On some ideals with linear free resolutions

TL;DR

This paper proves a conjecture that certain ideals generated by products of linear forms have linear free resolutions in specific cases, including linear codes and line arrangements, and explores their Betti numbers.

Contribution

It establishes the conjecture for cases involving linear codes with minimum distance and line arrangements in the projective plane, and analyzes Betti numbers in the latter case.

Findings

Proved the conjecture for linear codes with minimum distance d.

Confirmed the conjecture for line arrangements in P^2.

Analyzed the graded Betti numbers for line arrangements.

Abstract

Given $\Sigma\subset\mathbb K[x_1,\ldots,x_k]$, any finite collection of linear forms, some possibly proportional, and any $1\leq a\leq |\Sigma|$, it has been conjectured that $I_a(\Sigma)$, the ideal generated by all $a$-fold products of $\Sigma$, has linear graded free resolution. In this article we show the validity of this conjecture for two cases: the first one is when $a=d+1$ and $\Sigma$ is dual to the columns of a generating matrix of a linear code of minimum distance $d$; and the second one is when $k=3$ and $\Sigma$ defines a line arrangement in $\mathbb P^2$ (i.e., there are no proportional linear forms). For the second case we investigate what are the graded betti numbers of $I_a(\Sigma)$.

Equations

$$I_a(\Sigma ):=⟨\{{\ell }_{i_1}\cdots {\ell }_{i_a}|1\le i_1<\cdots <i_a\le n\}⟩.$$

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

$$I_a(\Sigma ):\ell =I_{a-1}(\Sigma \setminus \{\ell \}).$$

$$0⟶\frac{R(-1)}{I_{a-1}(\Sigma \setminus \{\ell \})}\stackrel{\cdot \ell }{⟶}\frac{R}{I_a(\Sigma )}⟶\frac{R}{⟨\ell ,I_a(\Sigma )⟩}⟶0.$$

$$J=J^{\mathrm{sat}}\cap {\mathfrak{m}}^a.$$

$$\mathrm{ht}(I_a(\mathcal{C}))=k+1-r.$$

$$I_a(\mathcal{C})={\mathfrak{m}}^a.$$

$$V(I_{d_1(\mathcal{C})+1}(\mathcal{C}))\subset {\mathbb{P}}^{k-1}.$$

$$(I_{d_1(\mathcal{C})+1}(\mathcal{C}){)}^{\mathrm{sat}}={\mathfrak{q}}_1\cap \cdots \cap {\mathfrak{q}}_m,$$

$$(I_{d^{\prime }+1}({\mathcal{C}}^{\prime }){)}^{\mathrm{sat}}=\bigcap _{\ell \notin {\mathfrak{q}}_i}{\mathfrak{q}}_i,$$

$$I_{d+1}(\mathcal{C}):\ell =I_d({\mathcal{C}}^{\prime }).$$

$$I_{d+1}(\mathcal{C}):\ell \subset {\mathfrak{m}}^{d+1}:\ell ={\mathfrak{m}}^d=I_d({\mathcal{C}}^{\prime }).$$

$$I_{d+1}(\mathcal{C}):\ell \subseteq ({\mathfrak{q}}_1:\ell )\cap \cdots \cap ({\mathfrak{q}}_m:\ell )\cap ({\mathfrak{m}}^{d+1}:\ell ).$$

$$I_{d+1}(\mathcal{C}):\ell \subseteq (I_{d^{\prime }+1}({\mathcal{C}}^{\prime }){)}^{\mathrm{sat}}\cap {\mathfrak{m}}^d.$$

$$I_{d+1}(\mathcal{C}):\ell \subseteq I_d({\mathcal{C}}^{\prime }).$$

$$0⟶\frac{R(-1)}{I_d({\mathcal{C}}^{\prime })}⟶\frac{R}{I_{d+1}(\mathcal{C})}⟶\frac{R}{⟨\ell ,I_{d+1}(\mathcal{C})⟩}⟶0.$$

$$\mathrm{reg}\left(\frac{R(-1)}{I_d({\mathcal{C}}^{\prime })}\right)=(d-1)+1=d.$$

$$⟨\overline{{\ell }_1},\dots ,\overline{{\ell }_u},x_k⟩=⟨{\ell }_1,\dots ,{\ell }_u,x_k⟩=⟨{\ell }_1,\dots ,{\ell }_u,{\ell }_{m+1},\dots ,{\ell }_n⟩.$$

$$n-d\ge u+n-m=n-\bar{d},$$

$$\mathrm{reg}\left(\frac{\bar{R}}{I_{d+1}(\bar{\mathcal{C}})}\right)=d.$$

$$(I_{n-b}(\mathcal{A}){)}^{\mathrm{sat}}={{\mathfrak{p}}_1}^{n_1-b}\cap \cdots \cap {{\mathfrak{p}}_s}^{n_s-b},$$

$$I_{n-b}(\mathcal{A}):z=I_{n-b-1}({\mathcal{A}}^{\prime })=⟨x,y{⟩}^{n-b-1}.$$

$$I_{n-b}(\mathcal{A}):\ell \subseteq ({{\mathfrak{p}}_1}^{n_1-b}:\ell )\cap \cdots \cap ({{\mathfrak{p}}_s}^{n_s-b}:\ell )\cap \underset{⟨x,y,z{⟩}^{n-b-1}}{\underbrace{⟨x,y,z{⟩}^{n-b}:\ell }}.$$

$$({{\mathfrak{p}}_1}^{n_1-b}:\ell )\cap \cdots \cap ({{\mathfrak{p}}_s}^{n_s-b}:\ell )={{\mathfrak{p}}_1}^{n_1-b-1}\cap \cdots \cap {{\mathfrak{p}}_r}^{n_r-b-1}\cap {{\mathfrak{p}}_{r+1}}^{n_{r+1}-b}\cap \cdots \cap {{\mathfrak{p}}_s}^{n_s-b}=(I_{n-1-b}({\mathcal{A}}^{\prime }){)}^{\mathrm{sat}}.$$

