# The facet ideals of chessboard complexes

Chengyao Jiang, Yakun Zhao, Hong Wang and Guangjun Zhu${}^{{}^{*}}$ Authors’
address: School of Mathematical Sciences, Soochow University, Suzhou 215006,
P. R. China  1562420606@qq.com(Chengyao Jiang),1768868280@qq.com(Yakun
Zhao),651634806@qq.com(Hong Wang),zhuguangjun@suda.edu.cn(Corresponding
author:Guangjun Zhu).

###### Abstract.

In this paper we describe the irreducible decomposition of the facet ideal
${\mathcal{F}}(\Delta_{m,n})$ of the chessboard complex $\Delta_{m,n}$ with
$n\geq m$. We also provide some lower bounds for depth and regularity of the
facet ideal ${\mathcal{F}}(\Delta_{m,n})$. When $m\leq 3$, we prove that these
lower bounds can be obtained.

∗ Corresponding author

2020 Mathematics Subject Classification. Primary 13D02; Secondary 13F55, 13C15

Keywords: Chessboard complex, facet ideal, irreducible decomposition, depth,
regularity

##  1\. Introduction

Graph complexes have provided an important link between combinatorics and
algebra, topology, and geometry (see [14, 22]). The two most important graph
complexes are the matching complex and independence complex; their algebraic
and topological properties have been widely studied by many authors (see [2,
8, 14, 17, 22, 23, 26]). The chessboard complex, which is the matching complex
$\mathcal{M}(K_{m,n})$ of a complete bipartite graph $K_{m,n}$, is first
introduced in the thesis of Garst [9]. He showed that $\mathcal{M}(K_{m,n})$
is Cohen–Macaulay if and only if $n\geq 2m-1$. Ziegler in [23] strengthened
this result by showing that $\mathcal{M}(K_{m,n})$ is shellable if $n\geq
2m-1$. Hence $\mathcal{M}(K_{m,n})$ has the homotopy type of a wedge of
$(m-1)$-spheres if $n\geq 2m-1$. Freidman and Hanlon studied Betti numbers of
chessboard complex in [8]. They showed that $b_{r-1}(\Delta_{m,n})=0$ if and
only if $(m-r)(n-r)>r$, and $b_{v-1}(\Delta_{m,n})>0$ if and only if $n\geq
2m-4$ or $(m,n)=(6,6),(7,7),(8,9)$, where $b_{i}(\Delta_{m,n})$ is the $i$-th
Betti number of $\Delta_{m,n}$, it equals the rank of the homology group
$H_{i}(\Delta_{m,n})$. The chessboard complex may be regarded as a simplicial
complex formed by all admissible rook configurations on an $m\times n$
chessboard, while an admissible rook configuration on a chessboard refers to a
subset of squares of the chessboard such that no two squares lie in the same
row or in the same column. It is for this reason that the name “chessboard
complex” is used.

Given a simplicial complex $\Delta$ on the vertex set
$\\{x_{1},\ldots,x_{n}\\}$, we consider the polynomial ring
$S=k[x_{1},\ldots,x_{n}]$ in $n$ variables over a field $k$. The Stanley-
Reisner ideal of $\Delta$, denoted by ${\mathcal{I}}_{\Delta}$, is the
squarefree monomial ideal of $S$ given by

${\mathcal{I}}_{\Delta}=(x_{i_{1}}\cdots
x_{i_{s}}\mid\\{x_{i_{1}},\ldots,x_{i_{s}}\\}\text{\ is not a face of \
}\Delta).$

The facet ideal of $\Delta$, denoted by ${\mathcal{F}}(\Delta)$, is the
squarefree monomial ideal of $S$ given by

${\mathcal{F}}(\Delta)=(x_{\ell_{1}}\cdots
x_{\ell_{s}}\mid\\{x_{\ell_{1}},\ldots,x_{\ell_{s}}\\}\text{\ is a facet of \
}\Delta).$

The Stanley-Reisner ring of $\Delta$ is $k[\Delta]=S/{\mathcal{I}}_{\Delta}$.
For any positive integers $m,n$, Björner et al. in [2] showed that a
chessboard complex $\Delta_{m,n}$ is $(v-2)$-connected and
$\operatorname{depth}\,(k[\Delta_{m,n}])=v$ for any positive integers $m,n$,
where $v=\min\\{m,n,\lfloor\frac{m+n+1}{3}\rfloor\\}$ and
$\left\lfloor\frac{m+n+1}{3}\right\rfloor$ is the largest integer
$\leq\frac{m+n+1}{3}$.

Hà and Van Tuyl in [11] studied edge ideals of $m$-uniform hypergraphs, they
showed that if $H$ is a properly-connected hypergraph with the edge ideal
$I(H)$ and $c$ is the maximal number of pairwise $(m+1)$-disjoint edges of
$H$, then $\operatorname{reg}\,(S/I(H))\geq c(m-1)$ and the equality holds if
$H$ is triangulated. Obviously, ${\mathcal{F}}(\Delta_{m,n})$ can be regarded
as the edge ideal of an $m$-uniform hypergraph, but this $m$-uniform
hypergraph is not properly-connected and has no pairwise $(m+1)$-disjoint
edges.

As far as we know, little is known about algebraic properties of the facet
ideal of a chessboard complex. In this article, we focus on properties
corresponding to the irreducible decomposition (see Theorem 3.3), depth and
regularity of the facet ideal of a chessboard complex. Our main results are as
follows:

######  Theorem 1.1.

Let $\Delta_{m,n}$ be a chessboard complex with $n\geq m\geq 1$. Let $V$ be
its vertex set and ${\mathcal{F}}(\Delta_{m,n})$ be its facet ideal. A prime
ideal $P$ of $S$ is a minimal prime of ${\mathcal{F}}(\Delta_{m,n})$ if and
only if $P$ is generated by a set $C$, which is obtained from $V$ by deleting
all elements in $s$ rows and all elements in $m-1-s$ columns for some $0\leq
s\leq m-1$.

######  Corollary 1.2.

Let $\Delta_{m,n}$ be a chessboard complex with $n\geq m\geq 1$, one has

  1. (a)

$\operatorname{height}\,({\mathcal{F}}(\Delta_{m,n}))=n$,

  2. (b)

$\dim\,(S/{\mathcal{F}}(\Delta_{m,n}))=(m-1)n$,

  3. (c)

$\operatorname{bight}\,({\mathcal{F}}(\Delta_{m,n}))=\left\\{\begin{array}[]{l@{\
\ }l}\lfloor\frac{(n+1)^{2}}{4}\rfloor&\text{if $n<2m-1$,}\\\
{(n-m+1)m}&\text{otherwise.}\end{array}\right.$

######  Theorem 1.3.

Let $\Delta_{m,n}$ be a chessboard complex with $n\geq m\geq 1$, then

  1. (a)

$\operatorname{reg}(S/{\mathcal{F}}(\Delta_{m,n}))\geq 2(m-1)$.

  2. (b)

$\operatorname{depth}\,(S/{\mathcal{F}}(\Delta_{m,n}))\geq 2(m-1)$.

The equalities hold if $m\leq 3$.

Moradi in [17] showed that if $G$ is a graph with $m$-clique ideal $K_{m}(G)$
and $c_{m}(G)$ is the minimum number of co-chordal subgraphs of $G$ required
to cover the $m$-cliques of $G$, then $\operatorname{reg}\,(S/K_{m}(G))\leq
c_{m}(G)(m-1)$. One can regard ${\mathcal{F}}(\Delta_{m,n})$ as the $m$-clique
ideal of the complement $G$ of the line graph of the complete bipartite graph
$K_{m,n}$ where $c_{m}(G)\geq 2$. This shows that the upper bound of
$\operatorname{reg}\,(S/K_{m}(G))$ given by Moradi can be strict.

Our paper is organized as follows. In the preliminary section, we collect the
necessary terminology and results from the literature. In Section $3$, we give
the irreducible decomposition of the facet ideal ${\mathcal{F}}(\Delta_{m,n})$
of a chessboard complex $\Delta_{m,n}$. In Section $4$, we give some exact
formulas for depth and regularity of powers of ${\mathcal{F}}(\Delta_{m,n})$
under the conditions that $n\geq m$ and $m\leq 2$. In Section $5$, we provide
some lower bounds for depth and regularity of the facet ideal
${\mathcal{F}}(\Delta_{m,n})$ with $n\geq m$. We also prove that these lower
bounds can be attained if $m=3$.

Throughout the paper, we assume that $x_{ij}$ is the square corresponding to
the $i$-th row and the $j$-th column in an $m\times n$ chessboard where $n\geq
m$, $1\leq i\leq m$ and $1\leq j\leq n$. By convention, we use the matrix
$(x_{ij})_{m\times n}$ for representing this chessboard. For example, we can
represent the chessboard on the left by the matrix on the right

|  $x_{11}$ |  $x_{12}$ |  $x_{13}$
---|---|---
$x_{21}$ |  $x_{22}$ |  $x_{23}$
$x_{31}$ |  $x_{32}$ |  $x_{33}$
$\begin{pmatrix}x_{11}&x_{12}&x_{13}\\\ \\\ x_{21}&x_{22}&x_{23}\\\ \\\
x_{31}&x_{32}&x_{33}\\\ \end{pmatrix}$

##  2\. Preliminaries

In this section, we gather together the needed notations and basic facts,
which will be used throughout this paper. However, for more details, we refer
the reader to [3, 6, 12, 13, 14].

Let $\Delta$ be a simplicial complex on the vertex set $V$. We denote the set
of its facets (maximal faces under inclusion) by $Facets\,(\Delta)$. If
$Facets\,(\Delta)=\\{F_{1},\ldots,F_{q}\\}$, we write $\Delta$ as
$\Delta=\langle F_{1},\ldots,F_{q}\rangle$. A subcomplex of $\Delta$ is a
simplicial complex whose facet set is a subset of the facet set of $\Delta$.
If $A\subseteq V$, the induced subcomplex of $\Delta$ over $A$ is the
subcomplex $\Delta|_{A}=\\{F\in\Delta\mid F\subseteq A\\}$.

Let $n$ be a positive integer, we set $[n]=\\{1,2,\ldots,n\\}$. Given a
chessboard complex $\Delta_{m,n}$ with a vertex set $\\{x_{ij}\mid
i\in[m],j\in[n]\\}$, we consider the polynomial ring $S=k[x_{ij}\mid
i\in[m],j\in[n]]$ in the entries of a generic matrix $(x_{ij})_{m\times n}$
over a field $k$. The facet ideal of $\Delta_{m,n}$ is the squarefree monomial
ideal of $S$ given by

${\mathcal{F}}(\Delta_{m,n})=(x_{1\ell_{1}}\cdots
x_{m\ell_{m}}\mid\ell_{1},\ldots,\ell_{m}\in[n]\text{\ are pairwise
different}).$

######  Example 2.1.

The facet ideal of the chessboard complex $\Delta_{3,3}$ is

${\mathcal{F}}(\Delta_{3,3})=(x_{11}x_{22}x_{33},x_{11}x_{23}x_{32},x_{12}x_{21}x_{33},x_{12}x_{23}x_{31},x_{13}x_{21}x_{32},x_{13}x_{22}x_{31}).$

Let $S=k[x_{1},\ldots,x_{n}]$ be a polynomial ring in $n$ variables over a
field $k$, $I\subset S$ a nonzero homogeneous ideal and

$0\rightarrow\bigoplus\limits_{j}S(-j)^{\beta_{p,j}(I)}\rightarrow\bigoplus\limits_{j}S(-j)^{\beta_{p-1,j}(I)}\rightarrow\cdots\rightarrow\bigoplus\limits_{j}S(-j)^{\beta_{0,j}(I)}\rightarrow
I\rightarrow 0$

is a minimal graded free resolution of $I$, and $S(-j)$ is an $S$-module
obtained by shifting the degrees of $S$ by $j$. The number $\beta_{i,j}(I)$,
the $(i,j)$-th graded Betti number of $I$, is an invariant of $I$ that equals
the minimal number of generators of degree $j$ in the $i$th syzygy module of
$I$. Of particular interest are the following invariants which measure the
size of the minimal graded free resolution of $I$. The projective dimension of
$I$, denoted by $\operatorname{pd}\,(I)$, is defined to be

$\mbox{pd}\,(I):=\mbox{max}\,\\{i\ |\ \beta_{i,j}(I)\neq 0\\}.$

The regularity of $I$, denoted by $\mbox{reg}\,(I)$, is defined by

$\mbox{reg}\,(I):=\mbox{max}\,\\{j-i\ |\ \beta_{i,j}(I)\neq 0\\}.$

By looking at the minimal free resolution and Auslander-Buchsbaum formula (see
Theorem 1.3.3 of [3]), it is easy to obtain the following lemma:

######  Lemma 2.2.

([13, Lemma 1.3]) Let $I\subset S$ be a nonzero proper homogeneous ideal. Then

  * (1)

$\operatorname{depth}\,(S/I)=n-\operatorname{pd}\,(S/I)$,

  * (2)

$\operatorname{depth}\,(S/I)=\operatorname{depth}\,(I)-1$,

  * (3)

$\operatorname{reg}\,(S/I)=\operatorname{reg}\,(I)-1$,

  * (4)

$\operatorname{pd}\,(S/I)=\operatorname{pd}\,(I)+1$.

Hence we shall work with $\operatorname{reg}\,(I)$ (resp.
$\operatorname{depth}\,(I)$) and $\operatorname{reg}\,(S/I)$ (resp.
$\operatorname{depth}\,(S/I)$) interchangeably.

For a monomial ideal $I$, $\mathcal{G}(I)$ denotes the unique minimal set of
monomial generators of $I$.

######  Definition 2.3.

Let $I\subset S$ be a squarefree monomial ideal with irreducible decomposition

$I=P_{1}\cap P_{2}\cap\cdots\cap P_{s},$

where $s\geq 1$ and $P_{i}$ is a prime ideal of $S$ for any $1\leq i\leq s$.
The Alexander dual of $I$, denoted by $I^{\,\vee}$, is a squarefree monomial
ideal

$I^{\,\vee}=({\bf x}_{P_{1}},{\bf x}_{P_{2}},\ldots,{\bf x}_{P_{s}}),$

where ${\bf x}_{P_{i}}=\prod\limits_{x_{j}\in\mathcal{G}(P_{i})}\\!\\!x_{j}$
for any $1\leq i\leq s$.

