On the Geramita-Harbourne-Migliore conjecture

TL;DR

This paper proves a conjecture about the linearity of resolutions for ideals generated by products of linear forms, leading to results on primary decompositions and resurgence of ideals of star configurations.

Contribution

It establishes the Geramita-Harbourne-Migliore conjecture on linear resolutions of certain ideals and applies this to primary decompositions and resurgence of star configuration ideals.

Findings

Ideals generated by all $a$-fold products have linear graded free resolutions.

Confirmed the Geramita-Harbourne-Migliore conjecture for these ideals.

Determined the resurgence of ideals of star configurations.

Abstract

Let $\Sigma$ be a finite collection of linear forms in $\mathbb K[x_0,\ldots,x_n]$, where $\mathbb K$ is a field. Denote ${\rm Supp}(\Sigma)$ to be the set of all nonproportional elements of $\Sigma$, and suppose ${\rm Supp}(\Sigma)$ is generic, meaning that any $n+1$ of its elements are linearly independent. Let $1\leq a\leq |\Sigma|$. In this article we prove the conjecture that the ideal generated by (all) $a$-fold products of linear forms of $\Sigma$ has linear graded free resolution. As a consequence we prove the Geramita-Harbourne-Migliore conjecture concerning the primary decomposition of ordinary powers of defining ideals of star configurations, and we also determine the resurgence of these ideals.

Equations

$$I_a(\Sigma ):=⟨\{L_{i_1}\cdots L_{i_a}|1\le i_1<\cdots <i_a\le N\}⟩,$$

$$\mathrm{ht}(I_a(\Sigma ))=n+2-r.$$

$$0\to R^{n_{b+1}}(-(d+b))\to \cdots \to R^{n_2}(-(d+1))\to R^{n_1}(-d)\to R\to R/I\to 0,$$

$$V_c(\mathcal{A}):=\bigcup _{1\le j_1<\cdots <j_c\le s}{H}_{j_1}\cap \cdots \cap H_{j_c}.$$

$$I(V_c(\mathcal{A}))=\bigcap _{1\le j_1<\cdots <j_c\le s}⟨{\ell }_{j_1},\dots ,{\ell }_{j_c}⟩,$$

$$I(V_c(\mathcal{A}){)}^{(m)}=\bigcap _{1\le j_1<\cdots <j_c\le s}⟨{\ell }_{j_1},\dots ,{\ell }_{j_c}{⟩}^m.$$

$$I(V_c(\mathcal{A}){)}^m=I(V_c(\mathcal{A}){)}^{(m)}\cap I(V_{c+1}(\mathcal{A}){)}^{(2m)}\cap \cdots \cap I(V_n(\mathcal{A}){)}^{((n-c+1)m)}\cap M^{(s-c+1)m}.$$

$$J=J^{\mathrm{sat}}\cap M^d.$$

$${\left(I(V_c(\mathcal{A}){)}^m\right)}^{\mathrm{sat}}=I(V_c(\mathcal{A}){)}^{(m)}\cap I(V_{c+1}(\mathcal{A}){)}^{(2m)}\cap \cdots \cap I(V_n(\mathcal{A}){)}^{((n-c+1)m)}.$$

$$I_a(\Sigma )=I_a({\ell }_1^{m_1}\cdots {\ell }_s^{m_s}),$$

$$I_a({\ell }_1^{m_1}\cdots {\ell }_s^{m_s})\subseteq \bigcap _{c=1}^s\left(\bigcap _{1\le i_1<\cdots <i_c\le s}⟨{\ell }_{i_1},\dots ,{\ell }_{i_c}{⟩}^{\mu (i_1,\dots ,i_c)}\right),$$

$$\mu (i_1,\dots ,i_u)=a-(N-{\nu }_{\Sigma }(\mathfrak{p})),$$

$$I\subseteq \bigcap _{c=1}^s\left(\bigcap _{1\le i_1<\cdots <i_c\le s}⟨{\ell }_{i_1},\dots ,{\ell }_{i_c}{⟩}^{\mu (i_1,\dots ,i_c)}\right)=\bigcap _{\mathfrak{p}\in \Gamma (\Sigma )}{\mathfrak{p}}^{a-N+{\nu }_{\Sigma }(\mathfrak{p})},$$

$$I^{\mathrm{sat}}=\bigcap _{\mathfrak{p}\in \Gamma (\Sigma )\setminus \{M\}}{\mathfrak{p}}^{a-N+{\nu }_{\Sigma }(\mathfrak{p})}.$$

$$I=I_a({\ell }_1^{m_1}\cdots {\ell }_{s-1}^{m_{s-1}}{\ell }_s^{m_s})={\ell }_s^{m_s}{I}_{a-m_s}({\ell }_1^{m_1}\cdots {\ell }_{s-1}^{m_{s-1}})+\cdots +{\ell }_s{I}_{a-1}({\ell }_1^{m_1}\cdots {\ell }_{s-1}^{m_{s-1}})+I_a({\ell }_1^{m_1}\cdots {\ell }_{s-1}^{m_{s-1}}).$$

$$I_{a-m_s}({\ell }_1^{m_1}\cdots {\ell }_{s-1}^{m_{s-1}})\supset \cdots \supset I_a({\ell }_1^{m_1}\cdots {\ell }_{s-1}^{m_{s-1}}),$$

$$I{R}_P=I_{a-m_s}({\ell }_1^{m_1}\cdots {\ell }_{s-1}^{m_{s-1}})R_P.$$

$$I{R}_P=I_{a-N+{\nu }_{\Sigma }(\mathfrak{p})}({\ell }_1^{m_1}\cdots {\ell }_c^{m_c})R_P,$$

$$I{R}_P=I_{a-N+{\nu }_{\Sigma }(\mathfrak{p})}(x_1^{m_1}\cdots x_c^{m_c})R_P,$$

$$\mathbb{I}{R}_P=\bigcap _{k=1}^c\left(\bigcap _{1\le j_1<\cdots <j_k\le c}⟨x_{j_1},\dots ,x_{j_k}{⟩}^{a-N+m_{j_1}+\cdots +m_{j_k}}\right)R_P.$$

$$a-N+m_{j_1}+\cdots +m_{j_k}=b-d+m_{j_1}+\cdots +m_{j_k}.$$

$$I_b(x_1^{m_1}\cdots x_c^{m_c})=\bigcap _{k=1}^c\left(\bigcap _{1\le j_1<\cdots <j_k\le c}⟨x_{j_1},\dots ,x_{j_k}{⟩}^{\mu (j_1,\dots ,j_k)}\right),$$

$$\xi =x_1^{t_1}\cdots x_c^{t_c}=\left(\prod _{i\in \{j_1,\dots ,j_k\}}{x}_i^{t_i}\cdot \prod _{i\in [c]\setminus \{j_1,\dots ,j_k\}}{x}_i^{m_i}\right)\cdot \prod _{i\in [c]\setminus \{j_1,\dots ,j_k\}}{x}_i^{t_i-m_i}.$$

$$\sum _{i\in \{j_1,\dots ,j_k\}}{t}_i\ge \mu (j_1,\dots ,j_k).$$

$$\sum _{i\in \{j_1,\dots ,j_k\}}{t}_i+\sum _{i\in [c]\setminus \{j_1,\dots ,j_k\}}{m}_i$$

$$I:{\ell }_1\subseteq (I^{\mathrm{sat}}:{\ell }_1)\cap \underset{M^{a-1}}{\underbrace{(M^a:{\ell }_1)}}.$$

