On the Castelnuovo-Mumford regularity of subspace arrangements

Aldo Conca Dipartimento di Matematica, Università di Genova,
Via Dodecaneso 35, 16146 Genova, Italy [email protected] and Manolis C. Tsakiris Academy of Mathematics and Systems Science,
Chinese Academy of Sciences, 100190, Beijing, China [email protected]

Abstract.

Let $X$ be the union of $n$ generic linear subspaces of codimension $>1$ in $\mathbb{P}^{d}$ . Improving an earlier bound due to Derksen and Sidman [DS02] we prove that the Castelnuovo-Mumford regularity of $X$ satisfies $\operatorname*{reg}(X)\leq n-\lfloor{n/(2d-1)\rfloor}$ .

Key words and phrases:

subspace arrangements, Castelnuovo-Mumford regularity, linear resolutions, Hilbert functions, flat degenerations

2020 Mathematics Subject Classification:

13D02,14M07,14B15

1. Introduction

The notion of Castelnuovo-Mumford regularity was introduced by Mumford in [MB66], to systematically control the vanishing of sheaf cohomology. Since then it has proved to be a fundamental notion throughout algebraic geometry and commutative algebra, both from the theoretical and computational perspectives. For instance, proofs for the existence of the Hilbert scheme rely on it [Kol99, MB66], while so do complexity bounds for Gröbner bases computations [BM93]. It even plays a role in the theory of subspace clustering [TV17] in machine learning.

In the present work we are concerned with the Castelnuovo-Mumford regularity $\operatorname*{reg}(X)$ of the reduced union $X$ of $n$ generic linear spaces $X_{1},\dots,X_{n}$ of arbitrary dimensions in a projective space $\mathbb{P}_{k}^{d}$ over an infinite field $k$ . When there are no pairwise intersections between the $X_{i}$ ’s, the regularity can be extracted from another important invariant; the Hilbert function. For points, i.e. when $\dim X_{i}=0$ , the Hilbert function is easy to compute; e.g. see [GMR83]. However, for higher dimensions determining the Hilbert function is very challenging. A landmark achievement regarding this problem is the work of Hartshorne & Hirschowitz [HH82] who settled the case of lines. Another important step forward was taken by Derksen [Der07], who gave a formula for the Hilbert polynomial of $X$ and proved that it agrees with the Hilbert function at degrees $\geq n$ . On the other hand, the Hilbert function is largely unknown at degrees $<n$ .

The hardest part of the proof of Hartshorne & Hirschowitz [HH82] was to treat the case $d=3$ , which was done by degeneration techniques via smooth quadric surfaces. Recently a new proof was given by Aladpoosh & Catalisano [AC21], where degeneration techniques via linear spaces replaced those via quadrics. These have the potential of generalizing to higher dimensions and have inspired us in this article.

For generic lines the formula for the regularity has been derived by Rice [Ric22] from [HH82]. Apart from this, there has been a single general result regarding $\operatorname*{reg}(X)$ , due to Derksen and Sidman [DS02], who proved that $\operatorname*{reg}(X)\leq n$ for any reduced union of linear spaces, and provided examples of (special) configurations showing that their result is sharp; for special subspace arrangements there exist various results such as [BDMV18] and [BPS05].

In this paper we prove:

Theorem 1.

For $n$ generic linear spaces $X_{1},\dots,X_{n}$ of codimension bigger than $1$ in $\mathbb{P}_{k}^{d}$ , the regularity of their reduced union $X$ satisfies

\operatorname*{reg}(X)\leq n-\lfloor{n/(2d-1)\rfloor}.

At the heart of our proof lies a surprising result:

Theorem 2.

For $n=2d-1$ generic linear spaces $X_{1},\dots,X_{n}$ of codimension $2$ in $\mathbb{P}_{k}^{d}$ , the regularity of their reduced union $X$ is equal to $n-1$ and the saturated ideal $I_{X}$ that defines $X$ has a linear graded minimal free resolution.

Another device that plays a role in the proof of Theorem 1, and of potential further significance to the problem of determining $\operatorname*{reg}(X)$ , is:

Theorem 3.

Let $J$ be any non-zero saturated homogeneous ideal of a polynomial ring $S$ of dimension $r$ over $k$ . For $c=1,\dots,r$ let $L_{c}$ be an ideal of $S$ generated by $c$ generic linear forms. Then

\operatorname*{reg}(J\cap L_{c})=\left\{\begin{matrix}&\operatorname*{reg}(J)+1,\,\,&\mbox{ if }c\leq\gamma\\ &\operatorname*{reg}(J),\,\,&\mbox{ if }c>\gamma\end{matrix}\right.,

where $\gamma=\max\{c:\operatorname*{reg}(J+L_{c})=\operatorname*{reg}(J)\}$ .

In Theorem 3 one always has $\gamma\geq\operatorname*{depth}(S/J)$ and in many cases the inequality is strict, e.g. if $J$ has a linear resolution then $\gamma\geq r-1$ .

The bound of Theorem 1 is in general not sharp. As asserted though by Theorem 2, it is sharp when $X$ is the union of $n=2d-1$ codimension- $2$ generic linear spaces in $\mathbb{P}_{k}^{d}$ . For that particular configuration, Theorem 2 implies that the Hilbert function of $X$ is fully determined: it coincides with the Hilbert function of $\mathbb{P}_{k}^{d}$ at degrees $<2d-2$ , while it agrees thereafter with the Hilbert polynomial of $X$ (described in [Der07]). Moreover, the ideal of $X$ has a linear resolution; we are not aware of other codimension- $2$ configurations where this phenomenon occurs. It further follows from our arguments that whenever the union of $n_{0}$ condimension- $2$ generic linear spaces in $\mathbb{P}_{k}^{d}$ has regularity $\leq n_{0}-\alpha$ , then a union of $n$ generic linear spaces of codimension $\geq 2$ in $\mathbb{P}_{k}^{d}$ has regularity $\leq n-\alpha\lfloor{n/n_{0}\rfloor}$ .

In §2 we set up notational conventions and in §3 we review preliminaries. In §4 we prove Theorem 2 and in §5 we prove Theorem 3. Finally we prove Theorem 1 in §6.

2. Conventions

In the rest of the paper we set $r=d+1$ and we use $\mathbb{P}^{r-1}_{k}=\operatorname*{Proj}S$ for $\mathbb{P}^{d}_{k}$ , where $S=k[x_{1},\dots,x_{r}]$ is the polynomial ring of dimension $r\geq 3$ over the infinite field $k$ , endowed with the standard $\mathbb{Z}$ -grading. We denote by $\mathfrak{m}=(x_{1},\dots,x_{r})$ the unique maximal homogeneous ideal of $S$ . For a finitely generated graded $S$ -module $M$ we denote by $[M]_{\nu}$ the component of $M$ of degree $\nu$ , while the Hilbert function $\operatorname*{HF}(M,\nu)$ gives the dimension of the $k$ -vector space $[M]_{\nu}$ .

For $Y$ a closed subscheme of $\mathbb{P}_{k}^{r-1}$ , we let $\mathcal{I}_{Y}$ be the ideal sheaf of $Y$ , and $I_{Y}$ the unique graded saturated ideal of $S$ for which $\mathcal{I}_{Y}=\tilde{I}_{Y}$ —here $\tilde{I}_{Y}$ is the coherent sheaf induced by $I_{Y}$ . When we say that a scheme $Y$ contains a scheme $Z$ , we will always mean an inclusion of ideals $I_{Y}\subset I_{Z}$ . Similarly, by the union $Y\cup Z$ of two schemes we will mean the scheme defined by the ideal $I_{Y}\cap I_{Z}$ .

With $b$ a non-negative integer and $n$ any integer, we denote the ordinary numerical binomial coefficient as ${n\choose b}=n!/((n-b)!b!)$ , with the convention that it is $0$ if $n<b$ . With $t$ an element in a commutative ring that contains $\mathbb{Q}$ , we denote the binomial polynomial coefficient as

\left[\begin{matrix}t\\ b\end{matrix}\right]=\frac{(t-b+1)\cdots(t-1)t}{b!}.

For any integer $a$ we have $\left[\begin{smallmatrix}a\\ b\end{smallmatrix}\right]={a\choose b}$ if and only if $a\geq 0$ .

Finally, for a positive integer $\ell$ we set $[\ell]=\{1,\dots,\ell\}$ .