######  Lemma 2.4.

([10, Theorem 2.18]) Let $I\subset S$ be a squarefree monomial ideal. Then

$\operatorname{pd}\,(S/I)=\operatorname{reg}\,(I^{\,\vee}).$

In this paper we shall use the following results.

######  Lemma 2.5.

([12, Lemmas 2.2 and 3.2]) Let $S_{1}=k[x_{1},\dots,x_{m}]$,
$S_{2}=k[x_{m+1},\dots,x_{n}]$ be two polynomial rings and
$S=S_{1}\otimes_{k}S_{2}$, let $I\subset S_{1}$, $J\subset S_{2}$ be two
nonzero homogeneous ideals. Then

  * (1)

$\operatorname{depth}\,(I+J)=\operatorname{depth}\,(I)+\operatorname{depth}\,(J)-1$,

  * (2)

$\operatorname{depth}\,(JI)=\operatorname{depth}\,(I)+\operatorname{depth}\,(J)$,

  * (3)

$\operatorname{reg}\,(I+J)=\operatorname{reg}\,(I)+\operatorname{reg}\,(J)-1$,

  * (4)

$\operatorname{reg}\,(JI)=\operatorname{reg}\,(I)+\operatorname{reg}\,(J)$.

######  Lemma 2.6.

([15, Proposition 4.1]) Let $I\subset S$ be a squarefree monomial ideal. If
each element of $\mathcal{G}(I)$ contains at least one variable not dividing
any other element of $\mathcal{G}(I)$. Then

$reg\,(I)=|\text{supp}\,(I)|-|\mathcal{G}(I)|+1$

where $\text{supp}\,(I)=\\{x_{i}:\,x_{i}|u\text{\ for some \
}u\in\mathcal{G}(I)\\}$.

######  Lemma 2.7.

([16, Corollary 2.12]) Let $I\subset S$ be a monomial ideal and $f$ a monomial
of degree $d$. Then

$\operatorname{reg}\,(S/(I,f))\leq\operatorname{reg}\,(S/I)+d-1.$

######  Lemma 2.8.

([19, Theorem 1.1]) Let $S_{1}=k[x_{1},\dots,x_{m}]$,
$S_{2}=k[x_{m+1},\dots,x_{n}]$ be two polynomial rings and
$S=S_{1}\otimes_{k}S_{2}$, let $I\subset S_{1}$, $J\subset S_{2}$ be two
nonzero monomial ideals. Then for all $t\geq 1$, there are equalities

  * (1)

$\operatorname{reg}\,(\frac{S}{(I+J)^{t}})=\max\limits_{\begin{subarray}{c}i\in[t-1]\\\
j\in[t]\\\
\end{subarray}}\\{\operatorname{reg}\,(\frac{S_{1}}{I^{t-i}})+\operatorname{reg}\,(\frac{S_{2}}{J^{i}})+1,\operatorname{reg}\,(\frac{S_{1}}{I^{t-j+1}})+\operatorname{reg}\,(\frac{S_{2}}{J^{j}})\\}$,

  * (2)

$\operatorname{depth}\,(\frac{S}{(I+J)^{t}})=\min\limits_{\begin{subarray}{c}i\in[t-1]\\\
j\in[t]\\\
\end{subarray}}\\{\operatorname{depth}\,(\frac{S_{1}}{I^{t-i}})+\operatorname{depth}\,(\frac{S_{2}}{J^{i}})+1,\operatorname{depth}\,(\frac{S_{1}}{I^{t-j+1}})+\operatorname{depth}\,(\frac{S_{2}}{J^{j}})\\}$.

In particular, If $t=1$, we obtain Lemma 2.5 $(1)$ and $(3)$.

######  Lemma 2.9.

[13, Lemmas 1.1 and 1.2] Let $0\rightarrow A\rightarrow B\rightarrow
C\rightarrow 0$ be a short exact sequence of finitely generated graded
$S$-modules. Then

  * (1)

$\operatorname{reg}\,(B)\leq\max\,\\{\operatorname{reg}\,(A),\operatorname{reg}\,(C)\\}$,
the equality holds if $\operatorname{reg}\,(A)-1\neq\operatorname{reg}\,(C)$,

  * (2)

$\operatorname{reg}\,(C)\leq\max\,\\{\operatorname{reg}\,(A)-1,\operatorname{reg}\,(B)\\}$,
the equality holds if $\operatorname{reg}\,(A)\neq\operatorname{reg}\,(B)$,

  * (3)

$\operatorname{depth}\,(B)\geq\min\,\\{\operatorname{depth}\,(A),\operatorname{depth}\,(C)\\}$,
the equality holds if
$\operatorname{depth}\,(A)-1\neq\operatorname{depth}\,(C)$.

######  Corollary 2.10.

Let $I\subset S$ be a nonzero proper homogenous ideal, and $f$ a homogenous
polynomial of degree $d$ in $S$. Then

  * (1)

$\operatorname{reg}\,(I)\leq\max\,\\{\operatorname{reg}\,(I\\!:\\!f)+d,\operatorname{reg}\,((I,f))\\}$,
the equality holds if
$\operatorname{reg}\,((I,f))\neq\operatorname{reg}\,(I\\!:\\!f)+d-1$,

  * (2)

$\operatorname{reg}\,(I,f)\leq\max\,\\{\operatorname{reg}\,(I\\!:\\!f)+d-1,\operatorname{reg}\,(I)\\}$,
the equality holds if
$\operatorname{reg}\,(I)\neq\operatorname{reg}\,(I\\!:\\!f)+d$,

  * (3)

$\operatorname{depth}\,(S/I)\geq\min\,\\{\operatorname{depth}\,(S/(I\\!:\\!f)),\operatorname{depth}\,(S/(I,f))\\}$,
the equality holds if
$\operatorname{depth}\,(S/(I\\!:\\!f))\neq\operatorname{depth}\,(S/(I,f))+1$.

###### Proof.

It follows from Lemma 2.9 by considering the following short exact sequence

$0\longrightarrow\frac{S}{I:f}(-d)\stackrel{{\scriptstyle\cdot
f}}{{\longrightarrow}}\frac{S}{I}\longrightarrow\frac{S}{(I,f)}\longrightarrow
0.$

######  Remark 2.11.

Let $I\subset S$ be a nonzero proper homogenous ideal and $f_{i}$ a homogenous
polynomial of degree $d_{i}$ in $S$ for $i=1,\ldots,m$. Set $J_{0}=I$,
$J_{i}=I+(f_{1},f_{2},\ldots,f_{i})$ for any $i\in[m]$. Applying the above
corollary, one has
$\operatorname{reg}\,(J_{i-1})\leq\max\\{\operatorname{reg}\,(J_{i-1}:f_{i})+d_{i},\
\operatorname{reg}\,(J_{i})\\}$ and the equality holds if
$\operatorname{reg}\,(J_{i})\geq\operatorname{reg}\,(J_{i-1}:f_{i})+d_{i}$.
Furthermore,

$\operatorname{reg}\,(I)\leq\max\\{\operatorname{reg}\,(J_{j-1}:f_{j})+d_{j},\
\operatorname{reg}\,(J_{m})\mid j\in[m]\\}$

and $\operatorname{reg}\,(I)=\operatorname{reg}\,(J_{m})$ if
$\operatorname{reg}\,(J_{m})\geq\max\\{\operatorname{reg}\,(J_{j-1}:f_{j})+d_{j}\mid
j\in[m]\\}$.

Similarly, we also have

$\operatorname{depth}\,(S/J_{i-1})\geq\min\\{\operatorname{depth}\,(S/(J_{i-1}:f_{i})),\
\operatorname{depth}\,(S/J_{i})\\}$

and the equality holds if
$\operatorname{depth}\,(S/(J_{i-1}:f_{i}))\leq\operatorname{depth}\,(S/J_{i})$.
Furthermore,

$\operatorname{depth}(S/I)\geq\min\\{\operatorname{depth}(S/(J_{j-1}:f_{j})),\
\operatorname{depth}(S/J_{m})\mid j\in[m]\\}$

and the equality holds if
$\operatorname{depth}(S/(I\\!:\\!f_{1}))\leq\min\\{\operatorname{depth}(S/(J_{j-1}\\!:\\!f_{j})),\operatorname{depth}(S/J_{m})\mid
2\leq j\leq m\\}.$

Applying the above remark iteratively, we can obtain

######  Corollary 2.12.

Let $I\subset S$ be a monomial ideal and $u_{i}$ a monomial for
$i=1,\ldots,r$. Set $V=\\{u_{1},u_{2},\ldots,u_{r}\\}$. Then

  * (1)

$\operatorname{reg}\,(I)\leq\max\\{\operatorname{reg}((I:\prod\limits_{u\in
W}u)+I_{W^{c}})+d_{W}\mid\ W\subseteq V\\}$,

  * (2)

$\operatorname{depth}\,(S/I)\geq\min\\{\operatorname{depth}\,(\frac{S}{(I:\prod\limits_{u\in
W}u)+I_{W^{c}}})\mid\ W\subseteq V\\}$,

where $W^{c}=V\setminus W$, $I_{W^{c}}$ is an ideal generated by $\\{v\mid
v\in W^{c}\\}$ and $d_{W}$ is the degree of $\prod\limits_{u\in W}u$.

######  Remark 2.13.

Let $I$ be a monomial ideal of $S$ with
$\mathcal{G}(I)=\\{u_{1},u_{2},\ldots,u_{r}\\}$, where $u_{i}$ is a monomial
of degree $d_{i}$ for any $i\in[r]$. Set $J_{0}=I$,
$J_{i}=(u_{i+1},u_{i+2},\ldots,u_{r})$. Thus we have the following short exact
sequence

$0\longrightarrow\frac{S}{J_{i}:u_{i}}(-d_{i})\stackrel{{\scriptstyle\cdot
u_{i}}}{{\longrightarrow}}\frac{S}{J_{i}}\longrightarrow\frac{S}{J_{i-1}}\longrightarrow
0.$

Therefore,
$\operatorname{reg}\,(J_{i-1})\leq\max\\{\operatorname{reg}\,(J_{i}:u_{i})+d_{i}-1,\
\operatorname{reg}\,(J_{i})\\}$ by Lemma 2.9 (2), and the equality holds if
$\operatorname{reg}\,(J_{i}:u_{i})+d_{i}>\operatorname{reg}\,(J_{i})$.
Furthermore,

$\operatorname{reg}\,(I)=\operatorname{reg}\,(J_{0})\leq\max\\{\operatorname{reg}\,(J_{j}:u_{j})+d_{j}-1,\
\operatorname{reg}\,(J_{r})\mid j\in[r]\\}$

and the equality holds if
$\operatorname{reg}\,(J_{1}:u_{1})+d_{1}>\max\\{\operatorname{reg}\,(J_{j}:u_{j})+d_{j}-1,\
\operatorname{reg}\,(J_{r})\mid 2\leq j\leq r\\}.$

##  3\. Irreducible decomposition of the facet ideal

In this section, we provide the irreducible decomposition of the facet ideal
${\mathcal{F}}(\Delta_{m,n})$ of a chessboard complex $\Delta_{m,n}$ with
$n\geq m$. We first give a definition of the vertex cover.

######  Definition 3.1.

Let $\Delta$ be a simplicial complex on the vertex set $V$. A vertex cover for
$\Delta$ is a subset $A$ of $V$ that intersects every facet of $\Delta$. If
$A$ is a minimal element (under inclusion) of the set of vertex covers of
$\Delta$, it is called a minimal vertex cover.

We need the following lemma.

######  Lemma 3.2.

([6, Proposition 1.8]) Let $\Delta$ be a simplicial complex,
${\mathcal{F}}(\Delta)$ its facet ideal. Then an ideal
$P=(x_{i_{1}},\ldots,x_{i_{s}})$ is a minimal prime of ${\mathcal{F}}(\Delta)$
if and only if $\\{x_{i_{1}},\ldots,x_{i_{s}}\\}$ is a minimal vertex cover of
$\Delta$.

Now, we describe the irreducible decomposition of
${\mathcal{F}}(\Delta_{m,n})$.

######  Theorem 3.3.

Let $\Delta_{m,n}$ be a chessboard complex with $n\geq m\geq 1$. Let $V$ be
its vertex set and ${\mathcal{F}}(\Delta_{m,n})$ be its facet ideal. A prime
ideal $P$ of $S$ is a minimal prime of ${\mathcal{F}}(\Delta_{m,n})$ if and
only if $P$ is generated by a set $C$, which is obtained from $V$ by deleting
all elements in $s$ rows and all elements in $m-1-s$ columns for some $0\leq
s\leq m-1$.

###### Proof.

The case $m=1$ is trivial. Now we assume that $m\geq 2$. It is enough to show
that the set $C$ is exactly a minimal vertex cover of $\Delta_{m,n}$ by Lemma
3.2.

First, we show that the set $C$ is a vertex cover of $\Delta_{m,n}$.

For any $F\in Facets(\Delta_{m,n})$, we know that $|F|=m$ and all elements in
$F$ are taken from different rows and columns. We may write $F$ as
$F=\\{x_{1t_{1}},\ldots,x_{mt_{m}}\\}$, where $t_{1},\ldots,t_{m}\in[n]$ are
different. Now, we show that there exists some $x_{kt_{k}}\in F$ such that
$x_{kt_{k}}\in C$, i.e., $F\cap C\neq\emptyset$. Assume that the set $C$ is
obtained from $V$ by deleting all elements in $i_{1},\ldots,i_{s}$ rows and
all elements in $j_{1},\ldots,j_{m-1-s}$ columns, where
$\\{i_{1},\ldots,i_{s}\\}\subsetneq[m]$ and
$\\{j_{1},\ldots,j_{m-1-s}\\}\subsetneq[n]$. By convention,
$\\{i_{1},\ldots,i_{s}\\}=\emptyset$ if $s=0$, and
$\\{j_{1},\ldots,j_{m-1-s}\\}=\emptyset$ if $s=m-1$. Note that there exist
$m-s$ elements in $F$, which are not in the $i_{1},\ldots,i_{s}$ rows, and at
least one of such $m-s$ elements is not in the $j_{1},\ldots,j_{m-1-s}$
columns since $m-s>m-1-s$. Say $x_{kt_{k}}$. Hence $F\cap C\neq\emptyset$, as
desired.

