Linear resolution of products of monomial ideals related to maximal minors

TL;DR

This paper proves that products of certain monomial ideals related to maximal minors have linear resolutions, extending previous results with a new elementary and combinatorial proof approach.

Contribution

It introduces a new elementary and combinatorial proof showing that products of specific monomial ideals have linear resolutions, generalizing earlier theorems.

Findings

Products of the defined monomial ideals have linear free resolutions.

The proof is self-contained, elementary, and combinatorial.

The result extends a theorem by Bruns and Conca.

Abstract

Let $ X $ be an $ m \times n $ matrix of distinct indeterminates over a field $ K $, where $ m \le n $. Set the polynomial ring $ K[X] := K[X_{ij} : 1 \le i \le m, 1 \le j \le n] $. Let $ 1 \le k < l \le n $ be such that $ l - k + 1 \ge m $. Consider the submatrix $ Y_{kl} $ of consecutive columns of $ X $ from $ k $th column to $ l $th column. Let $ J_{kl} $ be the ideal generated by `diagonal monomials' of all $ m \times m $ submatrices of $ Y_{kl} $, where the diagonal monomial of a square matrix means product of its main diagonal entries. We show that $ J_{k_1 l_1} J_{k_2 l_2} \cdots J_{k_s l_s} $ has a linear free resolution, where $ k_1 \le k_2 \le \cdots \le k_s $ and $ l_1 \le l_2 \le \cdots \le l_s $. This result is a variation of a theorem due to Bruns and Conca. Moreover, our proof is self-contained, elementary and combinatorial.

Equations

$$X_{11}>X_{12}>\cdots >X_{1n}>X_{21}>X_{22}>\cdots >X_{2n}>\cdots >X_{m1}>\cdots >X_{mn}.$$

$$X_t(a):=(X_{ij}:1⩽i⩽t,a⩽j⩽n).$$

$$I=⟨X_1{X}_2,X_1{X}_3,X_2^2,X_2{X}_3⟩=J.$$

$$\left[\begin{array}{cccccccc}{X}_{11}& {\mathbf{X}}_{12}& {\mathbf{X}}_{13}& {\mathbf{X}}_{14}& {\mathbf{X}}_{15}& {\mathbf{X}}_{16}& X_{17}& X_{18}\\ X_{21}& {\mathbf{X}}_{22}& {\mathbf{X}}_{23}& {\mathbf{X}}_{24}& {\mathbf{X}}_{25}& {\mathbf{X}}_{26}& X_{27}& X_{28}\\ X_{31}& {\mathbf{X}}_{32}& {\mathbf{X}}_{33}& {\mathbf{X}}_{34}& {\mathbf{X}}_{35}& {\mathbf{X}}_{36}& X_{37}& X_{38}\end{array}\right].$$

$$\displaystyle J_{2\,6}=\big{(}$$

$$\displaystyle X_{12}X_{25}X_{36},\,X_{13}X_{24}X_{35},\,X_{13}X_{24}X_{36},\,X_{13}X_{25}X_{36},\,X_{14}X_{25}X_{36}\big{)}.$$

$$(⟨f_1⟩:f_2)=(X_{34}),(⟨f_1,f_2⟩:f_3)=(X_{34},X_{35}),(⟨f_1,f_2,f_3⟩:f_4)=(X_{23}),$$

$$(⟨f_1,\dots ,f_4⟩:f_5)=(X_{23},X_{35}),(⟨f_1,\dots ,f_5⟩:f_6)=(X_{23},X_{24}),$$

$$(⟨f_1,\dots ,f_6⟩:f_7)=(X_{12}),(⟨f_1,\dots ,f_7⟩:f_8)=(X_{12},X_{35}),$$

$$(⟨f_1,\dots ,f_8⟩:f_9)=(X_{12},X_{24}),(⟨f_1,\dots ,f_9⟩:f_{10})=(X_{12},X_{13}),$$

$$(⟨f_1,\dots ,f_u⟩:f_{u+1})=⟨X_{1b_1},X_{2b_2},\dots ,X_{m{b}_m}:c_{i-1}<b_i<c_i,1⩽i⩽m⟩.$$

$$g:=X_{1c_1}\cdots X_{i-1c_{i-1}}{X}_{i{b}_i}{X}_{i+1c_{i+1}}\cdots X_{m{c}_m}.$$

$$X_{i{b}_i}\in \left(⟨g⟩:f_{u+1}\right)\subseteq (⟨f_1,\dots ,f_u⟩:f_{u+1}).$$

$$(⟨J_{k_1{l}_1}{J}_{k_2{l}_2}\cdots J_{k_s{l}_s},f_1,\dots ,f_u⟩:f_{u+1})=$$

$$⟨J_{k_2{l}_2}\cdots J_{k_s{l}_s},X_{1b_1},X_{2b_2},\dots ,X_{m{b}_m}:c_{i-1}<b_i<c_i,1⩽i⩽m⟩.$$

