either $h^0\left({\mathcal{I}}_{A\left(t\right)}(d_1,d_2,d_3)\right)=0$ or $h^1\left({\mathcal{I}}_{A\left(t\right)}(d_1,d_2,d_3)\right)=0$.

Proof. 

Since $h^0\left({\mathcal{O}}_X(d_1+1,d_2+1,d_3+1)\right)/8\notin \mathbb{Z}$, we need to check two different integers, t, $t_1:=\lfloor h^0\left({\mathcal{O}}_X(d_1+1,d_2+1,d_3+1)\right)/8\rfloor$ and $t_2:=\lceil h^0\left({\mathcal{O}}_X(d_1+1,d_2+1,d_3+1)\right)/8\rceil$ (Remark 4). Fix $t\in \{t_1,t_2\}$. We first perform a construction and see when it helps. Take a divisor $D\in |{\mathcal{O}}_X(0,0,1)|$. Recall that

$$h^0(D,{\mathcal{O}}_D(d_1,d_2,d_3))=h^0\left({\mathcal{O}}_X(d_1,d_2,d_3)\right)-h^0\left({\mathcal{O}}_X(d_1,d_2,d_3-1)\right)=(d_1+1)(d_2+1).$$

Since $(4,3)=1$ and $d_1+1)(d_2+1)\ge 6$, there is a unique pair $(u,v)\in {\mathbb{N}}^2$ such that $(d_1+1)(d_2+1)=4u+3v$ and $0\le v\le 3$.

(a) In this step, we prove that $t\ge u+v$. Since $8t\ge (d_1+1)(d_2+1)(d_3+1)-7$ and $d_3\ge 2$, it is sufficient to prove that $8u+8v\le 3(d_1+1)(d_2+1)-7$. We have $8u+6v=2(d_1+1)(d_2+1)$, $v\le 3$ and $(d_1+1)(d_2+1)\ge 9$. Moreover, either $d_1=d_2=2$ or $(d_1+1)(d_2+1)\ge 15$. In the latter case, Claim 1 is true. Assume $d_1=d_2=2$. We have $u=0$, $v=3$ and $t\ge 3$, concluding the proof of Claim 1.

(b) By step (a), we may assume $t\ge u+v$. Take a general $S\subset D$ such that $\#S=u$ and a general $S^{\prime }\subset D$ such that $\#S^{\prime }=v$ We degenerate $A\left(t\right)$ to $W:=A\cup B\cup A(t-u)$ with $A_{\mathrm{red}}=S$, each connected component of A has some type I, II or III with respect to D, $B_{\mathrm{red}}=S^{\prime }$, and each connected component of B is a residual sequence $(3,3,1,1,0)$ (Remark 6). Since $(A\cap B)\cap D$ is a general union of u 2-squares and v 2-points and $D\cong {\mathbb{P}}^1\times {\mathbb{P}}^1$, $h^i(D,{\mathcal{I}}_{D\cap W,D}(d_1,d_2,d_3))=0$, $i=0,1$. Hence, it is sufficient to prove that either $h^0\left({\mathcal{I}}_{{\mathrm{Res}}_D\left(W\right)}(d_1,d_2,d_3-1)\right)=0$ or $h^1\left({\mathcal{I}}_{{\mathrm{Res}}_D\left(W\right)}(d_1,d_2,d_3-1)\right)=0$. The scheme $D\cap {\mathrm{Res}}_D\left(W\right)$ is the union of $2S^{\prime }$ and u connected components of either degree four (if the corresponding connected component of A has type I) or two (if type II or III).

(b1) Assume $d_3\ge 4$ and that the proposition is true for all even integers $\le d_3-2$. We assume that all connected components of A have type I with respect to D. Hence, $h^i(D,{\mathcal{I}}_{{\mathrm{Res}}_D\left(W\right),D}(d_1,d_2,d_3-1))=0$, $i=0,1$. Hence, it is sufficient to prove that either $h^0\left({\mathcal{I}}_{{\mathrm{Res}}_{2D}\left(W\right)}(d_1,d_2,d_3-2)\right)=0$ or $h^1\left({\mathcal{I}}_{{\mathrm{Res}}_{2D}\left(W\right)}(d_1,d_2,d_3-2)\right)$ vanishes. We have ${\mathrm{Res}}_{2D}\left(W\right)={\mathrm{Res}}_{2D}\left(B\right)\cup A(t-u-v)$. By the inductive assumption either $h^0\left({\mathcal{I}}_{A(t-u-v)}(d_1,d_2,d_3-2)\right)=0$ or $h^1\left({\mathcal{I}}_{A(t-u-v)}(d_1,d_2,d_3-2)\right)=0$. If $h^0\left({\mathcal{I}}_{A(t-u-v)}(d_1,d_2,d_3-2)\right)=0$, then $h^0\left({\mathcal{I}}_{{\mathrm{Res}}_{2D}\left(W\right)}(d_1,d_2,d_3-2)\right)=0$. Thus, we may assume that $h^1\left({\mathcal{I}}_{A(t-u-v)}(d_1,d_2,d_3-2)\right)=0$. Since ${\mathrm{Res}}_{2D}\left(B\right)$ is a general union of v connected zero-dimensional schemes with the only restriction being that its reduction is v general points of D, if $d_3\ge 6$, then the inductive assumption gives that either $h^0\left({\mathcal{I}}_{A(t-u-v)}(d_1,d_2,d_3-4)\right)=0$ or $h^1\left({\mathcal{I}}_{A(t-u-v)}(d_1,d_2,d_3-4)\right)=0$. In this case, to prove $h^0\left({\mathcal{I}}_{{\mathrm{Res}}_{2D}\left(W\right)}(d_1,d_2,d_3-4)\right)=0$ or $h^1\left({\mathcal{I}}_{{\mathrm{Res}}_{2D}\left(W\right)}(d_1,d_2,d_3-4)\right)=0$, it is sufficient to use Lemma 2.

Now assume $d_3=4$. We have

$$h^0\left({\mathcal{I}}_{A(t-u-v)}(d_1,d_2,0)\right)=max\{0,(d_1+1)(d_2+1)-4(t-u-v)\}.$$

Hence, we may apply Lemma 2.

(b2) Assume $d_3=2$. Using $d_2$ instead of $d_3$ as in step (b1), we reduce to prove the case $d_2=2$. Using $d_1$ instead of $d_3$, we reduce to the case $(d_1,d_2,d_3)=(2,2,2)$. In this case, the proposition is true by Lemma 14. □

Proof of Theorem 2: 

All cases with $1\le d_i\le 2$ for all i are proved in Lemmas 11–14. Hence, we may assume that at least one among $d_1$, $d_2$ and $d_3$ is at least 3. If at least two of the integers $d_1$, $d_2$, and $d_3$ are odd, then we apply Proposition 5. If not all integers $d_1$, $d_2$, and $d_3$ have the same parity, then we apply Proposition 6. If $d_1$, $d_2$, and $d_3$ are even, then we apply Proposition 7. □

6. ${\mathbb{P}}^{\mathbf{2}}\times {\mathbb{P}}^{\mathbf{1}}$

In this section, we take $X={\mathbb{P}}^2\times {\mathbb{P}}^1$ and prove Theorem 3. Let $A\left(t\right)$ denote a general union of t 2cubes of X.

Let ${\pi }_1:X\to {\mathbb{P}}^2$ and ${\pi }_2:X\to {\mathbb{P}}^1$ denote the projection of X into its factor. We have $Pic\left(X\right)\cong {\mathbb{Z}}^2$, and we write ${\mathcal{O}}_X(a,b):={\pi }_1^{*}\left({\mathcal{O}}_{{\mathbb{P}}^2}\left(a\right)\right)\otimes {\pi }_2^{*}\left({\mathcal{O}}_{{\mathbb{P}}^1}\left(b\right)\right)$. The Künneth formula gives $h^0\left({\mathcal{O}}_X(a,b)\right)=0$ if either $a<0$ or $b<0$ and $h^0\left({\mathcal{O}}_X(a,b)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{a+2}{2}}\right)(b+1)$ for all $(a,b)\in {\mathbb{N}}^2$. The Künneth formula gives $h^1\left({\mathcal{O}}_X(a,b)\right)=0$ if either $b\ge -1$ or if $a<0$.