Next, we show that the vertex cover $C$ is minimal.

Let $C$ be a set obtained from $V$ by deleting all elements in
$i_{1},\ldots,i_{s}$ rows and all elements in $j_{1},\ldots,j_{m-1-s}$
columns, where $i_{1},\ldots,i_{s}\in[m]$ and $j_{1},\ldots,j_{m-1-s}\in[n]$
are different. Let $W$ be any proper subset of $C$. Then there exists some
$x_{pq}\in C\setminus W$. Hence $p\in[m]\setminus\\{i_{1},\ldots,i_{s}\\}$,
$q\in[n]\setminus\\{j_{1},\ldots,j_{m-1-s}\\}$. Choose $(m-1-s)$ different
elements $k_{1},\ldots,k_{m-1-s}\in[m]\setminus\\{p,i_{1},\ldots,i_{s}\\}$ and
$s$ different elements
$\ell_{1},\ldots,\ell_{s}\in[n]\setminus\\{q,j_{1},\ldots,j_{m-1-s}\\}$. Set
$F=\\{x_{i_{1}\ell_{1}},\ldots,x_{i_{s}\ell_{s}},x_{k_{1}j_{1}},\ldots,x_{k_{m-1-s}j_{m-1-s}},x_{pq}\\}$.
Note that $|F|=m$ and all elements in $F$ are taken from different rows and
columns, it follows that $F\in Facets(\Delta_{m,n})$ and
$C\cap\\{x_{i_{1}\ell_{1}},\ldots,x_{i_{s}\ell_{s}},x_{k_{1}j_{1}},\ldots,x_{k_{m-1-s}j_{m-1-s}}\\}=\emptyset$.
By the choice of $x_{pq}$, one has $F\cap W=\emptyset$. This implies that $W$
is not a vertex cover of $\Delta_{m,n}$.

Given an ideal $I\subset S$, we set

$\operatorname{bight}\,(I)=\sup\\{\operatorname{height}\,(P)\mid P\text{ is a
minimal prime ideal of }S\text{ over }I\\}.$

As a consequence of the above theorem, we have

######  Corollary 3.4.

Let $\Delta_{m,n}$ be a chessboard complex with $n\geq m\geq 1$, one has

  1. (a)

$\operatorname{height}\,({\mathcal{F}}(\Delta_{m,n}))=n$,

  2. (b)

$\dim\,(S/{\mathcal{F}}(\Delta_{m,n}))=(m-1)n$,

  3. (c)

$\operatorname{bight}\,({\mathcal{F}}(\Delta_{m,n}))=\left\\{\begin{array}[]{l@{\
\ }l}\lfloor\frac{(n+1)^{2}}{4}\rfloor&\text{if $n<2m-1$,}\\\
{(n-m+1)m}&\text{otherwise.}\end{array}\right.$

###### Proof.

(a) follows from Theorem 3.3.

(b) follows from a fact that
$\operatorname{height}({\mathcal{F}}(\Delta_{m,n}))+\dim(S/{\mathcal{F}}(\Delta_{m,n}))=\dim(S)=mn$.

(c) The case $m=1$ is trivial. Now suppose $m\geq 2$. Let $P$ be a minimal
prime of ${\mathcal{F}}(\Delta_{m,n})$, then its generators have forms as the
above theorem. It follows that

$\operatorname{height}\,(P)=[n-(m-1-s)](m-s)=-[s+\frac{n-(2m-1)}{2}]^{2}+\frac{(n+1)^{2}}{4}$

for some nonnegative integer $s\leq m-1$. Therefore,

$\operatorname{bight}\,({\mathcal{F}}(\Delta_{m,n}))=\sup\\{-[s+\frac{n-(2m-1)}{2}]^{2}+\frac{(n+1)^{2}}{4}\mid\,0\leq
s\leq m-1\\}.$

Let $f(s)=-[s+\frac{n-(2m-1)}{2}]^{2}+\frac{(n+1)^{2}}{4}$, then $f(s)$ is a
quadratic function in $s$. Hence, if $n\geq 2m-1$, then

$\operatorname{bight}\,({\mathcal{F}}(\Delta_{m,n}))=f(0)=-[\frac{n-(2m-1)}{2}]^{2}+\frac{(n+1)^{2}}{4}=(n-m+1)m;$

if $n<2m-1$, then

$\displaystyle\operatorname{bight}\,({\mathcal{F}}(\Delta_{m,n}))$
$\displaystyle=$ $\displaystyle f(\lfloor\frac{(2m-1)-n}{2}\rfloor)$
$\displaystyle=$
$\displaystyle-[\lfloor\frac{(2m-1)-n}{2}\rfloor+\frac{n-(2m-1)}{2}]^{2}+\frac{(n+1)^{2}}{4}=\lfloor\frac{(n+1)^{2}}{4}\rfloor,$

where the last equality holds because
$\operatorname{bight}\,({\mathcal{F}}(\Delta_{m,n}))=\frac{(n+1)^{2}}{4}$ if
$n$ is odd, otherwise,
$\operatorname{bight}\,({\mathcal{F}}(\Delta_{m,n}))=\frac{(n+1)^{2}}{4}-\frac{1}{4}$.
The proof is complete.

##  4\.  Powers of facet ideals of small chessboard complexes

In this section, we give some formulas for depth and regularity of powers of
the facet ideal ${\mathcal{F}}(\Delta_{m,n})$ of a chessboard complex
$\Delta_{m,n}$ under the condition that $n\geq m\geq 1$ and $m\leq 2$.

Let $G=(V(G),E(G))$ be a finite simple graph with the vertex set $V(G)$ and
edge set $E(G)$. We say that a graph $H$ is an induced subgraph of $G$ on a
subset $V^{\prime}$ of $V(G)$ if $V(H)=V^{\prime}$ and $E(H)=\\{uv\in
E(G):u,v\in V^{\prime}\\}$. The complement of $G$ is the graph $G^{c}$ with
$V(G^{c})=V(G)$ and $E(G^{c})=\\{uv:u,v\in V(G)\text{\ and\ }uv\notin
E(G)\\}$. Let $u\in V(G)$, the degree of $u$, denoted by $\deg\,(u)$, is the
number of vertices of $G$ adjacent to $u$. Let $G=(V(G),E(G))$ be a connected
graph with $n=|V(G)|$ vertices, it is called an $n$-cycle if $\deg\,(u)=2$ for
any $v\in V(G)$, denoted by $C_{n}$, where $n$ is referred to as the length of
the cycle.

A bipartite graph $G$ is a graph whose vertex set $V(G)$ is the disjoint union
$X\sqcup Y$ of $X$ and $Y$ and edge set $E(G)$ satisfies
$E(G)\subseteq\\{xy\mid x\in X,y\in Y\\}$. The bipartite complement of a graph
$H$ is the bipartite graph $H^{bc}$ with $V(H^{bc})=X\sqcup Y$ and
$E(H^{bc})=\\{xy:x\in X,y\in Y\text{\ and\ }xy\notin E(H)\\}$.

######  Lemma 4.1.

([1, Theorem 3.6]) Let $G$ be a connected bipartite graph and $I(G)$ its edge
ideal. If $\operatorname{reg}\,(I(G))=3$, then
$\operatorname{reg}\,(I(G)^{t})=2t+1$ for any $t\geq 1$.

######  Theorem 4.2.

Let $\Delta_{1,n}$ be a chessboard complex with $n\geq 1$, then for any $t\geq
1$,

  1. (a)

$S/{\mathcal{F}}(\Delta_{1,n})^{t}$ is Cohen-Macaulay of dimension $0$.

  2. (b)

$\operatorname{reg}\,(S/{\mathcal{F}}(\Delta_{1,n})^{t})=t-1$.

In particular, ${\mathcal{F}}(\Delta_{1,n})^{t}$ has a linear resolution.

###### Proof.

(a) is trivial, since ${\mathcal{F}}(\Delta_{1,n})=(x_{11},\ldots,x_{1n})$ is
the maximal graded ideal of $S$. (b) follows from Lemmas 2.2 and 2.8, since
${\mathcal{F}}(\Delta_{1,n})=(x_{11},\ldots,x_{1n})$ is a complete
intersection ideal.

Let $I$ be a squarefree monomial ideal in a polynomial ring
$k[x_{1},\ldots,x_{n}]$. The Stanley-Reisner complex $\delta(I)$ of $I$ is a
simplicial complex on the vertex set $\\{x_{1},\ldots,x_{n}\\}$, where
$\\{x_{i_{1}},\ldots,x_{i_{s}}\\}$ is a face of $\delta(I)$ if and only if
$x_{i_{1}}\cdots x_{i_{s}}\notin I$.

Let $\Delta_{2,n}$ be a chessboard complex with $n\geq 3$, then its facet
ideal ${\mathcal{F}}(\Delta_{2,n})=(x_{1i}x_{2j}\mid 1\leq i,j\leq n\text{\
and }i\neq j)$ can be regarded as the edge ideal of a connected bipartite
graph with bipartition
$\\{x_{11},\ldots,x_{1n}\\}\sqcup\\{x_{21},\ldots,x_{2n}\\}$, and the Stanley-
Reisner complex $\delta({\mathcal{F}}(\Delta_{2,n}))$ of
${\mathcal{F}}(\Delta_{2,n})$ is

$\delta({\mathcal{F}}(\Delta_{2,n})=\big{\langle}\,\\{x_{11},\ldots,x_{1n}\\},\\{x_{21},\ldots,x_{2n}\\},\\{x_{11},x_{21}\\},\\{x_{12},x_{22}\\},\ldots,\\{x_{1n},x_{2n}\\}\,\big{\rangle}$

from Theorem 3.3 and [6, Remark 1.13]. By [21, Lemma 3.1], we have

######  Lemma 4.3.

Let $\Delta_{2,n}$ be a chessboard complex with $n\geq 3$. Then
$\operatorname{depth}\,(S/{\mathcal{F}}(\Delta_{2,n})^{t})\\\ =1$ if and only
if $\Delta_{\alpha}({\mathcal{F}}(\Delta_{2,n})^{t})=\langle
F_{1},F_{2}\rangle$ for some
$\alpha=(\alpha_{1},\ldots,\alpha_{2n})\in{\mathbb{N}}^{2n}$,
$F_{1}=\\{x_{11},\ldots,x_{1n}\\}$, $F_{2}=\\{x_{21},\ldots,x_{2n}\\}$ and
$\Delta_{\alpha}({\mathcal{F}}(\Delta_{2,n})^{t})=\big{\langle}\,F\in
Facets\,(\delta({\mathcal{F}}(\Delta_{2,n})))\mid\sum\limits_{i\notin
F}\alpha_{i}\leq t-1\,\big{\rangle}$. Moreover, if
$t=\min\,\\{k\mid\operatorname{depth}\,(S/{\mathcal{F}}(\Delta_{2,n})^{k})=1\\}$,
then such $\alpha$ must satisfy $\sum\limits_{i\notin
F_{1}}\alpha_{i}=\sum\limits_{i\notin F_{2}}\alpha_{i}=t-1.$

######  Theorem 4.4.

Let $\Delta_{2,n}$ be a chessboard complex with $n\geq 2$, then for any $t\geq
1$, we have

  1. (a)

$\operatorname{reg}\,(S/{\mathcal{F}}(\Delta_{2,n})^{t})=2t$.

  2. (b)

$\operatorname{depth}\,(S/{\mathcal{F}}(\Delta_{2,n})^{t})=1$ or $2$.
Moreover, $\operatorname{depth}\,(S/{\mathcal{F}}(\Delta_{2,n})^{t})=1$ if and
only if $n=3$ and $t\geq 4$, or $n\geq 4$ and $t\geq 3$.

  3. (c)

$S/{\mathcal{F}}(\Delta_{2,2}^{t})$ is Cohen-Macaulay of dimension $2$.

###### Proof.

(a) ${\mathcal{F}}(\Delta_{2,n})$ can be regarded as the edge ideal of a
connected bipartite graph $G$ with bipartition
$\\{x_{11},\ldots,x_{1n}\\}\sqcup\\{x_{21},\ldots,x_{2n}\\}$. The bipartite
complement $G^{bc}$ of $G$ contains an induced cycle $C_{4}$ whose edges are
$x_{11}x_{12},x_{11}x_{21},x_{21}x_{22},x_{12}x_{22}$ and does not contain the
induced circle of length $\geq 6$, since $x_{1i}x_{2i}\in E(G^{c})$ for any
$1\leq i\leq n$. Hence $\operatorname{reg}\,({\mathcal{F}}(\Delta_{2,n}))=3$
by [7, Theorem 3.1], the desired result follows from Lemmas 4.1 and 2.2.

(b) ${\mathcal{F}}(\Delta_{2,n})$ is normally torsion free and
$\min\\{\operatorname{depth}\,(S/{\mathcal{F}}(\Delta_{2,n})^{t})\mid t\geq
1\\}=1$ by [20, Theorem 5.9] and [21, Theorem 4.4], since
${\mathcal{F}}(\Delta_{2,n})$ is the edge ideal of a connected bipartite
graph. By [3, Proposition 1.2.13] and Theorem 3.3, we obtain

$\displaystyle\operatorname{depth}(S/{\mathcal{F}}(\Delta_{2,n})^{t})$
$\displaystyle\leq$
$\displaystyle\min\\{\dim(S/P):P\in\operatorname{Ass}({\mathcal{F}}(\Delta_{2,n})^{t})\\}$
$\displaystyle=$
$\displaystyle\min\\{\dim(S/P):P\in\operatorname{Ass}({\mathcal{F}}(\Delta_{2,n}))\\}=2.$

Therefore, $\operatorname{depth}\,(S/{\mathcal{F}}(\Delta_{2,n})^{t})=1$ or
$2$.