3. Preliminaries

3.1. Regularity

The Castelnuovo-Mumford regularity of a closed scheme $Y\subset\mathbb{P}_{k}^{r-1}$ can be defined as the Castelunuovo-Mumford regularity $\operatorname*{reg}(I_{Y})$ of the saturated ideal $I_{Y}\subset S$ that defines $Y$ . Here more generally we review the Castelnuovo-Mumford regularity of a finitely generated graded module $M$ over the polynomial ring $S$ [Eis94, BH98, Pee10], and quote basic properties that we will need.

In terms of local cohomology, the regularity of $M$ is defined as

\operatorname*{reg}(M)=\max\{i+j:\,[H_{\mathfrak{m}}^{i}(M)]_{j}\neq 0\}.

In terms of the minimal graded free resolution of $M$ , $\operatorname*{reg}(M)$ is the maximum among the numbers $b_{i}-i$ , where $b_{i}$ is the largest degree of a minimal generator of the $i$ -th syzygy module of $M$ ; that is

\operatorname*{reg}(M)=\max\{b-i:\,[{\operatorname*{Tor}}_{i}^{S}(M,k)]_{b}\neq 0\}.

The following fact is known as the regularity lemma:

Proposition 4.

Let $0\rightarrow M^{\prime}\rightarrow M\rightarrow M^{\prime\prime}\rightarrow 0$ be a short exact sequence of finitely generated graded $S$ -modules and maps of degree zero. Then

	$\displaystyle\operatorname*{reg}(M)$	$\displaystyle\leq\max\{\operatorname{reg}(M^{\prime}),\operatorname{reg}(M^{\prime\prime})\}$
	$\displaystyle\operatorname*{reg}(M^{\prime})$	$\displaystyle\leq\max\{\operatorname{reg}(M),\operatorname{reg}(M^{\prime\prime})+1\}$
	$\displaystyle\operatorname*{reg}(M^{\prime\prime})$	$\displaystyle\leq\max\{\operatorname{reg}(M),\operatorname{reg}(M^{\prime})-1\}.$

Moreover, if $M^{\prime}$ has finite length, then

\operatorname*{reg}(M)=\max\{\operatorname*{reg}(M^{\prime}),\operatorname*{reg}(M^{\prime\prime})\}.

A linear form $x\in[S]_{1}$ is called almost regular on $M$ , if it is a non-zero divisor on $M/H_{\mathfrak{m}}^{0}(M)$ . When $M=S/J$ for some homogeneous ideal $J$ , this condition translates to $x$ being regular on $S/J^{\operatorname*{sat}}$ ; here $J^{\operatorname*{sat}}=J:\mathfrak{m}^{\infty}$ is the saturation of $J$ with respect to $\mathfrak{m}$ . We recall also that the saturation index $\operatorname*{sat}(J)$ of $J$ is the smallest integer $s$ for which $[J^{\operatorname*{sat}}]_{\nu}=[J]_{\nu}$ for every $\nu\geq s$ , i.e. $\operatorname*{sat}(J)=\operatorname*{reg}(J^{\operatorname*{sat}}/J)+1$ . With that, the behavior of the regularity upon taking quotients by almost regular elements is extremely useful for inductive arguments on the dimension:

Proposition 5.

Let $J$ be a homogeneous ideal of $S$ and $x\in S$ a linear form almost regular on $S/J$ . Then

\operatorname*{reg}(J)=\max\{\operatorname*{reg}(J+(x)),\,\operatorname*{sat}(J)\},

or equivalently

\operatorname*{reg}(S/J)=\max\{\operatorname*{reg}(S/J+(x)),\,\operatorname*{reg}(J^{\operatorname*{sat}}/J)\}.

For a proof of the above folklore fact we refer to [CH03].

3.2. Subspace Arrangements

Given an arrangement of $n$ $k$ -vector subspaces $V_{i},\,i\in[n]$ of $[S]_{1}$ , and the ideals $I_{i}$ that the $V_{i}$ ’s generate, we may consider the intersection $\cap_{i\in[n]}I_{i}$ or the product $\prod_{i\in[n]}I_{i}$ of the $I_{i}$ ’s. The intersection ideal is more geometric, because it is the radical ideal that defines the schematic union $X$ of the corresponding arrangement of linear spaces $X_{i}=\operatorname*{Proj}(S/I_{i})$ of $\mathbb{P}_{k}^{r-1}$ . But it is a much harder object to understand than the product ideal $\prod_{i\in[n]}I_{i}$ ; already describing a set of generators of $\cap_{i\in[n]}I_{i}$ is a difficult task [Idà90]. On the other hand, many properties of $\prod_{i\in[n]}I_{i}$ are by now well understood, including a complete description of its minimal graded free resolution and of a minimal primary decomposition [CT22].

Definition 6.

We call the subspace arrangement $V_{1},\dots,V_{n}$ of $[S]_{1}$ linearly general, if for every $A\subseteq[n]$ we have

\dim_{k}\sum_{i\in A}V_{i}=\min\left\{r,\,{\sum}_{i\in A}\dim_{k}V_{i}\right\}.

We call the subspace arrangement $X_{1},\dots,X_{n}$ of $\mathbb{P}_{k}^{r-1}$ linearly general, if the $V_{i}$ ’s are linearly general.

The following result allows valuable access from the product ideal to the intersection ideal.

Proposition 7 (Conca & Herzog, [CH03]).

Suppose that the subspace arrangement $V_{i},\,i\in[n]$ , is linearly general. Then for every $\nu\geq n$

\textstyle[\cap_{i\in[n]}I_{i}]_{\nu}=\big{[}\prod_{i\in[n]}I_{i}\big{]}_{\nu}.

Using a formula for the primary decomposition that they developed, Conca & Herzog [CH03] proved that

\textstyle\operatorname*{reg}\big{(}\prod_{i\in[n]}I_{i}\big{)}=n,

while Derksen & Sidman proved:

Proposition 8 (Derksen & Sidman, [DS02]).

\operatorname*{reg}(\cap_{i\in[n]}I_{i})\leq n.

Later on, Derksen gave a formula for the Hilbert polynomial and proved that it agrees with the Hilbert function from degrees $n$ and on:

Proposition 9 (Derksen, [Der07]).

Suppose that the subspace arrangement $V_{i},\,i\in[n]$ is linearly general. For any subset $A\subseteq[n]$ set $d_{A}=\dim_{k}\sum_{i\in A}V_{i}$ . Then for every $\nu\geq n$ we have

\operatorname*{HF}\big{(}S/\cap_{i\in[n]}I_{i},\nu\big{)}=\sum_{\emptyset\neq A\subseteq[n]}(-1)^{\#A+1}{\nu+r-1-d_{A}\choose r-1-d_{A}},

where all terms with $r-1-d_{A}<0$ are zero by convention.

Remark 10.

The notion of a linearly general subspace arrangement in projective space given in Definition 6 is equivalent to the notion of a transversal subspace arrangement as defined in [Der07].

3.3. Flat Degenerations and Hilbert functions

A typical technique that is employed in the study of families of schemes —such as unions of generic lines in $\mathbb{P}_{k}^{3}$ [HH82, AC21]— is that of a flat degeneration. This is a powerful tool due to the following fact:

Proposition 11.

Let $S[\lambda]=S\otimes_{k}k[\lambda]$ be the extended graded polynomial ring in $r+1$ variables over $k$ , where the $x_{i}$ ’s have degree $1$ and $\lambda$ has degree $0$ . Let $J_{\lambda}$ be a homogeneous ideal of $S[\lambda]$ and let $J_{c}$ be the ideal of $S$ obtained by replacing $\lambda$ with $c\in k$ . If $S[\lambda]/J_{\lambda}$ is a flat $k[\lambda]$ -module, then the Hilbert function of $J_{c}$ does not depend on $c$ .

For a proof see the arguments given in Exercise 20.14 of [Eis94]. We will use Proposition 11 later on, in the proof of Lemma 17, in the following fashion: under the flatness hypothesis, $J_{c}$ does not contain a form of degree $\nu$ if and only if $J_{0}$ does not contain a form of degree $\nu$ .

4. Proof of Theorem 2

4.1. First Hilbert Function Lemma

We begin by giving a formula for the Hilbert function at degree $\ell-1$ of a linearly general subspace arrangement of $\ell$ codimension- $2$ planes in $\mathbb{P}_{k}^{r-1}$ .

Lemma 12.