Take $H\in |{\mathcal{O}}_X(0,1)|$ and $D\in |{\mathcal{O}}_X(1,0)|$. Note that $H\cong {\mathbb{P}}^2$ and that any such isomorphism identifies ${\mathcal{O}}_H(a,b)$ and ${\mathcal{O}}_{{\mathbb{P}}^2}\left(a\right)$. Note that $D\cong {\mathbb{P}}^1\times {\mathbb{P}}^1$ and that any such isomorphism identifies ${\mathcal{O}}_D(a,b)$ and ${\mathcal{O}}_{{\mathbb{P}}^1\times {\mathbb{P}}^1}(a,b)$. For all $(a,b)\in {\mathcal{Z}}^2$, we have the exact sequences

$$0\to {\mathcal{O}}_X(a,b-1)\to {\mathcal{O}}_X(a,b)\to {\mathcal{O}}_H(a,b)\to 0$$

(2)

$$0\to {\mathcal{O}}_X(a-1,b)\to {\mathcal{O}}_X(a,b)\to {\mathcal{O}}_D(a,b)\to 0$$

(3)

Now assume $(a,b)\in {\mathbb{N}}^2$. The long cohomology exact sequence of (2) gives that the restriction map $H^0{\mathcal{O}}_X(a,b))\to H^0(H,{\mathcal{O}}_H(a,b))$ is surjective. The long cohomology exact sequence of (3) gives that the restriction map $H^0{\mathcal{O}}_X(a,b))\to H^0(D,{\mathcal{O}}_D(a,b))$ is surjective.

Remark 14.

Fix $(a,b)\in {\mathbb{N}}^2$. By Remark 4, to prove Theorem 3 for ${\mathcal{O}}_X(a,b)$ and all positive integers t, it is sufficient to prove it for the integers $t\in \{\lfloor \left({\displaystyle \genfrac{}{}{0pt}{}{a+2}{2}}\right)(b+1)\rfloor ,\lceil \left({\displaystyle \genfrac{}{}{0pt}{}{a+2}{2}}\right)(b+1)\rceil \}$.

Lemma 15.

We have $h^0\left({\mathcal{I}}_{A\left(1\right)}(1,1)\right)=0$.

Proof. 

It is sufficient to observe that $\mathbb{K}{[x_0,x_1,x_2,y_0,y_1]}_{(1,1)}$ contains no monomial of bidegree $(2,0)$, while there is a 2cube $A\subset 2D$, while obviously $A⊈D$. □

Lemma 16.

We have $h^1\left({\mathcal{I}}_{A\left(1\right)}(2,1)\right)=0$ and $h^0\left({\mathcal{I}}_{A\left(2\right)}(2,1)\right)=0$.

Proof. 

Recall that $h^0\left({\mathcal{O}}_X(2,1)\right)=12$, $h^0(D,{\mathcal{O}}_D(2,1))=6$ and $h^0(H,{\mathcal{O}}_H(2,1))=6$. We specialize $A\left(1\right)$ to a general 2cube of type III with respect to D and use Proposition 2, the residual exact sequence of D, and Lemma 2. We specialize $A\left(2\right)$ to a general union B of two 2cubes of type I with respect to sequence H. Recall that we proved (Proposition 1) that $h^0(H,{\mathcal{I}}_{B\cap ,H}(2,1))=h^0(H,{\mathcal{I}}_{{\mathrm{Res}}_H\left(B\right)\cap H,H}(2,0))=0$. Thus, it is sufficient to use twice the residual exact sequence of H and that $h^0\left({\mathcal{O}}_X(2,-1)\right)=h^1\left({\mathcal{O}}_X(2,-1)\right)=0$. □

Lemma 17.

We have $h^1\left({\mathcal{I}}_{A\left(2\right)}(2,2)\right)=0$ and $h^0\left({\mathcal{I}}_{A\left(3\right)}(2,2)\right)=0$.

Proof. 

Recall that $h^0\left({\mathcal{O}}_X(2,2)\right)=18$, $h^0(D,{\mathcal{O}}_D(2,2))=9$, $h^0(D,{\mathcal{O}}_D(2,1))=6$ and $h^0(D,{\mathcal{O}}_D(2,0))=3$. We specialize $A\left(2\right)$ to a general union A of a 2cube of type I and a 2cube of type III with respect to D. We apply twice the residual exact sequences of D and Proposition 2 and then use Remark 2. Hence, $h^0\left({\mathcal{I}}_{A\left(2\right)}(2,2)\right)=2$. Since a general 2cube contains a general degree-two scheme, Remark 2 gives $h^0\left({\mathcal{I}}_{A\left(3\right)}(2,2)\right)=0$. □

Lemma 18.

We have $h^1\left({\mathcal{I}}_{A\left(1\right)}(1,2)\right)=0$ and $h^0\left({\mathcal{I}}_{A\left(2\right)}(1,2)\right)=0$.

Proof. 

We degenerate $A\left(1\right)$ to one 2cube E with residual sequence $(3,3,1,1,0)$ with respect to A. Since the schemes $A\cap H$ and ${\mathrm{Res}}_H\left(A\right)\cap H$ are 2-points of H, we have $h^i(H,{\mathcal{I}}_{A\cap H,H}(1,2))=h^i(H,{\mathcal{I}}_{{\mathrm{Res}}_H\left(A\right)\cap H,H})=0$, $i=0,1$. We use twice the residual exact sequence of H and then use Remark 2 for the degree-two scheme ${\mathrm{Res}}_{2H}\left(A\right)$. Degenerate $A\left(2\right)$ to $A\left(1\right)\cup E$ and use that $h^0\left({\mathcal{I}}_{A\left(1\right)\cup E}(1,0)\right)=0$. □

Lemma 19.

We have $h^1\left({\mathcal{I}}_{A\left(3\right)}(3,2)\right)=0$ and $h^0\left({\mathcal{I}}_{A\left(4\right)}(3,2)\right)=0$.

Proof. 

Recall that $h^0\left({\mathcal{O}}_X(3,2)\right)=30$, $h^0(D,{\mathcal{O}}_D(3,2))=12$, $h^0(D,{\mathcal{O}}_D(2,2))=9$. We degenerate $A\left(3\right)$ to a general union of one 2cube of type l with respect to D and two 2cubes of type III with respect to D. Then, we use twice the residual exact sequence of D, Proposition 2 and Remark 2 to obtain that $h^1\left({\mathcal{I}}_{A\left(3\right)}(3,2)\right)=0$. We degenerate $A\left(4\right)$ to $A\left(1\right)\cup E$, where E is a general union of two 2cubes of type I with respect to D and one 2cube of type III with respect to D. We have $h^i(D,{\mathcal{I}}_{E\cap D,D}(3,2))=0$, $i=0,1$, and $h^0(D,{\mathcal{I}}_{{\mathrm{Res}}_D\left(E\right)\cap D,D}(2,3))=0$. Since $h^1\left({\mathcal{I}}_{A\left(1\right)}(1,2)\right)=0$ (Lemma 18), $h^1\left({\mathcal{I}}_{A\left(1\right)}(1,2)\right)=1$. Hence, Remark 2 gives $h^0\left({\mathcal{I}}_{A\left(1\right)\cup {\mathrm{Res}}_{2D}\left(E\right)}(1,2)\right)=0$. □

Lemma 20.

We have $h^i\left({\mathcal{I}}_{A\left(3\right)}(2,3)\right)=0$, $i=0,1$.

Proof. 

Recall that $h^0\left({\mathcal{O}}_H(2,x)\right)=6$ for all x. We degenerate $A\left(3\right)$ to $A\left(1\right)\cup E$, with E being a general union of two 2cubes with residual sequence $(3,3,1,1,0)$ with respect to H. We reduce to prove that $h^i\left({\mathcal{I}}_{A\left(1\right)\cup {\mathrm{Res}}_{2H}\left(E\right)}(2,1)\right)=0$. Since $h^1\left({\mathcal{I}}_{A\left(1\right)}(2,1)\right)=0$ (Lemma 16), $h^0\left({\mathcal{I}}_{A\left(1\right)}(2,1)\right)=4$. Hence, it is sufficient to use Lemma 2. □

Lemma 21.

We have $h^1\left({\mathcal{I}}_{A\left(2\right)}(3,1)\right)=0$ and $h^0\left({\mathcal{I}}_{A\left(3\right)}(3,1)\right)=0$.

Proof. 

We have $h^0(D,{\mathcal{O}}_D(3,1))=8$, $h^0(D,{\mathcal{O}}_D(2,1))=6$ and $h^0(D,{\mathcal{O}}_D(1,1))=4$. We degenerate $A\left(2\right)$ to a general union of one 2cube of type I and one 2cube of type III with respect to D. We degenerate $A\left(3\right)$ to a general union of three 2cubes with residual sequence of $(3,3,1,1,0)$ with respect to D. □

Lemma 22.