If $n=2$, then ${\mathcal{F}}(\Delta_{2,2})=(x_{11}x_{22},x_{12}x_{21})$ is a
complete intersection ideal. By Lemma 2.8, we have
$\operatorname{depth}\,(S/{\mathcal{F}}(\Delta_{2,2})^{t})=2$ for any $t\geq
1$; If $n=3$, then ${\mathcal{F}}(\Delta_{2,n})$ is the edge ideal of a cycle
graph of length $6$. It follows that
$\operatorname{depth}\,(S/{\mathcal{F}}(\Delta_{2,n})^{t})=1$ if and only if
$t\geq 4$ by [21, Lemma 5.3].

Now suppose $n\geq 4$. Choose
$\alpha=(\alpha_{1},\ldots,\alpha_{2n})\in{\mathbb{N}}^{2n}$, where

$\alpha_{i}=\left\\{\begin{array}[]{l@{\ \ }l}{1}&\text{if $i=1,2,n+3,n+4$}\\\
{0}&\text{otherwise},\\\ \end{array}\right.$

one has $\Delta_{\alpha}({\mathcal{F}}(\Delta_{2,n})^{t})=\langle
F_{1},F_{2}\rangle$ for all $t\geq 3$, where
$F_{1}=\\{x_{11},\ldots,x_{1n}\\}$ and $F_{2}=\\{x_{21},\ldots,x_{2n}\\}$.

If $\Delta_{\beta}({\mathcal{F}}(\Delta_{2,n})^{2})=\langle
F_{1},F_{2}\rangle$ for some
$\beta=(\beta_{1},\ldots,\beta_{2n})\in{\mathbb{N}}^{2n}$, then
$\sum\limits_{i=1}^{n}\beta_{i}\leq 1$ because of
$F_{2}\in\Delta_{\beta}({\mathcal{F}}(\Delta_{2,n})^{2})$. Similarly,
$\sum\limits_{i=n+1}^{2n}\beta_{i}\leq 1$. Therefore, there exists at most one
$\beta_{i}=1$ for $1\leq i\leq n$. Thus
$\\{x_{1i},x_{2i}\\}\in\Delta_{\beta}({\mathcal{F}}(\Delta_{2,n})^{2})$,
contradicting with the supposition
$\Delta_{\beta}({\mathcal{F}}(\Delta_{2,n})^{2})=\langle F_{1},F_{2}\rangle$.
It follows from Lemma 4.3 that
$\operatorname{depth}\,(S/{\mathcal{F}}(\Delta_{2,n})^{t})=1$ if and only if
$t\geq 3$.

(c) Since ${\mathcal{F}}(\Delta_{2,2})=(x_{11}x_{22},x_{12}x_{21})$ can be
regarded as the edge ideal of a bipartite graph, and its symbolic powers are
consistent with ordinary powers. Hence
${\mathcal{F}}(\Delta_{2,2})^{t}=(x_{11}x_{22},x_{12}x_{21})^{t}=(x_{11},x_{12})^{t}\cap(x_{21},x_{22})^{t}\cap(x_{11},x_{21})^{t}\cap(x_{21},x_{22})^{t}$
for any $t\geq 1$. It follows that
$\dim\,(S/{\mathcal{F}}(\Delta_{2,2})^{t})=2$. The result follows from (b).

##  5\. The facet ideal ${\mathcal{F}}(\Delta_{m,n})$ with $m\geq 3$

In this section, we provide some lower bounds for depth and regularity of the
facet ideal ${\mathcal{F}}(\Delta_{m,n})$ of a chessboard complex
$\Delta_{m,n}$ under the condition that $n\geq m\geq 3$, we also prove that
these lower bounds can be attained if $m=3$.

Let $\Delta$ be a simplicial complex on the vertex set $V$, $Facets\,(\Delta)$
the set of its facets. A subset $M$ of $Facets\,(\Delta)$ is called a matching
of $\Delta$ if the facets in $M$ are pairwise disjoint. Moreover, if
$Facets\,(\Delta|_{A})=M$ for $A=\cup_{F\in M}F$, then $M$ is called an
induced matching of $\Delta$.

######  Lemma 5.1.

([5, Corollary 3.5] and [18, Corollary 3.9]) Let $\Delta$ be a simplicial
complex with facet ideal ${\mathcal{F}}(\Delta)$. Then

$\operatorname{reg}\,(S/{\mathcal{F}}(\Delta))\geq\Big{\\{}|\bigcup\limits_{i=1}^{k}F_{i}|-k\mid\\{F_{1},\ldots,F_{k}\\}\text{\
is an induced matching in }\Delta\Big{\\}}.$

######  Corollary 5.2.

Let $\Delta_{m,n}$ be a chessboard complex with $n\geq m\geq 3$, then

$\operatorname{reg}(S/{\mathcal{F}}(\Delta_{m,n}))\geq 2(m-1).$

###### Proof.

Let $F_{1}=\\{x_{11},x_{22},\ldots,x_{mm}\\}$ and
$F_{2}=\\{x_{12},x_{23},\ldots,x_{(m-1)m},x_{m1}\\}$, then $\\{F_{1},F_{2}\\}$
is an induced matching of $\Delta_{m,n}$. The result follows from Lemma 5.1.

######  Theorem 5.3.

Let $\Delta_{3,n}$ be a chessboard complex with $n\geq 3$, then

$\operatorname{reg}\,(S/{\mathcal{F}}(\Delta_{3,n}))=4.$

###### Proof.

It is clear that $\operatorname{reg}\,(S/{\mathcal{F}}(\Delta_{3,n}))\geq 4$
by Corollary 5.2, so it is enough to show that
$\operatorname{reg}\,({\mathcal{F}}(\Delta_{3,n}))\leq 5$ by Lemma 2.2 (3). We
sort all the elements of $\mathcal{G}({\mathcal{F}}(\Delta_{2,n}))$ in
lexicographic order with
$x_{11}>x_{12}>\cdots>x_{1n}>x_{21}>x_{22}>\cdots>x_{2n}$, that is :

$x_{11}x_{22}>x_{11}x_{23}>\cdots>x_{11}x_{2n}>x_{12}x_{21}>x_{12}x_{23}>\cdots>x_{1n}x_{21}>\cdots>x_{1n}x_{2,n-1}.$

Let $u_{p}$ be the $p$-th element in
$\mathcal{G}({\mathcal{F}}(\Delta_{2,n}))$, where $1\leq p\leq r$ and
$r=|\mathcal{G}({\mathcal{F}}(\Delta_{2,n}))|$. Put
$J_{0}={\mathcal{F}}(\Delta_{3,n})$, $Q_{p}=(u_{1},u_{2},\ldots,u_{p})$ and
$J_{p}=J_{0}+Q_{p}$ for any $p\in[r]$. By Remark 2.11, we have

$None$
$\operatorname{reg}\,({\mathcal{F}}(\Delta_{3,n}))=\operatorname{reg}\,(J_{0})\leq\max\\{\operatorname{reg}\,(J_{p}:u_{p+1})+2,\
\operatorname{reg}\,(J_{r})\mid 0\leq p<r\\}.$

Note that
$\operatorname{reg}\,(J_{r})=\operatorname{reg}\,({\mathcal{F}}(\Delta_{2,n}))=3$
by Theorem 4.4. It is enough to show that
$\operatorname{reg}\,(J_{p}:u_{p+1})\leq 3$ for any $0\leq p<r$ by formula
(1). Let $u_{p+1}=x_{1i}x_{2j}$ with $1\leq i,j\leq n$ and $i\neq j$. By a
direct calculation, we obtain that

$None$
$J_{0}:u_{p+1}={\mathcal{F}}(\Delta_{3,n}):x_{1i}x_{2j}=K_{3}+(x_{3i})(K_{1}+(x_{1j})K_{2})+(x_{3j})(K_{2}+(x_{2i})K_{1}),$

and

$None$
$Q_{p}:u_{p+1}=Q_{p}:x_{1i}x_{2j}=\left\\{\begin{array}[]{ll}L_{2}&\text{if}\
\ i=1\ \text{and}\ j>1,\\\ L_{1}+L_{2}&\text{if}\ \ j>i\geq 2,\\\
L_{1}+(x_{11})L_{3}&\text{if}\ \ j=1\ \text{and}\ i>1,\\\
L_{1}+L_{2}+(x_{1j})L_{3}&\text{if}\ \ i>j\geq 2,\end{array}\right.$

where $\mathcal{G}(K_{\ell})=\\{x_{\ell 1},\ldots,x_{\ell
m}\\}\setminus\\{x_{\ell i},x_{\ell j}\\}$ for $\ell\in[3]$,
$\mathcal{G}(L_{1})=\\{x_{11},\ldots,x_{1(i-1)}\\}\setminus\\{x_{1j}\\}$,
$\mathcal{G}(L_{2})=\\{x_{21},\ldots,x_{2(j-1)}\\}\setminus\\{x_{2i}\\}$ and
$\mathcal{G}(L_{3})=\\{x_{2(j+1)},\ldots,x_{2n}\\}$. By Lemma 2.7 and formulas
(2), (3), one has

$\displaystyle\hskip 28.45274pt\mbox{reg}\,(J_{p}:u_{p+1})$ $\displaystyle=$
$\displaystyle\mbox{reg}\,((J_{0}:u_{p+1})+(Q_{p}:u_{p+1}))$
$\displaystyle\leq$
$\displaystyle\left\\{\begin{array}[]{ll}\mbox{reg}\,(J_{0}:u_{p+1})&\text{if}\
\ i<j,\\\ \mbox{reg}\,((J_{0}:u_{p+1})+(x_{1j})L_{3})&\text{if}\ \
i>j.\end{array}\right.\hskip 56.9055pt(4)$

We distinguish into the following two cases:

(i) If $i<j$. By formula (2), one has

$J_{0}:u_{p+1}x_{3j}=K_{3}+(x_{3i})(K_{1}+(x_{1j})K_{2})+K_{2}+(x_{2i})K_{1}=K_{3}+K_{2}+(x_{2i},x_{3i})K_{1}$

and $((J_{0}:u_{p+1}),x_{3j})=K_{3}+(x_{3j})+(x_{3i})(K_{1}+(x_{1j})K_{2})$.
By Lemma 2.5 (3), (4), we get that
$\operatorname{reg}\,(J_{0}:u_{p+1}x_{3j})=\operatorname{reg}\,((x_{2i},x_{3i})K_{1})=\operatorname{reg}\,((x_{2i},x_{3i}))+\operatorname{reg}\,(K_{1})=2$
and

$\displaystyle\operatorname{reg}\,((J_{0}:u_{p+1}),x_{3j})$ $\displaystyle=$
$\displaystyle\operatorname{reg}\,(K_{3}+(x_{3j})+(x_{3i})(K_{1}+(x_{1j})K_{2}))$
$\displaystyle=$
$\displaystyle\operatorname{reg}\,((x_{3i})(K_{1}+(x_{1j})K_{2}))=\operatorname{reg}\,((x_{3i}))+\operatorname{reg}\,(K_{1}+(x_{1j})K_{2})$
$\displaystyle=$ $\displaystyle 1+\operatorname{reg}\,((x_{1j})K_{2})=3.$

Hence, using Corollary 2.10, we have

$\operatorname{reg}\,(J_{p}\\!:\\!u_{p+1})\leq\operatorname{reg}\,(J_{0}\\!:\\!u_{p+1})\leq\max\\{\operatorname{reg}\,(J_{0}\\!:u_{p+1}x_{3j})+1,\operatorname{reg}\,((J_{0}\\!:\\!u_{p+1}),x_{3j})\\}=3.$

(ii) If $i>j$. Let $L=(J_{0}:u_{p+1})+(x_{1j})L_{3}$ for any $0\leq p\leq
r-1$. Using Corollary 2.10 repeatedly, we obtain

$\displaystyle\operatorname{reg}\,(J_{p}\\!:\\!u_{p+1})$ $\displaystyle\leq$
$\displaystyle\operatorname{reg}(L)\leq\max\\{\operatorname{reg}(L:x_{3i})+1,\operatorname{reg}(L,x_{3i})\\}$
$\displaystyle\leq$
$\displaystyle\max\\{\operatorname{reg}(L:x_{3i}x_{1j})+2,\operatorname{reg}(L:x_{3i},x_{1j})+1,\operatorname{reg}((L:x_{3j}),x_{3i})+1,$
$\displaystyle\ \ \ \ \ \ \operatorname{reg}(L,x_{3i},x_{3j})\\}.\hskip
199.16928pt(5)$

Note that

$\displaystyle(L:x_{3i}x_{1j})$ $\displaystyle=$ $\displaystyle
K_{3}+K_{1}+K_{2}+L_{3},$ $\displaystyle((L:x_{3i}),x_{1j})$ $\displaystyle=$
$\displaystyle K_{3}+K_{1}+(x_{3j})K_{2}+(x_{1j}),$
$\displaystyle((L:x_{3j}),x_{3i})$ $\displaystyle=$ $\displaystyle
K_{3}+K_{2}+(x_{2i})K_{1}+(x_{1j})L_{3}+(x_{3i})$ $\displaystyle=$
$\displaystyle K_{3}+K_{2}+(x_{2i})(K_{1},x_{1j})+(x_{3i}),$
$\displaystyle(L,x_{3i},x_{3j})$ $\displaystyle=$ $\displaystyle
K_{3}+(x_{1j})L_{3}+(x_{3i},x_{3j}).$

By Lemma 2.5 (3) and (4), we obtain $\mbox{reg}\,(L:x_{3i}x_{1j})=1$ and

$\mbox{reg}\,(((L:x_{3i}),x_{1j}))=\mbox{reg}\,(((L:x_{3j}),x_{3i}))=\mbox{reg}\,((L,x_{3i},x_{3j}))=2.$

Thus, by formula (5), we obtain that $\operatorname{reg}\,(J_{p}:u_{p+1})\leq
3$.