$$I^{\mathrm{sat}}=\bigcap _{\mathfrak{p}\in \Gamma (\Sigma )\setminus \{M\}}{\mathfrak{p}}^{a-N+{\nu }_{\Sigma }(\mathfrak{p})},$$

$$J^{\mathrm{sat}}=\bigcap _{\mathfrak{q}\in \Gamma ({\Sigma }^{\prime })\setminus \{M\}}{\mathfrak{q}}^{(a-1)-(N-1)+{\nu }_{{\Sigma }^{\prime }}(\mathfrak{q})},$$

$$I^{\mathrm{sat}}:{\ell }_1\subseteq \bigcap _{\mathfrak{q}\in \Gamma ({\Sigma }^{\prime })\setminus \{M\}}({\mathfrak{q}}^{a-N+{\nu }_{\Sigma }(\mathfrak{q})}:{\ell }_1).$$

$${\mathfrak{q}}^{a-N+{\nu }_{\Sigma }(\mathfrak{q})}:{\ell }_1={\mathfrak{q}}^{a-N+{\nu }_{\Sigma }(\mathfrak{q})}={\mathfrak{q}}^{(a-1)-(N-1)+{\nu }_{{\Sigma }^{\prime }}(\mathfrak{q})}.$$

Full text

Toh$\breve{{\rm a}}$neanu’s address: Department of Mathematics, University of Idaho, Moscow, Idaho 83844-1103, USA, Email: tohaneanu@uidaho.edu.

Xie’s address: Department of Mathematics, Widener University, Chester, Pennsylvania 19013, USA, Email: yxie@widener.edu.

1. Introduction

Let $R:=\mathbb{K}[x_{0},\ldots,x_{n}]$ be the ring of polynomials with coefficients in an arbitrary field $\mathbb{K}$, thought of as a graded ring with the standard grading. Denote $M:=\langle x_{0},\ldots,x_{n}\rangle$ the irrelevant ideal in $R$.

Let $\Sigma$ be a finite collection of linear forms $L_{1},\ldots,L_{N}$ in $R$, some possibly proportional. Denote $\Sigma=(L_{1},\ldots,L_{N})$. For any $1\leq a\leq N$, define

[TABLE]

the ideal generated by (all) $a$-fold products of linear forms of   $\Sigma$. For convention, $I_{0}(\Sigma)=R$, and $I_{a}(\Sigma)=0$, for any $a\geq N+1$. Often we will denote $I_{a}(\Sigma)$ with $I_{a}(L_{1}\cdots L_{N})$.

The rank of $\Sigma$, denoted ${\rm rk}(\Sigma)$, is ${\rm ht}(\langle L_{1},\ldots,L_{N}\rangle)$.

1.1. Linear Codes.

The ideals generated by $a$-fold products of linear forms originally occurred in coding theory context as a nice tool to determine the minimum distance of a linear code (see [5]).

Let $L=c_{0}x_{0}+\cdots+c_{n}x_{n}$ be an arbitrary element of $\Sigma$. Dually we get a column vector $(c_{0},\ldots,c_{n})^{T}\in\mathbb{K}^{n+1}$. This way we build an $(n+1)\times N$ matrix $G_{\Sigma}$, and consequently a linear code $\mathcal{C}_{\Sigma}$ which is the image of the linear map $\displaystyle\mathbb{K}^{n+1}\stackrel{{\scriptstyle\cdot G_{\Sigma}}}{{\longrightarrow}}\mathbb{K}^{N}$. This is a linear code of block-length $N$ and dimension ${\rm rk}(G_{\Sigma})={\rm rk}(\Sigma)$; a generating matrix of $\mathcal{C}_{\Sigma}$ is $G_{\Sigma}$. Conversely, to any linear code we can associate a collection of linear forms dual to the columns of some generating matrix.

Suppose ${\rm rk}(\Sigma)=n+1$. For any $1\leq r\leq n+1$ one can define the $r$-th generalized Hamming weight, $d_{r}(\mathcal{C}_{\Sigma})$, which by classical results in coding theory (see for example [1, Corollary 1.3 and Proposition 1.7]) has the following description: $N-d_{r}(\mathcal{C}_{\Sigma})$ is the maximum number of columns of $G_{\Sigma}$ that span a $n+1-r$ dimensional vector space. Observe $d_{1}(\mathcal{C}_{\Sigma})$ is the usual minimum distance of $\mathcal{C}_{\Sigma}$. Moreover, $N-d_{n+1}(\mathcal{C}_{\Sigma})=0$, since $G_{\Sigma}$ has no zero columns, and $N-d_{n}(\mathcal{C}_{\Sigma})$ is the maximum number of columns of $G_{\Sigma}$ that are proportional to each-other (i.e., the maximum number of linear forms of $\Sigma$ that are proportional to each-other).

The generalized Hamming weights help determine the heights of ideals generated by $a$-fold products of linear forms. By [1, Proposition 2.2], for $r=1,\ldots,n+1$, with the convention that $d_{0}(\mathcal{C}_{\Sigma})=0$, and for any $d_{r-1}(\mathcal{C}_{\Sigma})<a\leq d_{r}(\mathcal{C}_{\Sigma})$, one has

[TABLE]

For example, if $1\leq a\leq d_{1}(\mathcal{C}_{\Sigma})$, then ${\rm ht}(I_{a}(\Sigma))=n+1$, the maximum possible value (see also [5]).

The connections between homological properties of ideals generated by $a$-fold products of linear forms and linear codes and their combinatorics is further transparent in [1, Theorem 2.8], that presents a formula for the degree of $I_{a}(\Sigma)$, or in [1, Proposition 2.10], that presents a formula for the minimum number of generators of $I_{a}(\Sigma)$, both in terms of the coefficients of the Tutte polynomial of the matroid of $G_{\Sigma}$.

Let $I\subset R$ be a homogeneous ideal generated in degree $d$. We say that the $R$-module $I$, or $R/I$, has linear (minimal) graded free resolution, if one has the graded free resolution

[TABLE]

for some positive integer $b$.

The following is a conjecture regarding to the minimal free resolution of ideals generated by $a$-fold products of linear forms.

Conjecture 1.1**.**

(See for example [1, Conjecture 1])   For any $\Sigma=(L_{1},\ldots,L_{N})$ and any $1\leq a\leq N$, the ideal $I_{a}(\Sigma)$ has linear graded free resolution.

This conjecture was made six years ago and some consistent progress has been done towards proving this conjecture. Now in this article we prove it when the support of $\Sigma$ is generic (Theorem 2.3).

• (a)

First evidence of the validity of this conjecture was observed in [17, Theorem 3.1], where it is shown that for any $1\leq a\leq d_{1}(\mathcal{C}_{\Sigma})$, one has $I_{a}(\Sigma)=M^{a}$, and powers of the irrelevant ideal (more generally, of any linear prime) have linear graded free resolution (see for example [9, Corollary 1.5]).

• (b)

If $n=1$, so $\Sigma\subset\mathbb{K}[x_{0},x_{1}]$, [15, Theorem 2.2] proves that for any $1\leq a\leq N$, $I_{a}(\Sigma)$ has linear graded free resolution.

• (c)

If $a=N$, then $I_{N}(\Sigma)=\langle L_{1}\cdots L_{N}\rangle$, which has linear graded free resolution as it is a principal ideal.

• (d)

If $a=N-1$, then [15, Section 2.1] shows that $I_{N-1}(\Sigma)$ has linear graded free resolution.

• (e)

If $a=N-2$, then [15, Theorem 2.5] shows that $I_{N-2}(\Sigma)$ has linear graded free resolution.