Suppose that the subspace arrangement $V_{i},\,i\in[\ell]$ , is linearly general and $\dim_{k}V_{i}=2$ for every $i\in[\ell]$ . Then

\operatorname*{HF}(\cap_{i\in[\ell]}I_{i},\ell-1)=\sum_{j\geq 0}(-1)^{j}{\ell\choose j}{\ell+r-2-2j\choose\ell-1}.

Proof.

Consider the short exact sequence

0\rightarrow\frac{S}{\cap_{i\in[\ell]}I_{i}}\rightarrow\frac{S}{\cap_{i\in[\ell-1]}I_{i}}\oplus\frac{S}{I_{\ell}}\rightarrow\frac{S}{I_{\ell}+\cap_{i\in[\ell-1]}I_{i}}\rightarrow 0.

By Proposition 9

	$\displaystyle\operatorname*{HF}\left(\frac{S}{\cap_{i\in[\ell-1]}I_{i}},\ell-1\right)=$
	$\displaystyle\sum_{\emptyset\neq A\subseteq[\ell-1]}(-1)^{\|A\|+1}{\ell-1\choose\|A\|}{\ell-1+r-1-2\|A\|\choose r-1-2\|A\|}$
	$\displaystyle=\sum_{j\geq 1}(-1)^{j+1}{\ell-1\choose j}{\ell+r-2-2j\choose r-1-2j}.$

By Proposition 7

\left[\bigcap_{i\in[\ell-1]}I_{i}\right]_{\ell-1}=\left[\prod_{i\in[\ell-1]}I_{i}\right]_{\ell-1}.

Thus, with $\bar{S}=S/I_{\ell}$ and $\bar{I}_{i}=I_{i}+I_{\ell}/I_{\ell}$ , we have

\left[\frac{S}{I_{\ell}+\bigcap_{i\in[\ell-1]}I_{i}}\right]_{\ell-1}=\left[\frac{\bar{S}}{{\prod}_{i\in[\ell-1]}\bar{I}_{i}}\right]_{\ell-1}.

Since the ideals $\bar{I}_{i},\,i\in[\ell-1]$ are generated by a linearly general subspace arrangement of $[\bar{S}]_{1}$ , where $\bar{S}$ is a polynomial ring over $k$ in $r-2$ variables, we can compute the Hilbert function of the last term of the short exact sequence at degree $\ell-1$ again by Proposition 9:

	$\displaystyle\operatorname*{HF}\left(\frac{S}{\prod_{i\in[\ell-1]}\bar{I}_{i}},\ell-1\right)=$
	$\displaystyle\sum_{\emptyset\neq A\subseteq[r-2]}(-1)^{\|A\|+1}{\ell-1\choose\|A\|}{\ell-1+r-3-2\|A\|\choose r-3-2\|A\|}$
	$\displaystyle=\sum_{j\geq 1}(-1)^{j+1}{\ell-1\choose j}{\ell+r-4-2j\choose r-3-2j}.$

Putting everything together, we arrive at

	$\displaystyle\operatorname{HF}(\cap_{i\in[\ell]}I_{i},\ell-1)=\operatorname{HF}(\cap_{i\in[\ell-1]}I_{i},\ell-1)-\operatorname*{HF}(\cap_{i\in[\ell-1]}\bar{I}_{i},\ell-1)$
	$\displaystyle=\sum_{j\geq 0}(-1)^{j}{\ell-1\choose j}{\ell+r-2-2j\choose r-1-2j}$
	$\displaystyle-\sum_{j\geq 0}(-1)^{j}{\ell-1\choose j}{\ell+r-4-2j\choose r-3-2j}$

	$\displaystyle=\sum_{j\geq 0}(-1)^{j}{\ell-1\choose j}{\ell+r-2-2j\choose r-1-2j}$
	$\displaystyle+\sum_{j\geq 1}(-1)^{j}{\ell-1\choose j-1}{\ell+r-2-2j\choose r-1-2j}$
	$\displaystyle=\sum_{j\geq 0}(-1)^{j}\bigg{[}{\ell-1\choose j}+{\ell-1\choose j-1}\bigg{]}{\ell+r-2-2j\choose r-1-2j}$
	$\displaystyle=\sum_{j\geq 0}(-1)^{j}{\ell\choose j}{\ell+r-2-2j\choose r-1-2j}.\qed$

∎

4.2. Second Hilbert Function Lemma

We recall a well-known binomial identity:

Lemma 13.

Let $p(t)$ be a polynomial in $t$ and $d>\deg(p)$ . Then

\sum_{j=0}^{d}(-1)^{j}{d\choose j}p(j)=0.

We have the following interesting fact:

Lemma 14.

Through $r-1$ linearly general codimension- $2$ linear spaces of $\mathbb{P}_{k}^{r-1}$ passes a unique hypersurface of degree $r-2$ , i.e.

\operatorname*{HF}(\cap_{i\in[r-1]}I_{i},r-2)=1.

Proof.

Lemma 12 gives us

	$\displaystyle\operatorname*{HF}(\cap_{i\in[r-1]}I_{i},r-2)=$
	$\displaystyle=\sum_{j\geq 0}(-1)^{j}{r-1\choose j}{2r-3-2j\choose r-2}$

	$\displaystyle=\sum_{j=0}^{\lfloor(2r-3)/2\rfloor}(-1)^{j}{r-1\choose j}\begin{bmatrix}2r-3-2j\\ r-2\end{bmatrix}$
	$\displaystyle=\sum_{j=0}^{r-2}(-1)^{j}{r-1\choose j}\begin{bmatrix}2r-3-2j\\ r-2\end{bmatrix}.$

As a polynomial in $t$ , the degree of

p(t)=\begin{bmatrix}2r-3-2t\\ r-2\end{bmatrix}

is $r-2$ . Hence Lemma 13 gives

\sum_{j=0}^{r-1}(-1)^{j}{r-1\choose j}\begin{bmatrix}2r-3-2j\\ r-2\end{bmatrix}=0.

Now we are done by observing that the last term in the summation above for $j=r-1$ is equal to $-1$ . ∎

4.3. Sundials

The notion of a sundial played a crucial role in the proof of [HH82]. It was also used in [Hir81] and has more recently been generalized in various ways; e.g. see [AC21]. Here we discuss yet another generalization that we will employ in the proof of Lemma 17.

Definition 15.

In $\mathbb{P}^{r-1}_{k}$ consider two linear spaces $Y$ and $Z$ , of codimensions $2$ and $i+2$ respectively, which lie in a hyperplane $H$ , and intersect at a codimension $i+3$ linear space $W$ . A $(2,i+2)$ -sundial is a scheme defined by an ideal sheaf of the form $\mathcal{I}_{Y}\cap\mathcal{I}_{Z}\cap\mathcal{I}_{W}^{2}$ . We say that the hyperplane $H$ supports the sundial.

Lemma 16.

Let $Y,Z$ be linearly general linear spaces in $\mathbb{P}^{r-1}_{k}$ of codimensions $2$ and $2+i$ , such that $4+i\leq r$ . Then there exists a flat family $E_{c},\,c\in k$ of closed subschemes of $\mathbb{P}^{r-1}_{k}$ , such that $E_{c\neq 0}$ is isomorphic to $Y\cup Z$ , while $E_{0}$ is a $(2,i+2)$ -sundial.

Proof.

We may assume that $I_{Y}=(x_{1},x_{2})$ and $I_{Z}=(x_{3},\dots,x_{i+4})$ . With $\lambda$ as in Section 3.3, we define in the polynomial ring $S[\lambda]$ the ideal

J_{\lambda}=\big{(}x_{1},\lambda(x_{2}-x_{3})+x_{3}\big{)}\cap(x_{3},\dots,x_{i+4}).

Since $x_{1},\lambda(x_{2}-x_{3})+x_{3}$ is a regular sequence in $S[\lambda]$ and $S[\lambda]/(x_{3},\dots,x_{r})$ , we have

J_{\lambda}=\big{(}x_{1},\lambda(x_{2}-x_{3})+x_{3}\big{)}(x_{3},\dots,x_{i+4}).

Moreover, the canonical ring homomorphism $\varphi:k[\lambda]\rightarrow S[\lambda]/J_{\lambda}$ is flat if and only if $S[\lambda]/J_{\lambda}$ is torsion-free as a $k[\lambda]$ -module —now this follows immediately once one notes that $\big{(}x_{1},\lambda(x_{2}-x_{3})+x_{3}\big{)}$ and $(x_{3},\dots,x_{i+4})$ are prime ideals of $S[\lambda]$ , whose contraction in $k[\lambda]$ is zero.