$$I_{n-b}(\mathcal{A}):\ell \subseteq I_{n-1-b}({\mathcal{A}}^{\prime }).$$

$$0⟶R(-1)/I_{n-1-b}({\mathcal{A}}^{\prime })⟶R/I_{n-b}(\mathcal{A})⟶R/⟨\ell ,I_{n-b}(\mathcal{A})⟩⟶0.$$

$$I_a(\Sigma )={\ell }_1^{d_1}\cdots {\ell }_t^{d_t}\cdot ⟨x,y{⟩}^e,$$

$$0⟶{\overline{R}}^e(-(a+1))⟶{\overline{R}}^{e+1}(-a)⟶\overline{R};$$

$$0⟶R^{b_3(a,\mathcal{A})}(-(a+2))⟶R^{b_2(a,\mathcal{A})}(-(a+1))⟶R^{b_1(a,\mathcal{A})}(-a)⟶R$$

$$b_i(a,\mathcal{A})=\left(\genfrac{}{}{0ex}{}{a+2}{a+i-1}\right)\cdot \left(\genfrac{}{}{0ex}{}{a+i-2}{a-1}\right),i=1,2,3.$$

Full text

Tohaneanu’s address: Department of Mathematics, University of Idaho, Moscow, Idaho 83844-1103, USA, Email: tohaneanu@uidaho.edu.

1. Introduction

Let $R:=\mathbb{K}[x_{1},\ldots,x_{k}]$ be the ring of (homogeneous) polynomials with coefficients in a field $\mathbb{K}$, with the standard grading. Denote ${\mathfrak{m}}:=\langle x_{1},\ldots,x_{k}\rangle$ to be the irrelevant maximal ideal of $R$. Let $\ell_{1},\ldots,\ell_{n}$ be linear forms in $R$, some possibly proportional, and denote this collection by $\Sigma=(\ell_{1},\ldots,\ell_{n})\subset R$; we need a notation to take into account the fact that some of these linear forms are proportional. For $\ell\in\Sigma$, by $\Sigma\setminus\{\ell\}$ we will understand the collection of linear forms of $\Sigma$ from which $\ell$ has been removed. Also, we denote $|\Sigma|=n$, and ${\rm rk}(\Sigma):={\rm ht}(\langle\ell_{1},\ldots,\ell_{n}\rangle)$.

Let $1\leq a\leq n$ be an integer and define the ideal generated by $a$-fold products of $\Sigma$ to be the ideal of $R$

[TABLE]

We also make the convention $I_{0}(\Sigma):=R$, and $I_{b}(\Sigma)=0$, for all $b>n$. Also, if $\Sigma=\emptyset$, $I_{a}(\Sigma)=0$, for any $a\geq 1$.

When the linear forms $\ell_{1},\ldots,\ell_{n}$ define a hyperplane arrangement ${\mathcal{A}}\subset\mathbb{P}^{k-1}$ (so for all $i\neq j$, $\gcd(\ell_{i},\ell_{j})=1$), we will denote $\Sigma={\mathcal{A}}=\{\ell_{1},\ldots,\ell_{n}\}$.

A homogeneous ideal $I\subset R$ generated in degree $d$ it is said to have linear (minimal) graded free resolution, if one has the graded free resolution

[TABLE]

for some positive integer $b$. The integers $n_{j}\geq 1$ are called the betti numbers of $R/I$, and since we have exactly one shift at each step in the resolution, they match the graded betti numbers. By convention, the zero ideal has linear graded free resolution. Also we say that $R/I$ has linear graded free resolution if and only if $I$ has linear graded free resolution.

[1, Conjecture 1] states that for any collection of linear forms $\Sigma$, and any $1\leq a\leq|\Sigma|$, the ideals $I_{a}(\Sigma)$ have linear graded free resolution. In [9] is presented the current state of this conjecture, as well as new instances when the conjecture is true: Theorem 2.2 shows the validity for $k=2$, and Theorem 2.5 shows the validity for $a=|\Sigma|-2$.

In this article we check the validity of the conjecture when $a=d+1$, where $d$ is the minimum distance of the linear code dual to $\Sigma$; and for $k=3$, but when $\Sigma={\mathcal{A}}=\{\ell_{1},\ldots,\ell_{n}\}$.

The strategy to prove these two cases is the following.

• •

The proof is by induction on pairs $(|\Sigma|,{\rm rk}(\Sigma))$.

• •

Often we appeal to the simple observation that if $\ell\in\Sigma$, then for all $a$, $I_{a}(\Sigma)=\ell\cdot I_{a-1}(\Sigma\setminus\{\ell\})+I_{a}(\Sigma\setminus\{\ell\})$.

• •

By knowing how $I_{a}(\Sigma)^{\rm sat}$ looks like, from inductive hypotheses coupled with Remark 1.1 below, we are able to show that there exists an $\ell\in\Sigma$ such that

[TABLE]

• •

Consider the short exact sequence of graded $R$-modules:

[TABLE]

• •

Most of the time due to inductive hypotheses, the leftmost and the rightmost nonzero modules have linear graded free resolutions, equivalently, by [5, Theorem 1.2], their Castelnuovo-Mumford regularities are both $\leq a-1$.

• •

Apply the inequalities of regularity under short exact sequence (see [4, Corollary 20.19 b.]), to conclude that ${\rm reg}(R/I_{a}(\Sigma))\leq a-1$, and therefore that $R/I_{a}(\Sigma)$ has linear graded free resolution.

Remark 1.1**.**

Let $J\subset R$ be an ideal generated in degree $a$. Then $J\subseteq J^{\rm sat}\cap{\mathfrak{m}}^{a}$.111If $I\subset R$ is a homogeneous ideal, by definition $I^{\rm sat}:=\{f\in R|\exists n(f)\geq 0\mbox{ such that }{\mathfrak{m}}^{n(f)}\cdot f\subset I\}$. If $R/J$ has linear graded free resolution (equivalently, ${\rm reg}(R/J)=a-1$), since ${\rm H}_{\mathfrak{m}}^{0}(R/J)=J^{\rm sat}/J$, by [3, Theorem 4.3], we have $(J^{\rm sat}/J)_{e}=0,\mbox{ for any }e\geq a$. This means that $J^{\rm sat}\cap{\mathfrak{m}}^{a}\subseteq J$, and therefore

[TABLE]

In the last part we present an addition-deletion technique to find the graded betti numbers for $\mathbb{K}[x,y,z]/I_{a}({\mathcal{A}})$ for any $1\leq a\leq|{\mathcal{A}}|$, where ${\mathcal{A}}$ is a line arrangement in $\mathbb{P}^{2}$.