• (f)

Suppose $\Sigma$ has no proportional linear forms, i.e., $\Sigma$ defines a hyperplane arrangement in $\mathbb{P}^{n}$. If $\Sigma$ is generic (meaning any ${\rm rk}(\Sigma)$ linear forms of $\Sigma$ are linearly independent), then for any $1\leq a\leq N$, $I_{a}(\Sigma)$ has linear graded free resolution. Indeed, after a change of variables one can assume that ${\rm rk}(\Sigma)=n+1$. Then $d_{1}(\mathcal{C}_{\Sigma})=N-(n+1)+1=N-n$, and part (a) above shows the result for $1\leq a\leq N-n$. For $N-n+1\leq a\leq N$, via Lemma 3.1 below, $I_{a}(\Sigma)$ is the defining ideal of a star configuration, and these have been shown to have linear graded free resolution (given by the Eagon-Northcott complex): see [11, Remark 2.11], [12, Example 3.4 and Corollary 3.5], or [14, Corollary 3.7]. Independently, one can also obtain this result by applying the proof of [10, Theorem 2.5] and [1, Proposition 3.5].

• (f’)

More generally, whenever $R/I_{a}(\Sigma)$ is Cohen-Macaulay, then $I_{a}(\Sigma)$ has a linear graded free resolution. This can be seen from the discussions at the end of the proof of [16, Proposition 2.1].

• (g)

Generalizing the main result in [2], in [18] it is shown that for $a=d_{1}(\mathcal{C}_{\Sigma})+1$, $I_{a}(\Sigma)$ has linear graded free resolution, for any $\Sigma$, a collection of linear forms.

• (h)

If $\Sigma$ defines a line arrangement in $\mathbb{P}^{2}$, then by [18], for any $1\leq a\leq N$, $I_{a}(\Sigma)$ has linear graded free resolution.

In our first main result, Theorem 2.3, we add one more item to this list. Suppose $\Sigma=(\underbrace{\ell_{1},\ldots,\ell_{1}}_{m_{1}},\ldots,\underbrace{\ell_{s},\ldots,\ell_{s}}_{m_{s}})$ with $\gcd(\ell_{i},\ell_{j})=1$ if $i\neq j$. The support of $\Sigma$ is ${\rm Supp}(\Sigma):=\{\ell_{1},\ldots,\ell_{s}\}$. If the support of $\Sigma$ is generic, i.e., any ${\rm rk}(\Sigma)$ elements of ${\rm Supp}(\Sigma)$ are linearly independent, then we show that $I_{a}(\Sigma)$ has linear graded free resolution, for any $1\leq a\leq m_{1}+\cdots+m_{s}$.

1.2. Star Configurations.

Let ${\mathcal{A}}=\{H_{1},\ldots,H_{s}\}$ be a collection of $s\geq n+1$ hyperplanes in $\mathbb{P}^{n}$, and suppose $\ell_{1},\ldots,\ell_{s}\in R$ are defining linear forms of these hyperplanes: i.e., $H_{i}=V(\ell_{i}),i=1,\ldots,s$. Suppose ${\mathcal{A}}$ is generic, meaning that any $n+1$ of the defining linear forms are linearly independent, or in the language of [11, Definition 2.1], the hyperplanes $H_{1},\ldots,H_{s}$ meet properly.

Let $1\leq c\leq n$ be an integer and define the star configuration of codimension $c$ with support ${\mathcal{A}}$ (in $\mathbb{P}^{n}$) to be

[TABLE]

It is clear that the defining ideal is

[TABLE]

and the $m$-th symbolic power of this ideal is

[TABLE]

The following is a conjecture about the ordinary powers of $I(V_{c}({\mathcal{A}}))$.

Conjecture 1.2**.**

(See [11, Conjecture 4.1])   For any $m\geq 1$ one has

[TABLE]

Remark 1.3**.**

Let $J\subset R$ be an ideal generated in degree $d$. Then $J\subseteq J^{\rm sat}\cap M^{d}$.111If $I\subset R$ is a homogeneous ideal, by definition $I^{\rm sat}:=\{f\in R|\exists\,n(f)\geq 0\mbox{ such that }M^{n(f)}\cdot f\subset I\}$. If $R/J$ has linear graded free resolution (equivalently, ${\rm reg}(R/J)=d-1$), since ${\rm H}_{\mathfrak{m}}^{0}(R/J)=J^{\rm sat}/J$, by [7, Theorem 4.3], we have $(J^{\rm sat}/J)_{e}=0,\mbox{ for any }e\geq d$. This means that $J^{\rm sat}\cap M^{d}\subseteq J$, and therefore

[TABLE]

It is shown in [11, Corrolary 4.9] that for any $m\geq 1$,

[TABLE]

Also by [11, Proposition 2.9 (4)], one has that $I(V_{c}({\mathcal{A}}))$ is generated in degree $s-c+1$, and therefore, $I(V_{c}({\mathcal{A}}))^{m}$ is generated in degree $(s-c+1)m$. So in order to prove Conjecture 1.2, it is enough to show that $I(V_{c}({\mathcal{A}}))^{m}$ has linear graded free resolution.

Conjecture 1.2 has been verified to be true in the following cases (See [11, Remark 4.4]):

• (i)

$m=1$. This is true because of Remark 1.3 and item (f) above.

• (ii)

$c=1$. This is true because $I(V_{c}({\mathcal{A}}))$ is a principal ideal.

• (iii)

$c=n$. This is true by [3, Lemma 2.3.3(c), Lemma 2.4.2].

• (iv)

$n=s-1$. This is true by [11, Theorem 4.8].

We are not aware at this moment of any other consistent progress in proving this conjecture, other than the observations we make in Remark 3.4. Nonetheless, by identifying ordinary powers of defining ideals of star configurations with ideals generated by $a$-fold products of linear forms (see Lemma 3.1), then Theorem 2.3, Remark 1.3, and [11, Corrolary 4.9] will prove Conjecture 1.2 in its full generality (see Theorem 3.2).

In the end, again by using the identification provided by Lemma 3.1, we give a positive answer to [11, Question 4.12] and prove a result that calculates the resurgence of the defining ideal of any star configuration (see Theorem 3.6), which generalizes [11, Theorem 4.11].

2. Ideals generated by $a$-fold products of linear forms have linear graded free resolution

Let $\Sigma:=(\underbrace{\ell_{1},\ldots,\ell_{1}}_{m_{1}},\ldots,\underbrace{\ell_{s},\ldots,\ell_{s}}_{m_{s}})$ be a collection of $N:=m_{1}+\cdots+m_{s}$ linear forms in $R:=\mathbb{K}[x_{0},\ldots,x_{n}]$, with $s,m_{1},\ldots,m_{s}\geq 1$, and $\gcd(\ell_{i},\ell_{j})=1$ if $i\neq j$. For any $a\geq 0$, consider the ideal generated by $a$-fold products of linear forms of $\Sigma$

[TABLE]

with the convention that if $a=0$ this ideal equals the entire ring $R$, and if $a>N$ this ideal equals the zero ideal.

One of the most delicate issue is to find a primary decomposition for $I_{a}(\ell_{1}^{m_{1}}\cdots\ell_{s}^{m_{s}})$ (and therefore generalizing [1, Proposition 2.3]). We have a first lemma.

Lemma 2.1**.**

For any $1\leq a\leq N$, one has

[TABLE]

where $\displaystyle\mu(i_{1},\ldots,i_{c}):=a-\sum_{j\in[s]\setminus\{i_{1},\ldots,i_{c}\}}m_{j}$ and any ideal with power $\leq 0$ is replaced by $R$.

