# CLASSIFICATION OF COHEN–MACAULAY $t$–SPREAD LEXSEGMENT IDEALS VIA SIMPLICIAL
COMPLEXES

Marilena Crupi, Antonino Ficarra

###### Abstract

We study the minimal primary decomposition of completely $t$–spread lexsegment
ideals via simplicial complexes. We determine some algebraic invariants of
such a class of $t$–spread ideals. Hence, we classify all $t$–spread
lexsegment ideals which are Cohen–Macaulay. †† _Keywords:_ Betti numbers,
$t$–spread ideals, primary decomposition, simplicial complexes, Cohen–Macaulay
rings.
_2020 Mathematics Subject Classification:_ 05E40, 13B25, 13D02, 16W50.

Department of Mathematics and Computer Sciences, Physics and Earth Sciences,
University of Messina, Viale Ferdinando Stagno d’Alcontres 31, 98166 Messina,
Italy

E-mail addresses<EMAIL_ADDRESS><EMAIL_ADDRESS>
## Introduction

In combinatorial commutative algebra, one often associates to a combinatorial
object $X$ a suitable monomial ideal $I_{X}$ in a polynomial ring
$S=K[x_{1},\dots,x_{n}]$ in finitely many variables with coefficients in a
field $K$. Distinguished combinatorial properties of the object $X$ are
reflected by algebraic properties of the ideal $I_{X}$, and conversely. One
says that $X$ is Cohen–Macaulay if the ring $S/I_{X}$ is Cohen–Macaulay. A
typical task is to classify all Cohen–Macaulay objects $X$ of a given class.
For instance, in combinatorics, to a simplicial complex $\Delta$ on the vertex
set $[n]$ one associates the Stanley–Reisner ideal $I_{\Delta}$, _i.e._ , the
ideal generated by all squarefree monomials $x_{i_{1}}\cdots x_{i_{r}}$,
$r\leq n$, such that $\\{i_{1},\ldots,i_{r}\\}\notin\Delta$. A simplicial
complex $\Delta$ is said Cohen–Macaulay if the ideal $I_{\Delta}$ is
Cohen–Macaulay, _i.e._ , $S/I_{\Delta}$ is a Cohen–Macaulay ring. To classify
all Cohen–Macaulay simplicial complexes is an hopeless task. However, many
results can be found, for instance, in [18].

The aim of this article is to classify all Cohen–Macaulay $t$–spread
lexsegment ideals for $t\geq 1$.