$$(J:f_{u+1})\subseteq L,\text{mbox}whereJ=J_{k_1{l}_1}{J}_{k_2{l}_2}\cdots J_{k_s{l}_s}.$$

$$f\in ⟨X_{1b_1},X_{2b_2},\dots ,X_{m{b}_m}:c_{i-1}<b_i<c_i,1⩽i⩽m⟩\text{mbox}or$$

$$g_1{g}_2\cdots g_s=h_1{h}_2\cdots h_s\text{mbox}forsomediagonalmonomials$$

$$\displaystyle\quad\quad\quad\quad h_{j}\in J_{k_{j}l_{j}}\;(1\leqslant j\leqslant s)\mbox{ such that $h_{1}$ is divisible by $g$}.$$

$$g_j=X_{1b_1}{X}_{2b_2}\cdots X_{i_{l-1}{b}_{i_{l-1}}}\cdots X_{i_l{b}_{i_l}}\cdots X_{m{b}_m}$$

$${\mathcal{A}}_j$$

$${\mathcal{B}}_j$$

$${\mathcal{C}}_j$$

$${\mathcal{A}}_j={\mathcal{B}}_j\cap {\mathcal{C}}_j;{\mathcal{B}}_1\subseteq {\mathcal{B}}_2\subseteq \cdots \subseteq {\mathcal{B}}_s\text{mbox}and{\mathcal{C}}_1\supseteq {\mathcal{C}}_2\supseteq \cdots \supseteq {\mathcal{C}}_s.$$

$$\mbox{the circles correspond to $g_{j}\in J_{k_{j}l_{j}}$ are placed in $\mathcal{A}_{j}\;(=\mathcal{B}_{j}\cap\mathcal{C}_{j})$}.$$

$$\mbox{the $j$th circle of every row of $X$ is placed in $\mathcal{A}_{j}$}.$$

$$f_{u+1}$$

$$g_2$$

$$g_4$$

$$h_1$$

$$h_3$$

$$h_5$$

$$J_{28}=(X_{12}{X}_{23}{X}_{34},X_{12}{X}_{23}{X}_{35},\dots ,X_{14}{X}_{25}{X}_{38},X_{14}{X}_{26}{X}_{37},\dots ,X_{16}{X}_{27}{X}_{38}),$$

$$(J_{28}{J}_{37},X_{12}{X}_{23}{X}_{34},\dots ,X_{14}{X}_{25}{X}_{38}:X_{14}{X}_{26}{X}_{37})\ne J_{37}+(X_{12},X_{13},X_{25}).$$

$$0⟶\frac{K[X]}{(J:f_1)}(-m)\stackrel{f_1}{⟶}$$

$$0⟶\frac{K[X]}{(⟨J,f_1⟩:f_2)}(-m)\stackrel{f_2}{⟶}$$

$$⋮$$

$$0⟶\frac{K[X]}{(⟨J,f_1,\dots ,f_{r-1}⟩:f_r)}(-m)\stackrel{f_r}{⟶}$$

Full text

The third named author would like to thank the Institute of Mathematical Sciences, Chennai for providing postdoctoral fellowship for this study.

1. Introduction

The study of determinantal ideals, rings and varieties is a classical topic in commutative algebra, algebraic geometry and invariant theory (see [8]). One important method to study determinantal ideals is to understand their initial ideals via Gröbner basis. Let $K$ be a field, and $X$ be an $m\times n$ matrix of distinct indeterminates over $K$, where $m\leqslant n$. Set the polynomial ring $K[X]:=K[X_{ij}:1\leqslant i\leqslant m,1\leqslant j\leqslant n]$. Let $I_{t}(X)$ be the ideal generated by all $t\times t$ minors of $X$. The homological properties of these ideals as well as Gröbner bases and initial ideals with respect to diagonal (or antidiagonal) monomial orders are well understood. Among these ideals of minors the best-behaved is the ideal of maximal minors, namely the ideal $I_{m}(X)$. Let $\tau$ be a diagonal monomial order. For example, $\tau$ can be the lexicographic term order on $K[X]$ induced by the order

[TABLE]

It is well known that all the $m\times m$ minors of $X$ form a Gröbner basis of $I_{m}(X)$ with respect to $\tau$. Indeed, it is proved in [4, Theorem 0] and [23, Corollary 7.6], and generalized in [10, Theorem 2.1] that the maximal minors of $X$ form a universal Gröbner basis (i.e., a Gröbner basis with respect to every monomial order). Regarding powers of $I_{m}(X)$, in [9, Theorem 2.1], Conca proved that the natural generators of $I_{m}(X)^{s}$ form a Gröbner basis with respect to $\tau$, and $\operatorname{in}_{\tau}(I_{m}(X)^{s})=\operatorname{in}_{\tau}(I_{m}(X))^{s}$ for every $s\geqslant 1$. For every $s\geqslant 1$; see also [5, Theorem 3.10] for the similar result on arbitrary determinantal ideals $I_{t}(X)$. But, for $m>2$ and $s>1$, the natural generators of $I_{m}(X)^{s}$ do not necessarily form a universal Gröbner basis, due to Speyer and Sturmfels [22, Corollary 5.6]. In other words, there are monomial orders $<$ such that $\operatorname{in}_{<}(I_{m}(X)^{s})$ is strictly larger than $\operatorname{in}_{<}(I_{m}(X))^{s}$.