We have $h^i\left({\mathcal{I}}_{A\left(5\right)}(3,3)\right)=0$, $i=0,1$.

Proof. 

Recall that $h^0\left({\mathcal{O}}_H(3,x)\right)=10$ for all x. We specialize $A\left(5\right)$ to $A\left(2\right)\cup E$ with E a general union of one 2cube of type I and two 2cubes with residual sequence $(3,3,1,1,0)$ with respect to H. We reduce to prove that $h^i\left({\mathcal{I}}_{A\left(2\right)\cup {\mathrm{Res}}_{2H}\left(E\right)}(3,1)\right)=0$, $i=0,1$. Since $h^1\left({\mathcal{I}}_{A\left(2\right)}(3,1)\right)=0$ (Lemma 21), $h^0\left({\mathcal{I}}_{A\left(2\right)}(3,1)\right)=4$. Since ${\mathrm{Res}}_{2H}\left(E\right)$ is a general union of three degree-two connected zero-dimensional schemes with their reduction general in H, it is sufficient to quote Lemma 2. □

Lemma 23.

For all integers $t>0$ and $b\ge 4$, either $h^0\left({\mathcal{I}}_{A\left(t\right)}(3,b)\right)=0$ or $h^1\left({\mathcal{I}}_{A\left(t\right)}(3,b)\right)=0$.

Proof. 

We may assume $t\ge 3$ (Remark 14). Recall that $h^0\left({\mathcal{O}}_H(3,x)\right)=10$ for all x. By Lemmas 19, 21, and 22 and induction on b, we may assume that the theorem holds for all ${\mathcal{O}}_X(3,x)$ such that $1\le x<b$. We degenerate $A\left(t\right)$ to $A(t-3)\cup E$ with E being a general union of one 2cube of type I and two 2cubes with residual sequence $(3,3,1,1,0)$ with respect to H. We reduce to prove that $h^i\left({\mathcal{I}}_{A(t-3)\cup {\mathrm{Res}}_{2H}\left(E\right)}(3,x-2)\right)=0$, $i=0,1$. By the inductive assumptions, we have $h^0\left({\mathcal{I}}_{A(t-3)}(3,x-2)\right)=10(x-1)-8(t-3)$. Since ${\mathrm{Res}}_{2H}\left(E\right)$ is a general union of three degree-two connected zero-dimensional schemes with their reduction being general in H, it is sufficient to quote Lemma 2. □

Proof of Theorem 3: 

If $a\le 3$ and $b\le 3$, then the theorem is true for ${\mathcal{O}}_X(a,b)$ by Remark 14 and Lemmas 15–22. Lemma 23 gives the theorem for ${\mathcal{O}}_X(3,b)$ for all $b\ge 4$. Thus, it is sufficient to prove it for all ${\mathcal{O}}_X(a,b)$, $a\ge 4$ and $b\ge 4$. We fix the integer $b\ge 4$. By Lemma 23 and induction on a, we may assume that the theorem is true for all ${\mathcal{O}}_X(x,b)$ with $3\le x<a$. Recall that for $h^0(D,{\mathcal{O}}_D(y,b))=(y+1)(b+1)$. Since 4 and 3 are coprime, there are integers u and v such that $(a+1)(b+1)=4u+3v$ and $0\le v\le 3$. Since $(a+1)(b+1)\ge 25$, $u\ge 0$. Since $8t\ge \left({\displaystyle \genfrac{}{}{0pt}{}{a+2}{2}}\right)(b+1)-7$ (Remark 14), $a\ge 4$ and $v\le 3$, $t\ge u+v$. We degenerate the zero-dimensional scheme $A\left(t\right)$ to $A(t-u-v)\cup A$, with A being a general union of u 2cubes of type III with respect to D and v 2cubes with residual sequence $(3,3,1,1,0)$ with respect to D. Since $A(t-u-v)\cap D=\varnothing$ and $h^i(D,{\mathcal{I}}_{A\cap D,D}(a,b))=0$, $i=0,1$ (Proposition 2), it is sufficient to prove that either $h^0\left({\mathcal{I}}_{A(t-u-v)\cup {\mathrm{Res}}_D\left(A\right)}(a-1,b)\right)=0$ or $h^1\left({\mathcal{I}}_{A(t-u-v)\cup {\mathrm{Res}}_D\left(A\right)}(a-1,b)\right)=0$. The scheme ${\mathrm{Res}}_D\left(A\right)\cap D$ is a general union of u degree-two connected schemes and v 2-points of D. Since four and three are coprime, there are integers w and z such that we have $0\le z\le 3$ and $4w+3z=a(b+1)-2u-3v$. Assume for the moment $w\ge 0$ and $t\ge u+v+w+z$. We degenerate the scheme $A(t-u-v)$ to $A(t-u-v-w-z)\cup F$, with F being a general union of w 2cubes of type III with respect to D and z 2cubes with residual sequence $(3,3,1,1,0)$ with respect to D. The scheme $D\cap (F\cup {\mathrm{Res}}_D\left(E\right))$ is a general union of v connected degree-two schemes, $v+z$ 2-points of D and w 2-squares of D. Since we have $a-1\ge 3$ and $b\ge 3$, $h^i(D,{\mathcal{I}}_{D\cap (F\cup {\mathrm{Res}}_D\left(E\right)),D}(a-1,b))=0$, $i=0,1$. By the residual exact sequence of D, it is sufficient to prove that either $h^0\left({\mathcal{I}}_{A(t-u-v-w-z)\cup {\mathrm{Res}}_{2D}\left(E\right)\cup {\mathrm{Res}}_D\left(F\right)}(a-2,b)\right)=0$ or $h^{0}1\left({\mathcal{I}}_{A(t-u-v-w-z)\cup {\mathrm{Res}}_{2D}\left(E\right)\cup {\mathrm{Res}}_D\left(F\right)}(a-2,b)\right)=0$. Since $a\ge 4$ and $h^1\left({\mathcal{O}}_X(-1,b)\right)=0$, it is sufficient to use Lemma 2.

Now assume $w\ge 0$ and $t\le u+v+w+z-1$. Set $z^{\prime }:=max\{0,t-u-v-w\}$. If $z^{\prime }>0$, then set $w^{\prime }:=w$. If $z^{\prime }:=0$, then set $w^{\prime }:=t-u-v$. Let $F^{\prime }\subset X$ be a general union of $w^{\prime }$ 2cubes of type III with respect to D and $z^{\prime }$ 2cubes with residual sequence $(3,3,1,1,0)$ with respect to D. We take ∅ instead of $A(t-u-v-w-z)$ and $F^{\prime }$ instead of F and then quote again Lemma 2.

Now assume $w<0$. Since $z\le 3$, we obtain $a(b+1)-2u-3v\le 5$. Recall that $a\ge 4$ and $b\ge 4$. Since $4u+3v=(a+1)(b+1)$ and $v\le 3$, we obtain $ab+a\le b+20$, which is false, unless $(a,b)\in \left\{\right(4,4),(4,5\left)\right\}$. If $a=b=4$, then $t\in \{9,10\}$, $u=4$, $v=3$, $w=0$ and $z=1$. If $(a,b)=(4,5)$, then $t\in \{11,12\}$, $u=6$, $v=2$, $w=0$, $z=2$. □

7. The 3-Dimensional Smooth Quadric

In this section, we take as X the smooth quadric 3-fold $Q_3\subset {\mathbb{P}}^4$ and prove Theorem 4. Let $A\left(t\right)$ denote a general union of t 2cubes of $Q_3$. We recall that any line bundle on $Q_3$ is of the form ${\mathcal{O}}_{Q_3}\left(x\right)$ for some $x\in \mathbb{Z}$ and that $h^0\left({\mathcal{O}}_{Q_3}\left(x\right)\right)=0$ for all $x<0$, while $h^0\left({\mathcal{O}}_{Q_3}\right)=1$. Hence, Theorem 4 computes all the possible Hilbert functions of general unions of 2cubes on $Q_3$. Consider the exact sequence

$$0\to {\mathcal{O}}_{{\mathbb{P}}^4}(x-2)\to {\mathcal{O}}_{{\mathbb{P}}^4}\left(z\right)\to {\mathcal{O}}_{Q_3}\left(x\right)\to 0$$

(4)