The ideal $J_{\lambda}$ induces a flat family of closed subschemes $E_{c},\,c\in k$ of $\mathbb{P}_{k}^{r-1}$ . The ideal of $S$ defining $E_{c}$ is

J_{c}=\big{(}x_{1},c(x_{2}-x_{3})+x_{3}\big{)}(x_{3},\dots,x_{i+4}).

For $c\neq 0$ we have that

J_{c}=\big{(}x_{1},c(x_{2}-x_{3})+x_{3}\big{)}\cap(x_{3},\dots,x_{i+4}),

defines two codimension $2$ and $i+2$ linear spaces of $\mathbb{P}_{k}^{r-1}$ , which intersect at codimension $i+4$ . On the other hand,

	$\displaystyle J_{0}$	$\displaystyle=\big{(}x_{1},x_{3}\big{)}(x_{3},\dots,x_{i+4})$
		$\displaystyle=(x_{1},x_{3})\cap(x_{3},\dots,x_{i+4})\cap(x_{1},x_{3},\dots,x_{i+4})^{2}$

defines a $(2,i+2)$ -sundial. ∎

4.4. A Variant of Castelnuovo’s Inequality

For a homogeneous ideal $J$ of $S$ and any linear form $x$ , the short exact sequences

0\rightarrow S/J:(x)(-1)\stackrel{{\scriptstyle x}}{{\rightarrow}}S/J\rightarrow S/J+(x)\rightarrow 0,

0\rightarrow S(-1)\stackrel{{\scriptstyle x}}{{\rightarrow}}S\rightarrow S/(x)\rightarrow 0,

give

\operatorname*{HF}(J,\nu)=\operatorname*{HF}\big{(}J+(x)/(x),\nu\big{)}+\operatorname*{HF}(J:(x),\nu-1).

Castelnuovo’s inequality —usually quoted in sheaf-theoretic form— is an easy consequence of the equality above:

\operatorname*{HF}(J,\nu)\leq\operatorname*{HF}\big{(}\big{(}J+(x)/(x)\big{)}^{\operatorname*{sat}},\nu\big{)}+\operatorname*{HF}(J:(x),\nu-1).

Call $Y$ the closed subscheme of $\mathbb{P}_{k}^{r-1}$ defined by $J$ and $H$ the hyperplane defined by the linear form $x$ . Then the closed subscheme of $\mathbb{P}_{k}^{r-1}$ defined by the ideal $J:(x)$ is called the residual scheme of $Y$ along $H$ , denoted as $\operatorname*{Res}_{H}(Y)$ . The closed subscheme of $\mathbb{P}_{k}^{r-2}$ defined by the ideal $J+(x)/(x)$ is called the trace of $Y$ along $H$ , and is denoted by $\operatorname*{Tr}_{H}(Y)$ —note $(J+(x)/(x))^{\operatorname*{sat}}$ also defines $\operatorname*{Tr}_{H}(Y)$ .

In the proof of Lemma 17 we will be concerned with showing $\operatorname*{HF}(J,\nu)=0$ for certain ideals $J$ and degrees $\nu$ . Often, the scheme $Y$ defined by the ideal $J$ will be the reduced union in $\mathbb{P}_{k}^{r-1}$ of linear spaces $Y_{i}$ , together with a $(2,j+\ell)$ -sundial $U$ , for some $j,\ell\geq 0$ . That is, $J=(\cap_{i}I_{Y_{i}})\cap I_{U}$ . The linear form $x$ will always be chosen to define a hyperplane that supports the sundial; e.g., in the notation of the proof of Lemma 16 $I_{U}=J_{0}$ and we would take $x=x_{3}$ . Instead of dealing with the trace scheme $\operatorname*{Tr}_{H}(Y)$ defined by the possibly non-saturated ideal of $S/(x)$

\frac{(\cap_{i}I_{Y_{i}})\cap I_{U}+(x)}{(x)},

it will be more convenient to work with its subscheme $\operatorname*{Tr}^{\dagger}_{H}(Y)$ , defined by the larger ideal

\left(\bigcap_{i}\frac{I_{Y_{i}}+(x)}{(x)}\right)\cap\left(\frac{I_{U}+(x)}{(x)}\right).

Since the sundial will always be of type $(2,j+\ell)$ and $(x)$ defines a hyperplane that supports the sundial, we see that $\operatorname*{Tr}_{H}(U)$ — defined by the ideal $(I_{U}+(x))/(x)$ — is the union in $\mathbb{P}_{k}^{r-2}$ of a hyperplane together with a codimension- $(j+\ell-1)$ linear space $Z$ . Thus $\operatorname*{Tr}_{H}(U)$ always contributes a hyperplane as an irreducible component of $\operatorname*{Tr}^{\dagger}_{H}(Y)$ , and possibly one more, whenever $j+\ell-1=1$ . Let $1\leq\alpha\leq 2$ be the number of the hyperplanes that appear as irreducible components of $\operatorname*{Tr}_{H}(U)$ . Denote by $\operatorname*{Tr}^{\dagger\dagger}_{H}(Y)$ the scheme obtained by removing these components from $\operatorname*{Tr}^{\dagger}_{H}(Y)$ . It becomes clear that to show no hypersurface of $\mathbb{P}_{k}^{r-1}$ of degree $\nu$ contains $Y$ , it suffices to show that i) there is no hypersurface of $\mathbb{P}_{k}^{r-1}$ of degree $\nu-1$ that contains $\operatorname*{Res}_{H}(Y)$ , and ii) there is no hypersurface of $\mathbb{P}_{k}^{r-2}$ of degree $(\nu-\alpha)$ that contains $\operatorname*{Tr}^{\dagger\dagger}_{H}(Y)$ . The arguments in Section 4.5 make repeated use of this idea.

4.5. Third Hilbert Function Lemma

The hardest part of proving Theorem 2 is establishing the following key fact:

Lemma 17.

With $n=2r-3$ , we have

\operatorname*{HF}(\cap_{i\in[n]}I_{i},n-2)=0.

Recalling Proposition 11 and the discussion in Section 3.3, it suffices to show that this holds for a special member of a flat family of schemes, whose general member is the union of $n$ codimension- $2$ generic linear spaces of $\mathbb{P}_{k}^{r-1}$ . We will be denoting the statement of Lemma 17 by

\mathscr{H}_{2r-3,\,2r-5,\,r}

—the first index refers to the number of linear spaces in the arrangement, the second index refers to the degree of interest of the ideal of the arrangement, and the last index is the number of ambient variables.

To prove Lemma 17, we proceed by induction on $r$ . As a base for the induction, we take $r=3$ —the statement is then trivial, since no line in $\mathbb{P}^{2}$ contains three generic points. In the sequel we assume that i) $r\geq 4$ , and ii) $\mathscr{H}_{2r^{\prime}-3,2r^{\prime}-5,r^{\prime}}$ is true for any $3\leq r^{\prime}<r$ .

Within this induction hypothesis, we prove an auxiliary statement:

Lemma 18.

Let $0\leq j\leq r-4$ . Consider the union in $\mathbb{P}_{k}^{r-1}$ of $2r-5-j$ codimension- $2$ generic linear spaces, together with a codimension- $(j+3)$ generic linear space. Then there is no hypersurface of degree $2r-6-j$ containing this union.

We denote the statement of Lemma 18 by

\mathscr{H}^{\prime}_{2r-5-j,\,j+3,\,2r-6-j,\,r}.

We prove it by ascending induction on $r$ and descending induction on $j$ . For the base of our induction on $r$ we take $r=4$ —the only possibility is $j=0$ and the statement follows from the well-known fact that there is a unique quadric surface in $\mathbb{P}^{3}$ through three generic lines (thus no quadric contains three generic lines and a generic point). For any $r$ , the base for the induction on $j$ is for $j=r-4$ . The statement then becomes $\mathscr{H}^{\prime}_{r-1,r-1,r-2,r}$ . This follows immediately from Lemma 14, because through $r-1$ generic codimension- $2$ linear spaces in $\mathbb{P}_{k}^{r-1}$ passes a unique hypersurface of degree $r-2$ , and the codimension- $(r-1)$ space is a generic point, which can be taken to be outside that hypersurface. In the sequel we will assume that i) $\mathscr{H}^{\prime}_{2r^{\prime}-5-j,\,j+3,\,2r^{\prime}-6-j,\,r^{\prime}}$ is true for any $r^{\prime}<r$ and any $j$ such that $0\leq j\leq r^{\prime}-4$ , and ii) $\mathscr{H}^{\prime}_{2r-5-j^{\prime},\,j^{\prime}+3,\,2r-6-j^{\prime},\,r}$ is true for any $j^{\prime}$ for which $j<j^{\prime}\leq r-4$ .