The study of Castelnuovo-Mumford regularity (or simply, regularity) of powers and products of ideals in polynomial rings has been a central problem in commutative algebra and algebraic geometry. One important result in this direction was given by Cutkosky, Herzog and Trung [12], and independently by Kodiyalam [19]. They proved that if $I$ is a homogeneous ideal of $K[x_{1},\ldots,x_{d}]$, then the regularity of $I^{s}$ is asymptotically a linear function in $s$. This linear function behaves in the simplest possible way when $I$ is generated in degree $m$, and all its powers have linear free resolution, i.e., $\operatorname{reg}(I^{s})=m\cdot s$ for all $s$. In general, powers of an ideal with a linear free resolution need not have linear free resolution (see [20, Counterexample 1.10]). Recently, a number of authors have been interested in classifying or identifying families of ideals whose powers and products have linear free resolution. For example, Fröberg [13] characterized all squarefree monomial ideals generated by quadratic monomials, which have linear free resolution. Later, Herzog, Hibi and Zheng [18] proved that a monomial ideal $I$ generated in degree $2$ has a linear free resolution if and only if every power of $I$ has a linear free resolution. If $I$ is a polymatroidal ideal, then all powers have linear free resolution, [17, Corollary 12.6.4]. It was proved by Akin, Buchsbaum and Weyman [1, Theorem 5.4] that all powers of determinantal ideals of maximal minors of a generic matrix have linear free resolution. Recently, in [21, Theorem 5.1], Raicu classified the determinantal ideals of a generic matrix with all powers having linear free resolution. We are interested to find a family of monomial ideals (related to determinantal ideals) whose powers and products have linear free resolution.

The asymptotic behavior of regularity also has been studied for powers of more than one ideals in [2], [7] and [14]. For determinantal ideals, Conca and Herzog in [11, Theorem 6.1] showed that products of ideals of minors of a Hankel matrix have linear free resolution. Berget, Bruns and Conca [3, Theorem 4.7] proved an extension of [1, Theorem 5.4] to arbitrary products of the ideals $I_{t}(X_{t})$, where $X_{t}$ is the submatrix of the first $t$ rows of $X$. They proved that $I$ and $\operatorname{in}_{\tau}(I)$ have linear free resolutions, where $I:=I_{t_{1}}(X_{t_{1}})\cdots I_{t_{w}}(X_{t_{w}})$. In [6, Theorems 1.3 and 5.3], Bruns and Conca generalized this result further to a class of ideals that are fixed by the Borel group. They defined the northeast ideals $I_{t}(a)$ of maximal minors: $I_{t}(a)$ is generated by the $t\times t$ minors of the $t\times(n-a+1)$ northeast submatrix

[TABLE]

They proved that, if $I_{t_{1}}(a_{1}),\ldots,I_{t_{w}}(a_{w})$ are northeast ideals of maximal minors, and $I:=I_{t_{1}}(a_{1})\cdots I_{t_{w}}(a_{w})$, then $I$ and $\operatorname{in}_{\tau}(I)$ have linear free resolutions. The aim of this article is to prove a variation of above results. To state our result, we introduce a few notations.

1.1**.**

Let $1\leqslant k<l\leqslant n$ be such that $l-k+1\geqslant m$. We denote the submatrix $Y_{kl}$ of consecutive columns of $X$ from $k$th column to $l$th column. Denote by $[c_{1},\ldots,c_{m}]$ the $m\times m$ minor $\det(X_{ic_{j}})_{1\leqslant i,j\leqslant m}$ of $Y_{kl}$, where $k\leqslant c_{1}<c_{2}<\cdots<c_{m}\leqslant l$. The diagonal of $[c_{1},\ldots,c_{m}]$ is defined to be the set $\{X_{1c_{1}},\ldots,X_{mc_{m}}\}$. We call the product $X_{1c_{1}}\cdots X_{mc_{m}}$ as a diagonal monomial of $Y_{kl}$. Note that by a diagonal monomial, we always mean product of main diagonal entries of some maximal minor, not any arbitrary minor. Let $I_{kl}$ be the ideal of $K[X]$ generated by all $m\times m$ minors of $Y_{kl}$. Let $J_{kl}$ be the ideal of $K[X]$ generated by all diagonal monomials (of maximal minors) of $Y_{kl}$; see Example 2.1.