Proof.

Let $\xi=\ell_{1}^{t_{1}}\cdots\ell_{s}^{t_{s}}$ be a minimal “monomial” generator in $I_{a}(\ell_{1}^{m_{1}}\cdots\ell_{s}^{m_{s}})$, so $\sum_{j=1}^{s}t_{j}=a$ and $t_{j}\leq m_{j}$ for $1\leq j\leq s$. We will show that $\xi\in\langle\ell_{i_{1}},\ldots,\ell_{i_{c}}\rangle^{\mu(i_{1},\ldots,i_{c})}$ for every $1\leq i_{1}<\cdots<i_{c}\leq s$ and $1\leq c\leq s$.

To do that we just need to show that $\sum_{j\in\{i_{1},\ldots,i_{c}\}}t_{j}\geq\mu(i_{1},\ldots,i_{c})=a-\sum_{j\in[s]\setminus\{i_{1},\ldots,i_{c}\}}m_{j}$. This follows by the fact that if it does not hold, then $\sum_{j\in[s]\setminus\{i_{1},\ldots,i_{c}\}}t_{j}>\sum_{j\in[s]\setminus\{i_{1},\ldots,i_{c}\}}m_{j}$, which forces $t_{j}>m_{j}$ for some $j\in[s]\setminus\{i_{1},\ldots,i_{c}\}$, a contradiction. ∎

Next, we review the beginning of Section 2 in [17]. Let $\Sigma=(L_{1},\ldots,L_{N})$ be a collection of linear forms in $R$. Let $1\leq a\leq N$ and consider $I:=I_{a}(L_{1}\cdots L_{N})$. Let $\mathfrak{p}$ be a prime ideal containing $I$, and let $L_{i_{1}}\cdots L_{i_{a}},1\leq i_{1}<\cdots<i_{a}\leq N$, be an arbitrary generator of $I$. For convenience, suppose $i_{1}=1,\ldots,i_{a}=a$. Then one of $L_{1},\ldots,L_{a}$ belongs to $\mathfrak{p}$; say $L_{1}\in\mathfrak{p}$. But $L_{2}\cdots L_{a+1}\in I$, and therefore one of $L_{2},\ldots,L_{a+1}$ belongs to $\mathfrak{p}$; say $L_{2}\in\mathfrak{p}$ (here it may happen that $L_{1}$ and $L_{2}$ are proportional; it doesn’t matter to the argument). So any prime ideal containing $I$ must contain at least $N-a+1$ linear forms from $\Sigma$ (counted with multiplicity), and conversely, any linear prime generated by $N-a+1$ linear forms of $\Sigma$ will contain $I$.

For a prime ideal $\mathfrak{p}$, consider the closure of $\mathfrak{p}$ in $\Sigma$, denoted $cl_{\Sigma}(\mathfrak{p})$, to be the set of all elements of $\Sigma$, considered with multiplicity/repetition, that belong to ${\mathfrak{p}}$. Denote $\nu_{\Sigma}(\mathfrak{p}):=|cl_{\Sigma}(\mathfrak{p})|$. Immediately, $\mathfrak{p}\supseteq I$ if and only if $\nu_{\Sigma}(\mathfrak{p})\geq N-a+1$.

Now we go back to $\Sigma=(\underbrace{\ell_{1},\ldots,\ell_{1}}_{m_{1}},\ldots,\underbrace{\ell_{s},\ldots,\ell_{s}}_{m_{s}})$. The linear primes showing up in the intersection in Lemma 2.1 may not be distinct. This justifies considering $\Gamma(\Sigma)$, the set of all of the pairwise distinct primes of the form $\langle\ell_{i_{1}},\ldots,\ell_{i_{c}}\rangle$, where $1\leq i_{1}<\cdots<i_{c}\leq s$ and $1\leq c\leq s$.

For any $\mathfrak{p}\in\Gamma(\Sigma)$ one has $\mathfrak{p}=\langle\ell_{i_{1}},\ldots,\ell_{i_{c}}\rangle$ for some linear forms $\ell_{i_{1}},\ldots,\ell_{i_{c}}$, where $\displaystyle\langle\ell_{i_{1}},\ldots,\ell_{i_{c}}\rangle^{\mu(i_{1},\ldots,i_{c})}$ is a factor of the intersection in Lemma 2.1. If $\langle\ell_{j_{1}},\ldots,\ell_{j_{c^{\prime}}}\rangle=\mathfrak{p}$, then in the intersection we will “only see” $\displaystyle{\mathfrak{p}}^{\max\{\mu(i_{1},\ldots,i_{c}),\mu(j_{1},\ldots,j_{c^{\prime}})\}}$. So the maximum power of $\mathfrak{p}$ that can occur is the maximum $\mu(i_{1},\ldots,i_{u})$, whenever $\langle\ell_{i_{1}},\ldots,\ell_{i_{u}}\rangle=\mathfrak{p}$.

Suppose the closure of $\mathfrak{p}$ consists of $\ell_{i_{1}},\ldots,\ell_{i_{u}}$ taken with their corresponding multiplicities. So ${\mathfrak{p}}=\langle\ell_{i_{1}},\ldots,\ell_{i_{u}}\rangle$, and $\nu_{\Sigma}(\mathfrak{p})=m_{i_{1}}+\cdots+m_{i_{u}}$. So

[TABLE]

which is maximal possible, since we cannot have more than $\nu_{\Sigma}(\mathfrak{p})$ linear forms that generate $\mathfrak{p}$. Hence we have

[TABLE]

where of course, $a-N+\nu_{\Sigma}(\mathfrak{p})\leq 0$ if and only if $I\nsubseteq\mathfrak{p}$. For this reason, from now on $\Gamma(\Sigma)$ will consist only of primes that contain $I$. It is worth mentioning that $M=\langle\ell_{1},\ldots,\ell_{s}\rangle=\langle x_{0},\ldots,x_{n}\rangle\in\Gamma(\Sigma)$, and $\nu_{\Sigma}(M)=N$.

The next result is very relevant for the situation when the support of $\Sigma$ is generic, i.e., any ${\rm rk}(\Sigma)$ linear forms from ${\rm Supp}(\Sigma):=\{\ell_{1},\ldots,\ell_{s}\}$ are linearly independent. Via a change of variables, and an embedding into a ring with fewer variables, one can suppose ${\rm rk}(\Sigma)=n+1$ (maximum rank).

Proposition 2.2**.**

Let $\Sigma:=(\underbrace{\ell_{1},\ldots,\ell_{1}}_{m_{1}},\ldots,\underbrace{\ell_{s},\ldots,\ell_{s}}_{m_{s}})$ be a collection of linear forms of rank $n+1$ in $R:=\mathbb{K}[x_{0},\ldots,x_{n}]$, with $s,m_{1},\ldots,m_{s}\geq 1$, and ${\rm Supp}(\Sigma)$ generic. Set $N:=m_{1}+\cdots+m_{s}$ and for any $1\leq a\leq N$, denote $I:=I_{a}(\ell_{1}^{m_{1}}\cdots\ell_{s}^{m_{s}})$. Then

[TABLE]

Proof.

Denote the big intersection on the righthand side with $\mathbb{I}$. It is clear that $\mathbb{I}$ is a saturated ideal. Also, $I\subseteq\mathbb{I}$ (from Lemma 2.1). So it is enough to show that $IR_{P}\supseteq\mathbb{I}R_{P}$, for any $P\in{\rm Ass}(I)\setminus\{M\}$.