With $0\leq j<r-4$ , let $Y$ be the scheme of Lemma 18. Since $2+(j+3)\leq r$ , we apply Lemma 16 to degenerate $Y$ into a scheme that consists of $2r-6-j$ codimension- $2$ generic linear spaces and a $(2,j+3)$ -sundial. Now we apply the variant of Castelnuovo’s inequality described in Section 4.4 —it is enough to show that i) there is no hypersurface of $\mathbb{P}_{k}^{r-1}$ of degree $2r-7-j$ that contains the residual scheme $\operatorname*{Res}_{H}(Y)$ of $Y$ with respect to a hyperplane $H$ that supports the sundial, and ii) there is no hypersurface of $\mathbb{P}_{k}^{r-2}$ of degree $2r-7-j$ that contains $\operatorname*{Tr}^{\dagger\dagger}_{H}(Y)$ .

The residual scheme $\operatorname*{Res}_{H}(Y)$ is the union in $\mathbb{P}_{k}^{r-1}$ of $2r-6-j$ codimension- $2$ generic linear spaces, together with a codimension- $(j+4)$ generic linear space. The fact that such a scheme lies in no hypersurface of degree $2r-7-j$ is the statement

\mathscr{H}^{\prime}_{2r-5-(j+1),\,(j+1)+3,\,2r-6-(j+1),\,r}.

As $j+1\leq r-4,$ this is true by our induction hypothesis on $j$ for $\mathscr{H}^{\prime}$ .

Next, $\operatorname*{Tr}^{\dagger\dagger}_{H}(Y)$ is the union in $\mathbb{P}_{k}^{r-2}$ of $2r-6-j$ codimension- $2$ generic linear spaces, together with a codimension- $(j+2)$ generic linear space. We claim that there is no degree- $(2r-7-j)$ hypersurface of $\mathbb{P}_{k}^{r-2}$ that contains $\operatorname*{Tr}^{\dagger\dagger}_{H}(Y)$ . If $j=0$ , this is the statement

\mathscr{H}_{2(r-1)-3,2(r-1)-5,r-1},

which is true by our induction hypothesis on $r$ for $\mathscr{H}$ . If $j>0$ , it is the statement

\mathscr{H}^{\prime}_{2(r-1)-5-(j-1),\,(j-1)+3,\,2(r-1)-6-(j-1),\,r-1},

which is true by our induction hypothesis on $r$ for $\mathscr{H}^{\prime}$ (note that $j-1<(r-1)-4$ ). This concludes the proof of Lemma 18.

We now finish the proof of Lemma 17. So let $Y=\bigcup_{i\in[2r-3]}Y_{i}$ be the union in $\mathbb{P}_{k}^{r-1}$ of $2r-3$ codimension- $2$ generic linear spaces $Y_{i}$ . Since $r\geq 4$ , by Lemma 16 we degenerate $Y_{2r-4}\cup Y_{2r-3}$ within a flat family to a $(2,2)$ -sundial. The resulting scheme $Y^{\prime}$ is the union in $\mathbb{P}_{k}^{r-1}$ of $2r-5$ codimension- $2$ generic linear spaces, together with a generic $(2,2)$ -sundial. By Proposition 11, it is enough to show that there is no hypersurface of degree $2r-5$ that contains $Y^{\prime}$ . We do this by applying the variant of Castelnuovo’s inequality described in Section 4.4. So let $H$ be a hyperplane that supports the $(2,2)$ -sundial. The residual scheme $\operatorname*{Res}_{H}(Y^{\prime})$ is the union in $\mathbb{P}_{k}^{r-1}$ of $2r-5$ codimension- $2$ generic linear spaces together with a codimension- $3$ generic linear space. Saying that there is no hypersurface of degree $2r-6$ that contains $\operatorname*{Res}_{H}(Y^{\prime})$ , is the same as saying that

\mathscr{H}^{\prime}_{2r-5,\,3,\,2r-6,\,r}

holds true —now this follows from Lemma 18 with $j=0$ . Next, $\operatorname*{Tr}^{\dagger\dagger}_{H}(Y^{\prime})$ is the union in $\mathbb{P}_{k}^{r-2}$ of $2r-5$ codimension- $2$ generic linear spaces. We must show that there is no hypersurface of degree $2r-7$ that contains $\operatorname*{Tr}^{\dagger\dagger}_{H}(Y^{\prime})$ (note that $\alpha=2$ in the notation of Section 4.4). But this is statement

\mathscr{H}_{2(r-1)-3,\,2(r-1)-5,\,r-1},

which is true by our induction hypothesis on $r$ for $\mathscr{H}$ .

4.6. Saturation

The last ingredient is:

Lemma 19.

Let $J_{i},\,i\in[\ell]$ , be ideals generated by linear forms; suppose $\dim S/J_{i}\geq 2$ for at least one $i$ . Then for a generic linear form $h$

\displaystyle\big{[}\cap_{i\in[\ell]}J_{i}+(h)\big{]}^{\operatorname*{sat}}=\cap_{i\in[\ell]}\big{(}J_{i}+(h)\big{)}

Proof.

Consider the short exact sequence

0\rightarrow\frac{S}{\cap_{i\in[\ell]}J_{i}}\stackrel{{\scriptstyle\varphi}}{{\rightarrow}}\prod_{i\in[\ell]}S/J_{i}\rightarrow C\rightarrow 0,

where $\varphi$ takes the class of $a\in S$ in $S/\cap_{i\in[\ell]}J_{i}$ to the classes of $a$ mod $J_{i}$ for $i\in[\ell]$ , and $C$ is the cokernel. Tensoring with $S/(h)$ and using the fact that $h$ is $S/J_{i}$ -regular for every $i$ , we get

0\rightarrow{\operatorname*{Tor}}_{1}^{S}\left(C,\frac{S}{(h)}\right)\rightarrow\frac{S}{\cap_{i\in[\ell]}J_{i}+(h)}\stackrel{{\scriptstyle\varphi}}{{\rightarrow}}\prod_{i\in[\ell]}\frac{S}{J_{i}+(h)}\rightarrow\frac{C}{hC}\rightarrow 0.

We conclude that

M:=\frac{\cap_{i\in[\ell]}\big{(}J_{i}+(h)\big{)}}{\cap_{i\in[\ell]}J_{i}+(h)}={\operatorname*{Tor}}_{1}^{S}\left(C,\frac{S}{(h)}\right).

Now, we can compute $\operatorname*{Tor}_{1}^{S}(C,S/(h))$ as the kernel of the multiplication map $C(-1)\stackrel{{\scriptstyle h}}{{\rightarrow}}C$ . Since $h$ is generic, it is almost regular on $C(-1)$ and so $M$ has finite length. Thus with $A=\cap_{i\in[\ell]}\big{(}J_{i}+(h)\big{)}$ and $B=\cap_{i\in[\ell]}J_{i}+(h)$ we have $B\subset A\subseteq B^{\operatorname*{sat}}$ . This gives $B^{\operatorname*{sat}}=A^{\operatorname*{sat}}$ . But $A$ is radical because the ideals $J_{i}+(h)$ are prime and at least one of them is properly contained in the maximal homogeneous ideal of $S$ . ∎

4.7. Finishing the Proof of Theorem 2

We proceed by induction on $r$ ; the case $r=3$ is a simple exercise, so we assume $r\geq 4$ .

Let $h\in[S]_{1}$ be a generic linear form. As such, $h$ is regular on $S/\cap_{i\in[n]}I_{i}$ , and so Proposition 5 gives

\operatorname*{reg}\left(\frac{S}{\cap_{i\in[n]}I_{i}}\right)=\operatorname*{reg}\left(\frac{S}{\cap_{i\in[n]}I_{i}+(h)}\right).

It thus suffices to show that

\operatorname*{reg}\left(\frac{S}{\cap_{i\in[n]}I_{i}+(h)}\right)=n-2.

Towards that end, we consider the short exact sequence

0\rightarrow\frac{\cap_{i\in[n]}\big{(}I_{i}+(h)\big{)}}{\cap_{i\in[n]}I_{i}+(h)}\rightarrow\frac{S}{\cap_{i\in[n]}I_{i}+(h)}\rightarrow\frac{S}{\cap_{i\in[n]}\big{(}I_{i}+(h)\big{)}}\rightarrow 0.