The main result of this paper is the following:

Theorem 2.10. Let $1\leqslant k_{1}\leqslant k_{2}\leqslant\cdots\leqslant k_{s}<n$ and $1<l_{1}\leqslant l_{2}\leqslant\cdots\leqslant l_{s}\leqslant n$ be such that $k_{i}<l_{i}$ and $l_{i}-k_{i}+1\geqslant m$. Along with the notations as in 1.1, we set $J:=J_{k_{1}l_{1}}J_{k_{2}l_{2}}\cdots J_{k_{s}l_{s}}$. Then $J$ has a linear free resolution.

In order to prove the main result, we study certain colon ideals related to $J$. We show that colon ideals take very interesting forms, and they behave nicely for regularity. Using these results and various short exact sequences, we obtain the main result. Our proof is self-contained, elementary and combinatorial. We finish this section by presenting a conjecture based on the above results and some computational evidence. In particular, this conjecture generalizes [1, Theorem 5.4] and [9, Theorem 2.1], since $I_{m}(X)=I_{1n}$.

Conjecture 1.2**.**

Let $1\leqslant k_{1}\leqslant k_{2}\leqslant\cdots\leqslant k_{s}<n$ and $1<l_{1}\leqslant l_{2}\leqslant\cdots\leqslant l_{s}\leqslant n$ be such that $k_{i}<l_{i}$ and $l_{i}-k_{i}+1\geqslant m$. Set $I:=I_{k_{1}l_{1}}I_{k_{2}l_{2}}\cdots I_{k_{s}l_{s}}$ and $J:=J_{k_{1}l_{1}}J_{k_{2}l_{2}}\cdots J_{k_{s}l_{s}}$. Then

1. (i)

The initial ideal $\operatorname{in}_{\tau}(I)=\operatorname{in}_{\tau}(I_{k_{1}l_{1}})\operatorname{in}_{\tau}(I_{k_{2}l_{2}})\cdots\operatorname{in}_{\tau}(I_{k_{s}l_{s}})=J$, and the natural generators of $I$ form a Gröbner basis with respect to $\tau$. 2. (ii)

It follows from (i) and Theorem 2.10 that $I$ has a linear free resolution.

Remark 1.3*.*

It should be observed that Theorem 2.10 and the statements in Conjecture 1.2 do not follow from the results of Bruns and Conca [6, Theorems 1.3 and 5.3]. For example, consider the generic matrix $[X_{1}\;X_{2}\;X_{3}]$. Then the product of ideals $I_{12}:=\langle X_{1},X_{2}\rangle=:J_{12}$ and $I_{23}:=\langle X_{2},X_{3}\rangle=:J_{23}$ is

[TABLE]

Clearly, $I$ cannot be written as product of northeast ideals of maximal minors considered in [6, Theorems 1.3 and 5.3].

2. Main Result

In order to prove Theorem 2.10, we first show that the monomial ideal $J_{kl}$ as described in 1.1 has a linear free resolution by proving that the ideal has linear quotients. Recall that an ideal $I$ is said to have linear quotients if there exists an ordered set of generators $f_{1},\ldots,f_{r}$ of $I$ such that $(\langle f_{1},\ldots,f_{u}\rangle:f_{u+1})$ is generated by linear forms for every $1\leqslant u\leqslant r-1$. Before proving that $J_{kl}$ has linear quotients, we illustrate an example for reader’s convenience.

Example 2.1**.**

Let $X$ be a $3\times 8$ matrix:

[TABLE]

Setting $k=2$ and $l=6$, the submatrix $Y_{k\,l}=Y_{2\,6}$ is given by the bold faced entries, and the corresponding ideal $J_{k\,l}$ has the form:

[TABLE]

Denoting the generators of $J_{2\,6}$ by $f_{1},f_{2},\ldots,f_{10}$ (in order), it can be verified in Macaulay2 [15] that

[TABLE]

which shows that $J_{2\,6}$ has linear quotients.

2.2**.**

Throughout, we use the following elementary facts without giving any references. Consider some monomials $g_{1},\ldots,g_{u},g$ in the polynomial ring $K[X]$.

1. (i)

The monomial $g\in\langle g_{1},\ldots,g_{u}\rangle$ if and only if it is divisible by one of $g_{1},\ldots,g_{u}$. 2. (ii)

The colon ideal $(\langle g_{1},\ldots,g_{u}\rangle:g)$ is a monomial ideal. 3. (iii)

Let $I_{1},\ldots,I_{s}$ be monomial ideals. Then \big{(}\big{(}\sum_{v=1}^{s}I_{v}\big{)}:g\big{)}=\sum_{v=1}^{s}(I_{v}:g).