Given a chessboard complex $\Delta_{m,n}$ with $n\geq m\geq 2$. For $1\leq
i\leq n$, we define $A_{m,i}$ to be a subcomplex of $\Delta_{m,n}$, the set of
its facets is $Facets\,(A_{m,i})=\\{F\in Facets\,(\Delta_{m,n})\mid
x_{ms}\notin F\text{\ \ for any\ }1\leq s\leq i\\}$ and $B_{m,i}$ to be a
chessboard complex obtained by removing the $m$-th row cells and $i$-th column
cells of of the original chessboard. It is obvious that $A_{m,n}=\emptyset$.

######  Example 5.4.

For a chessboard complex $\Delta_{2,3}$, $A_{2,2}$ is a subcomplex of
$\Delta_{2,3}$ with facets $\\{x_{11},x_{23}\\}$ and $\\{x_{12},x_{23}\\}$.
$B_{2,2}$ is also a chessboard complex with facets $\\{x_{11}\\}$ and
$\\{x_{13}\\}$.

######  Lemma 5.5.

Let $\Delta_{m,n}$ be a chessboard complex and
${\mathcal{F}}_{0}:={\mathcal{F}}(\Delta_{m,n})$ its facet ideal. For any
$1\leq i\leq n$, let ${\mathcal{F}}_{i}$ and $G_{i}$ be facet ideals of
$A_{m,i}$ and $B_{m,i}$, respectively. Then

  1. (a)

${\mathcal{F}}_{n}=(0)$ and
$\sum\limits_{i=1}^{n}G_{i}={\mathcal{F}}(\Delta_{m-1,n})$, where
$\Delta_{m-1,n}$ is a chessboard complex obtained by removing the $m$-th row
cells of $\Delta_{m,n}$,

  2. (b)

$({\mathcal{F}}_{i-1}:x_{mi})={\mathcal{F}}_{i}+\operatorname{G}_{i}$ and
$({\mathcal{F}}_{i-1},x_{mi})=({\mathcal{F}}_{i},x_{mi})$ for any
$i=1,\ldots,n$.

###### Proof.

(a) is obvious. Since
${\mathcal{F}}_{i-1}=\sum\limits_{j=i}^{n}x_{mj}\operatorname{G}_{j}$ for
$1\leq i\leq n$, (b) follows.

######  Remark 5.6.

Let $\Delta_{m,n}$ be a chessboard complex and
${\mathcal{F}}_{0}:={\mathcal{F}}(\Delta_{m,n})$ be its facet ideal. Let
$V=\\{x_{m1},x_{m2},\ldots,x_{mn}\\}$, $W\subseteq V$ and
${\mathcal{F}}_{W}=({\mathcal{F}}_{0}:\prod\limits_{x_{mi}\in
W}x_{mi})+P_{W^{c}}$. Using the lemma above repeatedly, we obtain

${\mathcal{F}}_{W}=\sum\limits_{x_{mi}\in W}G_{i}+P_{W^{c}}$

where $P_{W^{c}}$ is an ideal generated by $W^{c}$ and $W^{c}$ is the
complement set of $W$ in $V$.

Given a chessboard complex $\Delta_{m,n}$ with $n\geq m$. For $1\leq
i_{1}<\cdots<i_{m}\leq n$, we define a subcomplex
${\mathcal{D}}_{i_{1},\ldots,i_{m}}$ of $\Delta_{m,n}$, the set of its facets
is $Facets\,({\mathcal{D}}_{i_{1},\ldots,i_{m}})=\\{F\in
Facets\,(\Delta_{m,n})\mid
F\subseteq\bigcup\limits_{\ell=1}^{m}\\{x_{1i_{\ell}},x_{2i_{\ell}},\ldots,x_{mi_{\ell}}\\}\\}$.
In fact, it is a chessboard complex on an $m\times m$ chessboard composed of
the $i_{1}$-th, $\dots,i_{m}$-th columns of the original chessboard.

######  Theorem 5.7.

Let $\Delta_{m,n}$ be a chessboard complex with $n\geq m$ and

$\Omega=\\{{\mathcal{D}}_{i_{1},\ldots,i_{m}}\mid 1\leq i_{1}<\cdots<i_{m}\leq
n\\}.$

Then

$\operatorname{depth}\,\left(\frac{S}{\sum\limits_{{\mathcal{D}}_{i_{1},\ldots,i_{m}}\in\Omega^{\prime}}{\mathcal{F}}({\mathcal{D}}_{i_{1},\ldots,i_{m}})}\right)\geq
2(m-1)$

for any nonempty subcollection $\Omega^{\prime}$ of $\Omega$.

###### Proof.

We apply induction on $m$. Case $m=1$ is trivial. Now suppose $m\geq 2$. Let
$V=\\{x_{m1},x_{m2},\ldots,x_{mn}\\}$, $W\subseteq V$ and denote

${\mathcal{F}}_{W}=(\sum\limits_{{\mathcal{D}}_{i_{1},\ldots,i_{m}}\in\Omega^{\prime}}{\mathcal{F}}({\mathcal{D}}_{i_{1},\ldots,i_{m}}):\prod\limits_{x_{m\ell}\in
W}x_{m\ell})+P_{W^{c}}$

where $P_{W^{c}}$ is an ideal generated by $V\setminus W$. By Corollary 2.12,
it is enough to show that

$\operatorname{depth}\,(S/{\mathcal{F}}_{W})\geq 2(m-1).$

If $W=\emptyset$, then
${\mathcal{F}}_{W}=\sum\limits_{{\mathcal{D}}_{i_{1},\ldots,i_{m}}\in\Omega^{\prime}}{\mathcal{F}}({\mathcal{D}}_{i_{1},\ldots,i_{m}})+P_{W^{c}}=P_{W^{c}}=(x_{m1},\ldots,x_{mn})$.
In this case,
$\operatorname{depth}\,(S/{\mathcal{F}}_{W})=\operatorname{depth}\,(S/(x_{m1},\ldots,x_{mn}))=n(m-1)\geq
2(m-1).$

If $W\neq\emptyset$. We may assume
$W=\\{x_{ms_{1}},x_{ms_{2}},\ldots,x_{ms_{k}}\\}$ and

$\Omega^{\prime\prime}=\\{{\mathcal{D}}_{i_{1},\ldots,i_{m}}\in\Omega^{\prime}\mid\\{i_{1},\ldots,i_{m}\\}\cap\\{s_{1},\ldots,s_{k}\\}\neq\emptyset\\}.$

We consider the following two cases:

(a) If $\Omega^{\prime\prime}=\emptyset$, then
$\\{i_{1},\ldots,i_{m}\\}\cap\\{s_{1},\ldots,s_{k}\\}=\emptyset$ for any
${\mathcal{D}}_{i_{1},\ldots,i_{m}}\in\Omega^{\prime}$. It follows that

$({\mathcal{F}}({\mathcal{D}}_{i_{1},\ldots,i_{m}}):\prod\limits_{x_{m\ell}\in
W}x_{m\ell})={\mathcal{F}}({\mathcal{D}}_{i_{1},\ldots,i_{m}})\subset
P_{W^{c}}$

because of $\\{x_{mi_{1}},\ldots,x_{mi_{m}}\\}\subseteq W^{c}$. Thus

${\mathcal{F}}_{W}=(\sum\limits_{{\mathcal{D}}_{i_{1},\ldots,i_{m}}\in\Omega^{\prime}}{\mathcal{F}}({\mathcal{D}}_{i_{1},\ldots,i_{m}}):\prod\limits_{x_{m\ell}\in
W}x_{m\ell})+P_{W^{c}}=P_{W^{c}}.$

Thus, we have
$\operatorname{depth}\,(S/{\mathcal{F}}_{W})=\operatorname{depth}\,(S/P_{W^{c}})=n(m-1)+k>2(m-1)$.

(b) If $\Omega^{\prime\prime}\neq\emptyset$, then

$\displaystyle{\mathcal{F}}_{W}\\!\\!$ $\displaystyle=$
$\displaystyle\\!\\!(\sum\limits_{{\mathcal{D}}_{i_{1},\ldots,i_{m}}\in\Omega^{\prime}}{\mathcal{F}}({\mathcal{D}}_{i_{1},\ldots,i_{m}}):\prod\limits_{x_{m\ell}\in
W}x_{m\ell})+P_{W^{c}}$ $\displaystyle=$
$\displaystyle\\!\\!\\!\\!\sum\limits_{{\mathcal{D}}_{i_{1},\ldots,i_{m}}\in\Omega^{\prime\prime}}\\!\\!\\!({\mathcal{F}}({\mathcal{D}}_{i_{1},\ldots,i_{m}}):\\!\\!\prod\limits_{x_{m\ell}\in
W}\\!\\!x_{m\ell})+\\!\\!\\!\\!\sum\limits_{{\mathcal{D}}_{i_{1},\ldots,i_{m}}\in\Omega^{\prime}\setminus\Omega^{\prime\prime}}({\mathcal{F}}({\mathcal{D}}_{i_{1},\ldots,i_{m}}):\\!\\!\prod\limits_{x_{m\ell}\in
W}\\!\\!x_{m\ell})+P_{W^{c}}$ $\displaystyle=$
$\displaystyle\\!\\!\\!\sum\limits_{{\mathcal{D}}_{i_{1},\ldots,i_{m}}\in\Omega^{\prime\prime}}\\!\\!\\!({\mathcal{F}}({\mathcal{D}}_{i_{1},\ldots,i_{m}})\\!:\\!\\!\prod\limits_{x_{m\ell}\in
W}x_{m\ell})+P_{W^{c}}$ $\displaystyle=$
$\displaystyle\\!\\!\\!\sum\limits_{{\mathcal{D}}_{i_{1},\ldots,i_{m}}\in\Omega^{\prime\prime}}\\!\sum\limits_{s_{\ell}\in
A_{i_{1},\ldots,i_{m}}}\\!\\!\\!\\!\\!{\mathcal{F}}({\mathcal{B}}_{i_{1},\ldots,i_{m},s_{\ell}})\\!+\\!P_{W^{c}}$
$\displaystyle=$ $\displaystyle{\mathcal{F}}^{\prime}_{W}+P_{W^{c}},$

where the third and fourth equalities hold because of
$({\mathcal{F}}({\mathcal{D}}_{i_{1},\ldots,i_{m}}):\prod\limits_{x_{m\ell}\in
W}x_{m\ell})={\mathcal{F}}({\mathcal{D}}_{i_{1},\ldots,i_{m}})\subset
P_{W^{c}}$ for any
${\mathcal{D}}_{i_{1},\ldots,i_{m}}\in\Omega^{\prime}\setminus\Omega^{\prime\prime}$
and Remark 5.6, where
$A_{i_{1},\ldots,i_{m}}=\\{i_{1},\ldots,i_{m}\\}\cap\\{s_{1},\ldots,s_{k}\\}$,
${\mathcal{B}}_{i_{1},\ldots,i_{m},s_{\ell}}$ is a chessboard complex obtained
by removing the $m$-th row cells and $s_{\ell}$-th column cells of the
chessboard corresponding to ${\mathcal{D}}_{i_{1},\ldots,i_{m}}$ and

${\mathcal{F}}^{\prime}_{W}=\sum\limits_{{\mathcal{D}}_{i_{1},\ldots,i_{m}}\in\Omega^{\prime\prime}}\sum\limits_{s_{\ell}\in
A_{i_{1},\ldots,i_{m}}}{\mathcal{F}}({\mathcal{B}}_{i_{1},\ldots,i_{m},s_{\ell}}).$

We distinguish into the following two cases:

(i) If $k=1$, then $A_{i_{1},\ldots,i_{m}}=\\{s_{1}\\}$ for any
${\mathcal{D}}_{i_{1},\ldots,i_{m}}\in\Omega^{\prime\prime}$. Thus there
exists some $j\in[m]$ such that $i_{j}=s_{1}$. In this case,
${\mathcal{F}}^{\prime}_{W}=\sum\limits_{{\mathcal{D}}_{i_{1},\ldots,i_{m}}\in\Omega^{\prime\prime}}{\mathcal{F}}({\mathcal{B}}_{i_{1},\ldots,i_{m},s_{\ell}})$
and ${\mathcal{B}}_{i_{1},\ldots,i_{m},s_{1}}$ is a subcomplex of
$\Delta_{m-1,n-1}$ with
$Facets\,({\mathcal{B}}_{i_{1},\ldots,i_{m},s_{1}})=\\{F\in
Facets\,(\Delta_{m-1,n-1})\mid
F\subset\bigcup\limits_{\begin{subarray}{c}j=1\\\ i_{j}\neq s_{1}\\\
\end{subarray}}^{m}\\{x_{1i_{j}},x_{2i_{j}},\ldots,x_{mi_{j}}\\}\\}$, whereas
$\Delta_{m-1,n-1}$ is a chessboard complex obtained by deleting the $m$-th row
cells and the $s_{1}$-th column cells of the $m\times n$ chessboard. By
induction hypothesis, we obtain

$\operatorname{depth}\,(S/{\mathcal{F}}_{W})=\operatorname{depth}\,(S_{1}/{\mathcal{F}}^{\prime}_{W})+m\geq
2(m-2)+m\geq 2(m-1)$

where $S_{1}=k[x_{ij}\mid i\in[m-1]\text{ and\ }j\in[n]\setminus\\{s_{1}\\}]$.

(ii) If $k\geq 2$, then for any $s_{\ell}\in A_{i_{1},\ldots,i_{m}}$, all
${\mathcal{B}}_{i_{1},\ldots,i_{m},s_{\ell}}$ are subcomplexes of
$\Delta_{m-1,n}$ with
$Facets\,({\mathcal{B}}_{i_{1},\ldots,i_{m},s_{\ell}})=\\{F\in
Facets\,(\Delta_{m-1,n})\mid F\subset\bigcup\limits_{\begin{subarray}{c}j=1\\\
i_{j}\neq s_{\ell}\\\
\end{subarray}}^{m}\\{x_{1i_{j}},x_{2i_{j}},\ldots,x_{mi_{j}}\\}\\}$, where
$\Delta_{m-1,n}$ is a chessboard complex obtained by removing the $m$-th row
cells of the $m\times n$ chessboard. By induction hypothesis, we obtain

$\operatorname{depth}\,(S/{\mathcal{F}}_{W})=\operatorname{depth}\,(S_{2}/{\mathcal{F}}^{\prime}_{W})+k\geq
2(m-2)+k\geq 2(m-1)$

where $S_{2}=k[x_{ij}\mid i\in[m-1]\text{ and\ }j\in[n]]$.