The long cohomology exact sequence of (4) gives $h^i\left({\mathcal{O}}_{Q_3}\left(x\right)\right)=0$ for all $x\in \mathbb{Z}$ and $i=1,2$, $h^0\left({\mathcal{O}}_{Q_3}\left(x\right)\right)=0$ for $x<0$ and $h^0\left({\mathcal{O}}_{Q_3}\left(x\right)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{x+4}{4}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{x+2}{4}}\right)=(x+2)(x+1)(2x+3)/6$ for all $x\ge 0$. Let $H\subset {\mathbb{P}}^4$ be a hyperplane such that $Q_2:=H\cap Q_3$ is a smooth quadric surface of H. The isomorphism between $Q_2$ and ${\mathbb{P}}^1\times {\mathbb{P}}^1$ maps ${\mathcal{O}}_{{\mathbb{P}}^1\times {\mathbb{P}}^1}(x,x)$ to ${\mathcal{O}}_{Q_2}\left(x\right)$. Hence, $h^0(Q_2,{\mathcal{O}}_{Q_2}\left(x\right))=0$ for $x<0$ and $h^0(Q_2,{\mathcal{O}}_{Q_2}\left(x\right))={(x+1)}^2$ for all $x\in \mathbb{N}$. The long cohomology exact sequence of the exact sequence

$$0\to {\mathcal{O}}_{Q_3}(x-1)\to {\mathcal{O}}_{Q_3}\left(x\right)\to {\mathcal{O}}_{Q_2}\left(x\right)\to 0$$

(5)

gives the surjectivity of the restriction map $H^0\left({\mathcal{O}}_{Q_3}\left(x\right)\right)\to H^0(Q_2,{\mathcal{O}}_{Q_2}\left(x\right))$ and that $h^0\left({\mathcal{O}}_{Q_3}\left(x\right)\right)=h^0\left({\mathcal{O}}_{Q_3}(x-2)\right)+h^0(Q_2,{\mathcal{O}}_{Q_2}\left(x\right))$ for all $x\ge 0$.

Remark 15.

By Remark 4, to test all positive integers t for ${\mathcal{O}}_{Q_3}\left(d\right)$, it is sufficient to test the integers $t\in \{\lfloor {\displaystyle \frac{(d+2)(d+1)(2d+3)}{48}}\rfloor ,\lceil {\displaystyle \frac{(d+2)(d+1)(2d+3)}{48}}\rceil \}$.

Remark 16.

Since any 2cube $A\subset Q_3$ is not contained in a hyperplane section of $Q_3$, we have $h^0\left({\mathcal{I}}_A\left(x\right)\right)=0$ for all $x\le 1$.

Lemma 24.

We have $h^1\left({\mathcal{I}}_{A\left(1\right)}\left(2\right)\right)=0$.

Proof. 

Recall that $h^0\left({\mathcal{O}}_{Q_3}\left(2\right)\right)=14$ and $h^0(Q_2,{\mathcal{O}}_{Q_2}\left(2\right))=9$. We specialize $A\left(1\right)$ to a 2cube A with residual sequence $(4,2,1,1,0)$ with respect to $Q_2$. Proposition 2 gives $h^1(Q_2,{\mathcal{I}}_{Q_2\cap A,Q_2}\left(2\right))=0$. Hence, the long cohomology exact of (5) shows that it is sufficient to prove that $h^i\left({\mathcal{I}}_{{\mathrm{Res}}_{Q_2}\left(A\right)}\left(1\right)\right)=0$. Apply Lemma 2. □

Lemma 25.

We have $h^1\left({\mathcal{I}}_{A\left(3\right)}\left(3\right)\right)=0$ and $h^0\left({\mathcal{I}}_{A\left(4\right)}\left(3\right)\right)=0$.

Proof. 

Recall that $h^0(Q_2,{\mathcal{O}}_{Q_2}\left(3\right))=16$. We specialize $A\left(4\right)$ to B with B general with the condition that $B_{\mathrm{red}}\subset Q_2$ and that each connected component of B has residual sequence $(4,2,1,1,0)$ with respect to $Q_2$. Since $h^i\left({\mathcal{I}}_{B\cap Q_2,Q_2}\left(3\right)\right)=0$, $i=0,1$, to prove that $h^0\left({\mathcal{I}}_{A\left(4\right)}\left(3\right)\right)=0$, it is sufficient to prove that $h^0\left({\mathcal{I}}_{{\mathrm{Res}}_{Q_2}\left(B\right)}\left(2\right)\right)=0$. The scheme $E:=Q_2\cap {\mathrm{Res}}_{Q_2}\left(B\right)$ is a general union of four degree-two connected zero-dimensional schemes, $h^1(Q_2,{\mathcal{I}}_{E,Q_2}\left(2\right))=0$ and $h^0(Q_2,{\mathcal{I}}_{E,Q_2}\left(2\right))=1$ (Proposition 2). Since we have $deg\left({Res}_{Q_2}\left(B\right)\right)=2+h^0\left({\mathcal{O}}_{Q_3}\left(2\right)\right)$, it is sufficient to apply Lemma 2.

To prove that $h^1\left({\mathcal{I}}_{A\left(1\right)}\left(2\right)\right)=0$, it is sufficient to perform the same proof, taking the union of three of the connected components of B. □

Lemma 26.

We have $h^1\left({\mathcal{I}}_{A\left(6\right)}\left(4\right)\right)=0$ and $h^0\left({\mathcal{I}}_{A\left(7\right)}\left(4\right)\right)=0$.

Proof. 

We specialize $A\left(x\right)$, $x\in \{6,7\}$, to $A\cup A(x-6)$ with $A_{\mathrm{red}}$ general in $Q_2$, two of the connected components of A being of type I, another component of type III, and three components having residual sequence $(3,3,1,1,0)$ with respect to $Q_2$. Proposition 2 gives $h^i(Q_2,{\mathcal{I}}_{A\cap Q_2,Q_2}=h^i(Q_2,{\mathcal{I}}_{{\mathrm{Res}}_{Q_2}\left(A\right)\cap Q_2,Q_2}\left(3\right))=0$, $i=0,1$. Hence, it is sufficient to prove that $h^i\left({\mathcal{I}}_{{\mathrm{Res}}_{2Q_2}\left(A\right)}\left(2\right)\right)=0$ and $h^0\left({\mathcal{I}}_{A\left(1\right)\cup {\mathrm{Res}}_{2Q_2}\left(A\right)}\left(2\right)\right)=0$. Since ${\mathrm{Res}}_{2Q_2}\left(A\right)$ is a general union of four connected degree-two schemes with the only restriction being that their reduction is contained in $Q_2$, it is sufficient to quote Lemma 2. □

Lemma 27.

Fix an odd integer $d\ge 5$. If $d=5$, we assume that Theorem 4 holds for ${\mathcal{O}}_{Q_3}\left(3\right)$ and ${\mathcal{O}}_{Q_3}\left(4\right)$ and that $h^1\left({\mathcal{I}}_{A\left(1\right)}\left(2\right)\right)=0$. If $d\ge 7$, we assume that Theorem 4 is true for ${\mathcal{O}}_{Q_3}\left(x\right)$, $x\in \{d-3,d-2,d-1\}$. Then, Theorem 4 is true for ${\mathcal{O}}_{Q_3}\left(d\right)$.

Proof. 

Set $e:={(d+1)}^2/4$. By Remark 15, we may assume $t\ge e$. We degenerate $A\left(t\right)$ to $A\cup A(t-e)$ with $A\cap Q_2$ being a general union of e 2-squares of $Q_2$ and each connected component of A has type III with respect to $Q_2$. We have $h^i(Q_2,{\mathcal{I}}_{A\cap Q_2,Q_2}\left(d\right))=0$, $i=0,1$ (Proposition 2). By the long cohomology exact sequence of the residual sequence of $Q_2$, it is sufficient to prove that either $h^1\left({\mathcal{I}}_{{\mathrm{Res}}_{Q_2}\left(A\right)\cup A(t-e)}(d-1)\right)$ or $h^1\left({\mathcal{I}}_{{\mathrm{Res}}_{Q_2}\left(A\right)\cup A(t-e)}(d-1)\right)$ is 0. We have $deg(Q_2\cap {Res}_{Q_2}\left(A\right))=2e={(d+1)}^2/2$. Recall that $h^0{(Q_2,{\mathcal{O}}_{Q_2}(d-1))}^2=d^2$. Since $(4,3)=1$, there are unique integers u and v such that $4u+3v=d^2-{(d+1)}^2/2$ and $0\le v\le 3$. Since $d\ge 5$, we see that $u\ge 0$ ($u=v=1$ if $d=5$, $u=v=3$ if $d=7$).