Let $\mathfrak{p}:=\langle cl_{\Sigma}(P)\rangle$. Since $I\subset P$ and $P\neq M$, then $\mathfrak{p}\in\Gamma(\Sigma)\setminus\{M\}$.

Suppose ${\mathfrak{p}}=\langle\ell_{1},\ldots,\ell_{c}\rangle$, where, because ${\rm Supp}(\Sigma)$ is generic, we have ${\rm ht}(\mathfrak{p})=c\leq n$. Also because ${\rm Supp}(\Sigma)$ is generic, and since $c\leq n$, we have that $\ell_{c+1},\ldots,\ell_{s}\notin\mathfrak{p}$, and by the way $\mathfrak{p}$ was constructed, these linear forms are not in $P$ as well. So they become invertible under localization.

We have222For any collection $(L_{1},\ldots,L_{N})$ of linear forms, and for any $1\leq a\leq N$, one has $I_{a}(L_{1}\cdots L_{N})=L_{N}I_{a-1}(L_{1}\cdots L_{N-1})+I_{a}(L_{1}\cdots L_{N-1})$; then we apply this several times.

[TABLE]

Under localization, $\ell_{s}$ is invertible, and since

[TABLE]

we have

[TABLE]

Doing this for all $\ell_{c+1},\ldots,\ell_{s}$ we have

[TABLE]

since $\nu_{\Sigma}(\mathfrak{p})=m_{1}+\cdots+m_{c}$, which is $\geq N-a+1$.

Also, under localization in the intersection $\mathbb{I}$ we can “see” only $\mathfrak{q}\in\Gamma(\Sigma)$, with $\mathfrak{q}\subseteq\mathfrak{p}$. Furthermore, if there is an $\ell\in\Sigma$ such that $\ell\in\mathfrak{q}$, then $\ell\in\{\ell_{1},\ldots,\ell_{c}\}$. This is true because otherwise we will have $c+1\leq n+1$ or fewer elements of ${\rm Supp}(\Sigma)$ that are linearly dependent.

After a change of variables, suppose $\ell_{1}=x_{1},\ldots,\ell_{c}=x_{c}$. Everything put together gives

[TABLE]

and

[TABLE]

Let $b:=a-N+\nu_{\Sigma}(\mathfrak{p})=a-N+m_{1}+\cdots+m_{c}\geq 1$. Also let $d=m_{1}+\cdots+m_{c}$. Therefore,

[TABLE]

In order to show our equality under localization it is enough to show

[TABLE]

where $\mu(j_{1},\ldots,j_{k}):=b-d+m_{j_{1}}+\cdots+m_{j_{k}}$.

But in Lemma 2.1 we have seen that the inclusion $\subseteq$ is satisfied.

To prove the other inclusion, let $\xi$ be a monomial in the intersection of ideals of the right side333We can work just with monomials, since the righthand side is the intersection of monomial ideals..

By taking $k=c$ in the intersection, since $\xi\in\langle x_{1},\ldots,x_{c}\rangle^{b}$, one can write $\xi=x_{1}^{t_{1}}\cdots x_{c}^{t_{c}}$ with $\sum_{j=1}^{c}t_{j}\geq b$.

If $t_{j}\geq m_{j}$ for all $j=1,\ldots,c$, then $\xi=(x_{1}^{t_{1}-m_{1}}\cdots x_{c}^{t_{c}-m_{c}})x_{1}^{m_{1}}\cdots x_{c}^{m_{c}}$. This is an element of $I_{d}(x_{1}^{m_{1}}\cdots x_{c}^{m_{c}})$, which in turn is included in $I_{b}(x_{1}^{m_{1}}\cdots x_{c}^{m_{c}})$, as $b\leq d$.

Let $\{x_{j_{1}},\ldots,x_{j_{k}}\}\subseteq\{x_{1},\ldots,x_{c}\}$ such that for any $i\in\{j_{1},\ldots,j_{k}\}$, $t_{i}\leq m_{i}$ and for any $i\in[c]\setminus\{j_{1},\ldots,j_{k}\}$, $t_{i}>m_{i}$. Now

[TABLE]

Set $\xi^{\prime}=\prod_{i\in\{j_{1},\ldots,j_{k}\}}x_{i}^{t_{i}}\cdot\prod_{i\in[c]\setminus\{j_{1},\ldots,j_{k}\}}x_{i}^{m_{i}}$. Then, to prove $\xi\in I_{b}(x_{1}^{m_{1}}\cdots x_{c}^{m_{c}})$, we just need to show $\xi^{\prime}\in I_{b}(x_{1}^{m_{1}}\cdots x_{c}^{m_{c}})$.

Since the exponent of every $x_{i}$ in $\xi^{\prime}$ is less than or equal to $m_{i}$, we just need to show $\sum_{i\in\{j_{1},\ldots,j_{k}\}}t_{i}+\sum_{i\in[c]\setminus\{j_{1},\ldots,j_{k}\}}m_{i}\geq b$.

This follows by the fact that $\xi\in\langle x_{j_{1}},\ldots,x_{j_{k}}\rangle^{\mu(j_{1},\ldots,j_{k})}$, therefore

[TABLE]

Hence

[TABLE]

and the proof is completed. ∎

Now we are ready to prove Conjecture 1.1 when the support of $\Sigma$ is generic.

Theorem 2.3**.**

Let $\Sigma:=(\underbrace{\ell_{1},\ldots,\ell_{1}}_{m_{1}},\ldots,\underbrace{\ell_{s},\ldots,\ell_{s}}_{m_{s}})$ be a collection of linear forms in $R:=\mathbb{K}[x_{0},\ldots,x_{n}]$, with $s,m_{1},\ldots,m_{s}\geq 1$, and with ${\rm Supp}(\Sigma)$ generic. Set $N:=m_{1}+\cdots+m_{s}$. Then for any $1\leq a\leq N$, the ideal $I_{a}(\ell_{1}^{m_{1}}\ell_{2}^{m_{2}}\cdots\ell_{s}^{m_{s}})$ has linear graded free resolution.

Proof.

We are proving the result by induction on pairs $(N,{\rm rk}(\Sigma))$, with $N\geq{\rm rk}(\Sigma)\geq 1$.

Base Cases. If ${\rm rk}(\Sigma)=1$, then $s=1$, and therefore $I_{a}(\ell_{1}^{m_{1}})=\langle\ell_{1}^{a}\rangle$ (which has linear graded free resolution).

If $N={\rm rk}(\Sigma)$, then $s={\rm rk}(\Sigma)$ and $m_{1}=\cdots=m_{s}=1$. This is a particular case of item (f) in the Introduction, so this base case is also verified.

Inductive Step. Suppose $N>{\rm rk}(\Sigma)\geq 2$.

Let $\ell_{1}\in\Sigma$, and denote $I:=I_{a}(\ell_{1}^{m_{1}}\ell_{2}^{m_{2}}\cdots\ell_{s}^{m_{s}})$, and $J:=I_{a-1}(\ell_{1}^{m_{1}-1}\ell_{2}^{m_{2}}\cdots\ell_{s}^{m_{s}})$. First we want to show that $I:\ell_{1}=J$. Since $I=\ell_{1}J+I_{a}(\ell_{1}^{m_{1}-1}\ell_{2}^{m_{2}}\cdots\ell_{s}^{m_{s}})$, we obviously have $J\subseteq I:\ell_{1}$.

For the other inclusion, after a possible change of variables and embedding in a smaller ring, suppose ${\rm rk}(\Sigma)=n+1$.