2. The scheme of projective codewords of minimum weight

A linear code $\mathcal{C}$ of dimension $k$, and length $n$, is the image of a $\mathbb{K}$-linear map $\mathbb{K}^{k}\longrightarrow\mathbb{K}^{n}$, given by a (generating) matrix $G$ of size $k\times n$; most of the time one supposes that $G$ has no zero column, and we suppose this as well. The minimum distance, $d_{1}(\mathcal{C})$, is the minimum number of nonzero entries in a nonzero vector in $\mathcal{C}$. $\dim(\mathcal{C})={\rm rk}(G)$, $n$, and $d_{1}(\mathcal{C})$ are called the parameters of $\mathcal{C}$, and they are invariant under rescaling and permutation of the columns of $G$. The left-multiplication of $G$ by any $k\times k$ invertible matrix gives the same linear code $\mathcal{C}$. Because of these, one has the following duality:

To each column $(c_{1},\ldots,c_{k})^{T}$ of $G$ consider the dual linear form $c_{1}x_{1}+\cdots+c_{k}x_{k}$ in $R=\mathbb{K}[x_{1},\ldots,x_{k}]$. Consequently, to $G$ we can associate a collection of linear forms $\Sigma=(\ell_{1},\ldots,\ell_{n})\subset R$. Also, this process is reversible: to any collection of linear forms we can associate a generating matrix of a linear code $\mathcal{C}$. The dimension of $\mathcal{C}$ equals the rank of the collection of linear forms dual to some (any) generating matrix of the code. Because of this duality, we will replace $\Sigma$ by its dual code $\mathcal{C}$.

Suppose ${\rm rk}(G)=k$. Generalizing the notion of minimum distance, for any $1\leq r\leq k$ one can define the $r$-th generalized Hamming weight, $d_{r}(\mathcal{C})$, which by classical results in coding theory (see for example [1, Corollary 1.3 and Proposition 1.7]) has the following description: $n-d_{r}(\mathcal{C})$ is the maximum number of columns of $G$ that span a $k-r$ dimensional vector space. For example, $n-d_{k}(\mathcal{C})=0$, since we assumed that $G$ has no zero columns. Or, $n-d_{k-1}(\mathcal{C})$ is the maximum number of columns of $G$ that are proportional to each-other.

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

[TABLE]

For example, if $1\leq a\leq d_{1}(\mathcal{C})$, then ${\rm ht}(I_{a}(\mathcal{C}))=k$, the maximum possible value.

From [8, Theorem 3.1], we have that for any $1\leq a\leq d_{1}(\mathcal{C})$,

[TABLE]

This has been the starting point of [1, Conjecture 1], since powers of the irrelevant ideal have linear graded free resolution (see for example [5, Corollary 1.5]).

In this first part of the paper we will show that $I_{d_{1}(\mathcal{C})+1}(\mathcal{C})$ has also linear graded free resolution, generalizing one of the main results in [2]. As mentioned in there, such a result has very good applications in error-correcting messages, via saturation with respect with an extra variable.

2.1. Projective codewords of minimum weight.

The weight of a vector ${\bf v}\in\mathbb{K}^{n}$, denoted ${\rm wt}({\bf v})$, is the number of its nonzero entries. So the product of any ${\rm wt}({\bf v})+1$ entries of ${\bf v}$ is zero, and there exist a product of ${\rm wt}({\bf v})$ entries of ${\bf v}$ that is nonzero.

Elements of $\mathcal{C}$ are called codewords, and a codeword of minimum weight is a codeword of weight $d_{1}(\mathcal{C})$. A projective codeword of minimum weight is the equivalence class of a codeword of minimum weight, under multiplication by nonzero scalars. So we can think of a projective codeword of minimum weight as a point $[\beta_{1},\ldots,\beta_{n}]\in\mathbb{P}^{n-1}$. Suppose ${\rm rk}(G)=k$. Then, each projective codeword of minimum weight corresponds uniquely to a point $[\alpha_{1},\ldots,\alpha_{k}]\in\mathbb{P}^{k-1}$, via the left-multiplication $(\alpha_{1},\ldots,\alpha_{k})\cdot G=(\beta_{1},\ldots,\beta_{n})$.

These two observations put together give that the variety of projective codewords of minimum weight corresponds to

[TABLE]

We will abuse terminology by saying that the scheme of projective codewords of minimum weight is defined by $I_{d_{1}(\mathcal{C})+1}(\mathcal{C})$.

Remark 2.1**.**

By [8, Lemma 2.2], if $V(I_{d_{1}(\mathcal{C})+1}(\mathcal{C}))=\{Q_{1},\ldots,Q_{m}\}$, then

[TABLE]

where, for $1\leq i\leq m$, ${\mathfrak{q}_{i}}\subset R$ is the linear prime ideal of the point $Q_{i}$.

Remark 2.2**.**

Let $\ell\in\Sigma$, and denote by $\mathcal{C}^{\prime}$ the linear code dual to $\Sigma\setminus\{\ell\}$. Denote $d:=d_{1}(\mathcal{C})$, and $d^{\prime}:=d_{1}(\mathcal{C}^{\prime})$. Also denote by $G$, respectively $G^{\prime}$, the corresponding dual generating matrices; $G^{\prime}$ is the matrix obtained from $G$ by removing the column dual to $\ell$. The following classical results happen (see [6]):

• (i)

If $\dim(\mathcal{C}^{\prime})=k-1$, then all the $n-1$ columns of $G^{\prime}$ will span a $k-1$ dimensional vector space, so $d=1$.

• (ii)

If $d\geq 2$, then $\dim(\mathcal{C}^{\prime})=k$, and $d^{\prime}$ equals $d$ or $d-1$.