(a) Assume for the moment that $t-e\ge u+v$. We degenerate ${\mathrm{Res}}_{Q_2}\left(A\right)\cup A(t-e)$ to ${\mathrm{Res}}_{Q_2}\left(A\right)\cup B\cup A(t-e-u-v)$ with B being a general union of u 2cubes of type III with respect $Q_2$ and v 2cubes of type $(3,3,1,1,0)$ with respect to $Q_2$. Since $h^i(Q_2,{\mathcal{I}}_{B\cap Q_2,Q_2}(d-1))=0$, $i=0,1$, it is sufficient to prove that either $h^1\left({\mathcal{I}}_{A(t-e-u-v)\cup {\mathrm{Res}}_{2Q_2}\cup {\mathrm{Res}}_{Q_2}\left(B\right)}(d-2)\right)=0$. The scheme $Q_2\cap ({Res}_{2Q_2}\left(A\right)\cup {Res}_{Q_2}\left(B\right))$ is a general union of e points and v 2-points of $Q_2$. The scheme ${\mathrm{Res}}_{3Q_2}\left(A\right)\cup {\mathrm{Res}}_{2Q_2}\left(B\right)$ is a general union of e points and v connected degree-two schemes with the only restriction being that their reduction is contained in $Q_2$. Hence, it is sufficient to apply Lemma 2.

(b) By step (a), we may assume $t\le u+v+e-1$. Hence, $8t\le 8e+8u+8v-8=2{(d+1)}^2+2d^2-{(d+1)}^2+2v$. Assume for the moment $d\ge 7$. Since $v\le 3$, we obtain $8t\le (d+2)(d+1)(2d+3)/6-8$, contradicting our assumption on t. Now assume $d=5$. We have $u=v=1$ and $e=4$, while $\lfloor h^0\left({\mathcal{O}}_{Q_3}\left(5\right)\right)/8\rfloor =\lfloor 91/8\rfloor =11$. □

Lemma 28.

Fix an even integer $d\ge 6$ and assume that Theorem 4 is true for ${\mathcal{O}}_{Q_3}\left(x\right)$, $x\in \{d-3,d-2,d-1\}$. Then, Theorem 4 is true for ${\mathcal{O}}_{Q_3}\left(d\right)$.

Proof. 

Since 4 and 3 are coprime, there are integers e and f such that ${(d+1)}^2=4e+3f$ with $0\le f\le 3$. By Remark 15, we have $t\ge e+f$. We specialize $A\left(t\right)$ to $A\cup A(t-e-f)$ with A being a general union of e 2cubes of type III with respect to $Q_2$. Take integers u and v such that $0\le v\le 3$ and $2e+3f+4u+3v=d^2$. Since $d\ge 6$, we have $u\ge 0$. Indeed, for $d=6$ (respectively, $d=8$), we have $e=10$, $f=3$, $u=0$, and $v=2$ (respectively, $e=13$, $f=3$, $u=5$, and $v=3$). Then, we continue as in the proof of Lemma 27. □

Proof 

(Proof of Theorem 4:). The cases $d=3,4$ are covered by Lemmas 25 and 26. The cases $d\ge 5$ are covered by Lemmas 24–28. □

8. Bricks

In this section, we take $X={\mathbb{P}}^2$ and check the Hilbert function of a general union of a prescribed number of bricks. For all positive integers t, let $\mathcal{B}\left(t\right)$ be the set of all unions $Z\subset {\mathbb{P}}^2$ of t disjoint bricks. The set $\mathcal{B}\left(t\right)$ is an irreducible variety. Let $B\left(t\right)$ denote a general element of $\mathcal{B}\left(t\right)$. Fix $p\in {\mathbb{P}}^2$ and an integral curve $T\subset {\mathbb{P}}^2$ such that $p\in T_{\mathrm{reg}}$. We consider the set $\Delta$ of all $B\in \mathcal{B}\left(1\right)$ such that $B_{\mathrm{red}}=\left\{p\right\}$. The possible residual sequence does not depend on the choice of p and T with the only restriction that $p\in T_{\mathrm{reg}}$. We consider the following residual sequence $(a_1,\cdots ,a_s,0)$, where $a_1\ge \cdots \ge a_s>0$, with respect to T. Fix any $B\in \Delta$. Since $B\supset 2p$, $a_1\ge 2$, hence $s\le 5$. Since B is not curvilinear, $s\ge 2$ and hence $a_1\le 5$.

Remark 17.

Set $R:=\mathbb{K}\left[\right[x,y\left]\right]$ and $O:=(0,0)$. The ring R is a two-dimensional regular local ring with $(x,y)$ as its maximal ideal. Set $I:=(x^2,y^3)$ and ${\mathcal{O}}_Z:=R/I$. The degree-six vector space ${\mathcal{O}}_Z$ is the structural sheaf and the affine coordinate ring of the degree-six zero-dimensional scheme Z, which is a brick. The vector space ${\mathcal{O}}_Z$ has a basis given by the images of the six monomials 1, x, y, $y^2$, $xy$, and $x{y}^2$. For each $f\in (x,y)$ such that $f\notin (x^2,xy,y^2)$, the principal ideal $\left(f\right)$ is associated to a Cartier divisor $H_f$ of $Spec\left(R\right)$. For a few power series f, we compute the residual sequence of Z with respect to $H_f$.

Let $(a_1,\cdots ,a_s,0)$ with $a_1\ge \cdots \ge a_s>0$ be the residual sequence of Z with respect to $H_f$. Since Z is not curvilinear, $a_1\ge 2$ and hence $s\le 5$.

The residual sequence of Z with respect to $H_x$ is $(3,3,0)$. The residual sequence of Z with respect to $H_y$ is $(2,2,2,0)$.

(a) In this step, we prove that $(4,1,1,0)$ is the residual sequence of Z with respect to $H_{x+y}$. Since $(x+y)(x-y)=x^2-y^2$, the ideal $J:=(x+y,x^2,y^3)$ contains $y^2$, and hence it contains $x{y}^2$. Since $y(x+y)=y^2+xy$, J contains $xy$. Hence, $deg(R/J)=2$. Thus, $a_1=4$. Note that ${(x+y)}^2\notin I$. Since $deg\left(Z\right)=6$ and $a_2\ge a_3$, the residual sequence of Z with respect to $H_{x+y}$ is $(4,1,1,0)$.

(b) In this step, we prove that $(3,2,1,0)$ is the residual sequence of Z with respect to $H_{x+y^2}$. Set $J_1:=(x+y^2,x^2,y^3)$. Since $y(x+y^2)\in J_1$, we have $xy\in J_1$ and hence $x{y}^2\in J_1$. Moreover, $-x$ and $y^2$ have the same image in $R/J_1$. Hence, $deg(H_{x+y^2}\cap Z)=3$, i.e., $a_1=3$. We have ${(x+y^2)}^3\in I$. Since $x{y}^2\notin I$, ${(x+y^2)}^2\notin I$. Hence, the residual sequence with respect to $H_{x+y^2}$ is $(3,2,1,0)$. Since $x(x+y^2)\notin I$ and $y(x+y^2)\notin I$, ${Res}_{H_{x+y^2}}\left(Z\right)$ is the 2-point with $(0,0)$ as its reduction.

The same residual sequence is obtained using $H_{y+y^2}$.

(c) Take $f=x+y+y^2$ and set $J_2:=(x+y+y^2,x^2,y^3)$. Since $y^2(x+y+y^2)\in J_2$, $x{y}^2\in J_2$. Since $xf=x^2+xy+x{y}^2$, $xy\in J_2$. Since $yf=xy+y^2+y^3\in J_2$, $y^2\in J_2$. Hence, $x+y\in J_2$. Thus, the vector space $R/J_2$ has 1 and x as a basis. Thus, $a_1=4$. Since $f^3\notin I$, $(4,1,1,0)$ is the residual sequence of Z with respect to $H_{x+y+y^2}$.

(d) Take $f=x+y+xy$ and set $J_3:=(f,x^2,y^3)$. Since $y^{2}f\in J_3$, $x{y}^2\in J_3$. Since $xf\in J_3$, $xy\in J_3$. Hence, $x+y\in J_3$. Thus, we see that $J_3=J$ and $a_1=4$. Since $f^2\notin I$, Z has residual sequence $(4,1,1,0)$ with respect to $H_{x+y+xy}$.

(e) Take $f=x+y^2$. Since $f^3\in I$, $a_3=0$. Set $J_4:=(f,I)$. Since $y^{2}f\in J_4$, $x{y}^2\in J_4$. Moreover, $y^2$ and $-x$ have the same image in $R/J_4$. Hence, Z has residue sequence $(4,2,0)$.

Theorem 5.