Let $\Sigma^{\prime}:=(\underbrace{\ell_{1},\ldots,\ell_{1}}_{m_{1}-1},\underbrace{\ell_{2},\ldots,\ell_{2}}_{m_{2}},\ldots,\underbrace{\ell_{s},\ldots,\ell_{s}}_{m_{s}})$. Obviously, $\Sigma^{\prime}$ has generic support.

First suppose that ${\rm rk}(\Sigma^{\prime})=n$ (which necessarily implies that $m_{1}=1$). So, after a change of variables we can suppose $\ell_{1}=x_{0}$, and $\ell_{2},\ldots,\ell_{s}\in R^{\prime}:=\mathbb{K}[x_{1},\ldots,x_{n}]$.

Let $f\in I_{a}(x_{0}\ell_{2}^{m_{2}}\cdots\ell_{s}^{m_{s}}):x_{0}$. Then $x_{0}f=x_{0}g+h$, where $g\in I_{a-1}(\ell_{2}^{m_{2}}\cdots\ell_{s}^{m_{s}})$ and $h\in I_{a}(\ell_{2}^{m_{2}}\cdots\ell_{s}^{m_{s}})$. So $x_{0}(f-g)\in I_{a}(\ell_{2}^{m_{2}}\cdots\ell_{s}^{m_{s}})$, which, since $x_{0}$ is a nonzero divisor mod $I_{a}(\ell_{2}^{m_{2}}\cdots\ell_{s}^{m_{s}})$, leads to $f-g\in I_{a}(\ell_{2}^{m_{2}}\cdots\ell_{s}^{m_{s}})$. But this ideal is trivially included in $I_{a-1}(\ell_{2}^{m_{2}}\cdots\ell_{s}^{m_{s}})$, and therefore $f\in I_{a-1}(\ell_{2}^{m_{2}}\cdots\ell_{s}^{m_{s}})$.

Suppose ${\rm rk}(\Sigma^{\prime})=n+1$. As a reminder, $M:=\langle x_{0},\ldots,x_{n}\rangle$.

We have $I\subseteq I^{\rm sat}\cap M^{a}$, leading to

[TABLE]

By inductive hypotheses, since ${\rm Supp}(\Sigma^{\prime})$ remains generic, $J$ has linear graded free resolution, and so, from Remark 1.3, $J=J^{\rm sat}\cap M^{a-1}$. So in order to prove our desired inclusion, it is enough to show that $I^{\rm sat}:\ell_{1}\subseteq J^{\rm sat}$.

By Proposition 2.2 we have

[TABLE]

where $\Gamma(\Sigma)$ is the set of all linear primes that contain $I$, and that are generated by subsets of ${\rm Supp}(\Sigma)$; and

[TABLE]

where $\Gamma(\Sigma^{\prime})$ is the set of all linear primes that contain $J$, and that are generated by subsets of ${\rm Supp}(\Sigma^{\prime})$.

Since $I\subset J$ (because $I_{a}(\ell_{1}^{m_{1}-1}\ell_{2}^{m_{2}}\cdots\ell_{s}^{m_{s}})\subset J$), and ${\rm Supp}(\Sigma^{\prime})\subseteq{\rm Supp}(\Sigma)$, we have $\Gamma(\Sigma^{\prime})\subseteq\Gamma(\Sigma)$, and hence

[TABLE]

Let $\mathfrak{q}\in\Gamma(\Sigma^{\prime})\setminus\{M\}$.

• (1)

If $\ell_{1}\notin\mathfrak{q}$, then $\nu_{\Sigma^{\prime}}(\mathfrak{q})=\nu_{\Sigma}(\mathfrak{q})$. So

[TABLE]

• (2)

If $\ell_{1}\in\mathfrak{q}$, then $\nu_{\Sigma^{\prime}}(\mathfrak{q})=\nu_{\Sigma}(\mathfrak{q})-1$, so

[TABLE]

Therefore, from items (1) and (2) above, and by Proposition 2.2, we obtained the desired inclusion $I^{\rm sat}:\ell_{1}\subseteq J^{\rm sat}$, and therefore

[TABLE]

Now this equality gives the short exact sequence of $R$-graded modules:

[TABLE]

Since $J$ is generated in degree $a-1$, by inductive hypotheses and [9, Theorem 1.2], the Castelnuovo-Mumford regularity of the leftmost nonzero module is

[TABLE]

Considering the rightmost nonzero module $R/\langle\ell_{1},I\rangle$, after a change of variables, we may suppose $\ell_{1}=x_{0}$. For $j=2,\ldots,s$, let $\ell_{j}=c_{j}x_{0}+\bar{\ell}_{j}$, where $c_{j}\in\mathbb{K}$ and $\bar{\ell}_{j}\in S:=\mathbb{K}[x_{1},\ldots,x_{n}]$. Let $\overline{\Sigma}=(\underbrace{\bar{\ell}_{2},\ldots,\bar{\ell}_{2}}_{m_{2}},\ldots,\underbrace{\bar{\ell}_{s},\ldots,\bar{\ell}_{s}}_{m_{s}})$. Then $|\overline{\Sigma}|=m_{2}+\cdots+m_{s}\leq N-1$, and ${\rm rk}(\overline{\Sigma})=n$. Moreover, one has

[TABLE]

and therefore $R/\langle\ell_{1},I\rangle\cong S/I_{a}(\overline{\Sigma})$.

Suppose there exist $d_{1},\ldots,d_{n}\in\mathbb{K}$ such that $d_{1}\bar{\ell}_{i_{1}}+\cdots+d_{n}\bar{\ell}_{i_{n}}=0$, for some $2\leq i_{1}<\cdots<i_{n}\leq s$. Then we have $d_{1}\ell_{i_{1}}+\cdots+d_{n}\ell_{i_{n}}-(c_{i_{1}}d_{1}+\cdots+c_{i_{n}}d_{n})\ell_{1}=0$. But these $n+1$ linear forms are linearly independent because ${\rm Supp}(\Sigma)$ is generic, so $d_{1}=\cdots=d_{n}=0$. This leads to ${\rm Supp}(\overline{\Sigma})$ being generic as well.

If $a\leq|\overline{\Sigma}|$, then by induction hypotheses, ${\rm reg}(S/I_{a}(\overline{\Sigma}))=a-1$, and so via [7, Corollary 4.6], ${\rm reg}(R/\langle\ell_{1},I\rangle)=a-1$. On the other hand, if $a>|\overline{\Sigma}|$, then $I_{a}(\overline{\Sigma})=0$ by convention, so ${\rm reg}(R/\langle\ell_{1},I\rangle)={\rm reg}(R/\langle\ell_{1}\rangle)=0$.

Finally, [8, Corollary 20.19 b.] gives that ${\rm reg}(R/I)\leq a-1$, and since $I$ is generated in degree $a$, we obtain that $I$ has linear graded free resolution. ∎

With the notations and discussions at the beginning of this section, we have the following result.

Corollary 2.4**.**

Let $\Sigma:=(\underbrace{\ell_{1},\ldots,\ell_{1}}_{m_{1}},\ldots,\underbrace{\ell_{s},\ldots,\ell_{s}}_{m_{s}})$ be a collection of linear forms in $R:=\mathbb{K}[x_{0},\ldots,x_{n}]$, with $s,m_{1},\ldots,m_{s}\geq 1$, ${\rm rk}(\Sigma)=n+1$, and with ${\rm Supp}(\Sigma)$ generic. Set $N:=m_{1}+\cdots+m_{s}$. Then for any $1\leq a\leq N$, we have the decomposition

[TABLE]

Proof.