By Lemma 19, the first module in the exact sequence has finite length. Thus Proposition 4 gives

\operatorname*{reg}\left(\frac{S}{\cap_{i\in[n]}I_{i}+(h)}\right)=

\max\left\{\operatorname*{reg}\left(\frac{\cap_{i\in[n]}\big{(}I_{i}+(h)\big{)}}{\cap_{i\in[n]}I_{i}+(h)}\right),\operatorname*{reg}\left(\frac{S}{\cap_{i\in[n]}\big{(}I_{i}+(h)\big{)}}\right)\right\}.

Set $\bar{S}=S/(h)$ and $\bar{I}_{i}=I_{i}+(h)/(h)$ . Then

\frac{S}{\cap_{i\in[n-2]}(I_{i}+(h))}=\frac{\bar{S}}{\cap_{i\in[n-2]}\bar{I}_{i}},

and our induction hypothesis on $r$ gives

\operatorname*{reg}\left(\frac{\bar{S}}{\cap_{i\in[n-2]}\bar{I}_{i}}\right)=n-4.

Hence $\operatorname*{reg}(\cap_{i\in[n-2]}\bar{I}_{i})=n-3$ . Since taking quotient with a generic linear form can not increase the regularity (Proposition 5), we have

\operatorname*{reg}(\cap_{i\in[n-2]}\bar{I}_{i}+\bar{I}_{n-1})\leq n-3.

This, together with the short exact sequence

0\rightarrow\frac{\bar{S}}{\cap_{i\in[n-1]}\bar{I}_{i}}\rightarrow\frac{\bar{S}}{\cap_{i\in[n-2]}\bar{I}_{i}}\oplus\frac{\bar{S}}{\bar{I}_{n-1}}\rightarrow\frac{\bar{S}}{\cap_{i\in[n-2]}\bar{I}_{i}+\bar{I}_{n-1}}\rightarrow 0,

and the regularity lemma, give

\operatorname*{reg}(\cap_{i\in[n-1]}\bar{I}_{i})\leq n-2.

One more application of this argument with $\cap_{i\in[n-1]}\bar{I}_{i}$ and $\bar{I}_{n}$ gives

\operatorname*{reg}(\cap_{i\in[n]}\bar{I}_{i})\leq n-1.

We have thus shown that

\operatorname*{reg}\left(\frac{S}{\cap_{i\in[n]}(I_{i}+(h))}\right)\leq n-2,

and so it remains to prove

\operatorname*{reg}\left(\frac{\cap_{i\in[n]}\big{(}I_{i}+(h)\big{)}}{\cap_{i\in[n]}I_{i}+(h)}\right)\leq n-2.

Since $\operatorname*{reg}(\cap_{i\in[n]}I_{i}+(h))\leq n$ , we have that $\cap_{i\in[n]}I_{i}+(h)$ is saturated from degree $n$ and on. Hence in view of Proposition 19, $\cap_{i\in[n]}\big{(}I_{i}+(h)\big{)}$ and $\cap_{i\in[n]}I_{i}+(h)$ agree from degree $n$ and on, and so it is enough to prove that

\operatorname*{HF}(\cap_{i\in[n]}(I_{i}+(h)),n-1)=\operatorname*{HF}(\cap_{i\in[n]}I_{i}+(h),n-1).

For the Hilbert function on the left we note

\operatorname*{HF}\left(\frac{S}{\cap_{i\in[n]}(I_{i}+(h))},n-1\right)=\operatorname*{HF}\left(\frac{\bar{S}}{\cap_{i\in[n]}\bar{I}_{i}},n-1\right),

and now

\operatorname*{HF}(\cap_{i\in[n]}(I_{i}+(h)),n-1)=\operatorname*{HF}(S,n-2)+\operatorname*{HF}(\cap_{i\in[n]}\bar{I}_{i},n-1).

To get a handle on $\operatorname*{HF}(\cap_{i\in[n]}I_{i}+(h),n-1)$ , we work with the short exact sequence

0\rightarrow\frac{S}{\cap_{i\in[n]}I_{i}}(-1)\stackrel{{\scriptstyle h}}{{\rightarrow}}\frac{S}{\cap_{i\in[n]}I_{i}}\rightarrow\frac{S}{\cap_{i\in[n]}I_{i}+(h)}\rightarrow 0,

whose degree $n-1$ component gives

	$\displaystyle\operatorname*{HF}(\cap_{i\in[n]}I_{i}+(h),n-1)=$	$\displaystyle\operatorname{HF}(\cap_{i\in[n]}I_{i},n-1)-\operatorname{HF}(\cap_{i\in[n]}I_{i},n-2)$
		$\displaystyle+\operatorname*{HF}(S,n-2).$

By Lemma 17,

\operatorname*{HF}(\cap_{i\in[n]}I_{i},n-2)=0,

and so

\operatorname*{HF}(\cap_{i\in[n]}I_{i}+(h),n-1)=\operatorname*{HF}(S,n-2)+\operatorname*{HF}(\cap_{i\in[n]}I_{i},n-1).

To finish the proof we have to show that

\operatorname*{HF}(\cap_{i\in[n]}\bar{I}_{i},n-1)=\operatorname*{HF}(\cap_{i\in[n]}I_{i},n-1).

Both these Hilbert function values are accessible via Lemma 12. Indeed, they are equal:

Lemma 20.

We have that

\sum_{j\geq 0}(-1)^{j}{n\choose j}{n+r-2-2j\choose n-1}=\sum_{j\geq 0}(-1)^{j}{n\choose j}{n+r-3-2j\choose n-1}.

Proof.

The statement is equivalent to

\sum_{j\geq 0}(-1)^{j}{n\choose j}{n+r-3-2j\choose n-2}=0,

and recalling that $n=2r-3$ , equivalent to

\sum_{j\geq 0}(-1)^{j}{2r-3\choose j}{3r-6-2j\choose 2r-5}=0.

The degree of the polynomial

\begin{bmatrix}3r-6-2t\\ 2r-5\end{bmatrix}

is $2r-5$ , so Lemma 13 gives

\sum_{j=0}^{2r-3}(-1)^{j}{2r-3\choose j}\begin{bmatrix}3r-6-2j\\ 2r-5\end{bmatrix}=0.

In the summation above there are $2r-2$ terms. For $j=0,\dots,r-2$ , we claim that the terms corresponding to $j$ and $2r-3-j$ are equal. To see this, with $b$ a positive integer, first recall the polynomial identity

\begin{bmatrix}t\\ b\end{bmatrix}=(-1)^{b}\begin{bmatrix}b-t-1\\ b\end{bmatrix}.

Now, the $(2r-3-j)$ th term is

	$\displaystyle(-1)^{2r-3-j}{2r-3\choose 2r-3-j}\begin{bmatrix}3r-6-2(2r-3-j)\\ 2r-5\end{bmatrix}$
	$\displaystyle=(-1)^{-3-j}{2r-3\choose j}\begin{bmatrix}-r+2j\\ 2r-5\end{bmatrix}$
	$\displaystyle=(-1)^{-3-j}{2r-3\choose j}(-1)^{2r-5}\begin{bmatrix}2r-5+r-2j-1\\ 2r-5\end{bmatrix}$
	$\displaystyle=(-1)^{j}{2r-3\choose j}\begin{bmatrix}3r-6-2j\\ 2r-5\end{bmatrix},$

which is precisely the $j$ th term. Consequently,

\sum_{j=0}^{r-2}(-1)^{j}{2r-3\choose j}\begin{bmatrix}3r-6-2j\\ 2r-5\end{bmatrix}=0.

For those values of $j$ , the binomial polynomial coincides with the binomial coefficient, hence

\sum_{j=0}^{r-2}(-1)^{j}{2r-3\choose j}{3r-6-2j\choose 2r-5}=0.

Finally, the binomial coefficient

{3r-6-2j\choose 2r-5}

is already zero for any $j>r-2$ , so that

\sum_{j\geq 0}(-1)^{j}{2r-3\choose j}{3r-6-2j\choose 2r-5}=0.

∎

5. Proof of Theorem 3

Recalling that $L_{c}$ is an ideal of $S$ generated by $c$ generic linear forms $\underline{h}=h_{1},\dots,h_{c}$ , we begin by recording some basic observations:

Lemma 21.