Denote the ideal of $K[Y_{kl}]$ generated by all diagonal monomials of $Y_{kl}$ by the same notation $J_{kl}$. It can be observed that having linear quotients of $J_{kl}$ as an ideal of $K[X]$ or as an ideal of $K[Y_{kl}]$ are equivalent. So in order to show that $J_{kl}$ has linear quotients (as an ideal of $K[X]$), it suffices to prove the following:

Lemma 2.3**.**

Let $\{f_{1},\ldots,f_{r}\}$ be the collection of all diagonal monomials of $X$ ordered by $\tau$, i.e., $f_{1}>\cdots>f_{r}$ and $J=\langle f_{1},\ldots,f_{r}\rangle$. Then the colon ideal $(\langle f_{1},\ldots,f_{u}\rangle:f_{u+1})$ is generated by some variables for every $1\leqslant u\leqslant r-1$. Indeed, if $f_{u+1}=X_{1c_{1}}X_{2c_{2}}\cdots X_{mc_{m}}$, setting $c_{0}=0$, we have

[TABLE]

In the right hand side, $b_{i}$ is varying in between $c_{i-1}$ and $c_{i}$ for every $1\leqslant i\leqslant m$.

Proof.

Fix $i$ and $b_{i}$ such that $1\leqslant i\leqslant m$ and $c_{i-1}<b_{i}<c_{i}\;(<c_{i+1})$. Set

[TABLE]

Clearly, $g>f_{u+1}$, and hence $g$ is one of $f_{1},\ldots,f_{u}$. Therefore

[TABLE]

This proves ‘the containment $\supseteq$’. For another containment, it is enough to show that, for every $1\leqslant v\leqslant u$, $(\langle f_{v}\rangle:f_{u+1})\subseteq(X_{ja_{j}})$ for some $1\leqslant j\leqslant m$ and $c_{j-1}<a_{j}<c_{j}$. Let $f_{v}=X_{1a_{1}}X_{2a_{2}}\cdots X_{ma_{m}}$. Since $f_{v}$ is a diagonal monomial of $X$, we have that $1\leqslant a_{1}<\cdots<a_{m}\leqslant n$. Set $a_{0}:=0$. Since $f_{v}>f_{u+1}$, there exists $1\leqslant j\leqslant m$ such that $a_{j}<c_{j}$ and $a_{l}=c_{l}$ for all $1\leqslant l<j$. Note that $c_{j-1}=a_{j-1}<a_{j}<c_{j}$. It can be easily shown that any monomial in $(\langle f_{v}\rangle:f_{u+1})$ is divisible by $X_{ja_{j}}$. Hence $(\langle f_{v}\rangle:f_{u+1})\subseteq(X_{ja_{j}})$, which completes the proof. ∎

As an immediate consequence, we obtain the following:

Corollary 2.4**.**

The ideal $J_{kl}$ of $K[X]$ as described in 1.1, has linear quotients, and hence it has linear free resolution.

Proof.

The corollary follows from Lemma 2.3 and [11, Lemma 4.1]. ∎

The following lemma is technical and most crucial in our study. For better understanding, we refer Example 2.6, which illustrates the proof of the lemma.

Lemma 2.5**.**

Let $1\leqslant k_{1}\leqslant k_{2}\leqslant\cdots\leqslant k_{s}<n$ and $1<l_{1}\leqslant l_{2}\leqslant\cdots\leqslant l_{s}\leqslant n$ be such that $k_{i}<l_{i}$ and $l_{i}-k_{i}+1\geqslant m$. Suppose $s\geqslant 2$. Along with the notations as in 1.1, we set $J:=J_{k_{1}l_{1}}J_{k_{2}l_{2}}\cdots J_{k_{s}l_{s}}$, where $J_{k_{1}l_{1}}$ is generated by the diagonal monomials $\{f_{1},\ldots,f_{r}\}$ of $Y_{k_{1}l_{1}}$ ordered by $\tau$, i.e., $f_{1}>\cdots>f_{r}$. Then, for every $0\leqslant u\leqslant r-1$, the ideal $(\langle J,f_{1},\ldots,f_{u}\rangle:f_{u+1})$ is generated by $J_{k_{2}l_{2}}\cdots J_{k_{s}l_{s}}$ and some variables. Indeed, if $f_{u+1}=X_{1c_{1}}X_{2c_{2}}\cdots X_{mc_{m}}$, setting $c_{0}=k_{1}-1$, we have

[TABLE]

In the right hand side, $b_{i}$ is varying in between $c_{i-1}$ and $c_{i}$ for every $1\leqslant i\leqslant m$.

Proof.