Let $S=K[x_{1},\dots,x_{n}]$ be the polynomial ring in $n$ variables with
coefficients over a fixed field $K$. Let $t\geq 0$ be an integer. A monomial
$u=x_{i_{1}}x_{i_{2}}\cdots x_{i_{d}}$ is $t$–spread if $i_{j+1}-i_{j}\geq t$,
for all $j=1,\dots,d-1$. If $t\geq 1$, a $t$–spread monomial is a squarefree
monomial. A monomial ideal is $t$–spread if it is generated by $t$–spread
monomials. Such a topic introduced in [13] has immediately captured the
attention of many algebraists (see, for instance, [1, 2, 4, 3, 5, 6, 11, 15]).
Let $t\geq 1$. A $t$–spread ideal $I$ of $S$ is said a _$t$ –spread lexsegment
ideal_ if for all $t$–spread monomials $u\in I$ and all $t$–spread monomials
$v\in S$ with $\deg(u)=\deg(v)$ and such that
$v\geq_{{\operatorname{slex}}}u$, then $v\in I$ [6], where
$\geq_{{\operatorname{slex}}}$ is the squarefree lexicographic order [8, 18].
Therefore, for $t=1$, such a definition coincides with the _classical_
definition of squarefree lexsegment ideal [8]. Let $t\geq 1$ and let
$M_{n,d,t}$ be the set of all $t$–spread monomials of $S$ of degree $d$.
Assume that $M_{n,d,t}$ is endowed with the squarefree lexicographic order
$\geq_{{\operatorname{slex}}}$. Let $u,v\in M_{n,d,t}$. If
$u\geq_{{\operatorname{slex}}}v$, the set $\mathcal{L}_{t}(u,v)=\\{w\in
M_{n,d,t}:u\geq_{{\operatorname{slex}}}w\geq_{{\operatorname{slex}}}v\\}$ is
called an _arbitrary $t$–spread lexsegment_. If $u$ is the maximum monomial of
$M_{n,d,t}$, or $v$ is the minimum monomial of $M_{n,d,t}$, we say that
$\mathcal{L}_{t}(u,v)$ is an _initial_ or a _final $t$–spread lexsegment_,
respectively. The ideal generated by an arbitrary $t$–spread lexsegment is
called _arbitrary $t$–spread lexsegment ideal_. It is clear that the notion of
initial $t$–spread lexsegment ideal coincides with the notion of $t$–spread
lexsegment ideal in [6]. An arbitrary $t$–spread lexsegment ideal
$I=(\mathcal{L}_{t}(u,v))$ is said _completely $t$–spread lexsegment_ if
$I=(\mathcal{L}_{t}(\max(M_{n,d,t}),v))\cap(\mathcal{L}_{t}(u,\min(M_{n,d,t})))$,
[15]. One can easily observe that every initial and every final $t$–spread
lexsegment ideal is a completely $t$–spread lexsegment, but not all arbitrary
$t$–spread lexsegment ideals are completely $t$–spread lexsegment ideals. In
[15], the authors characterized all completely $t$–spread lexsegment ideals
and classified all completely $t$–spread lexsegment ideals with a linear
resolution. The case $t=0$ has been faced by De Negri and Herzog in [10].

In this article we determine the minimal primary decomposition of completely
$t$–spread lexsegment ideals and some algebraic invariants of such a class of
ideals by tools from the simplicial complex theory. Hence, we classify all
$t$–spread lexsegment ideals which are Cohen–Macaulay. The case $t=0$ has been
considered in [14] and the case $t=1$ has been analyzed in [9].

The article is organized as follows. Section 1 contains preliminary notions
and results. In Section 2, we explicitly compute the minimal primary
decomposition for initial and final $t$–spread lexsegment ideals (Theorems
2.1, 2.7). As a consequence we are able to compute the standard primary
decomposition for the class of completely $t$–spread lexsegments (Theorem
2.12). Moreover, we obtain formulas for the Krull dimension, the projective
dimension and the depth of $S/I$ when $I$ is an initial or a final $t$–spread
lexsegment ideal. In Section 3, we classify all $t$–spread lexsegment ideals
which are Cohen–Macaulay by the results stated in Section 2. The notion of
Betti splitting will be a crucial tool. Finally, Section 4 contains our
conclusions and perspectives.

All the examples are constructed by means of _Macaulay2_ packages one of which
developed in [1]. Furthermore, some functions described in [1] have allowed us
to test classes of Cohen–Macaulay $t$–spread lexsegment ideals whose
determination has been fundamental for the development of Section 3.

## 1\. Preliminaries

Let $S=K[x_{1},\dots,x_{n}]$ be the polynomial ring in $n$ indeterminates with
coefficients in a field $K$. $S$ is a ${\mathbb{N}}$–graded $K$–algebra with
$\deg(x_{i})=1$ for all $i$.

If $I$ is a monomial ideal of $S$, we denote by $G(I)$ the unique minimal set
of monomial generators of $I$. Moreover, we set $G(I)_{j}=\\{u\in
G(I):\deg(u)=j\\}$.