By Theorem 2.3, the ideal $I:=I_{a}(\ell_{1}^{m_{1}}\ell_{2}^{m_{2}}\cdots\ell_{s}^{m_{s}})$ has linear graded free resolution. Therefore, by Remark 1.3, $I=I^{\rm sat}\cap M^{a}$. Since $\nu_{\Sigma}(M)=N$, by Proposition 2.2, we get the first equality in the statement. The second equality has been addressed in the discussions right before presenting Proposition 2.2. ∎

It is worth noting that since the first equality presents a decomposition of $I$ that is irredundant and irreducible, we have $\Gamma(\Sigma)={\rm Ass}(I)$.

3. Star configurations

3.1. The Geramita-Harbourne-Migliore Conjecture.

Let ${\mathcal{A}}=\{H_{1},\ldots,H_{s}\}$ be a collection of $s\geq n+1$ hyperplanes in $\mathbb{P}^{n}$, and suppose $\ell_{1},\ldots,\ell_{s}\in R$ are defining linear forms of these hyperplanes: i.e., $H_{i}=V(\ell_{i}),i=1,\ldots,s$; we will abuse notation by saying ${\mathcal{A}}=\{\ell_{1},\ldots,\ell_{s}\}\subset R$. Suppose ${\mathcal{A}}$ is generic, and let $1\leq c\leq n$.

Lemma 3.1**.**

The defining ideal of the star configuration $V_{c}({\mathcal{A}})$ satisfies

[TABLE]

Furthermore, for any $m\geq 1$, we have

[TABLE]

Proof.

We prove the first part by induction on $|{\mathcal{A}}|=s\geq n+1$. The base case of induction is $s=n+1$. In this instance, after a change of variables we can suppose that ${\mathcal{A}}$ consists of the coordinate hyperplanes, i.e., $\ell_{i}=x_{i-1},i=1,\ldots,n+1$. So

[TABLE]

which by standard results on squarefree monomial ideals equals to

[TABLE]

Suppose $s>n+1$. In the proof of [11, Proposition 2.9 (4)] we have

[TABLE]

where ${\mathcal{A}}^{\prime}={\mathcal{A}}\setminus\{V(\ell_{s})\}$, and by inductive hypotheses we have

[TABLE]

But it is obvious that at the level of ideals generated by fold products of linear forms we have

[TABLE]

So the conclusion follows.

The second part follows immediately from the simple observation that $(I_{a}(\ell_{1}\cdots\ell_{s}))^{m}=I_{ma}(\ell_{1}^{m}\cdots\ell_{s}^{m})$. ∎

Theorem 3.2**.**

Conjecture 1.2 is true in its full generality.

Proof.

As we mentioned already in the Introduction, from Lemma 3.1 and Theorem 2.3, $I(V_{c}({\mathcal{A}}))^{m}$ have linear graded free resolution, and therefore via Remark 1.3 and from [11, Corollary 4.9], Conjecture 1.2 is true. ∎

Remark 3.3**.**

As an exercise, we will show that the decomposition of $(I_{m(s-c+1)}(\ell_{1}^{m}\cdots\ell_{s}^{m}))^{\rm sat}$ presented in Proposition 2.2 matches the one presented in [11, Corollary 4.9].

Let $\Sigma=(\underbrace{\ell_{1},\ldots,\ell_{1}}_{m},\ldots,\underbrace{\ell_{s},\ldots,\ell_{s}}_{m})$. So in Proposition 2.2 we have $N=ms$, and $a=m(s-c+1)$.

As we have seen, any associated prime of $I$ (so an element in $\Gamma(\Sigma)$) is generated by at least $N-a+1=m(c-1)+1$ linear forms from $\Sigma$. So we must pick at least $c$ linear forms from ${\mathcal{A}}:=\{\ell_{1},\ldots,\ell_{s}\}$ to be able to generate such a minimal prime (indeed picking $c-1$ elements from ${\mathcal{A}}$ each $m$ times it forces us to pick another linear form from the remaining $s-c$ elements of ${\mathcal{A}}$). Since any $n+1$ elements of ${\mathcal{A}}$ are linearly independent, since we don’t want to obtain $M$, the most number of linear forms from ${\mathcal{A}}$ we can pick is $n$.

So, any codimension $j$ associated prime $\mathfrak{p}$ over $I$, where $c\leq j\leq n$, is of the form $\mathfrak{p}=\langle\ell_{i_{1}},\ldots,\ell_{i_{j}}\rangle$, where $1\leq i_{1}<\cdots<i_{j}\leq s$. 444We can adjust conveniently how many times we pick each $\ell_{i_{1}},\ldots,\ell_{i_{j}}$, such that we have $m(c-1)+1$ linear forms chosen from $\Sigma$. But in this situation, $\nu_{\Sigma}(\mathfrak{p})=mj$.

With $a-N+\nu_{\Sigma}(\mathfrak{p})=m(j-c+1)$, we get that the codimension $j$ component of $I^{\rm sat}$ is

[TABLE]

which equals $I(V_{j}({\mathcal{A}}))^{(m(j-c+1))}$. As $j=c,\ldots,n$, we obtain indeed the decomposition in [11, Corollary 4.9].

We end this subsection with an observation on how to prove Conjecture 1.2 in a new case, without appealing to the technique we have developed in the previous section.

Remark 3.4**.**

With the identification in Lemma 3.1 established, the main result in [4] will give that Conjecture 1.2 is true for $c=2$ and also for the already proven case $c=n$. Here is how:

$c=2$. At the end of the proof of [16, Proposition 2.1], we can see that for any $\Sigma=(L_{1},\ldots,L_{N})$ and any $1\leq a\leq N$, $I_{a}(L_{1}\cdots L_{N})$ is generated by the maximal minors of an $a\times N$ matrix with linear entries. The matrix looks like

[TABLE]

where the constants $c_{uv}$ are generic (meaning that none of the $a\times a$ minors of $X$ vanish. In fact, for any $j=1,\ldots,a$, the ideal generated by the $j\times j$ minors of $X$, denoted $I_{j}(X)$, equals $I_{j}(L_{1}\cdots L_{N})$.

Let $c=2$, $N=s$, $a=s-c+1=s-1$, and $\Sigma=(\ell_{1},\ldots,\ell_{s})$, where $\ell_{i}$ define the generic arrangement ${\mathcal{A}}$. By Lemma 3.1, for $1\leq j\leq a=s-1$, we have

[TABLE]

and therefore

• •

${\rm ht}(I_{1}(X))=n+1$,

• •

${\rm ht}(I_{a}(X))=2$,

• •

${\rm ht}(I_{j}(X))=\min\{s-j+1,n+1\}$, for $j=2,\ldots,a-1=s-2$.

Since

[TABLE]

we satisfy the conditions of [4, Theorem 3.7], and therefore, any power of $I_{a}(X)=I(V_{2}({\mathcal{A}}))$ has linear graded free resolution, hence by Remark 1.3, the conjecture is true for $c=2$.

$c=n$. From Lemma 3.1 we have $I(V_{n}({\mathcal{A}}))=I_{s-n+1}(\ell_{1}\cdots\ell_{s})$, which also equals $I_{s-n+1}(X)$. So ${\rm ht}(I_{s-n+1}(X))=n$.