• (iii)

Let $d\geq 2$. Suppose $d^{\prime}=d-1$, and suppose $\ell=\ell_{n}$ dual to the last column of $G$. Let ${\bf v^{\prime}}=(\beta_{1},\ldots,\beta_{d-1},0,\ldots,0)\in\mathbb{K}^{n-1}$ be a projective codeword of minimum weight of $\mathcal{C}^{\prime}$, and suppose it corresponds to $\underbrace{(\alpha_{1},\ldots,\alpha_{k})}_{Q}\cdot G^{\prime}={\bf v^{\prime}}$. Obviously, ${\bf v}:=({\bf v^{\prime}},\ell(Q))\in\mathbb{K}^{n}$ is a codeword of $\mathcal{C}$. If $\ell(Q)=0$, then ${\bf v}$ is a projective codeword of weight $d-1$ of $\mathcal{C}$; impossible. So $\ell(Q)\neq 0$, and ${\bf v}$ is a projective codeword of minimum weight (equal to $d$) of $\mathcal{C}$. So, if $d^{\prime}=d-1$, then

[TABLE]

where $(I_{d+1}(\mathcal{C}))^{\rm sat}={\mathfrak{q}_{1}}\cap\cdots\cap{\mathfrak{q}_{m}}$.

Theorem 2.3**.**

Let $\Sigma$ be a collection of $n$ linear forms in $R:=\mathbb{K}[x_{1},\ldots,x_{k}]$. Let $d$ be the minimum distance of the linear code dual to $\Sigma$. Then, $I_{d+1}(\Sigma)$ has linear graded free resolution.

Proof.

We will prove the result by induction on the pairs $(|\Sigma|,{\rm rk}(\Sigma))$, where $|\Sigma|\geq{\rm rk}(\Sigma)\geq 2$.

Base Cases. If ${\rm rk}(\Sigma)=2$, after a change of variables we can suppose that $\Sigma\subset\mathbb{K}[x_{1},x_{2}]$. [9, Theorem 2.2] proves this case.

If $|\Sigma|={\rm rk}(\Sigma)$ (which we can suppose it equals $k\geq 2$), then after a change of variables we can assume that $\Sigma=(x_{1},\ldots,x_{k})$. In this case $d=1$. From results concerning star configurations (see part (3) in the Introduction of [9]), $I_{a}(\Sigma)$ has linear graded free resolution for any $1\leq a\leq k$.

Induction Step. Suppose $|\Sigma|>{\rm rk}(\Sigma)=k\geq 3$. Let $\mathcal{C}$ be the linear code dual to $\Sigma$. So the length of $\mathcal{C}$ is $n$, the dimension is $k$, and the minimum distance is $d\geq 1$. The inductive hypotheses say that if $\ell\in\mathcal{C}$, and if $\mathcal{C}^{\prime}:=\mathcal{C}\setminus\{\ell\}$ has minimum distance $d^{\prime}$, then $I_{d^{\prime}+1}(\mathcal{C}^{\prime})$ has linear graded free resolution.

CLAIM 1: There exists $\ell\in\Sigma$ such that, with the notations above,

[TABLE]

Proof.

Suppose $d=1$. Then, there exists a codeword of $\mathcal{C}$ with exactly one nonzero entry (say, the $n$-th entry). Since this is a vector in $\mathbb{K}^{n}$ which is a linear combination of the $k$ rows of $G$, we can replace the last row of $G$ by this vector, and after a change of variables we can assume $\ell_{n}=x_{k}$. So, after a left-multiplication of $G$ by an invertible matrix, we can assume that $\ell_{1},\ldots,\ell_{n-1}$ are linear forms in variables $x_{1},\ldots,x_{k-1}$, and $\ell_{n}=x_{k}$.

Take $\ell=\ell_{n}=x_{k}$. Let $f\in I_{2}(\mathcal{C}):x_{k}$. Since $I_{2}(\mathcal{C})=x_{k}I_{1}(\mathcal{C}^{\prime})+I_{2}(\mathcal{C}^{\prime})$, we have $x_{k}(f-g)\in I_{2}(\mathcal{C}^{\prime})$, for some $g\in I_{1}(\mathcal{C}^{\prime})$. But from the change of variables above, $x_{k}$ is a nonzero divisor mod $I_{2}(\mathcal{C}^{\prime})$, so $f-g\in I_{2}(\mathcal{C}^{\prime})$. Since $I_{2}(\mathcal{C}^{\prime})\subset I_{1}(\mathcal{C}^{\prime})$, we have $I_{2}(\mathcal{C}):x_{k}\subset I_{1}(\mathcal{C}^{\prime})$. The inclusion $\supseteq$ is obvious, so for $d=1$ the claim is shown.

Suppose $d\geq 2$. Let $\ell$ be any element of $\Sigma$. From Remark 2.2 (ii), $\dim(\mathcal{C}^{\prime})=k$, and $d^{\prime}=d$ or $d^{\prime}=d-1$.

If $d^{\prime}=d$, then $I_{d}(\mathcal{C}^{\prime})={\mathfrak{m}}^{d}$. Since $I_{d+1}(\mathcal{C})\subset{\mathfrak{m}}^{d+1}$, we have that

[TABLE]

Since the other inclusion is obvious, the claim is shown for this situation.

Suppose $d^{\prime}=d-1$. We have that $I_{d+1}(\mathcal{C})\subseteq(I_{d+1}(\mathcal{C}))^{\rm sat}\cap{\mathfrak{m}}^{d+1}$. Therefore, from Remark 2.1,

[TABLE]

If $\ell\in{\mathfrak{q}_{i}}$, then ${\mathfrak{q}_{i}}:\ell=R$, and if $\ell\notin{\mathfrak{q}_{j}}$, then ${\mathfrak{q}_{j}}:\ell={\mathfrak{q}_{j}}$. Also, ${\mathfrak{m}}^{d+1}:\ell={\mathfrak{m}}^{d}$. All together, via Remark 2.2 (iii), give

[TABLE]

But, by inductive hypotheses, $I_{d^{\prime}+1}(\mathcal{C}^{\prime})$ has linear graded free resolution, and therefore, by Remark 1.1, the last intersection above equals $I_{d^{\prime}+1}(\mathcal{C}^{\prime})$ itself. So in this case we obtained that

[TABLE]

Since the other inclusion is obvious, we proved the claim for this remaining case as well. ∎

Let $\ell\in\Sigma$ be such that via CLAIM 1, $I_{d+1}(\mathcal{C}):\ell=I_{d}(\mathcal{C}^{\prime})$. Suppose $\ell=\ell_{n}$.