Given a non empty subset $A\subseteq[n]$, we set ${\bf x}_{A}=\prod_{i\in
A}x_{i}$. Moreover, we set ${\bf x}_{\varnothing}=1$. Every monomial $w\in S$
can be written as $w=x_{i_{1}}x_{i_{2}}\cdots x_{i_{d}}$, with sorted indexes
$1\leq i_{1}\leq i_{2}\leq\dots\leq i_{d}\leq n$. We define the support of $w$
as follows:

${\operatorname{supp}}(w)=\\{j:x_{j}\,\mbox{divides
$w$}\\}=\\{i_{1},i_{2},\dots,i_{d}\\}.$

With the notation above,
$\min(w)=\min\big{\\{}i:i\in{\operatorname{supp}}(u)\big{\\}}=i_{1}$,
$\max(w)=\max\big{\\{}i:i\in{\operatorname{supp}}(u)\big{\\}}=i_{d}$.
Moreover, we set $\max(1)=\min(1)=0$. If $i_{1}\neq i_{2}\neq\dots\neq i_{d}$
we say that $u$ is squarefree. In such a case, $u$ can be written as $u={\bf
x}_{A}$ with $A={\operatorname{supp}}(u)$.

We quote next definition from [13].

###### Definition 1.1

Given $n\geq 1,t\geq 0$ and $u=x_{i_{1}}x_{i_{2}}\cdots x_{i_{d}}\in S$, with
$1\leq i_{1}\leq i_{2}\leq\ldots\leq i_{d}\leq n$, we say that $u$ is
$t$–spread if $i_{j+1}-i_{j}\geq t$, for all $j=1,\dots,d-1$. We say that a
monomial ideal $I$ of $S$ is $t$–spread if it is generated by $t$–spread
monomials.

We note that any monomial ideal of $S$ is a $0$–spread monomial ideal, and any
squarefree monomial ideal of $S$ is a $1$–spread monomial ideal.

Let $n,d,t\geq 1$. We denote by $M_{n,d,t}$ the set of all $t$–spread
monomials of $S$ of degree $d$. Throughout this paper we tacitly assume that
$n\geq 1+(d-1)t$, otherwise $M_{n,d,t}=\varnothing$. From [13], we have that
$|M_{n,d,t}|=\binom{n-(t-1)(d-1)}{d}$. The monomial ideal generated by
$M_{n,d,t}$ is denoted by $I_{n,d,t}$ and is called the $t$–spread Veronese
ideal of degree $d$ of the polynomial ring $S=K[x_{1},\dots,x_{n}]$ [13].

From now on we consider $t\geq 1$ and assume that $M_{n,d,t}$ is endowed with
the _squarefree lexicographic order_ , $\geq_{{\operatorname{slex}}}$, [18].
Recall that given $u=x_{i_{1}}x_{i_{2}}\cdots
x_{i_{d}},v=x_{j_{1}}x_{j_{2}}\cdots x_{j_{d}}\in M_{n,d,t}$, with $1\leq
i_{1}<i_{2}<\dots<i_{d}\leq n$, $1\leq j_{1}<j_{2}<\dots<j_{d}\leq n$, then
$u>_{{\operatorname{slex}}}v$ if

$i_{1}=j_{1},\ \dots,\ i_{s-1}=j_{s-1}\ \ \text{and}\ \ i_{s}<j_{s}$

for some $1\leq s\leq d$.

Let $U$ be a not empty subset of $M_{n,d,t}$. We denote by $\max(U)$
($\min(U)$, respectively) the maximum (minimum, respectively) monomial $w\in
U$, with respect to $\geq_{{\operatorname{slex}}}$. One has

$\displaystyle\max(M_{n,d,t})$ $\displaystyle=x_{1}x_{1+t}x_{1+2t}\cdots
x_{1+(d-1)t},$ $\displaystyle\min(M_{n,d,t})$
$\displaystyle=x_{n-(d-1)t}x_{n-(d-2)t}\cdots x_{n-t}x_{n}.$

In [15], the following definitions have been introduced.