With $J$ any homogeneous ideal, we have:

(i)

$\operatorname*{reg}(J\cap L_{c})\leq\operatorname*{reg}(J)+1$ .
(ii)

$(J\cap L_{c})/JL_{c}$ has finite length.
(iii)

If $\operatorname*{reg}(J+L_{c})<\operatorname*{reg}(J)$ then $\operatorname*{reg}(J\cap L_{c})=\operatorname*{reg}(J)$ .

Proof.

(i) We have a short exact sequence:

(1)

\displaystyle 0\to\frac{S}{J\cap L_{c}}\to\frac{S}{J}\oplus\frac{S}{L_{c}}\to\frac{S}{J+L_{c}}\to 0.

Since $L_{c}$ is generated by generic linear forms, Proposition 5 gives $\operatorname*{reg}(S/(J+L_{c}))\leq\operatorname*{reg}(S/J)$ . Furthermore $\operatorname*{reg}(S/L_{c})=0$ . Now the claim follows from the regularity lemma (Proposition 4).

(ii) The sequence $\underline{h}$ is almost regular on $S/J$ . Hence the Koszul homology $H_{i}(\underline{h},S/J)$ has finite length for all $i>0$ . In particular $H_{1}(\underline{h},S/J)=\operatorname*{Tor}^{S}_{1}(S/J,S/L_{c})=(J\cap L_{c})/JL_{c}$ has finite length.

(iii) This follows from the above exact sequence together with the regularity lemma. ∎

The next step, even though intuitively non-surprising, is the hardest part behind the proof of Theorem 3:

Proposition 22.

For every non-zero saturated homogeneous ideal $J\subset S$ and every $c\in[r-1]$ we have:

\operatorname*{reg}(J\cap L_{c})\geq\operatorname*{reg}(J\cap L_{c+1})

Proof.

We discuss first the case $c<r-1$ . Suppose, by contradiction that $\operatorname*{reg}(J\cap L_{c})<\operatorname*{reg}(J\cap L_{c+1})$ . Let us write $L_{c+1}=L_{c}+L_{1}^{\prime}$ , where $L_{1}^{\prime}=(h_{c+1})$ . Set

K=(J\cap L_{c})+(J\cap L_{1}^{\prime}).

Note that we have:

JL_{c+1}=JL_{1}^{\prime}+JL_{c}\subseteq K\subseteq J\cap L_{c+1}.

Since by Lemma 21 $(J\cap L_{c+1})/JL_{c+1}$ has finite length, $J\cap L_{c+1}/K$ also has finite length, and so by Proposition 4

\operatorname*{reg}\left(\frac{S}{J\cap L_{c+1}}\right)\leq\operatorname*{reg}\left(\frac{S}{K}\right).

We have an exact complex

\displaystyle 0\to\frac{S}{J\cap L_{c}\cap L_{1}^{\prime}}\to\frac{S}{J\cap L_{1}^{\prime}}\oplus\frac{S}{J\cap L_{c}}\stackrel{{\scriptstyle f}}{{\to}}\frac{S}{K}\to 0

and we further have a canonical projection

g:\frac{S}{K}\to\frac{S}{J\cap L_{c+1}}.

Note that the restriction of $f$ to $S/(J\cap L_{1}^{\prime})$ composed with $g$ gives the canonical projection

\pi_{1,c+1}:\frac{S}{J\cap L_{1}^{\prime}}\to\frac{S}{J\cap L_{c+1}}.

Now we have

\displaystyle\operatorname*{reg}\left(\frac{S}{J\cap L_{c}\cap L_{1}^{\prime}}\right)\leq\operatorname*{reg}\left(\frac{S}{J\cap L_{c}}\right)+1\leq\operatorname*{reg}\left(\frac{S}{J\cap L_{c+1}}\right),

where the first inequality follows from Lemma 21(i) applied to the ideal $J\cap L_{c}$ , and the second inequality is by hypothesis.

Let $i\in[r]$ and $a\in\mathbb{N}$ be such that

\left[H^{i}_{\mathfrak{m}}\left(\frac{S}{J\cap L_{c+1}}\right)\right]_{a}\neq 0\mbox{ and }i+a=\operatorname*{reg}\left(\frac{S}{J\cap L_{c+1}}\right).

Since $J$ is saturated, $J\cap L_{c+1}$ is as well saturated; thus $i>0$ .

Consider the maps induced in cohomology by the above short exact sequence:

\begin{matrix}\left[H^{i}_{\mathfrak{m}}\left(\frac{S}{J\cap L_{1}^{\prime}}\right)\right]_{a}\\ \oplus\\ \left[H^{i}_{\mathfrak{m}}\left(\frac{S}{J\cap L_{c}}\right)\right]_{a}\end{matrix}\to\left[H_{\mathfrak{m}}^{i}\left(\frac{S}{J\cap L_{c+1}}\right)\right]_{a}\to\left[H^{i+1}_{\mathfrak{m}}\left(\frac{S}{J\cap L_{c}\cap L_{1}^{\prime}}\right)\right]_{a},

where we have used the fact that

H_{\mathfrak{m}}^{i}\left(\frac{S}{K}\right)\simeq H_{m}^{i}\left(\frac{S}{J\cap L_{c+1}}\right)

via the map induced by the projection $f$ . Now

\left[H^{i}_{\mathfrak{m}}\left(\frac{S}{J\cap L_{c}}\right)\right]_{a}=0,

since

a+i=\operatorname*{reg}\left(\frac{S}{J\cap L_{c+1}}\right)>\operatorname*{reg}\left(\frac{S}{J\cap L_{c}}\right).

Moreover,

\left[H^{i+1}_{\mathfrak{m}}\left(\frac{S}{J\cap L_{c}\cap L_{1}^{\prime}}\right)\right]_{a}=0,

since

i+1+a=1+\operatorname*{reg}\left(\frac{S}{J\cap L_{c+1}}\right)>\operatorname*{reg}\left(\frac{S}{J\cap L_{c}\cap L_{1}^{\prime}}\right).

Therefore we have a surjective map

\displaystyle\varepsilon:\left[H^{i}_{\mathfrak{m}}\left(\frac{S}{J\cap L_{1}^{\prime}}\right)\right]_{a}\to\left[H_{\mathfrak{m}}^{i}\left(\frac{S}{J\cap L_{c+1}}\right)\right]_{a}\neq 0,

which, as observed above, is induced by the canonical projection $\pi_{1,c+1}$ . But $\pi_{1,c+1}$ is the composition of the two canonical projections

\frac{S}{J\cap L_{1}^{\prime}}\stackrel{{\scriptstyle\pi_{1,c}}}{{\to}}\frac{S}{J\cap L_{c}^{\prime}}\stackrel{{\scriptstyle\pi_{c,c+1}}}{{\to}}\frac{S}{J\cap L_{c+1}},

where $L_{c}^{\prime}$ is an ideal generated by $c$ generic linear forms, contained in $L_{c+1}$ and containing $L_{1}^{\prime}$ . The above composition induces maps

\displaystyle\left[H^{i}_{\mathfrak{m}}\left(\frac{S}{J\cap L_{1}^{\prime}}\right)\right]_{a}\to\left[H_{\mathfrak{m}}^{i}\left(\frac{S}{J\cap L_{c}^{\prime}}\right)\right]_{a}\to\left[H_{\mathfrak{m}}^{i}\left(\frac{S}{J\cap L_{c+1}}\right)\right]_{a},

whose composition is the above surjective map $\epsilon$ . But as we have already seen, the middle term is zero, which is a contradiction.

It remains to discuss the case $c=r-1$ . In this special case, since $L_{c+1}=L_{r}=\mathfrak{m}$ and $J\cap L_{c+1}=J$ , we have to prove that $\operatorname*{reg}(J\cap L_{r-1})\geq\operatorname*{reg}(J)$ . We have $J+L_{r-1}=(p)+L_{r-1}$ where $\deg(p)=\min\{a:J_{a}\neq 0\}$ . Note that $\operatorname*{reg}(J+L_{r-1})=\deg(p)$ . If $\deg(p)<\operatorname*{reg}(J)$ we conclude by Lemma 21iii) that $\operatorname*{reg}(J\cap L_{r-1})=\operatorname*{reg}(J)$ . Otherwise $\deg(p)=\operatorname*{reg}(J)$ (one has that $J$ has a linear resolution) and the short exact sequence in the proof of Lemma 21(i) gives:

0\to H^{0}_{\mathfrak{m}}\left(\frac{S}{J+L_{r-1}}\right)=\frac{S}{J+L_{r-1}}\to H^{1}_{\mathfrak{m}}\left(\frac{S}{J\cap L_{r-1}}\right)

It follows that

\operatorname*{reg}\left(\frac{S}{J\cap L_{r-1}}\right)\geq 1+\operatorname*{reg}\left(\frac{S}{J+L_{r-1}}\right)=\operatorname*{reg}(J),

that is $\operatorname*{reg}(J\cap L_{r-1})\geq\operatorname*{reg}(J)+1$ . ∎

We can now refine Lemma 21:

Proposition 23.