Set $L:=\left\langle J_{k_{2}l_{2}}\cdots J_{k_{s}l_{s}},X_{1b_{1}},X_{2b_{2}},\ldots,X_{mb_{m}}:c_{i-1}<b_{i}<c_{i},\;1\leqslant i\leqslant m\right\rangle$. In view of 2.2.(iii) and Lemma 2.3, we only need to prove that

[TABLE]

So we consider a monomial $f\in(J:f_{u+1})$. Hence, since $ff_{u+1}$ is a monomial, it can be written as $ff_{u+1}=g_{1}g_{2}\cdots g_{s}p$ for some diagonal monomials $g_{j}\in J_{k_{j}l_{j}}$ $(1\leqslant j\leqslant s)$ and a monomial $p$. Suppose $g:=\text{gcd}(f_{u+1},(g_{1}g_{2}\cdots g_{s}))=X_{i_{1}c_{i_{1}}}X_{i_{2}c_{i_{2}}}\cdots X_{i_{k}c_{i_{k}}}$, where $i_{1}<i_{2}<\cdots<i_{k}$. Then $f_{u+1}=gg^{\prime}$ for some $g^{\prime}$ which must divide $p$. We show that either

[TABLE]

From (2.5.3), it follows that $fgg^{\prime}=ff_{u+1}=g_{1}g_{2}\cdots g_{s}p=h_{1}h_{2}\cdots h_{s}p$, which yields that $f=(h_{2}\cdots h_{s})q$ for some monomial $q$, and hence $f\in J_{k_{2}l_{2}}\cdots J_{k_{s}l_{s}}$. Thus, in both cases (2.5.2) and (2.5.3), $f\in L$, which proves (2.5.1). Suppose (2.5.2) does not hold true. It remains to prove (2.5.3).

Each $g_{j}$ is a product of $m$ many variables. Draw circle around each of these variables in the matrix $X$. We do the process for every $1\leqslant j\leqslant s$, and every time we draw a new circle if there is a repetition of variable. So there might be more than one circle around a variable. Moreover, in each row, there are total $s$ many circles listed as 1st, 2nd, 3rd etc., from left to right. (Possibly, $j$th and $(j+1)$st circles of a row are at the same point). For $1\leqslant j\leqslant s$, we construct $h_{j}$ as the product of $m$ many variables corresponding to $j$th circles of all rows of $X$. It follows from the construction that $g_{1}g_{2}\cdots g_{s}=h_{1}h_{2}\cdots h_{s}$. We make the following statements.

Claim 1. The $1$st circle in $(i_{l})$th row of $X$ is at $(i_{l},c_{i_{l}})$ for every $1\leqslant l\leqslant k$.

Proof of Claim 1. Since each $X_{i_{l}\,c_{i_{l}}}$ divides one of $g_{j}$, $1\leqslant j\leqslant s$, there is a circle around $(i_{l},c_{i_{l}})$ for every $1\leqslant l\leqslant k$. If possible, suppose that the $1$st circle in $(i_{l})$th row is around $(i_{l},v)$ for some $1\leqslant l\leqslant k$, where $v<c_{i_{l}}$ and $l$ is the minimum such possible number, i.e., the $1$st circle in $(i_{j})$th row of $X$ is around $(i_{j},c_{i_{j}})$ for all $1\leqslant j<l$. Since $(i_{l},v)$ contains one circle, there is $g_{j}$ for some $1\leqslant j\leqslant s$ such that $X_{i_{l}\,v}$ divides $g_{j}$. We claim that there exists $X_{i\,b_{i}}$ with $i_{l-1}<i\leqslant i_{l}$ and $c_{i-1}<b_{i}<c_{i}$ such that $X_{i\,b_{i}}$ divides $g_{j}$. Assuming the claim, and observing that $X_{i\,b_{i}}$ does not divide $f_{u+1}$, it follows from $ff_{u+1}=g_{1}g_{2}\cdots g_{s}p$ that $X_{i\,b_{i}}$ divides $f$, which contradicts the assumption that (2.5.2) does not hold true. It remains to prove the claim. We may write

[TABLE]

for some $b_{i}$, $1\leqslant i\leqslant m$, where $b_{i_{l}}=v$. Note that $v<c_{i_{l}}$ and $b_{1}<b_{2}<\cdots<b_{m}$. If possible, assume that $b_{i}$ does not lie in between $c_{i-1}$ and $c_{i}$ for every $i_{l-1}<i\leqslant i_{l}$. Hence, since $b_{i_{l}}=v<c_{i_{l}}$, it follows that $b_{i_{l}}\leqslant c_{i_{l}-1}$. Then, starting from $i_{l}$, by using backward induction, one obtains that $b_{i}\leqslant c_{i-1}$ for all $i_{l-1}<i\leqslant i_{l}$. In particular, $b_{i_{l-1}}<b_{1+i_{l-1}}\leqslant c_{i_{l-1}}$. Since $X_{i_{l-1}b_{i_{l-1}}}$ divides $g_{j}$, there is a circle at $(i_{l-1},b_{i_{l-1}})$. So the $1$st circle in $(i_{l-1})$th row of $X$ does not lie at $(i_{l-1},c_{i_{l-1}})$, which contradicts the minimality of $l$. Therefore there is some $i$ with $i_{l-1}<i\leqslant i_{l}$ for which $b_{i}$ is lying in between $c_{i-1}$ and $c_{i}$. This completes the proof of Claim 1.