###### Definition 1.2

Let $u,v\in M_{n,d,t}$, $u\geq_{{\operatorname{slex}}}v$. The set

$\mathcal{L}_{t}(u,v)=\big{\\{}w\in
M_{n,d,t}:u\geq_{{\operatorname{slex}}}w\geq_{{\operatorname{slex}}}v\big{\\}}$

is called an _arbitrary $t$–spread lexsegment set_, or simply a _$t$ –spread
lexsegment set_. The set

$\mathcal{L}_{t}^{i}(v)=\big{\\{}w\in
M_{n,d,t}:w\geq_{{\operatorname{slex}}}v\big{\\}}=\mathcal{L}_{t}(\max(M_{n,d,t}),v)$

is called an _initial $t$–spread lexsegment_ and the set

$\mathcal{L}_{t}^{f}(u)=\big{\\{}w\in
M_{n,d,t}:w\leq_{{\operatorname{slex}}}u\big{\\}}=\mathcal{L}_{t}(u,\min(M_{n,d,t}))$

is called a _final $t$–spread lexsegment_.

One can observe that, if $u,v\in M_{n,d,t}$, with
$u\geq_{{\operatorname{slex}}}v$, then
$\mathcal{L}_{t}(u,v)=\mathcal{L}_{t}^{i}(v)\cap\mathcal{L}_{t}^{f}(u)$.

###### Definition 1.3

A $t$–spread monomial ideal $I$ of $S$ is called an arbitrary $t$–spread
lexsegment ideal, or simply a $t$–spread lexsegment ideal if it is generated
by an (arbitrary) $t$–spread lexsegment.
A $t$–spread lexsegment ideal $I=(\mathcal{L}_{t}(u,v))$ is a completely
$t$–spread lexsegment ideal if

$I=J\cap T,$

where $J=(\mathcal{L}_{t}^{i}(v))$ and $T=(\mathcal{L}_{t}^{f}(u))$.

In [15], all completely $t$–spread lexsegment ideals have been characterized.

Now, we recall some important notions that will be useful in the sequel.

A $t$–spread monomial ideal $I$ is a $t$–spread strongly stable ideal if for
all $t$–spread monomial $u\in I$, all $j\in{\operatorname{supp}}(u)$ and all
$1\leq i<j$ such that $x_{i}(u/x_{j})$ is $t$–spread, it follows that
$x_{i}(u/x_{j})\in I$.

Every initial $t$–spread lexsegment ideal $I$ of $S$ is a $t$–spread strongly
stable ideal. Hence, in order to compute the graded Betti numbers of $I$ one
may use the Ene, Herzog, Qureshi’s formula ([13, Corollary 1.12]):

$\beta_{i,i+j}(I)\ =\ \sum_{u\in G(I)_{j}}\binom{\max(u)-t(j-1)-1}{i}.$ (1)

A $t$–spread final lexsegment ideal $I$ of $S$ is also a $t$–spread strongly
stable ideal, but with the order of the variables reversed, _i.e._ ,
$x_{n}>x_{n-1}>\cdots>x_{1}$. Hence, in order to compute their graded Betti
numbers one may use the following _modified_ Ene, Herzog, Qureshi’s formula:

$\beta_{i,i+j}(I)\ =\ \sum_{u\in G(I)_{j}}\binom{n-\min(u)-t(j-1)}{i}.$ (2)

Let $>_{{\operatorname{lex}}}$ be the usual lexicographic order on $S$, with
$x_{1}>x_{2}>\cdots>x_{n}$, [18]. The next theorem collects some results from
[15] that will be pivotal for the aim of this article.

###### Theorem 1.4

[15]. Given $n,d,t\geq 1$, let $u=x_{i_{1}}x_{i_{2}}\cdots x_{i_{d}}$ and
$v=x_{j_{1}}x_{j_{2}}\cdots x_{j_{d}}$ be $t$–spread monomials of degree $d$
of $S$ such that $u\geq_{{\operatorname{slex}}}v$.

  1. (1)

Let $I=(\mathcal{L}_{t}(u,v))$. The following conditions are equivalent:

    1. (a)

$I$ is a completely $t$–spread lexsegment ideal, _i.e._ , $I=J\cap T$;

    2. (b)

for every $w\in M_{n,d,t}$ with $w<_{{\operatorname{slex}}}v$ there exists an
integer $s>i_{1}$ such that $x_{s}$ divides $w$ and
$x_{i_{1}}(w/x_{s})\leq_{{\operatorname{lex}}}u$.