Fix positive integers t and d. Then, either $h^0\left({\mathcal{I}}_{B\left(t\right)}\left(d\right)\right)=0$ or $h^1\left({\mathcal{I}}_{B\left(t\right)}\left(d\right)\right)=0$.

Proof. 

To prove that for all t either $h^0\left({\mathcal{I}}_{B\left(t\right)}\left(d\right)\right)=0$ or $h^1\left({\mathcal{I}}_{B\left(t\right)}\left(d\right)\right)=0$, it is sufficient to prove it for $t\in \{\lfloor (d+2\left)\right(d+1)/12\rfloor ,\lceil (d+2\left)\right(d+1)/12\rceil \}$ (Remark 4). By the semicontinuity theorem for cohomology to prove that either $h^0\left({\mathcal{I}}_{B\left(t\right)}\left(d\right)\right)=0$ or $h^1\left({\mathcal{I}}_{B\left(t\right)}\left(d\right)\right)=0$ for a fixed $(t,d)$ it is sufficient to prove the existence of $B\in \mathcal{B}\left(t\right)$ such that either $h^0\left({\mathcal{I}}_B\left(d\right)\right)=0$ or $h^1\left({\mathcal{I}}_B\left(d\right)\right)=0$. Take a line L, a smooth conic D, and a smooth plane cubic C.

(a) In this step, we prove that $h^i\left({\mathcal{I}}_{B\left(1\right)}\left(2\right)\right)=0$, $i=0,1$. Since $h^0\left({\mathcal{O}}_{{\mathbb{P}}^2}\left(2\right)\right)=6$, it is sufficient to prove that $h^0\left({\mathcal{I}}_{B\left(1\right)}\left(2\right)\right)=0$. There are bricks A such that $h^0\left({\mathcal{I}}_A\left(2\right)\right)\ne 0$, e.g., take a line $R\ne L$ and take the brick $2L\cap 3R$.

Take any brick E such that $h^0\left({\mathcal{I}}_E\left(2\right)\right)\ne 0$ and set $\left\{p\right\}:=E_{\mathrm{red}}$. Since $E\supset 2p$, any conic containing E is singular at p. Hence, there is a line R such that $p\in R$ and $deg(R\cap E)\ge 3$. Fix $p\in D$. Take a brick B with residual sequence $(4,1,1,0,0)$ with respect D and assume the existence of a conic $D^{\prime }$ containing B. Call $R\subset D^{\prime }$ a line such that $deg(B\cap R)\ge 3$. The theorem of Bezout gives that R is contained in the smooth conic D, which is a contradiction.

(b) In this step, we prove that $h^1\left({\mathcal{I}}_{B\left(1\right)}\left(3\right)\right)=0$ and $h^0\left({\mathcal{I}}_{B\left(2\right)}\left(3\right)\right)=0$. We specialize $B\left(1\right)$ to B with residual sequence $(4,2,0)$ with respect to L. Since $h^i(L,{\mathcal{I}}_{B\cap L,L}\left(3\right))=0$, $i=0,1$, it is sufficient to use that $h^1\left({\mathcal{I}}_{{\mathrm{Res}}_L\left(B\right)}\left(2\right)\right)=0$. We degenerate $B\left(2\right)$ to $B\cup B\left(1\right)$ and use that $h^0\left({\mathcal{I}}_{B\left(1\right)}\left(2\right)\right)=0$ (step (a)).

(c) In this step, we prove that $h^1\left({\mathcal{I}}_{B\left(2\right)}\left(4\right)\right)=0$ and $h^0\left({\mathcal{I}}_{B\left(3\right)}\left(4\right)\right)=0$. We specialize $B\left(2\right)$ to a general union of 2 bricks with residual sequence $(4,2,0)$ with respect to D. To conclude the proof for $B\left(2\right)$, it is sufficient to observe that $h^1\left({\mathcal{I}}_E\left(2\right)\right)=0$ if E is the union of two connected degree-two subschemes of D. We degenerate $B\left(3\right)$ to a general union Z of 3 bricks of residual sequence $(4,2,0)$ with respect to C. We have $deg(Z\cap C)=12$ and (since the elliptic curve C has finite 4-torsion) for a general $Z_{\mathrm{red}}\subset C$, this scheme is not an element of $|{\mathcal{O}}_C\left(4\right)|$. Hence, $h^i(C,{\mathcal{I}}_{Z\cap C,C}\left(4\right))=0$, $i=0,1$. By the residual exact sequence of C, it is sufficient to use $h^0\left({\mathcal{I}}_E\left(1\right)\right)=0$, where E is a general union of three connected degree-two subschemes of C.

(d) In this step, we prove that $h^1\left({\mathcal{I}}_{B\left(3\right)}\left(5\right)\right)=0$ and $h^0\left({\mathcal{I}}_{B\left(4\right)}\left(5\right)\right)=0$. We specialize $B\left(3\right)$ to a general union B of three bricks of type $(4,1,1,0)$ with respect to C. Set $E:={\mathrm{Res}}_C\left(B\right)$. Since $h^1(C,{\mathcal{I}}_{C\cap B,C}\left(5\right))=0$, the residual exact sequence of C shows that it is sufficient to prove that $h^i\left({\mathcal{I}}_E\left(2\right)\right)=0$, $i=0,1$. The scheme E is a general union of three connected degree-two schemes with reduction contained in C. Apply Lemma 2. We specialize $B\left(4\right)$ to a general union $B_1$ of 4 bricks with residual sequence $(4,2,0)$ with respect to C. Since $h^0(C,{\mathcal{I}}_{C\cap B_1,C}\left(5\right))=0$, it is sufficient to prove that $h^0\left({\mathcal{I}}_G\left(2\right)\right)=0$, where G is a general union of four connected degree-two subschemes of C. Apply Remark 2 to G.

From now on, we assume $d\ge 6$ and that the theorem is true for all line bundles of degree $<d$.

(e) Assume $d\equiv 3mod4$. We may assume $t\ge (d+1)/4$. We specialize the scheme $B\left(t\right)$ to $B\cup B(t-(d+1)/4)$, with B being a general union of $(d+1)/4$ bricks of residual sequence $(4,1,1,0)$ with respect to L. Set $E:={\mathrm{Res}}_L\left(B\right)$. The scheme E is a general union of $(t+1)/4$ connected degree-two schemes with reduction contained in L. Note that we have $L\cap B(t-(d+1)/4))=\varnothing$ and $h^i(L,{\mathcal{I}}_{B\cap L,L}\left(d\right))=0$. Thus, by the residual exact sequence of L, it is sufficient to prove that either $h^0\left({\mathcal{I}}_{E\cup B(t-(d+1)/4)}(d-1)\right)=0$ or $h^1\left({\mathcal{I}}_{E\cup B(t-(d+1)/4)}(d-1)\right)=0$. Apply Lemma 2 and the inductive assumption in degree $d-1$ and $d-2$.

(f) Assume $d\equiv 0(mod4)$. Note that $d\ge 8$. We may assume $t\ge d/4$. We specialize $B\left(t\right)$ to $B\cup B(t-d/4)$ with $B_{\mathrm{red}}$ being general in C and each connected component of B with residual sequence $(4,1,1,0)$ with respect to C. We have $h^i(C,{\mathcal{I}}_{B\cap C,C}\left(d\right))=0$, because $B_{\mathrm{red}}$ is general in C and the elliptic curve C has finite 4-torsion. We use Lemma 2 and the inductive assumption in degrees $d-3$ and $d-6$.

(g) Assume $d\equiv 2(mod4)$. We may assume $t\ge (d+2)/4$. Take $S\subset L$ such that $\#S=(d-2)/4$ and $P\in L\setminus L$. We degenerate $B\left(t\right)$ to $B\cup B^{\prime }\cup B(t-(d+2)/4)$ with B a general union of $(d-2)/4$ bricks such that $B_{\mathrm{red}}=S$ and each connected component of B has residual sequence $(4,1,1,0)$ with respect to L, $B_{\mathrm{red}}^{\prime }=\left\{P\right\}$ and $B^{\prime }$ has residual sequence $(3,2,1,0)$ with respect to L. Set $E:={\mathrm{Res}}_L(B\cup B^{\prime })$. The scheme E is the union of $2P$ and a general union of connected degree-two schemes with S as their reduction. Since $h^i(L,{\mathcal{I}}_{(B\cup B^{\prime })\cap L,L}\left(d\right))=0$, $i=0,1$, the residual exact sequence of L shows that it is sufficient to prove that either $h^0\left({\mathcal{I}}_{E\cup B(t-(d+2)/4)}(d-1)\right)=0$ or $h^{0}1\left({\mathcal{I}}_{E\cup B(t-(d+2)/4)}(d-1)\right)=0$. Use Lemma 2 and the inductive assumption in degrees $d-1$, $d-2$ and $d-3$.