######  Corollary 5.8.

Let $\Delta_{m,n}$ be a chessboard complex with $n\geq m\geq 1$, then

$\operatorname{depth}\,(S/{\mathcal{F}}(\Delta_{m,n}))\geq 2(m-1).$

###### Proof.

Since ${\mathcal{F}}(\Delta_{m,n})$ can be expressed as
${\mathcal{F}}(\Delta_{m,n})=\sum\limits_{{\mathcal{D}}_{i_{1},\ldots,i_{m}}\in\Omega}{\mathcal{F}}({\mathcal{D}}_{i_{1},\ldots,i_{m}})$
where $\Omega=\\{{\mathcal{D}}_{i_{1},\ldots,i_{m}}\mid 1\leq
i_{1}<\cdots<i_{m}\leq n\\}$. The desired result follows from Theorem 5.7.

Let $G$ be a simple graph on the set $V=\\{x_{1},\ldots,x_{n}\\}$, and let
$m\leq n$ be an integer. A path of length $m$ of $G$ is a sequence
$x_{i_{1}},\ldots,x_{i_{m}}$ of vertices with the property that
$(x_{i_{k}},x_{i_{k+1}})$ is a directed edge of $G$ from $x_{i_{k}}$ to
$x_{i_{k+1}}$ for $k=1,\ldots,m-1$. The path ideal of $G$ of length $m$ is a
monomial ideal

$P_{m}=(\\{x_{i_{1}}\cdots x_{i_{m}}\mid x_{i_{1}},\ldots,x_{i_{m}}\text{\ is
a path of length\ }m\text{\ in\ }G\\}).$

In particular, if $G$ is a cycle, then

$P_{m}=(x_{i+1}\cdots x_{i+m}\mid i=1,2,\ldots,n),$

where $x_{i+j}=x_{k}$ if $i+j\equiv k$ mod $n$ for $i\in[n]$, $j\in[m]$. The
following lemma holds from [24, Corollary 5.1] and [25, Theorem 3.3].

######  Lemma 5.9.

Let $G$ be a path or a cycle with $n$ vertices and $P_{m}$ be its path ideal
of length $m$. Then
$\operatorname{reg}\,(P_{2})=\left\lfloor\frac{n+1}{3}\right\rfloor+1$ and
$\operatorname{reg}\,(P_{n-1})=n-1$.

######  Lemma 5.10.

Let $S=k[x_{1},\ldots,x_{6}]$ be the polynomial ring over a field $k$ and let
$I=(x_{1}x_{2},x_{1}x_{3},x_{2}x_{3},x_{4}x_{5},x_{4}x_{6},x_{5}x_{6},x_{1}x_{4},x_{2}x_{5},x_{3}x_{6})\subset
S$ be a monomial ideal. Then we obtain by using CoCoA [4] that
$\operatorname{reg}\,(I)=3$.

######  Lemma 5.11.

Let $n\geq 3$ be an integer and let

$I=\left(\prod\limits_{k=1}^{n}x_{1k},\prod\limits_{k=1}^{n}x_{2k}\right)+\sum\limits_{\begin{subarray}{c}i,j\in[n]\\\
i<j\end{subarray}}\left(\prod\limits_{k=1}^{2}\frac{x_{k1}\cdots
x_{kn}}{x_{ki}x_{kj}}\right)$

be a monomial ideal of $S=k[x_{11},\ldots,x_{1n},x_{21},\ldots,x_{2n}]$. Then
$\operatorname{reg}\,(I)=2n-3.$

###### Proof.

We may write $I$ as $I=J+K$, where
$K=(\prod\limits_{k=1}^{n}x_{1k},\prod\limits_{k=1}^{n}x_{2k},\prod\limits_{k=3}^{n}x_{1k}x_{2k})$
and $J$ is a monomial ideal with
$\mathcal{G}(J)=\mathcal{G}(I)\setminus\mathcal{G}(K)$. We sort all the
elements of $\mathcal{G}(J)$ in lexicographic order with
$x_{11}>x_{12}>\cdots>x_{1n}>x_{21}>x_{22}>\cdots>x_{2n}$, that is:

$\prod\limits_{k=1}^{2}\frac{x_{k1}\cdots
x_{kn}}{x_{k(n-1)}x_{kn}}>\prod\limits_{k=1}^{2}\frac{x_{k1}\cdots
x_{kn}}{x_{k(n-2)}x_{kn}}>\cdots>\prod\limits_{k=1}^{2}\frac{x_{k1}\cdots
x_{kn}}{x_{k1}x_{k4}}>\prod\limits_{k=1}^{2}\frac{x_{k1}\cdots
x_{kn}}{x_{k1}x_{k3}}.$

Let $u_{s}$ be the $s$-th element in $\mathcal{G}(J)$ for $s\in[r]$, where
$r=|\mathcal{G}(J)|$. Put $J_{0}=I$, $J_{s}=K+(u_{s+1},\ldots,u_{r})$ for
$s\in[r-1]$ and $J_{r}=K$.

Note that the degree of $u_{s}$ equals to $2n-4$ for any $s\in[r]$, by Remark
2.13, we obtain that

$\operatorname{reg}(I)=\operatorname{reg}(J_{0})\leq\max\\{\operatorname{reg}(J_{s}:u_{s})+2n-5,\operatorname{reg}(J_{r})\mid\
s\in[r]\\}$

and the equality holds if
$\operatorname{reg}(J_{1}:u_{1})+2n-4>\max\\{\operatorname{reg}\,(J_{s}:u_{s})+2n-5,\
\operatorname{reg}\,(J_{r})\mid 2\leq s\leq r\\}$.

First, we prove that $\operatorname{reg}\,(J_{s}:u_{s})=2$ for any $s\in[r]$.

Since $u_{s}=\prod\limits_{k=1}^{2}\frac{x_{k1}\cdots x_{kn}}{x_{kp}x_{kq}}$
with $p,q\in[n]$, $p<q$ and $(p,q)\neq(1,2)$, one has

$J_{s}=\left(\prod\limits_{k=1}^{n}x_{1k},\prod\limits_{k=1}^{n}x_{2k}\right)+\sum\limits_{\begin{subarray}{c}1\leq
i<p,\,i<j\leq n\\\ \text{or\
}i=p,\,p<j<q\end{subarray}}\left(\prod\limits_{k=1}^{2}\frac{x_{k1}\cdots
x_{kn}}{x_{ki}x_{kj}}\right).$

By a direct calculation, we obtain

$\displaystyle J_{s}:u_{s}$ $\displaystyle=$
$\displaystyle\left(\prod\limits_{k=1}^{n}x_{1k},\prod\limits_{k=1}^{n}x_{2k}\right):u_{s}+\sum\limits_{\begin{subarray}{c}1\leq
i<p,\,i<j\leq n\\\ \text{or\
}i=p,\,p<j<q\end{subarray}}\left(\prod\limits_{k=1}^{2}\frac{x_{k1}\cdots
x_{kn}}{x_{ki}x_{kj}}\right):u_{s}$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ll}(x_{11}x_{1q},x_{21}x_{2q},x_{1q}x_{2q})&\text{if}\
\ p=1,\\\ (x_{1p}x_{1q},x_{2p}x_{2q},x_{1p}x_{2p},x_{1q}x_{2q})&\text{if}\ \
p\geq 2.\\\ \end{array}\right.$

It can be regarded as the edge ideal of a path or cycle with $4$ vertices, it
follows from Lemma 5.9 (1) that $\operatorname{reg}\,(J_{s}:u_{s})=2$.

Next, we prove that $\operatorname{reg}\,(J_{r})=2n-3$.

Let $J_{r}=J_{r+1}+(\prod\limits_{k=1}^{n}x_{2k})$ and
$J_{r+1}=(\prod\limits_{k=1}^{n}x_{1k},\prod\limits_{k=3}^{n}x_{1k}x_{2k})$.
By Lemma 2.6, one has

$\operatorname{reg}\,(J_{r+1})=\operatorname{reg}\,((\prod\limits_{k=1}^{n}\\!x_{1k},\prod\limits_{k=3}^{n}\\!x_{1k}x_{2k}))=|\text{supp}\,(J_{r+1})|-|\mathcal{G}(J_{r+1})|+1=2n-3.$

Moreover, $\operatorname{reg}\,(J_{r+1}:\prod\limits_{k=1}^{n}x_{2k})=n-2$
because of
$J_{r+1}:\prod\limits_{k=1}^{n}x_{2k}=(\prod\limits_{k=3}^{n}x_{1k})$. Thus,
by Corollary 2.10, we have

$\operatorname{reg}\,(J_{r})=\max\\{\operatorname{reg}\,(J_{r+1}:\prod\limits_{k=1}^{n}x_{2k})+n-1,\operatorname{reg}\,(J_{r+1})\\}=2n-3.$

This concludes the proof.

######  Lemma 5.12.

Let $n\geq 4$ be an integer and let

$I=\sum\limits_{j=1}^{n}\left(\frac{x_{11}\cdots
x_{1n}}{x_{1j}},\frac{x_{21}\cdots
x_{2n}}{x_{2j}}\right)+\sum\limits_{\begin{subarray}{c}i,j\in[n-1],\\\
i<j\end{subarray}}\left(\prod\limits_{k=1}^{2}\frac{x_{k1}\cdots
x_{k(n-1)}}{x_{kj}x_{kj}}\right)$

be a monomial ideal of $S=k[x_{11},\ldots,x_{1n},x_{21},\ldots,x_{2n}]$. Then
$\operatorname{reg}\,(I)=2n-5.$

###### Proof.

Let $I=J+K$, where $J=\sum\limits_{j=1}^{n-1}\left(\frac{x_{11}\cdots
x_{1n}}{x_{1j}},\frac{x_{21}\cdots x_{2n}}{x_{2j}}\right)$ and $K$ is a
monomial ideal with $\mathcal{G}(K)=\mathcal{G}(I)\setminus\mathcal{G}(J)$. We
sort all the elements of $\mathcal{G}(J)$ in lexicographic order with
$x_{11}>x_{12}>\cdots>x_{1n}>x_{21}>x_{22}>\cdots>x_{2n}$, that is:

$\frac{x_{11}\cdots x_{1n}}{x_{1(n-1)}}>\cdots>\frac{x_{11}\cdots
x_{1n}}{x_{11}}>\frac{x_{21}\cdots
x_{2n}}{x_{2(n-1)}}>\cdots>\frac{x_{21}\cdots x_{2n}}{x_{21}}.$

Let $u_{s}$ be the $s$-th element in $\mathcal{G}(J)$ and $J_{s}$ be a
monomial ideal with
$\mathcal{G}(J_{s})=\mathcal{G}(I)\setminus\\{u_{1},\ldots,u_{s}\\}$ for
$1\leq s\leq 2n-2$. Then $J_{s}=K+(u_{s+1},\ldots,u_{2n-2})$ for any
$s\in[2n-3]$ and
$J_{2n-2}=K=\left(\prod\limits_{k=1}^{n-1}x_{1k},\prod\limits_{k=1}^{n-1}x_{2k}\right)+\sum\limits_{\begin{subarray}{c}i,j\in[n-1],\\\
i<j\end{subarray}}\left(\prod\limits_{k=1}^{2}\frac{x_{k1}\cdots
x_{k(n-1)}}{x_{kj}x_{kj}}\right)$. It follows from Lemma 5.11 that
$\operatorname{reg}\,(J_{2n-2})=2n-5$. Note that the degree of $u_{s}$ is
$n-1$ for any $s\in[2n-2]$, by Remark 2.13, we obtain that

$\operatorname{reg}(I)\leq\max\\{\operatorname{reg}(J_{s}:u_{s})+n-2,\operatorname{reg}(J_{2n-2})\mid\
s\in[2n-2]\\}$

and the equality holds if
$\operatorname{reg}(J_{1}:u_{1})+n-1>\max\\{\operatorname{reg}\,(J_{s}:u_{s})+n-2,\
\operatorname{reg}\,(J_{2n-2})\mid 2\leq j\leq 2n-2\\}$.