As we explained in the Introduction, we have the short exact of graded $R$-modules

[TABLE]

First we deal with the leftmost nonzero module.

• (a)

If $d=1$, then $I_{1}(\mathcal{C}^{\prime})=\langle\ell_{1},\ldots,\ell_{n-1}\rangle$, which is a linear prime ideal, hence it has a linear graded free resolution.

• (b)

If $d\geq 2$ (so $\dim(\mathcal{C}^{\prime})=k$), and $d^{\prime}=d$, then $I_{d}(\mathcal{C}^{\prime})={\mathfrak{m}}^{d}$, which has linear graded free resolution.

• (c)

If $d\geq 2$ and $d^{\prime}=d-1$, then $I_{d}(\mathcal{C}^{\prime})=I_{d^{\prime}+1}(\mathcal{C}^{\prime})$, which by inductive hypotheses has linear graded free resolution.

In conclusion,

[TABLE]

Now we deal with the rightmost nonzero module. We can suppose, after a change of variables, that $\ell=\ell_{n}=x_{k}$. Suppose that for some $m=0,\ldots,n-1$, after a permutation of the elements of $\Sigma$, and a rescaling, we have $\ell_{m+1}=\cdots=\ell_{n}=x_{k}$, and that for all $1\leq i\leq m$, $\ell_{i}=\bar{\ell_{i}}+c_{i}x_{k},c_{i}\in\mathbb{K}$, where $\bar{\ell_{i}}$ are linear forms in variables $x_{1},\ldots,x_{k-1}$.

Consider $\bar{\mathcal{C}}$ the linear code dual to $\bar{\Sigma}:=(\bar{\ell_{1}},\ldots,\bar{\ell_{m}})\subset\bar{R}:=\mathbb{K}[x_{1},\ldots,x_{k-1}]$. Let $\bar{G}$ be the corresponding generating matrix. Since we assumed that ${\rm rk}(\Sigma)=k$, then ${\rm rk}(\bar{\Sigma})={\rm rk}(\bar{G})=\dim(\bar{\mathcal{C}})=k-1$. The length of $\bar{\mathcal{C}}$ is $m$. Let $\bar{d}:=d_{1}(\bar{\mathcal{C}})$ be the minimum distance of $\bar{\mathcal{C}}$.

We have that $u:=m-\bar{d}$ is the maximum number of columns of $\bar{G}$ that span an $(k-1)-1=k-2$ vector space. WLOG, suppose these are the first $u$ columns of $\bar{G}$. Then ${\rm ht}(\langle\bar{\ell_{1}},\ldots,\bar{\ell_{u}}\rangle)=k-2$. Since $x_{k}$ is a nonzero divisor mod $\langle x_{1},\ldots,x_{k-1}\rangle$, we have ${\rm ht}(\langle\bar{\ell_{1}},\ldots,\bar{\ell_{u}},x_{k}\rangle)=k-1$. But

[TABLE]

So the first $u$ and the last $n-m$ columns of $G$ span a $k-1$ dimensional vector space. Therefore

[TABLE]

leading to $\bar{d}\geq d$.

• (1)

If $\bar{d}\geq d+1$, then $I_{d+1}(\bar{\mathcal{C}})=\langle x_{1},\ldots,x_{k-1}\rangle^{d+1}$, which has a linear graded free resolution.

• (2)

If $\bar{d}=d$, then $I_{d+1}(\bar{\mathcal{C}})=I_{\bar{d}+1}(\bar{\mathcal{C}})$, which by inductive hypotheses has a linear graded free resolution.

To sum up we got

[TABLE]

But $R/\langle\ell,I_{d+1}(\mathcal{C})\rangle$ and $\bar{R}/I_{d+1}(\bar{\mathcal{C}})$ are isomorphic as $R$-modules, so they have the same regularity (see [3, Corollary 4.6]).

As we explained in the Introduction, we obtain that $I_{d+1}(\mathcal{C})$ has linear graded free resolution. ∎

3. Line arrangements in $\mathbb{P}^{2}$

In this section we assume that $\Sigma=(\ell_{1},\ldots,\ell_{n})\subset R:=\mathbb{K}[x,y,z]$, with $\gcd(\ell_{i},\ell_{j})=1$, for $i\neq j$. So $\ell_{1},\ldots,\ell_{n}$ define a line arrangement ${\mathcal{A}}\subset\mathbb{P}^{2}$; we will write $\Sigma={\mathcal{A}}=\{\ell_{1},\ldots,\ell_{n}\}\subset R$. Suppose ${\rm rk}({\mathcal{A}})=3$.

We will show that for any $1\leq a\leq n$, the ideal $I_{a}({\mathcal{A}})$ has linear graded free resolution. This is known to be true in various instances:

• (a)

If $a=n$, then $I_{n}({\mathcal{A}})=\langle\ell_{1}\cdots\ell_{n}\rangle$, which has linear graded free resolution.

• (b)

If $a=n-1$, then, by [7], $0\longrightarrow R(-n)^{n-1}\longrightarrow R(-n+1)^{n}\longrightarrow I_{n-1}({\mathcal{A}})\longrightarrow 0$ is a linear graded free resolution.

• (c)

If $a=n-2$, then [9, Theorem 2.4] shows that $I_{n-2}({\mathcal{A}})$ has linear graded free resolution.

• (d)

Let $\mathcal{C}$ be the linear code dual to ${\mathcal{A}}$. Let $d:=d_{1}(\mathcal{C})$. Then, as we observed, $n-d$ is the maximum number of columns of the generating matrix $G$ that span a $3-1=2$ dimensional vector space. So $m:=n-d$ is the maximum number of concurrent lines of ${\mathcal{A}}$. If $1\leq a\leq n-m$, then $I_{a}({\mathcal{A}})=\langle x,y,z\rangle^{a}$, which has linear graded free resolution.