Suppose that $J$ is saturated. Then $\operatorname*{reg}(J\cap L_{c})=\operatorname*{reg}(J)+1$ if and only if $\operatorname*{reg}(J+L_{c})=\operatorname*{reg}(J)$ .

Proof.

According to Lemma 21(i) and Proposition 22, the regularity of $J\cap L_{c}$ can either be $\operatorname*{reg}(J)+1$ or $\operatorname*{reg}(J)$ .

By Lemma 21(iii) we know that $\operatorname*{reg}(J\cap L_{c})=\operatorname*{reg}(J)+1$ implies $\operatorname*{reg}(J+L_{c})=\operatorname*{reg}(J)$ . We argue for the reverse direction. For the sake of a contradiction, suppose

\operatorname*{reg}(J+L_{c})=\operatorname*{reg}(J)=\operatorname*{reg}(J\cap L_{c}).

We may assume $\operatorname*{reg}(J+L_{c+1})<\operatorname*{reg}(J)$ ; otherwise we simply replace $L_{c}$ with $L_{c+1}$ , noting that $\operatorname*{reg}(J\cap L_{c})=\operatorname*{reg}(J)$ implies $\operatorname*{reg}(J\cap L_{c+1})=\operatorname*{reg}(J)$ by Proposition 22. Set $M=S/(J+L_{c})$ . The generic hyperplane section $M/xM$ of $M$ can be realized by taking $x=h_{c+1}$ , i.e. $M/xM=S/J+L_{c+1}$ . Since $x$ is almost regular on $M$ , the kernel $0:_{M}x$ of the multiplication map $M\stackrel{{\scriptstyle x}}{{\rightarrow}}M$ has finite length; thus Proposition 20.20 in [Eis94] gives

\operatorname*{reg}(M)=\max\{\operatorname*{reg}(M/xM),\operatorname*{reg}(0:_{M}x)\}.

In our setting, by hypothesis we have $\operatorname*{reg}(M/xM)<\operatorname*{reg}(M)$ , so $\operatorname*{reg}(M)=\operatorname*{reg}(0:_{M}x).$ Now the short exact sequence in the proof of Lemma 21(i), together with the assumption that $J$ is saturated, implies that $H_{\mathfrak{m}}^{0}(M)$ is a submodule of $H_{\mathfrak{m}}^{1}(S/J\cap L_{c})$ . Since $0:_{M}x$ is a submodule of $H_{\mathfrak{m}}^{0}(M)$ , we have

	$\displaystyle\operatorname*{reg}(J)-1$	$\displaystyle=\operatorname{reg}(S/J\cap L_{c})\geq 1+\operatorname{reg}(H_{\mathfrak{m}}^{0}(M))$
		$\displaystyle\geq 1+\operatorname{reg}(0:_{M}x)=1+\operatorname{reg}(M)=\operatorname*{reg}(J),$

which is a contradiction. ∎

Theorem 3 now immediately follows from Proposition 23 and Lemma 21(iii), together with the remark that if $c\leq\operatorname*{depth}(S/J)$ , then $L_{c}$ is generated by an $S/J$ -regular sequence and so $\operatorname*{reg}(S/J)=\operatorname*{reg}(S/J+L_{c})$ .

6. Proof of Theorem 1

For $i\in[n]$ we let $I_{i}$ be generated by $c_{i}\geq 2$ generic linear forms. We first prove:

Lemma 24.

Suppose that $n\geq 2r-3$ . Then $\operatorname*{reg}(\cap_{i\in[n]}I_{i})\leq n-1$ .

Proof.

For each $i\in[n]$ let $I_{i}^{\prime}\subset I_{i}$ be an ideal generated by two generic linear forms. By Theorem 2 and Lemma 21(i), we have $\operatorname*{reg}(\cap_{i\in[n]}I_{i}^{\prime})\leq n-1$ . If $c_{i}=2$ for every $i\in[n]$ , we are done. So we may assume $c_{n}>2$ . We may write $I_{n}=(h_{1},h_{2},\dots,h_{c_{n}})$ and $I_{n}^{\prime}=(h_{1},h_{2})$ . Set $J=\cap_{i\in[n-1]}I_{i}^{\prime}$ , $L_{2}=I_{n}^{\prime}$ and $L_{c_{n}}=I_{n}$ . Then Proposition 22 gives

n-1\geq\operatorname*{reg}(\cap_{i\in[n]}I_{i}^{\prime})=\operatorname*{reg}(J\cap L_{2})\geq\operatorname*{reg}(J\cap L_{c_{n}})=\operatorname*{reg}((\cap_{i\in[n-1]}I_{i}^{\prime})\cap I_{n}).

Continuing in a similar fashion by inductively applying Proposition 22 to the ideal $(\cap_{i\in[n-1]}I_{i}^{\prime})\cap I_{n}$ , proves the statement. ∎

Theorem 1 now follows immediately from Lemma 24 and the following sub-additivity property of ideals of subspace arrangements:

Lemma 25.

Let $I_{a}\subsetneq\mathfrak{m},\,a\in\mathcal{A}=[\ell]$ and $J_{b}\subsetneq\mathfrak{m},\,b\in\mathcal{B}=[m]$ be ideals generated by generic linear forms, where $\ell$ and $m$ are positive integers. Then

\operatorname*{reg}\left((\cap_{a\in\mathcal{A}}I_{a})\cap(\cap_{b\in\mathcal{B}}J_{b})\right)\leq\operatorname*{reg}(\cap_{a\in\mathcal{A}}I_{a})+\operatorname*{reg}(\cap_{b\in\mathcal{B}}J_{b}).

Proof.

For convenience, we set $I_{\mathcal{A}}=\cap_{a\in\mathcal{A}}I_{a}$ and $J_{\mathcal{B}}=\cap_{b\in\mathcal{B}}J_{b}$ . We have inclusions of ideals

\textstyle(\prod_{a\in\mathcal{A}}I_{a})(\prod_{b\in\mathcal{B}}I_{b})\subset I_{\mathcal{A}}J_{\mathcal{B}}\subset I_{\mathcal{A}}\cap J_{\mathcal{B}}.

By Proposition 3.4 in [CH03], we have

\textstyle\left[(\prod_{a\in\mathcal{A}}I_{a})(\prod_{b\in\mathcal{B}}I_{b})\right]^{\operatorname*{sat}}=I_{\mathcal{A}}\cap J_{\mathcal{B}},

and hence also

\textstyle(I_{\mathcal{A}}J_{\mathcal{B}})^{\operatorname*{sat}}=I_{\mathcal{A}}\cap J_{\mathcal{B}}.

Thus the $S$ -module

{\operatorname*{Tor}}_{1}^{S}\left(\frac{S}{I_{\mathcal{A}}},\frac{S}{J_{\mathcal{B}}}\right)=\frac{I_{\mathcal{A}}\cap J_{\mathcal{B}}}{I_{\mathcal{A}}J_{\mathcal{B}}},

is zero-dimensional, and so [Cav07] give

\operatorname*{reg}\left(\frac{S}{I_{\mathcal{A}}+J_{\mathcal{B}}}\right)=\operatorname*{reg}\left(\frac{S}{I_{\mathcal{A}}}\otimes_{S}\frac{S}{J_{\mathcal{B}}}\right)\leq\operatorname*{reg}\left(\frac{S}{I_{\mathcal{A}}}\right)+\operatorname*{reg}\left(\frac{S}{J_{\mathcal{B}}}\right).

With this, the short exact sequence

0\rightarrow\frac{S}{I_{\mathcal{A}}\cap J_{\mathcal{B}}}\rightarrow\frac{S}{I_{\mathcal{A}}}\oplus\frac{S}{J_{\mathcal{B}}}\rightarrow\frac{S}{I_{\mathcal{A}}+J_{\mathcal{B}}}\rightarrow 0

and the regularity lemma (Proposition 4) give the assertion. ∎

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