(h) Assume $d\equiv 1(mod4)$. Hence, $d\ge 9$. We may assume $t\ge (d+3)/4$. Take $S\subset L$ such that $\#S=(d-5)/4$ and $S^{\prime }\subset L\setminus L$ such that $\#S^{\prime }=2$. We degenerate $B\left(t\right)$ to $B\cup B^{\prime }\cup B(t-(d+3)/4)$ with B a general union of $(d-5)/4$ bricks such that $B_{\mathrm{red}}=S$ and each connected component of B has residual sequence $(4,1,1,0)$ with respect to L, $B_{\mathrm{red}}^{\prime }=S^{\prime }$ and each connected component of $B^{\prime }$ has residual sequence $(3,2,1,0)$ with respect to L. Then, we continue as in step (g). □

Proposition 8.

Fix an integer $t\ge 2$. For any $B\in \mathcal{B}\left(t\right)$ and any integer $d\ge 2$, we have $h^0\left({\mathcal{I}}_B\left(d\right)\right)\le \left({\displaystyle \genfrac{}{}{0pt}{}{d}{2}}\right)$ and equality holds for all d if and only if it holds for $d=2$, i.e., if and only if there is a line L such that $B\subset 2L$.

Proof. 

For any line $L\subset {\mathbb{P}}^2$, there are bricks $E\subset 2L$. For any such scheme E, we have $h^0(2L,{\mathcal{I}}_{E,2L}\left(2\right))=0$ by the theorem of Bezout. Hence, the residual exact sequence of $2L$ gives $h^0\left({\mathcal{I}}_E\left(d\right)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{d}{2}}\right)$ for all $d\ge 2$. Take a brick B such that $h^0\left({\mathcal{I}}_B\left(2\right)\right)\ne 0$ and fix $Q\in |{\mathcal{I}}_B\left(2\right)|$. Since any brick contains a 2-point, $B_{\mathrm{red}}\subseteq Sing\left(Q\right)$. Since $t\ge 2$, Q is a double line. □

Lemma 29.

Let M be a surface and $P\in M_{\mathrm{reg}}$. Let $B\subset M$ such that $B_{\mathrm{red}}=\left\{P\right\}$. There is no curve $T\subset M$ such that $P\in T_{\mathrm{reg}}$, and B has a residual sequence $(a_1,\cdots ,a_s,0)$ with $a_1\ge 5$.

Proof. 

It is sufficient to observe that $B\subset (4P,M)$ and that $deg\left(\right(4P,M)\cap T)=4$ because P is a smooth point of T. □

Example 1.

Fix a positive integer t, a line $L\subset {\mathbb{P}}^2$ and a set $S\subset L$ such that $\#S=t$. By parts (a) or (e) of Remark 17, there is $B\in \mathcal{B}\left(t\right)$ such that $B_{\mathrm{red}}=S$ and $deg(L\cap B)=4t$. Since $h^1(L,{\mathcal{I}}_{B\cap L,L}(4t-2))=1$ and $h^1(L,{\mathcal{I}}_{B\cap L,L}\left(x\right))=0$ for all $x\ge 4t-1$, Remark 1 and the residual exact sequence of L give $h^1\left({\mathcal{I}}_B(4t-2)\right)=1$ and $h^1\left({\mathcal{I}}_B\left(x\right)\right)=0$ for all $x\ge 4t-1$.

Proposition 9.

Let t be a positive integer. For every $B\in \mathcal{B}\left(t\right)$, we have $h^1\left({\mathcal{I}}_B\left(x\right)\right)=0$ for all $x\ge 4t-1$. Moreover, $h^1\left({\mathcal{I}}_B(4t-2)\right)\ne 0$ if and only if B is as in Example 1.

Proof. 

Set $B_0:=B$. Take a line $L_1\subset {\mathbb{P}}^2$ such that $b_1:=deg(L_1\cap B)$ is maximal and set $B_1:={\mathrm{Res}}_{L_1}\left(B\right)$. For all integers $i\ge 2$, define the line $L_i$, the integer $b_i\ge 0$ and the scheme $B_i$ in the following way. Let $L_i\subset {\mathbb{P}}^2$ be a line such that $b_i:=deg(L_i\cap B_{i-1})$ is maximal and set $B_i:={\mathrm{Res}}_{L_i}\left(B_{i-1}\right)$. Note that we allow the case $L_i=L_j$ for some $i\ne j$. Since $B_i\subseteq B_{i-1}$, we have $b_{i-1}\ge b_i$. Since $h^0\left({\mathcal{O}}_{{\mathbb{P}}^2}\left(1\right)\right)=3$, if $deg\left(B_{i-1}\right)\le 2$, then $B_{i-1}\subset L_i$, $B_i=\varnothing$ and $b_j=0$ for all $j>0$. Hence, there is an integer $i\le 4t-1$ such that $b_i=0$. Lemma 29 gives that $b_1\le 4t$ and that $b_1=4t$ if and only if B is as in Example 1. Fix an integer $y\ge 4t-2$ and assume $h^1\left({\mathcal{I}}_B\left(y\right)\right)>0$. Using $i-1$ residual exact sequence, first with $L_1$, then with $L_2$ and so on, we obtain the existence of a minimal integer $e\le 2t-1$ such that $h^1(L_e,{\mathcal{I}}_{L_e\cap B_{e-1},L_e}(y+1-e))>0$. Thus, $b_e\ge y+3-e$. Since $b_i\ge b_e$ for $i<e$, we obtain $6t\ge e(y+3-e)$. Set $f_y\left(z\right)=z(y+3-z)$. By assumption, $2e\le y+3)$. The function $f_y\left(z\right)$ is strictly increasing in the closed interval $1\le z\le (y+3)/2$. If $e=1$, then we conclude. Now assume $e\ge 2$. We plug in $e=2$ and $e=(y+3)/2$. We obtain $6t\ge 2(y+1)\ge 2(4t-1)$, contradicting the inequality $6t\ge e(y+3-e)$. □

Ref. [6] described in many cases the Hilbert function of general unions $Z\subset X_{\mathrm{reg}}$ of connected zero-dimensional schemes with prescribed isomorphic type and degree at most 4 if $X\subset {\mathbb{P}}^r$ is an integral surface. It is natural to ask if similar results may be proved for embedded varieties $X\subset {\mathbb{P}}^r$ with $n:=dimX>2$. We will see that the situation is easier for $n=3$ and much easier for $n>3$. Fix $P\in X_{\mathrm{reg}}$ and a zero-dimensional scheme $Z\subset X$ such that $Z_{\mathrm{red}}=\left\{P\right\}$. Set $z:=deg\left(Z\right)$. If $z\le 2$, then Remark 2 and Lemmas 2 and 9 work for any n. Thus, we may assume $z\ge 3$. In this case, another invariant of the isomorphism type of Z is its embedding dimension. Let e be the embedding dimension of Z. The case $e=1$, i.e., the curvilinear case, is handled by Remark 2. Thus, we may assume $e\ge 2$. We have $e\le z-1$ and equality holds if and only if there is a germ $Y\subseteq X$ at P of a smooth e-dimensional variety such that $Z\subset Y$ and $Z=(2P,Y)$. Call $\mathcal{Z}(X,Z)$ the set of all zero-dimensional schemes $W\subset X$ such that $W_{\mathrm{red}}=\left\{P\right\}$ and $W\cong Z$ as an abstract scheme.

J. Briançon gave the classification of all connected zero-dimensional schemes of a smooth surface with degree at most 6 ([8], pp. 74–80). The case $deg\left(Z\right)=4$ was used in [5,6]. In the case $deg\left(Z\right)=5$, there are two cases with $deg(Z\cap L)=3$ and $Z\subset 2L$ for some line L[8] ($K^2$ and $S^2\setminus K^2$ at p. 78) and one of them was used to study the secant varieties of the tangential variety of an embedded surface ([2]). The proofs performed to compute the Hilbert function of a general union of them ([2,3,4]) work for both of them. Indeed, they work by specialization and the Horace Lemma for general union of any of the two types of schemes. As in the case $deg\left(Z\right)=4$ and $2p\subset Z\subset 3p$, both cases are parametrized by irreducible varieties and the one called $K^2$ is a flat limit of the one called $S^2$ in [8].

Remark 18.

Assume $e=n$. Since $e\le z-1\le 3$ and $n\ge 3$, this is possible only if $n=3$ and $Z=2P$. The Hilbert function of general unions of 2-points is a very important topic of research ([1]).