  2. (2)

Let $I=(\mathcal{L}_{t}(u,v))$ be a completely $t$–spread lexsegment ideal
with $\min(v)>\min(u)=1$. $I$ has a linear resolution if and only if one of
the following conditions hold:

    1. (i)

$i_{2}=1+t$;

    2. (ii)

$i_{2}>1+t$ and for the largest $w\in M_{n,d,t}$,
$w<_{{\operatorname{lex}}}v$, we have
$x_{1}(w/x_{\max(w)})\leq_{{\operatorname{lex}}}x_{1}x_{i_{2}-t}x_{i_{3}-t}\cdots
x_{i_{d}-t}$.

  3. (3)

Let $I=(\mathcal{L}_{t}(u,v))$ be a completely $t$–spread lexsegment ideal and
suppose $I=(\mathcal{L}_{t}(u,v))$ has a linear resolution. Then, for all
$i\geq 0$,

$\beta_{i}(I)=\sum_{w\in\mathcal{L}_{t}^{f}(u)}\binom{n-\min(w)-(d-1)t}{i}-\sum_{\begin{subarray}{c}w\in\mathcal{L}_{t}^{f}(v)\\\
w\neq v\end{subarray}}\binom{\max(w)-(d-1)t-1}{i}.$

We close the section recalling some notions about simplicial complexes.

Given $n\in\mathbb{N}$, we set $[n]=\\{1,2,\dots,n\\}$. We recall that a
simplicial complex on the vertex set $[n]$ is a family of subsets of $[n]$
such that

  1. -

$\\{i\\}\in\Delta$ for all $i\in[n]$, and

  2. -

if $F\subseteq[n]$, $G\subseteq F$, we have $G\in\Delta$.

The dimension of $\Delta$ is the number $d=\max\\{|F|-1:F\in\Delta\\}$. Any
$F\in\Delta$ is called a face and $|F|-1$ is the dimension of $F$. A facet of
$\Delta$ is a maximal face with respect to the inclusion. The set of facets of
$\Delta$ is denoted by $\mathcal{F}(\Delta)$.

It is well known that for any squarefree ideal ($1$–spread ideal) $I$ of $S$
there exists a unique simplicial complex $\Delta$ on $[n]$ such that
$I=I_{\Delta}$, where

$I_{\Delta}=({\bf x}_{F}:F\subseteq[n],F\notin\Delta)$

is the Stanley–Reisner ideal of $\Delta$ [18]. In the sequel when we say that
a simplicial complex $\Delta$ is associated to a squarefree ideal $I$ of $S$,
we mean that $I_{\Delta}=I$.

A simplicial complex $\Delta$ on $[n]$ is called pure if its facets have the
same dimension. We say that $\Delta$ is Cohen–Macaulay if
$K[\Delta]=S/I_{\Delta}=K[x_{1},\dots,x_{n}]/I_{\Delta}$ is a Cohen–Macaulay
ring. If $\Delta$ is Cohen–Macaulay then $\Delta$ is pure (see, for instance,
[18]).

It is well known that any monomial ideal (not necessarily squarefree) has a
unique minimal primary decomposition. We refer to it as the standard primary
decomposition. Moreover, for a squarefree monomial ideal, its standard primary
decomposition is $I=\bigcap_{\mathfrak{p}\in\textup{Min}(I)}\mathfrak{p}$,
[18, Corollary 1.3.6], where $\text{Min}(I)$ is the set of minimal primes of
$I$. Every minimal prime is a monomial prime ideal, _i.e._ ,
$\mathfrak{p}=(x_{i_{1}},x_{i_{2}},\dots,x_{i_{s}})$, for some
$A=\\{i_{1}<i_{2}<\dots<i_{s}\\}\subseteq[n]$. To denote such an ideal we also
use the notation $\mathfrak{p}_{A}$.