• (e)

If $a=n-m+1$, by Theorem 2.3, $I_{n-m+1}({\mathcal{A}})$ has linear graded free resolution.

• (f)

If ${\mathcal{A}}$ is generic (i.e., if any three linear forms of ${\mathcal{A}}$ are linearly independent), then for any $1\leq a\leq n$, $I_{a}({\mathcal{A}})$ has linear graded free resolution. This is true from parts (a), (b), and (d) above, since $m=2$.

Suppose $Sing({\mathcal{A}}):=\{P_{1},\ldots,P_{s}\}\subset\mathbb{P}^{2}$ is the set of all the intersection points of the lines of ${\mathcal{A}}$. For $i=1,\ldots,s$, let ${\mathfrak{p}_{i}}\subset R$ be the defining ideal of $P_{i}$; also, let $n_{i}=n(P_{i})$ be the number of lines of ${\mathcal{A}}$ that intersect at the point $P_{i}$. Note that $m=\max\{n_{1},\ldots,n_{s}\}$, and $n_{i}\geq 2$ for all $i=1,\ldots,s$. If ${\mathcal{A}}$ is generic, then $n_{i}=2$ for all $i=1,\ldots,s$.

Lemma 3.1**.**

For any $1\leq b\leq m-1$ we have

[TABLE]

where if $n_{j}-b\leq 0$, then ${\mathfrak{p}_{j}}^{n_{j}-b}$ is by convention equal to $R$.

Proof.

Since $1\leq b\leq m-1$, then $n-m+1\leq n-b\leq n-1$. If $\mathcal{C}$ is the linear code dual to ${\mathcal{A}}$, we have $d_{1}(\mathcal{C})=n-m$, and $d_{2}(\mathcal{C})=n-1$. So $d_{1}(\mathcal{C})<n-b\leq d_{2}(\mathcal{C})$, which leads to ${\rm ht}(I_{n-b}({\mathcal{A}}))=2$. Since for all $j=1,\ldots,s$, $(n-b)-n+n_{j}=n_{j}-b$, and since $I_{n-b}({\mathcal{A}})=(I_{n-b}({\mathcal{A}}))^{\rm sat}\cap J$, where $J$ is an ideal which is $\langle x,y,z\rangle$-primary, [1, Proposition 2.3] proves the result. ∎

Theorem 3.2**.**

Let ${\mathcal{A}}=\{\ell_{1},\ldots,\ell_{n}\}\subset R:=\mathbb{K}[x,y,z]$ be a line arrangement of rank 3. Then, for any $1\leq a\leq n$, the ideal $I_{a}({\mathcal{A}})$ has linear graded free resolution.

Proof.

If $m=\max\{n_{1},\ldots,n_{s}\}$, from parts (a) and (d) above we can suppose $n-m+1\leq a\leq n-1$. Let $b:=n-a$, so $1\leq b\leq m-1$. Since ${\rm rk}({\mathcal{A}})=3$, then $m<n$, so $b\leq n-2$.

We will show that $I_{n-b}({\mathcal{A}})$ has linear graded free resolution, by induction on $n=|{\mathcal{A}}|\geq 3$.

Base Case. If $n=3$, then part (f) above proves the claim.

Inductive Step. Suppose $n>3$. Let $\ell\in{\mathcal{A}}$, and let ${\mathcal{A}}^{\prime}={\mathcal{A}}\setminus\{\ell\}$.

CLAIM 2: We have $I_{n-b}({\mathcal{A}}):\ell=I_{n-b-1}({\mathcal{A}}^{\prime})$.

Proof.

If ${\rm rk}({\mathcal{A}}^{\prime})=2$, after a change of variables we can suppose $\ell=\ell_{n}=z$, and ${\mathcal{A}}^{\prime}=\{\ell_{1},\ldots,\ell_{n-1}\}\subset\mathbb{K}[x,y]$. Same argument as in the beginning of proof of CLAIM 1, we have

[TABLE]

The last equality comes from $1\leq n-b-1\leq n-2$, and from the fact that the minimum distance of the linear code dual to ${\mathcal{A}}^{\prime}$ is $n-1-1=n-2$.

Suppose ${\rm rk}({\mathcal{A}}^{\prime})=3$. Suppose $\ell$ passes through $P_{1},\ldots,P_{r}$, and avoids $P_{r+1},\ldots,P_{s}$. Since ${\rm rk}({\mathcal{A}})=3$, then $r\geq 2$. We have $I_{n-b}({\mathcal{A}})\subseteq(I_{n-b}({\mathcal{A}}))^{\rm sat}\cap\langle x,y,z\rangle^{n-b}$. Therefore

[TABLE]

If $\ell\in{\mathfrak{p}_{j}}$, then ${\mathfrak{p}_{j}}^{n_{j}-b}:\ell={\mathfrak{p}_{j}}^{n_{j}-b-1}$; and if $\ell\notin{\mathfrak{p}_{j}}$, then ${\mathfrak{p}_{j}}^{n_{j}-b}:\ell={\mathfrak{p}_{j}}^{n_{j}-b}$.

For $i=1,\ldots,r$, there are $n_{i}-1$ lines of ${\mathcal{A}}^{\prime}$ passing through $P_{i}$, and for $j=r+1,\ldots,s$, there are $n_{j}$ lines of ${\mathcal{A}}^{\prime}$ passing through $P_{j}$. If for some $i\in\{1,\ldots,r\}$, $n_{i}=2$, then $P_{i}$ doesn’t show up in $Sing({\mathcal{A}}^{\prime})$. But at the same time, $n_{i}-b-1\leq 0$, so by Lemma 3.1 we can conclude

[TABLE]

By inductive hypotheses and Remark 1.1, $I_{n-1-b}({\mathcal{A}}^{\prime})=(I_{n-1-b}({\mathcal{A}}^{\prime}))^{\rm sat}\cap\langle x,y,z\rangle^{n-1-b}$, so we obtained

[TABLE]

Since the other inclusion is obvious, the claim is shown. ∎

The equality in CLAIM 2 gives the short exact sequence:

[TABLE]

We can suppose $\ell=\ell_{n}=z$. Then $\langle\ell,I_{n-b}({\mathcal{A}})\rangle=\langle z,I_{n-b}(\bar{{\mathcal{A}}})\rangle$, where $\bar{{\mathcal{A}}}=(\bar{\ell_{1}},\ldots,\bar{\ell_{n-1}})\subset\mathbb{K}[x,y]$, and $\ell_{i}\equiv\bar{\ell_{i}}\mbox{ mod }z$. Some of the linear forms of $\bar{{\mathcal{A}}}$ may be proportional. Nonetheless, by [9, Theorem 2.2], $I_{n-b}(\bar{{\mathcal{A}}})$ has linear graded free resolution, and so does the rightmost nonzero module in the short exact sequence above.