(a) Assume $n=3$ and $e=2$. Let $T\subset X$ be a surface such that $P\in T_{\mathrm{reg}}$. There is $W\in \mathcal{Z}(X,Z)$ such that $W\subset T$. There is $A\in \mathcal{Z}(X,Z)$ such that $A⊈T$ and $A\cap T$ is curvilinear. Hence, we control the Hilbert function inside T of general unions of schemes like A. Now assume $A⊈T$ and $A\cap T$ not curvilinear. This case happens only if $z=4$. In this case $A\cap T=(2P,T)$ and ${Res}_T\left(A\right)=\left\{P\right\}$. This case may occur both if Z is a 2-square and if Z is a tile (with the terminology of [6]).

(b) Assume $n>3$ and $e=2$.

(b1) Assume $z=3$. In this case Z is the germ at P of $(2P,Y)$, where Y is a surface with $P\in Y_{\mathrm{reg}}$. Take $W\in \mathcal{Z}(X,Z)$. Let $T\subset X$ be a hypersurface. If $P\in \mathrm{Sing}\left(T\right)$, then $W\subset T$. Now assume $P\in T_{\mathrm{reg}}$. We have $deg(W\cap T)\ge 2$. There are hypersurfaces $F,G$ such that $P\in F_{\mathrm{reg}}\cap G_{\mathrm{reg}}$, $deg(W\cap F)=2$ and $W\subset G$.

(b2) Assume $z=4$. In this case, there are two isomorphic types for Z (called $Z_2$ and K in [8], pp. 76-77, called 2-squares and tiles in [6]). The following remarks are the same for both isomorphism types. Fix $W\in \mathcal{Z}(X,Z)$. Since $Z\supset (2P,Y)$ for some surface Y smooth at P, $deg(W\cap T)\ge 2$ for all hypersurfaces $T\subset X$ such that $P\in T_{\mathrm{reg}}$. Moreover, if $deg(W\cap T)=2$, then $deg\left({Res}_T\left(W\right)\right)=2$ and both $T\cap W$ and ${Res}_T\left(W\right)$ are curvilinear. There is a hypersurface $T\subset X$ such that $P\in T_{\mathrm{reg}}$ and $W\subset T$. There is a hypersurface $J\subset X$ such that $P\in J_{\mathrm{reg}}$, $W⊈J$ and $J\cap W$ is the 2-point of a smooth surface contained in J. In this case ${Res}_J\left(W\right)=\left\{P\right\}$. There is a hypersurface $F\subset X$ such that $P\in F_{\mathrm{reg}}$ and $deg(F\cap W)=2$. If Z (and hence W) is a tile, and there is a hypersurface $G\subset X$ such that $deg(W\cap G)=3$ and $W\cap G$ is curvilinear. No such hypersurface exists if Z is a 2-square ([7] (Prop. 2.15)). Conversely, for all isomorphic types, 2-squares or tiles, of Z and any hypersurface $E\subset X$ with $P\in E_{\mathrm{reg}}$ there is $A\in \mathcal{Z}(X,Z)$ such that $Z\subset E$.

(c) Assume $e=3$ and hence $z=4$ and Z the 2-point of the germ of a 3-fold smooth at P. Assume $n>3$. Fix any $W\in \mathcal{Z}(X,Z)$. Let $E\subset X$ be the germ at P of a smooth hypersurface of X. Either $W\subset E$ or $W\cap E$ is the 2-point of the germ of a smooth surface contained in E and ${Res}_E\left(W\right)=\left\{P\right\}$. We have $W\subset E$ if and only if W is contained in the Zariski tangent space of E at P.

9. Over the Rational and the Real Numbers

Up to now we assumed that the base field $\mathbb{K}$ is algebraically closed and of characteristic zero. For several reasons we need to assume that the base field $\mathbb{K}$ has characteristic 0, i.e., $\mathbb{Q}\subseteq \mathbb{K}$. The 3-dimensional Segre varieties ${\mathbb{P}}^3$, ${\mathbb{P}}^2\times {\mathbb{P}}^1$ and ${\mathbb{P}}^1\times {\mathbb{P}}^1\times {\mathbb{P}}^1$ are defined over $\mathbb{Q}$ and hence over $\mathbb{R}$. However, ${\mathbb{P}}^1$ has 2 real structures, the usual one with ${\mathbb{P}}^1\left(\mathbb{R}\right)$ diffeomorphic to the circle and another one with ${\mathbb{P}}^1\left(\mathbb{R}\right)=\varnothing$. The latter real structure is realized by the real smooth conic associated to a positive-definite quadratic form in 3 variables. Hence, ${\mathbb{P}}^1\times {\mathbb{P}}^1\times {\mathbb{P}}^1$ and ${\mathbb{P}}^2\times {\mathbb{P}}^1$ have a real structure with no real point. The same is true for $Q_3$. In this case it is obvious that over $\mathbb{Q}$, there even more possibilities, equivalent to the classification over $\mathbb{Q}$ of the rank 4 quadratic forms in 4 variables. From now on we take over $\mathbb{Q}$ the usual structure of ${\mathbb{P}}^3$, ${\mathbb{P}}^2\times {\mathbb{P}}^1$ and ${\mathbb{P}}^1\times {\mathbb{P}}^1\times {\mathbb{P}}^1$ associated to the polynomial rings $\mathbb{Q}[x_0,x_1,x_2,x_3]$, $\mathbb{Q}[x_0,x_1,x_2,y_0,y_1]$, $\mathbb{Q}[x_0,x_1,y_0,y_1,z_0,z_1]$. For $Q_3$ we take the usual structure. With these assumptions for these 3-folds $X\left(\mathbb{Q}\right)$ is Zariski dense in $X\left(\mathbb{C}\right)$ and $X\left(\mathbb{Q}\right)$ is dense in $X\left(\mathbb{R}\right)$ for the euclidean topology. Moreover, for each $p\in X\left(\mathbb{Q}\right)$ the $\mathbb{Q}$-points $T_{p}X\left(\mathbb{Q}\right)$ of the tangent space $T_p\left(X\right)$ are dense in $T_{p}X\left(\mathbb{C}\right)$. For each $P\in Q_3\left(\mathbb{Q}\right)$ we find a hyperplane section $Q_2\subset Q_3$ containing p and with $Q_2\left(\mathbb{Q}\right)\cong {\mathbb{P}}^1\left(\mathbb{Q}\right)\times {\mathbb{P}}^1\left(\mathbb{Q}\right)$. With these assumptions, we may define the residual sequences and residual sequence needed to say that in Theorems 1–4 we may find $A\left(t\right)$ whose connected components are defined over $\mathbb{Q}$ (we use that, with the same assumptions, we also have Propositions 1 and 2). The same holds for ${\mathbb{P}}^2$ and Theorem 5 on bricks. Indeed, we were very explicit in Remarks 5 and 17 and in Proposition 1 to see why the residual sequences and residual sequence may be defined over $\mathbb{Q}$. For Theorem 5 on the bricks, we also use that there is a smooth elliptic curve C defined over $\mathbb{Q}$ and with $C\left(\mathbb{Q}\right)$ infinite ([15], Ch. VIII). We are not claiming that the classification in [8] holds over $\mathbb{Q}$, only that the classes over $\mathbb{C}$ have representative defined over $\mathbb{Q}$. This is obvious by the equations given in [8], Ch. IV.

10. Discussion

We evaluate general unions of a class of degree-eight zero-dimensional schemes at the elements of certain multivariate polynomials. These sets of multivariate polynomials are the finite vector spaces which are addenda of the multigraded rings obtained from Segre–Veronese embeddings of 3-dimensional multi-projective spaces. The Segre–Veronese embeddings of multiprojective space are the one used to study partially symmetric tensors with a fixed partially symmetric tensor rank and evaluations of there addenda at general unions of some zero-dimensional schemes, called double points.

Our main results are for general unions of certain degree-eight zero-dimensional schemes on smooth 3-folds, 2cubes with degree-two sides. We also study unions (even non general) of other low degree schemes. Our proofs use an inductive procedure, called the Horace Method. The tools used in the proofs comes from Algebraic Geometry and some elementary parts of Commutative Algebra. We would be happy if somebody else carry over our suggestions with practical implementations.

Our main contribution is to obtain for each type of connected zero-dimensional scheme discrete invariants (called here “residual sequence”) to use in this inductive method.

Although not needed in this paper, for low-degree cases needed to start the inductive process, the computer algebra system should be extremely important and fast.

In the introduction, we pose several open projects and a question for the interested readers.