Claim 2. For every $1\leqslant j\leqslant s$ and every $1\leqslant i\leqslant m-1$, the $j$th circle in $i$th row is on the left of the $j$th circle in $(i+1)$st row.

Proof of Claim 2. If possible, suppose the $j$th circle in some $i$th row is either on the same column of the $j$th circle in $(i+1)$st row or on the right side of it. Consider $g_{v}$ corresponding to each circle of the first $j$ many circles in $(i+1)$st row. Each such $g_{v}$ corresponds to one circle in $i$th row which is situated on the left of the $j$th circle of same row. Therefore, in $i$th row, there are at least $j$ many circles before the $j$th circle, which is a contradiction. This completes the proof of Claim 2.

Claim 3. The monomial $h_{j}\in J_{k_{j}l_{j}}$ for every $1\leqslant j\leqslant s$.

Proof of Claim 3. For every $1\leqslant j\leqslant s$, we set

[TABLE]

It follows from the constructions of $\mathcal{A}_{j}$, $\mathcal{B}_{j}$ and $\mathcal{C}_{j}$ that

[TABLE]

Moreover, it can be observed that $(i,v)\in\mathcal{A}_{j}$ if and only if the corresponding entry $X_{iv}$ of $X$ belongs to the submatrix $Y_{k_{j}l_{j}}$. Therefore, for every $1\leqslant j\leqslant s$,

[TABLE]

Since $h_{j}$ is a diagonal monomial of $X$ (by Claim 2), in view of the construction of $h_{j}$, it is enough to show that

[TABLE]

In view of (2.5.4) and (2.5.5), we obtain that the circles correspond to $g_{1},g_{2},\ldots,g_{j}$ belong into $\mathcal{B}_{j}$. Thus, in every row of $X$, at least $j$ many circles are there in $\mathcal{B}_{j}$. Therefore the $j$th circle of every row of $X$ must be placed in $\mathcal{B}_{j}$. Hence, in order to prove (2.5.6), it remains to show that the $j$th circle of every row of $X$ is in $\mathcal{C}_{j}$. If possible, suppose that the $j$th circle of some $i$th row does not belong into $\mathcal{C}_{j}$. It follows that the first $j$ circles of the $i$th row do not belong into $\mathcal{C}_{j}$. So at most $(s-j)$ many circles of the $i$th row can be there in $\mathcal{C}_{j}$. On the other hand, in view of (2.5.4) and (2.5.5), since the circles correspond to $g_{j},g_{j+1},\ldots,g_{s}$ are placed in $\mathcal{C}_{j}$, at least $(s-j+1)$ many circles of the $i$th row are there in $\mathcal{C}_{j}$, which is a contradiction. This completes the proof of (2.5.6), and hence Claim 3.

By Claim 3, one obtains that each $h_{j}\in J_{k_{j}l_{j}}$ $(1\leqslant j\leqslant s)$ is a diagonal monomial; while Claim 1 yields that $h_{1}$ is divisible by $g=X_{i_{1}c_{i_{1}}}X_{i_{2}c_{i_{2}}}\cdots X_{i_{k}c_{i_{k}}}$. This completes the proof of (2.5.3), and hence the lemma. ∎

We now give an example describing the procedure that we follow in Lemma 2.5.

Example 2.6**.**

Let $X$ be a $6\times 16$ matrix of distinct indeterminates over a field $K$. We describe the matrix $X$ by the figure below, where the horizontal lines represent the rows of $X$, the vertical lines represent the columns of $X$, and the entries of $X$ are situated on the intersections of horizontal and vertical lines. We follow the notations used in the proof of Lemma 2.5. Set $(k_{1},l_{1})=(1,12)$, $(k_{2},l_{2})=(3,13)$, $(k_{3},l_{3})=(7,15)$, $(k_{4},l_{4})=(9,16)$ and $(k_{5},l_{5})=(10,16)$, which correspond five submatrices $Y_{k_{j}l_{j}}$ of $X$ and five ideals $J_{k_{j}l_{j}}$ of $K[X]$ for $1\leqslant j\leqslant 5$ respectively.

\bigstar$$\bigstar$$\bigstar$$\bigstar$$\bigstar$$\bigstar$$6$$5$$4$$3$$2$$1$$16$$15$$14$$13$$12$$11$$10$$9$$8$$7$$6$$5$$4$$3$$2$$1

Following the proof of Lemma 2.5, in a particular situation, let $f_{u+1}$, $g_{1},~{}g_{2},~{}g_{3},~{}g_{4}$ and $g_{5}$ denote the diagonal monomials which are the product of entries marked as $\blacklozenge$, $\bigstar$, $\diamond$, $\times$, $\bullet$ and $\blacktriangle$ respectively, i.e.,

[TABLE]