About the leftmost nonzero module in the short exact sequence, if ${\rm rk}({\mathcal{A}}^{\prime})=2$, we saw at the beginning of the proof of CLAIM 2, that $I_{n-1-b}({\mathcal{A}}^{\prime})$ has linear graded free resolution. If ${\rm rk}({\mathcal{A}}^{\prime})=3$, then by inductive hypotheses, $I_{n-1-b}({\mathcal{A}}^{\prime})$ has linear graded free resolution. Therefore, by the strategy mentioned in the Introduction, $I_{n-b}({\mathcal{A}})$ also has linear graded free resolution. ∎

4. Graded betti numbers

4.1. The case $k=2$.

First suppose $\Sigma=(\underbrace{\ell_{1},\ldots,\ell_{1}}_{m_{1}},\ldots,\underbrace{\ell_{t},\ldots,\ell_{t}}_{m_{t}})\subset\overline{R}:=\mathbb{K}[x,y]$, with $t\geq 1$, and $\gcd(\ell_{i},\ell_{j})=1$. Let $1\leq a\leq v:=m_{1}+\cdots+m_{t}$, and for $i=1,\ldots,t$, let $d_{i}:=\max\{m_{i}+a-v,0\}$. Then, by [1, Proposition 2.3] and [9, Lemma 2.1]222In the pre-published version, this lemma states that if $n_{1}=\max\{n_{1},\ldots,n_{m}\}$, then, for $1\leq a\leq n-n_{1}$, $I_{a}(\Sigma)=\langle x,y\rangle^{a}$, and for $n\geq a\geq n-n_{1}+1$, $I_{a}(\Sigma)=\ell_{1}I_{a-1}(\Sigma\setminus\{\ell_{1}\})$.,

[TABLE]

where $e=\max\{a-(d_{1}+\cdots+d_{t}),0\}$. Then, the minimal graded free resolution of $\overline{R}/I_{a}(\Sigma)$ is

[TABLE]

so the graded betti numbers are $b_{1}(a,\Sigma)=e+1$, $b_{2}(a,\Sigma)=e$, and $b_{3}(a,\Sigma)=0$.

4.2. The case $k=3$.

Let ${\mathcal{A}}=\{\ell_{1},\ldots,\ell_{n}\}\subset R:=\mathbb{K}[x,y,z]$ be a line arrangement in $\mathbb{P}^{2}$, of rank 3. Let ${\mathfrak{m}}:=\langle x,y,z\rangle$.

Suppose $Sing({\mathcal{A}}):=\{P_{1},\ldots,P_{s}\}\subset\mathbb{P}^{2}$ is the set of all the intersection points of the lines of ${\mathcal{A}}$. For $i=1,\ldots,s$, let ${\mathfrak{p}_{i}}\subset R$ be the defining ideal of $P_{i}$; also, let $n_{i}=n(P_{i})\geq 2$ be the number of lines of ${\mathcal{A}}$ that intersect at the point $P_{i}$, and let $m:=\max\{n_{1},\ldots,n_{s}\}$.

For $1\leq a\leq n$, let

[TABLE]

be the minimal (linear) graded free resolution of $R/I_{a}({\mathcal{A}})$.

• (i)

As we already discussed, if $1\leq a\leq n-m$, then $I_{a}({\mathcal{A}})={\mathfrak{m}}^{a}$, and by [5, Proposition 1.7(c)], the graded betti numbers of $R/I_{a}({\mathcal{A}})$ are

[TABLE]

• (ii)

If $a=n$, then $I_{n}({\mathcal{A}})=\langle\ell_{1}\cdots\ell_{n}\rangle$, and the graded betti numbers of $R/I_{n}({\mathcal{A}})$ are

[TABLE]

• (iii)

If $a=n-1$, by [7, Lemma 3.2], the graded betti numbers of $R/I_{n-1}({\mathcal{A}})$ are

[TABLE]

• (iv)

If $a=n-2$, [9, Theorem 2.4] presents the minimal free resolution of $R/I_{n-2}({\mathcal{A}})$, and hence its graded betti numbers.

• (v)

If ${\mathcal{A}}$ is generic, then $m=2$, and therefore items (i), (ii), and (iii) give the graded betti numbers of $R/I_{a}({\mathcal{A}})$ for any $1\leq a\leq n$.

Remark 4.1**.**

From the itemized list above, what is left to analyze are the graded betti numbers of $R/I_{a}({\mathcal{A}})$, for $n-m+1\leq a\leq n-2$, and when ${\mathcal{A}}$ is not generic. In this case we have ${\rm ht}(I_{a}({\mathcal{A}}))=2$. We also have that $R/I_{a}({\mathcal{A}})$ is not Cohen-Macaulay. To show this we use the same trick as in proof of [8, Proposition 2.4]: let $\ell_{1},\ell_{2},\ell_{3}\in{\mathcal{A}}$ be such that $\langle\ell_{1},\ell_{2},\ell_{3}\rangle={\mathfrak{m}}$. Let $V(\ell)$ be some line that does not pass through $P_{1},\ldots,P_{s}$. So $\ell$ is a nonzero divisor mod $(I_{a}({\mathcal{A}}))^{\rm sat}$. Also $\ell=c_{1}\ell_{1}+c_{2}\ell_{2}+c_{3}\ell_{3}$, for some constants $c_{1},c_{2},c_{3}\in\mathbb{K}$.