Now, we compute $\operatorname{reg}\,(J_{s}:u_{s})$ for any $s\in[2n-2]$. We
divide into the following two cases:

(i) If $u_{s}=\frac{x_{11}\cdots x_{1n}}{x_{1p}}$ for some $p\in[n-1]$, then

$\displaystyle K:u_{s}\\!\\!\\!$ $\displaystyle=$
$\displaystyle\\!\\!\\!\left(\prod\limits_{k=1}^{n-1}x_{1k},\prod\limits_{k=1}^{n-1}x_{2k}\right):u_{s}+\sum\limits_{\begin{subarray}{c}i,j\in[n-1],\\\
i<j\end{subarray}}\left(\prod_{k=1}^{2}\frac{x_{k1}\cdots
x_{k(n-1)}}{x_{kj}x_{kj}}\right):u_{s}$ $\displaystyle=$
$\displaystyle\\!\\!\\!(x_{1p},\prod\limits_{k=1}^{n-1}x_{2k})+\sum\limits_{j\in[n-1]\setminus\\{p\\}}\left(\prod_{k=1}^{2}\frac{x_{k1}\cdots
x_{k(n-1)}}{x_{kp}x_{kj}}\right):\frac{x_{11}\cdots x_{1n}}{x_{1p}}$
$\displaystyle=$
$\displaystyle\\!\\!\\!(x_{1p},\prod\limits_{k=1}^{n-1}x_{2k})+\\!\\!\sum\limits_{j\in[n-1]\setminus\\{p\\}}\\!\\!\\!\left(\frac{x_{21}\cdots
x_{2(n-1)}}{x_{2p}x_{2j}}\right)$

$\displaystyle=$
$\displaystyle\\!(x_{1p})+\\!\\!\sum\limits_{j\in[n-1]\setminus\\{p\\}}\\!\\!\\!\left(\frac{x_{21}\cdots
x_{2(n-1)}}{x_{2p}x_{2j}}\right)$

where the second equality holds because $x_{1p}$ is a factor of
$\prod\limits_{k=1}^{2}\frac{x_{k1}\cdots x_{k(n-1)}}{x_{kj}x_{kj}}$ when
$i,j\neq p$. Therefore, one has

$J_{s}:u_{s}=K:u_{s}+(u_{s+1},\ldots,u_{2n-2}):u_{s}=(x_{1p})+\sum\limits_{j\in[n-1]\setminus\\{p\\}}\left(\frac{x_{21}\cdots
x_{2(n-1)}}{x_{2p}x_{2j}}\right).$

Since $\sum\limits_{j\in[n-1]\setminus\\{p\\}}\left(\frac{x_{21}\cdots
x_{2(n-1)}}{x_{2p}x_{2j}}\right)$ can be regarded as the path ideal of length
$(n-3)$ of a cycle on the set
$\\{x_{21},\ldots,x_{2(n-1)}\\}\setminus\\{x_{2p}\\}$. Thus, by Lemmas 2.5 and
5.9, we obtain

$\operatorname{reg}\,(J_{s}:u_{s})=\operatorname{reg}\,(\sum\limits_{j\in[n-1]\setminus\\{p\\}}\left(\frac{x_{21}\cdots
x_{2(n-1)}}{x_{2p}x_{2j}}\right))=n-3.$

(ii) If $u_{s}=\frac{x_{21}\cdots x_{2n}}{x_{2p}}$ for some $p\in[n-1]$,
similar to (i), we obtain
$J_{s}:u_{s}=(x_{2p})+\sum\limits_{j\in[n-1]\setminus\\{p\\}}\left(\frac{x_{11}\cdots
x_{1(n-1)}}{x_{1p}x_{1j}}\right)$ and $\operatorname{reg}\,(J_{s}:u_{s})=n-3$.
This finishes the proof.

######  Theorem 5.13.

Let $\Delta_{3,n}$ be a chessboard complex with $n\geq 3$, then

$\operatorname{depth}\,(S/{\mathcal{F}}(\Delta_{3,n}))=4.$

###### Proof.

Let ${\mathcal{F}}^{\,\vee}$ be the Alexander dual of the facet ideal
${\mathcal{F}}(\Delta_{3,n})$. Then, by Lemma 2.4 and Auslander-Buchsbaum
formula, we have

$\operatorname{depth}\,(S/{\mathcal{F}}(\Delta_{3,n}))=3n-\text{pd}\,(S/{\mathcal{F}}(\Delta_{3,n}))=3n-\operatorname{reg}\,({\mathcal{F}}^{\,\vee}).$

We will prove that $\operatorname{reg}\,({\mathcal{F}}^{\,\vee})=3n-4$, the
desired result follows.

By Theorem 3.3, one has

$\displaystyle{\mathcal{F}}^{\,\vee}$ $\displaystyle=$
$\displaystyle\left(\prod\limits_{k=1}^{n}x_{1k},\prod\limits_{k=1}^{n}x_{2k},\prod\limits_{k=1}^{n}x_{3k}\right)+\sum\limits_{\begin{subarray}{c}i,j\in[n],\\\
i<j\end{subarray}}\left(\prod\limits_{k=1}^{3}\frac{x_{k1}\cdots
x_{kn}}{x_{ki}x_{kj}}\right)+$
$\displaystyle\sum\limits_{j=1}^{n}\left(\prod\limits_{k=1,2}\frac{x_{k1}\cdots
x_{kn}}{x_{kj}},\prod\limits_{k=1,3}\frac{x_{k1}\cdots
x_{kn}}{x_{kj}},\prod\limits_{k=2,3}\frac{x_{k1}\cdots
x_{kn}}{x_{kj}}\right).$

Let $J$ be a monomial ideal with
$\mathcal{G}(J)=\mathcal{G}({\mathcal{F}}^{\,\vee})\setminus\\{\prod\limits_{k=1}^{n}x_{3k}\\}$.
Claim: $\operatorname{reg}\,(J:\prod\limits_{k=1}^{n}x_{3k})=2n-3$ and
$\operatorname{reg}(J)\leq 3n-4$. Thus the asserted results hold by Corollary
2.10.

First, we compute $\operatorname{reg}\,(J:\prod\limits_{k=1}^{n}x_{3k})$. Note
that

$J:\prod\limits_{k=1}^{n}x_{3k}=\sum\limits_{j=1}^{n}\left(\frac{x_{11}\cdots
x_{1n}}{x_{1j}},\frac{x_{21}\cdots
x_{2n}}{x_{2j}}\right)+\sum\limits_{\begin{subarray}{c}i,j\in[n],\\\
i<j\end{subarray}}\left(\prod\limits_{k=1}^{2}\frac{x_{k1}\cdots
x_{kn}}{x_{ki}x_{kj}}\right).$

If $n=3$, then
$J:\prod\limits_{k=1}^{3}x_{3k}=\sum\limits_{k=1,2}(x_{k1}x_{k2},x_{k1}x_{k3},x_{k2}x_{k3})+(x_{11}x_{21},x_{12}x_{22},x_{13}x_{23})$.
Thus $\operatorname{reg}\,(J:\prod\limits_{k=1}^{3}x_{3k})=3=2n-3$ by Lemma
5.10.

If $n\geq 4$. Set $W=\left\\{\prod\limits_{k=1}^{2}\frac{x_{k1}\cdots
x_{k(n-1)}}{x_{kp}x_{kq}}\mid p,q\in[n-1],p<q\right\\}$, we sort all the
elements of $W$ in lexicographic order with
$x_{11}>x_{12}>\cdots>x_{1n}>x_{21}>x_{22}>\cdots>x_{2n}$, that is:

$\prod\limits_{k=1}^{2}\frac{x_{k1}\cdots
x_{k(n-1)}}{x_{k(n-2)}x_{k(n-1)}}>\prod\limits_{k=1}^{2}\frac{x_{k1}\cdots
x_{k(n-1)}}{x_{k(n-3)}x_{k(n-1)}}>\cdots>\prod\limits_{k=1}^{2}\frac{x_{k1}\cdots
x_{k(n-1)}}{x_{k1}x_{k3}}>\prod\limits_{k=1}^{2}\frac{x_{k1}\cdots
x_{k(n-1)}}{x_{k1}x_{k2}}.$

Let $u_{s}$ be the $s$-th element in $W$ and
$I_{s}=(J:\prod\limits_{k=1}^{n}x_{3k})+(u_{1},\ldots,u_{s})$ for $s\in[r]$.
Then

$I_{r}=(J:\prod\limits_{k=1}^{n}x_{3k})+(W)=\sum\limits_{j=1}^{n}\left(\frac{x_{11}\cdots
x_{1n}}{x_{1j}},\frac{x_{21}\cdots
x_{2n}}{x_{2j}}\right)+\sum\limits_{\begin{subarray}{c}i,j\in[n-1],\\\
i<j\end{subarray}}\left(\prod\limits_{k=1}^{2}\frac{x_{k1}\cdots
x_{k(n-1)}}{x_{ki}x_{kj}}\right).$

It follows from Lemma 5.12 that $\operatorname{reg}\,(I_{r})=2n-5.\hskip
142.26378pt(1)$

Now, we compute $\operatorname{reg}\,(I_{s-1}:u_{s})$ for any $s\in[r]$.

Set $\prod\limits_{k=1}^{2}\frac{x_{k1}\cdots x_{k(n-1)}}{x_{kp}x_{kq}}$ for
some $p,q\in[n-1]$ and $p<q$. By simple calculations, one has

$\displaystyle(J:\prod\limits_{k=1}^{n}x_{3k}):u_{s}\\!\\!$ $\displaystyle=$
$\displaystyle\\!\\!\sum\limits_{j=1}^{n}\left(\frac{x_{11}\cdots
x_{1n}}{x_{1j}},\frac{x_{21}\cdots
x_{2n}}{x_{2j}}\right):u_{s}+\sum\limits_{\begin{subarray}{c}i,j\in[n],\\\
i<j\end{subarray}}\left(\prod\limits_{k=1}^{2}\frac{x_{k1}\cdots
x_{kn}}{x_{ki}x_{kj}}\right):u_{s}$ $\displaystyle=$
$\displaystyle\sum\limits_{k=1,2}(x_{kp}x_{kq},x_{kp}x_{kn},x_{kq}x_{kn})+(x_{1p}x_{2p},x_{1q}x_{2q},x_{1n}x_{2n}).$

Hence, it is easy to check that

$\displaystyle I_{s-1}:u_{s}$ $\displaystyle=$
$\displaystyle((J:\prod\limits_{k=1}^{n}x_{3k})+(u_{i}\mid i\in[s-1])):u_{s}$
$\displaystyle=$
$\displaystyle(x_{1p}x_{1q},x_{1p}x_{1n},x_{1q}x_{1n},x_{2p}x_{2q},x_{2p}x_{2n},x_{2q}x_{2n},x_{1p}x_{2p},x_{1q}x_{2q},x_{1n}x_{2n}),$

where $(u_{i}\mid i\in[0])=(0)$. It follows from Lemma 5.10 that

$None$ $\operatorname{reg}\,(I_{s-1}:u_{s})=3.$

Therefore, by Remark 2.11 and formulas (1), (2), one has
$\operatorname{reg}\,(J:\prod\limits_{k=1}^{3}x_{3k})=2n-3$.

Finally, we prove $\operatorname{reg}\,(J)\leq 3n-4$.

We may write $J$ as
$J=\left(\prod\limits_{k=1}^{n}x_{1k},\prod\limits_{k=1}^{n}x_{2k}\right)+K_{1}+K_{2}$
where $K_{2}=\sum\limits_{\begin{subarray}{c}i,j\in[n],\\\
i<j\end{subarray}}\left(\prod\limits_{k=1}^{3}\frac{x_{k1}\cdots
x_{kn}}{x_{ki}x_{kj}}\right)$ and
$K_{1}=\sum\limits_{j=1}^{n}\left(\prod\limits_{k=1,2}\\!\\!\frac{x_{k1}\cdots
x_{kn}}{x_{kj}},\prod\limits_{k=1,3}\\!\\!\frac{x_{k1}\cdots
x_{kn}}{x_{kj}},\prod\limits_{k=2,3}\\!\\!\frac{x_{k1}\cdots
x_{kn}}{x_{kj}}\right)$. We sort all the elements of $\mathcal{G}(K_{1})$ and
$\mathcal{G}(K_{2})$ in lexicographic order with
$x_{11}>x_{12}>\cdots>x_{1n}>x_{21}>x_{22}>\cdots>x_{2n}>x_{31}>x_{32}>\cdots>x_{3n}$
respectively. Let $w_{p}$ and $w_{r_{1}+q}$ be the $p$-th element of
$\mathcal{G}(K_{1})$ and the $q$-th element of $\mathcal{G}(K_{2})$
respectively, where $p\in[r_{1}]$, $r_{1}=|\mathcal{G}(K_{1})|$, $q\in[r_{2}]$
and $r_{2}=|\mathcal{G}(K_{2})|$. Put
$J_{s}=\left(\prod\limits_{k=1}^{n}x_{1k},\prod\limits_{k=1}^{n}x_{2k}\right)+(w_{s+1},\ldots,w_{r})$
for any $s\in[r-1]$, where $r=r_{1}+r_{2}$ and
$J_{r}=\left(\prod\limits_{k=1}^{n}x_{1k},\prod\limits_{k=1}^{n}x_{2k}\right)$.
It follows that $\operatorname{reg}\,(J_{r})=2n-1\leq 3n-4$ from Lemma 2.5
(3).

Claim: $\operatorname{reg}\,(J_{s}:w_{s})+d_{s}\leq 3n-3$ for any $s\in[r]$,
where $d_{s}$ is the degree of $w_{s}$. Thus we obtain that
$\operatorname{reg}\,(J)\leq 3n-4$ by Remark 2.13.

We prove this claim by considering the following two cases:

(i) If $w_{s}\in\mathcal{G}(K_{1})$, then we can assume that
$w_{s}=\prod\limits_{k=\ell_{1},\ell_{2}}\frac{x_{k1}\cdots x_{kn}}{x_{kp}}$
for $1\leq\ell_{1}<\ell_{2}\leq 3$ and $p\in[n]$. Thus

$\displaystyle J_{s}:w_{s}$ $\displaystyle=$
$\displaystyle\left((\prod\limits_{k=1}^{n}x_{1k},\prod\limits_{k=1}^{n}x_{2k})+(w_{s+1},\ldots,w_{r})\right):w_{s}$
$\displaystyle=$
$\displaystyle\left(\prod\limits_{k=1}^{n}x_{1k},\prod\limits_{k=1}^{n}x_{2k}\right)\\!:\\!w_{s}\\!+\\!(w_{s+1},\ldots,w_{r_{1}})\\!:\\!w_{s}\\!+\\!\sum\limits_{\begin{subarray}{c}i,j\in[n],\\\
i<j\end{subarray}}\left(\prod\limits_{k=1}^{3}\\!\frac{x_{k1}\cdots
x_{kn}}{x_{ki}x_{kj}}\right)\\!:\\!u_{s}$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ll}(x_{1p},x_{2p})+J^{\prime}_{s}&\text{if}\
\ell_{1}=1\text{\ and\ }\ell_{2}=2,\\\
(x_{\ell_{1}p})+J^{\prime}_{s}&\text{otherwise.}\end{array}\right.$

where $\\{\ell_{3}\\}=[3]\setminus\\{\ell_{1},\ell_{2}\\}$,
$J^{\prime}_{s}=\sum\limits_{j\in[n]\setminus\\{p\\}}\left(\frac{x_{\ell_{31}}\cdots
x_{\ell_{3n}}}{x_{\ell_{3p}}x_{\ell_{3j}}}\right)$.