Set $g:=\text{gcd}(f_{u+1},(g_{1}g_{2}g_{3}g_{4}g_{5}))=X_{36}X_{6\,11}$. Since $g$ does not divide $g_{1}$, we rearrange $g_{1},\ldots,g_{5}$ as follows. Draw circle around each of the symbols $\bigstar$, $\diamond$, $\times$, $\bullet$ and $\blacktriangle$ correspond to $g_{j}$ for all $1\leqslant j\leqslant 5$. Note that there are more than one circles around the entries $X_{2\,11}$, $X_{5\,10}$ and $X_{6\,16}$. In each row, there are total $5$ circles listed as 1st, 2nd, 3rd etc., from left to right. For every $1\leqslant j\leqslant 5$, let $h_{j}$ be the product of six many variables corresponding to $j$th circles of all six rows of $X$, i.e.,

[TABLE]

Clearly, $g_{1}\cdots g_{5}=h_{1}\cdots h_{5}$. The first circles of 3rd and 6th rows of $X$ are at $(3,6)$ and $(6,11)$ respectively as shown in Claim 1. Hence $g=X_{36}X_{6\,11}$ divides $h_{1}$. Moreover, $h_{j}$ ($1\leqslant j\leqslant 5$) are diagonal monomials of $X$ as shown in Claim 2. Also $h_{1}\in J_{1\,12}$, $h_{2}\in J_{3\,13}$, $h_{3}\in J_{7\,15}$, $h_{4}\in J_{9\,16}$ and $h_{5}\in J_{10\,16}$ as shown in Claim 3.

The remarks below show that the hypotheses of Lemma 2.5 cannot be weakened.

Remark 2.7*.*

Note that Lemma 2.5 is not necessarily true for different order of the ideals $J_{k_{1}l_{1}},J_{k_{2}l_{2}},\ldots,J_{k_{s}l_{s}}$. For example, let $X$ be a $3\times 9$ generic matrix. Following the notations as in 1.1, consider $J_{15}$ and $J_{37}$. In this case, $J_{37}=(X_{13}X_{24}X_{35},X_{13}X_{24}X_{36},\ldots)$ and $(J_{37}J_{15}:X_{13}X_{24}X_{35})\neq J_{15}$.

Remark 2.8*.*

In Lemma 2.5, the orders $k_{1}\leqslant k_{2}\leqslant\cdots\leqslant k_{s}$ and $l_{1}\leqslant l_{2}\leqslant\cdots\leqslant l_{s}$ cannot be omitted. As in Remark 2.7, consider $J_{28}$ and $J_{37}$. Writing

[TABLE]

we note that

[TABLE]

We recall the following well-known facts for later use.

2.9**.**

Let $I$ be a homogeneous ideal of $K[X]$. Then

1. (i)

$\operatorname{reg}(I)=\operatorname{reg}(K[X]/I)+1$. 2. (ii)

Suppose $I$ is generated by homogeneous elements all of degree $d$. Then $\operatorname{reg}(I)\geqslant d$, and equality occurs if and only if $I$ has a linear free resolution. 3. (iii)

If $I$ is generated by some variables, then $\operatorname{reg}(K[X]/I)=0$.

Now we are in a position to prove our main result.

Theorem 2.10**.**

Let $1\leqslant k_{1}\leqslant k_{2}\leqslant\cdots\leqslant k_{s}<n$ and $1<l_{1}\leqslant l_{2}\leqslant\cdots\leqslant l_{s}\leqslant n$ be such that $k_{i}<l_{i}$ and $l_{i}-k_{i}+1\geqslant m$. Along with the notations as in 1.1, we set $J:=J_{k_{1}l_{1}}J_{k_{2}l_{2}}\cdots J_{k_{s}l_{s}}$. Then $J$ has a linear free resolution.

Proof.

We prove the assertion by induction on $s$. By virtue of Corollary 2.4, $J_{k_{1}l_{1}}$ has a linear free resolution. Thus, if $s=1$, then there is nothing to prove. So we may assume that $s\geqslant 2$, and $J^{\prime}:=J_{k_{2}l_{2}}J_{k_{3}l_{3}}\cdots J_{k_{s}l_{s}}$ has a linear free resolution, i.e., $\operatorname{reg}(J^{\prime})=(s-1)m$. Setting $J_{k_{1}l_{1}}=(f_{1},\ldots,f_{r})$ as in Lemma 2.5, we consider the following short exact sequences:

[TABLE]

It follows from these short exact sequences that

[TABLE]

By Corollary 2.4, one obtains that $\operatorname{reg}\left(K[X]/J_{k_{1}l_{1}}\right)=m-1$. Moreover, for every $1\leqslant i\leqslant r$, we have

[TABLE]

Thus it follows from the above inequalities that $\operatorname{reg}(K[X]/J)\leqslant sm-1$, i.e., $\operatorname{reg}(J)\leqslant sm$, and hence $\operatorname{reg}(J)=sm$. So $J$ has a linear free resolution. ∎