Since $a-1\leq n-3$, there exist $\ell_{i_{1}},\ldots,\ell_{i_{a-1}}\in{\mathcal{A}}\setminus\{\ell_{1},\ell_{2},\ell_{3}\}$. Then

[TABLE]

Since $\ell$ is a nonzero divisor, we obtain $\ell_{i_{1}}\cdots\ell_{i_{a-1}}\in(I_{a}({\mathcal{A}}))^{\rm sat}$, which is an element of degree $a-1$. So $I_{a}({\mathcal{A}})\subsetneq(I_{a}({\mathcal{A}}))^{\rm sat}$.

To conclude the remark we have that in these conditions $b_{3}(a,{\mathcal{A}})\geq 1$.

Let $\ell\in{\mathcal{A}}$, and consider ${\mathcal{A}}^{\prime}:={\mathcal{A}}\setminus\{\ell\}$. If ${\rm rk}({\mathcal{A}}^{\prime})=3$, by Theorem 3.2, $I_{a-1}({\mathcal{A}}^{\prime})$ has linear graded free resolution, CLAIM 2 in the proof of that theorem is satisfied:

[TABLE]

If ${\rm rk}({\mathcal{A}}^{\prime})=2$, then CLAIM 2 is satisfied as we can see at the beginning of its proof.

As we have seen over and over again, this leads to the short exact sequence

[TABLE]

$\bullet$ The leftmost nonzero module has minimal graded free resolution

[TABLE]

$\bullet$ Concerning the rightmost nonzero module, suppose $\ell=z$ and suppose $\ell(P_{1})=\cdots=\ell(P_{t})=0$ and $\ell(P_{j})\neq 0,j=t+1,\ldots,s$. Then

[TABLE]

where $\displaystyle\overline{{\mathcal{A}}}:=(\underbrace{\bar{\ell_{1}},\ldots,\bar{\ell_{1}}}_{n_{1}-1},\ldots,\underbrace{\bar{\ell_{t}},\ldots,\bar{\ell_{t}}}_{n_{t}-1})\subset\overline{R}:=\mathbb{K}[x,y]$. Then, the rightmost nonzero module has minimal free resolution

[TABLE]

where $b_{1}(a,\overline{{\mathcal{A}}})$ and $b_{2}(a,\overline{{\mathcal{A}}})$ are determined according to Subsection 4.1.

Applying mapping cone technique, after appropriate cancellations (or by using loosely [5, Lemma 1.13]), we obtain

• (A)

$b_{3}(a,{\mathcal{A}})=b_{3}(a-1,{\mathcal{A}}^{\prime})+b_{2}(a,\overline{{\mathcal{A}}})$.

• (B)

$b_{2}(a,{\mathcal{A}})=b_{2}(a-1,{\mathcal{A}}^{\prime})+b_{2}(a,\overline{{\mathcal{A}}})+b_{1}(a,\overline{{\mathcal{A}}})$.

• (C)

$b_{1}(a,{\mathcal{A}})=b_{1}(a-1,{\mathcal{A}}^{\prime})+b_{1}(a,\overline{{\mathcal{A}}})$.

Example 4.2**.**

Consider the line arrangement ${\mathcal{A}}=\{x,x-z,x+z,z,y,y-z\}$. We have $n=6$ and $m=4$, so we are left to analyze the graded betti numbers when $a=3$, and $a=4$.

To be as explicit as possible, for each of these two case we will pick different sets of lines that will be recursively deleted.

$\boxed{a=3}$

$\bullet$ Take $\ell=y$. Then ${\mathcal{A}}^{\prime}=\{x,x-z,x+z,z,y-z\}$, and $\overline{{\mathcal{A}}}\approx(z,z,x+z,x,x-z)$ with $b_{1}(3,\overline{{\mathcal{A}}})=4$, $b_{2}(3,\overline{{\mathcal{A}}})=3$. From (A), (B), (C) we have

[TABLE]

$\bullet$ Take $\ell^{\prime}=y-z$. Then ${\mathcal{A}}^{\prime\prime}:={\mathcal{A}}^{\prime}\setminus\{\ell^{\prime}\}=\{x,x-z,x+z,z\}$, and $\overline{{\mathcal{A}}^{\prime}}\approx(x,x+y,x-y,y)$. Then we have $b_{1}(1,{\mathcal{A}}^{\prime\prime})=2$, $b_{2}(1,{\mathcal{A}}^{\prime\prime})=1$, $b_{3}(1,{\mathcal{A}}^{\prime\prime})=0$, and $b_{1}(2,\overline{{\mathcal{A}}^{\prime}})=3$, $b_{2}(2,\overline{{\mathcal{A}}^{\prime}})=2$. From (A), (B), (C) we have

[TABLE]

leading to

[TABLE]

$\boxed{a=4}$

$\bullet$ Take $\ell=z$. Then ${\mathcal{A}}^{\prime}=\{x,x-z,x+z,y-z,y\}$, and $\overline{{\mathcal{A}}}\approx(x,x,x,y,y)$ with $b_{1}(4,\overline{{\mathcal{A}}})=2$, $b_{2}(4,\overline{{\mathcal{A}}})=1$. From (A), (B), (C) we have

[TABLE]

$\bullet$ Take $\ell^{\prime}=x$. Then ${\mathcal{A}}^{\prime\prime}:={\mathcal{A}}^{\prime}\setminus\{\ell^{\prime}\}=\{x-z,x+z,y-z,y\}$, and $\overline{{\mathcal{A}}^{\prime}}\approx(z,z,y-z,y)$. We have that ${\mathcal{A}}^{\prime\prime}$ is generic, with $1\leq 2\leq 4-2$, so from (i) above we have $b_{1}(2,{\mathcal{A}}^{\prime\prime})=6$, $b_{2}(2,{\mathcal{A}}^{\prime\prime})=8$, $b_{3}(2,{\mathcal{A}}^{\prime\prime})=3$. Also we have $b_{1}(3,\overline{{\mathcal{A}}^{\prime}})=3$, $b_{2}(3,\overline{{\mathcal{A}}^{\prime}})=2$. From (A), (B), (C) we have

[TABLE]

leading to

[TABLE]

Observe that we obtained, as expected, the same graded betti numbers in [9, Theorem 2.4] (with the notations of that theorem, we have $n=6,a=n-2=4,m=15,p({\mathcal{A}})=3+1=4$).