Note that $J^{\prime}_{s}$ may be regarded as the path ideal of length $(n-2)$
of a cycle on the vertex set
$\\{x_{\ell_{3}1},\ldots,x_{\ell_{3}n}\\}\setminus\\{x_{\ell_{3}p}\\}$. By
Lemmas 2.5 and 5.9 (2), one has

$None$
$\operatorname{reg}(J_{s}:w_{s})+d_{s}=\operatorname{reg}(J^{\prime}_{s})+d_{s}=(n-2)+2(n-1)=3n-4<3n-3.$

(ii) If $w_{s}\in\mathcal{G}(K_{2})$, then
$w_{s}=\prod\limits_{k=1}^{3}\frac{x_{k1}\cdots x_{kn}}{x_{kp}x_{kq}}$ for
some $p,q\in[n]$, where $p<q$. By direct calculations, we obtain

$\displaystyle J_{s}:w_{s}\\!\\!$ $\displaystyle=$
$\displaystyle\\!\\!\left\\{\begin{array}[]{ll}\left(\prod\limits_{k=1}^{n}x_{1k},\prod\limits_{k=1}^{n}x_{2k}\right):w_{s}&\text{if}\
\ s=r\\\
\left(\prod\limits_{k=1}^{n}x_{1k},\prod\limits_{k=1}^{n}x_{2k}\right):w_{s}+(w_{s+1},\ldots,w_{r}):w_{s}&\text{if}\
\ s<r\\\ \end{array}\right.$ $\displaystyle=$
$\displaystyle\\!\\!\left\\{\begin{array}[]{ll}(x_{11}x_{12},x_{21}x_{22})&\text{if}\
s=r\\\
(x_{1p}x_{1q},x_{2p}x_{2q})\\!+\\!\sum\limits_{\begin{subarray}{c}i=p,\,j<q\\\
\text{or}\ 1\leq i<p,j\leq
n\end{subarray}}\\!\\!\\!\left(\prod\limits_{k=1}^{3}\frac{x_{k1}\cdots
x_{kn}}{x_{ki}x_{kj}}\right):w_{s}&\text{if}\ s<r\\\ \end{array}\right.$
$\displaystyle=$
$\displaystyle\\!\\!\left\\{\begin{array}[]{ll}(x_{11}x_{12},x_{21}x_{22})&\text{if}\
\ p=1\text{\ and\ }q=2\\\
(x_{1p}x_{1q},x_{2p}x_{2q},x_{1q}x_{2q}x_{3q})&\text{if}\ \ p=1\text{\ and\
}q\geq 3\\\
(x_{1p}x_{1q},x_{2p}x_{2q},x_{1p}x_{2p}x_{3p},x_{1q}x_{2q}x_{3q})&\text{otherwise}.\end{array}\right.$

Hence $\operatorname{reg}\,(J_{s}:w_{s})=3$ if $p=1$ from the expression of
$J_{s}:w_{s}$ and Lemma 2.6; Otherwise, let
$J^{\prime\prime}_{s}=(x_{1p}x_{1q},x_{2p}x_{2q},x_{1p}x_{2p}x_{3p})$, then
$J^{\prime\prime}_{s}:x_{1q}x_{2q}x_{3q}=(x_{1p},x_{2p})$. From Lemma 2.6,
$\operatorname{reg}\,(J^{\prime\prime}_{s})=3$ and
$\operatorname{reg}\,(J^{\prime\prime}_{s}:x_{1q}x_{2q}x_{3q})=1$. Hence
$\operatorname{reg}\,(J_{s}:w_{s})=3$ by Corollary 2.10. This implies
$\operatorname{reg}\,(J_{s}:w_{s})+d_{s}=3+3(n-2)=3n-3$. This proof is
completed.

Let $I\subset S$ be a monomial ideal. The $a$-invariant of $S/I$, denoted by
$a(S/I)$, is the degree (as a rational function) of the Hilbert series of
$S/I$.

######  Corollary 5.14.

Let $\Delta_{m,n}$ be a chessboard complex with $n\geq m$. If $m\leq 3$, then

$a(S/{\mathcal{F}}(\Delta_{m,n}))=0.$

###### Proof.

By [18, Theorem 3.20], we have $0\leq
a(S/{\mathcal{F}}(\Delta_{m,n}))\leq\operatorname{reg}\,(S/{\mathcal{F}}(\Delta_{m,n}))-\operatorname{depth}\,(S/{\mathcal{F}}(\Delta_{m,n}))=0$,
as desired.

We conclude this paper with the following natural conjectures.

######  Conjecture 5.15.

Let $\Delta_{m,n}$ be a chessboard complex with $n\geq m\geq 1$, then

$\operatorname{reg}\,(S/{\mathcal{F}}(\Delta_{m,n}))=\operatorname{depth}\,(S/{\mathcal{F}}(\Delta_{m,n}))=2(m-1).$

Of course, the above conjecture is true if $m\leq 3$ by Theorem 5.3 and
Theorem 5.13. By using CoCoA [4], we get that
$\operatorname{reg}\,(S/{\mathcal{F}}(\Delta_{4,4}))=\operatorname{depth}\,(S/{\mathcal{F}}(\Delta_{4,4}))=6$,
which is consistent with the above conjecture.

Acknowledgments

The authors are grateful to the computer algebra system CoCoA [4] for
providing us with a large number of examples. This research is supported by
the Natural Science Foundation of Jiangsu Province (No. BK20221353) and by
foundation of the Priority Academic Program Development of Jiangsu Higher
Education Institutions. Finally, the authors would like to thank referees who
read carefully the manuscript and gave very helpful comments, which improved
the paper both in mathematics and presentation.

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# Encoding large information structures in linear algebra and statistical
models

David Banh [1]  Alan Huang [2]
AskExplain
The University of Queensland
Corresponding email<EMAIL_ADDRESS>
###### Abstract

Large information sizes in samples and features can be encoded to speed up the
learning of statistical models based on linear algebra and remove unwanted
signals. Encoding information can reduce both sample and feature dimension to
a smaller representational set. Here two examples are shown on linear mixed
models and mixture models speeding up the run time for parameter estimation by
a factor defined by the user’s choice on dimension reduction (can be linear,
quadratic or beyond based on dimension specification).

GitHub at: https://github.com/AskExplain/encoding_information

##  1 Introduction

Large sizes in the samples and features in statistics creates large matrix
objects in linear algebra. Large information structures generally occur when
the dimensional size is large enough such that the (generalised) inverse of
the cross product of two matrices is computationally infeasible to calculate
quickly (this generally occurs for sizes greater than n = 10,000).

Here, an encoding of the information is proposed to reduce the matrix
dimensions to a tractable size such that after inverting and decoding, gives a
representational structure similar to when the operation runs on the full
matrix. Examples are used for mixture models [1] and linear mixed models [2].

Encoding information can be done with Singular Value Decomposition (SVD), or,
a similar method based on SVD by same authors called Generative Encoding [3]
[4].

##  2 Methods

###  2.1 Linear Mixed Model

For a linear mixed model to be encoded, first consider the mixed model
equation:

$Y=X\beta+Zu+e$

To encode sample information, a function is introduced as a parameter $\alpha$
to re-weight the samples into a reduced dimension:

$\alpha Y=\alpha X\beta+\alpha Zu+e$

Provided each of $Y$, $X$ and $Z$ are of $n$ samples, and $\alpha$ is a
parameter that transforms the $n$ samples into $m$ samples (where $n>m$), then
the final model will be learned via the covariance of a smaller size.

####  2.1.1 Genetic Relatedness Matrix

For example, the general model in genetics to measure the heritability of a
trait is given as:

$Y\sim N(X\beta,G\sigma_{g}+D_{1}\sigma_{e})$

Reducing the sample size would introduce to this model, the following:

$\alpha Y\sim N(\alpha X\beta,(\alpha G\alpha^{T})\sigma_{g}+D_{2}\sigma_{e})$

Notice that $\alpha$ is of $m$ dimensions rather than $n$, enabling the linear
mixed model to learn with a smaller Genetic Relatedness Matrix, yet still
retaining the ability to learn $\sigma_{g}$ the heritability of the phenotype.

###  2.2 Mixture Model

Rather than encoding sample information, the dimensions of the feature
structure can be encoded. For example for a mixture model to be encoded, first
consider the expression:

$Y\sim\pi_{i}N(\mu_{i},\Sigma_{i})$

To encode sample structure, a function is introduced as a parameter $\alpha$
to re-weight the samples into a reduced dimension:

$Y\beta\sim\pi_{i}N(\mu_{i}\beta,\beta^{T}\Sigma_{i}\beta)$

By reducing the dimensions of the mixture model to learn a model from $r$
features rather than $p$ features (where $p>r$) the model can be learned
faster as the feature information $p$ is learned in the reduced dimensions
$r$.

####  2.2.1 Factor analytic models

For example, the general model for mixtures of factor analyses in psychology
or economic studies used to measure the scores of particular behaviours or
events is given as:

$X\sim\pi_{i}N(\mu_{i}+\Lambda_{i}z,D_{i})$
$X|z\sim\pi_{i}N(\mu_{i},\Lambda_{i}\Lambda_{i}^{T}+D_{i})$

where the conditional expression based on the latent features $z$ is found in
the second expression above.

Taking into account the feature encoding to reduce the dimensional size of $p$
takes the dimensions of the features to $r$ dimensions

$X\beta\sim\pi_{i}N(\mu_{i}\beta+\beta\Lambda_{i}z_{\beta},D_{i_{\beta}})$

$X\beta|z_{\beta}\sim\pi_{i}N(\mu_{i}\beta,(\beta\Lambda_{i}\Lambda_{i}^{T}\beta^{T})+D_{i_{\beta}})$

##  3 Results

###  3.1 Linear Mixed Model

To test the results on a linear mixed model, 1000 permutations were run on a
simulated dataset with a heritability of 0.5.

The package GMMAT in R was used to run the analysis with a Genetic Relatedness
Matrix. The table below shows the results comparing the encoded and original
linear mixed models comparing the heritability estimates and runtime of the
full mixed model with the encoded model (including the time to learn the
encoding).

This is for 100 permutation runs of a simulation with 1000 samples and 100
SNPs simulated according to

$Z\sim Binom(m=2,p=0.5)$ $\lambda_{s}\sim N(0,\sqrt{(h_{2}/p)})$ $Y\sim
N(Z\lambda_{s},\sqrt{(1-h_{2})})$

Where the Genetic Relatedness Matrix (1000 by 1000 dimensions) is given by:

$GRM=(ZZ^{T})/p$

Figure 1:  Comparing an encoded and standard linear mixed model for a
heritability estimate distributed around 0.5 (true is at $h^{2}=0.5$), and
runtime in seconds

###  3.2 Mixture Model

To test the results on a linear mixed model, approximately 1000 permutations
were run on a test dataset from the pdfCluster [5] package using the OliveOil
dataset only on the numerical dataset.

Given the OliveOil dataset has two categorical features structured
hierarchically, one as a subset of the other, the categorical feature with the
fewest number of categories was used. This equates to 3 known clusters to
label.

The features were encoded ranging from 2 to 8 (the original number of
numerical features in the OliveOil dataset).

Figure 2:  Mixture model with an encoding to 2 to 8 features, compared with
the original features (p = 8).

##  4 Discussion

Given information is being encoded, it is expected for there to be information
loss leading to higher variability compared to the standard linear mixed
model. However, due to reduced sample size via an encoding the runtime is
faster - almost half the speed of the original mixed model according to the R
package GMMAT [6].

Notice that with an encoding of the features into a lower dimensional space -
mixture model clustering accuracy has more flexibility due to an increase the
degrees of freedom leading to fit worse or better models. On average, the
larger the encoding - up to the original number of features, the higher the
accuracy of the final clustering.

##  5 Conclusion

Information in a vector, matrix, or tensor object can be encoded by learning
representational features that simultaneously: encodes the object and
represents the object via factorisation. Through considering functions that
encodes and re-represents the information via factorisation, an optimal model
is learned that extracts relevant signals from the data object to manipulate
the feature and, or sample structure.

This has implications on linear algebra and statistical methods - encoding the
samples can reduce the computational run time for mixed models when a sample
covariance matrix is used. Alternatively, features can be encoded to reduce
the computational run time of a feature covariance in mixture models.

##  6 Acknowledgements

Professor Geoff McLachlan has been a tremendous help in the guidance of past
work on mixture models (see Deep Gaussian Mixture Models). Professor Jian Yang
has also been an inspiration for the mixed model work. Also a thanks to Yuna
Zhang for jump starting the work by providing preliminary scripts on Average
Information with mixed models.

## References

  * [1] Cinzia Viroli and Geoffrey J. McLachlan.  Deep gaussian mixture models.  Statistics and Computing, 29(1):43–51, Jan 2019.
  * [2] Jian Yang, S. Hong Lee, Michael E. Goddard, and Peter M. Visscher.  Gcta: a tool for genome-wide complex trait analysis.  American journal of human genetics, 88(1):76–82, Jan 2011.  21167468[pmid].
  * [3] David Banh and Alan Huang.  Scalable parametric encoding of multiple modalities.  bioRxiv, 2022.
  * [4] David Banh.  Sample summary with generative encoding.  CoRR, abs/2201.08233, 2022.
  * [5] Adelchi Azzalini and Giovanna Menardi.  Clustering via nonparametric density estimation: The r package pdfcluster.  Journal of Statistical Software, 57(11):1–26, 2014.
  * [6] Han Chen, Chaolong Wang, Matthew P. Conomos, Adrienne M. Stilp, Zilin Li, Tamar Sofer, Adam A. Szpiro, Wei Chen, John M. Brehm, Juan C. Celedón, Susan Redline, George J. Papanicolaou, Timothy A. Thornton, Cathy C. Laurie, Kenneth Rice, and Xihong Lin.  Control for population structure and relatedness for binary traits in genetic association studies via logistic mixed models.  American journal of human genetics, 98(4):653–666, Apr 2016.  27018471[pmid].