Let $1\leq j\leq s-n$. We still have $I_{j}(X)=I_{j}(\ell_{1}\cdots\ell_{s})$. Because ${\mathcal{A}}$ is generic, the linear code $\mathcal{C}_{{\mathcal{A}}}$ has minimum distance $d_{1}(\mathcal{C}_{{\mathcal{A}}})=s-(n+1)+1=s-n$. By [1, Proposition 2.2], for all $1\leq j\leq s-n$, we have ${\rm ht}(I_{j}(\ell_{1}\cdots\ell_{s}))=n+1$. Therefore the conditions in [4, Theorem 3.7] are satisfied and the conjecture is also true for $c=n$.

3.2. Resurgence.

We finish the paper by determining the resurgence of defining ideals of star configurations.

Let $0\neq I\subset R:=\mathbb{K}[x_{0},\ldots,x_{n}]$ be a homogeneous ideal. The resurgence of $I$ is defined as

[TABLE]

The resurgence of ideals is an important invariant describing the containment of symbolic powers and regular powers. We always have $\rho(I)\geq 1$ since if $m<r$, then $I^{(m)}\nsubseteq I^{r}$. By a result of Ein-Lazarsfeld-Smith [6] and Hochster-Huneke [13] that $I^{(Nr)}\subseteq I^{r}$ for any $r\geq 1$, one can see that $\rho(I)\leq N$.

Let ${\mathcal{A}}=\{H_{1},\ldots,H_{s}\}$ be a set of generic $s\geq n+1$ hyperplanes in $\mathbb{P}^{n}$ with defining linear forms $\ell_{1},\ldots,\ell_{s}$, i.e., $H_{i}=V(\ell_{i})$ for $1\leq i\leq s$. Let $N=s-1$ and ${\mathcal{A}}^{\prime}=\{H^{\prime}_{0},H^{\prime}_{1},\ldots,H^{\prime}_{N}\}$ be the set of coordinate hyperplanes in $\mathbb{P}^{N}$, i.e., $H^{\prime}_{i}=V(z_{i})$, where $z_{0},\ldots z_{N}$ are coordinate variables.

Define

[TABLE]

by $\phi:z_{i}\longmapsto\ell_{i+1}$ for $0\leq i\leq N$.

Set $I:=I\left(V_{c}({\mathcal{A}},\mathbb{P}^{n})\right)$ (resp. $I^{\prime}:=I\left(V_{c}({\mathcal{A}}^{\prime},\mathbb{P}^{N})\right)$) be the defining ideal of the star configuration of codimension $c$ with support in ${\mathcal{A}}$ (resp. in ${\mathcal{A}}^{\prime}$). By [11, Theorem 3.1], one has that $\phi({I^{\prime}}^{(m)})=I^{(m)}$ for every $m\geq 1$.

In the following, we first give a positive answer to Question 4.12 in [11], that asks whether $\rho(I)=\rho(I^{\prime})$. Then we prove a result that calculates the resurgence of the defining ideal of any star configuration, which generalizes [11, Theorem 4.11].

Proposition 3.5**.**

With the above setting, one has $\rho(I)=\rho(I^{\prime})$ for $1\leq c\leq n$.

Proof.

By the discussion after [11, Question 4.12], one has that $\rho\left(I\right)\leq\rho\left(I^{\prime}\right)$, so we only need to prove $\rho\left(I\right)\geq\rho\left(I^{\prime}\right)$.

Assume $I^{(m)}\subseteq I^{r}$. We want to show ${I^{\prime}}^{(m)}\subseteq{I^{\prime}}^{r}$, which yields the desired inequality.

By [11, Theorem 3.1], the symbolic power ${I^{\prime}}^{(m)}$ is generated by monomials in $z_{i}$. Let $\xi^{\prime}\in{I^{\prime}}^{(m)}$ be a monomial generator, then one can write $\xi^{\prime}=z_{0}^{t_{1}}z_{1}^{t_{2}}\cdots z_{N}^{t_{s}}$. Hence $\xi:=\phi(\xi^{\prime})=\ell_{1}^{t_{1}}\ell_{2}^{t_{2}}\cdots\ell_{s}^{t_{s}}\in\phi({I^{\prime}}^{(m)})=I^{(m)}$. Since $I^{(m)}\subseteq I^{r}=I_{ra}(\ell_{1}^{r}\cdots\ell_{s}^{r})$, where $a=s-c+1$ (see Lemma 3.1). Hence

[TABLE]

where $k_{j}\leq r,t_{j}-k_{j}\geq 0$ for $1\leq j\leq s$, and $\sum_{j=1}^{s}k_{j}=ra$.

In the same manner,

[TABLE]

Since $k_{j}\leq r$ for $1\leq j\leq s$ and $\sum_{j=1}^{s}k_{j}=ra$, one has $z_{0}^{k_{1}}z_{1}^{k_{2}}\cdots z_{N}^{k_{s}}\in I_{ra}(z_{0}^{r}z_{1}^{r}\cdots z_{N}^{r})={I^{\prime}}^{r}$. Therefore $\xi^{\prime}\in{I^{\prime}}^{r}$. ∎

Theorem 3.6**.**

$I:=I\left(V_{c}({\mathcal{A}},\mathbb{P}^{n})\right)$* be the defining ideal of the star configuration of codimension $c$ with support in ${\mathcal{A}}$. Then*

[TABLE]

Proof.

By Proposition 3.5, setting $s=N+1$, it is enough to show $\displaystyle\rho(I^{\prime})=\frac{c(N-c+2)}{N+1}$, where $I^{\prime}:=I\left(V_{c}({\mathcal{A}}^{\prime},\mathbb{P}^{N})\right)$ is the defining ideal of the star configuration of codimension $c$ with support in ${\mathcal{A}}^{\prime}=\{V(z_{0}),\ldots,V(z_{N})\}$, the coordinate hyperplane arrangement.

As mentioned after [11, Definition 4.10], citing [3], we have that $\displaystyle\rho(I^{\prime})\geq\frac{c(N-c+2)}{N+1}$, so we have to show the other inequality; i.e., if $m/r\geq c(N-c+2)/(N+1)$, then ${I^{\prime}}^{(m)}\subseteq{I^{\prime}}^{r}$.

Let $\xi=z_{0}^{t_{0}}\cdots z_{N}^{t_{N}}\in{I^{\prime}}^{(m)}$ be an arbitrary monomial generator. Then for every $0\leq i_{1}<\cdots<i_{c}\leq N$, $\xi\in\langle z_{i_{1}},\ldots,z_{i_{c}}\rangle^{m}.$ So for every $0\leq i_{1}<\cdots<i_{c}\leq N$, we have

[TABLE]

By Lemma 3.1 and Corollary 2.4, one has

[TABLE]

so we need to show $\xi\in\langle z_{i_{1}},\ldots,z_{i_{c^{\prime}}}\rangle^{r(c^{\prime}-c+1)}$, for every $0\leq i_{1}<\cdots<i_{c^{\prime}}\leq N$, where $c\leq c^{\prime}\leq N+1$.

Let $0\leq i_{1}<\cdots<i_{c^{\prime}}\leq N$, with $c\leq c^{\prime}\leq N+1$. From above, for any $U\subset\{i_{1},\ldots,i_{c^{\prime}}\}$ with $|U|=c$ we have

[TABLE]

Summing up all of these inequalities over all such subsets $U$ we have

[TABLE]

leading to

[TABLE]

But the last quantity is $\geq r(c^{\prime}-c+1)$, since $c^{\prime}\leq N+1$, concluding the proof. ∎

Acknowledgment.   The authors would like to thank Kuei-Nuan Lin and Yi-Huang Shen for careful reading of an earlier version of this manuscript where we claimed we are proving Conjecture 1.1 in its full generality. Their corrections made us restrict the hypotheses to collections of linear forms with generic support; consequently, that conjecture is still open in the general case.