###### Proposition 1.5

[18, Lemma 1.5.4] The standard primary decomposition of $I_{\Delta}$ is

$I_{\Delta}=\bigcap_{F\in\mathcal{F}(\Delta)}\mathfrak{p}_{[n]\setminus F}.$

Given two positive integers $j\geq k$, we set
$[j,k]=\\{\ell\in\mathbb{N}:j\leq\ell\leq k\\}$ and we associate to a
$t$–spread monomial two special sets of integers.

Let $t\geq 1$ and let $w=x_{\ell_{1}}x_{\ell_{2}}\cdots x_{\ell_{d}}$ be a
$t$–spread monomial of $S$ of degree $d$.

If $\max(w)\leq n+1-t$, we define the $t$–spread support of $w$ as follows:

${\operatorname{supp}}_{t}(w)=\bigcup_{r=1}^{d}[\ell_{r},\ell_{r}+(t-1)];$ (3)

whereas, if $\min(w)\geq t$, we define the $t$–spread cosupport of $w$ as
follows:

${\operatorname{cosupp}}_{t}(w)=\bigcup_{r=1}^{d}[\ell_{r}-(t-1),\ell_{r}].$
(4)

Moreover, we set
${\operatorname{supp}}_{t}(1)={\operatorname{cosupp}}_{t}(1)=\varnothing$.

###### Remark 1.6

Let $w=x_{2}x_{5}x_{10}x_{13}\in S=K[x_{1},\ldots,x_{13}]$ be a 3–spread
monomial. One can note that $\max(w)=13>13+1-3=11$. Furthermore,
${\operatorname{supp}}_{3}(w/x_{\max(w)})=\\{2,3,4,5,6,7,10,11,12\\}$ and
${\operatorname{cosupp}}_{3}(w/x_{\min(w)})=\\{3,4,5,8,9,10,11,12,13\\}$.

Observe that for all $w\in M_{n,d,t}$, we can always compute
${\operatorname{supp}}_{t}(w/x_{\max(w)})$ and
${\operatorname{cosupp}}_{t}(w/x_{\min(w)})$.

In [13, Theorem 2.3], Ene, Herzog, Qureshi computed the standard primary
decomposition of the $t$–spread Veronese ideal of degree $d$, $I_{n,d,t}$,

$I_{n,d,t}=\bigcap_{D\in\mathcal{D}}\mathfrak{p}_{[n]\setminus D},$ (5)

where

$\mathcal{D}=\Big{\\{}\bigcup_{r=1}^{d-1}[\ell_{r},\ell_{r}+(t-1)]\subseteq[n]\
:\ \ell_{r+1}-\ell_{r}\geq t,\ r=1,\dots,d-2\Big{\\}}.$

Using our notation, each set $D\in\mathcal{D}$ may be written as
${\operatorname{supp}}_{t}(x_{\ell_{1}}\cdots x_{\ell_{d-1}})$, with
$x_{\ell_{1}}\cdots x_{\ell_{d-1}}\in M_{n+1-t,d-1,t}$. Hence, the primary
decomposition (5) becomes

$I_{n,d,t}=\bigcap_{w\in
M_{n+1-t,d-1,t}}\mathfrak{p}_{[n]\setminus{\operatorname{supp}}_{t}(w)}.$ (6)

In particular, $I_{n,d,t}$ is an height–unmixed ideal, _i.e._ , all its
minimal primes have the same height, $n-(d-1)t$ [13, Theorem 2.3].
Furthermore, $I_{n,d,t}$ is a completely $t$–spread lexsegment ideal [15].

## 2\. The primary decomposition of
completely $t$–spread lexsegment ideals

In this section we study the primary decomposition of a completely $t$–spread
lexsegment ideal of $S=K[x_{1},\ldots,x_{n}]$. The case $t=1$ has been
analyzed in [19]. The sets defined in (3) and (4) will be pivotal for our aim.

We start the section with some comments and remarks.

Let $n,d,t\geq 1$, $u=x_{i_{1}}x_{i_{2}}\cdots x_{i_{d}}$ and
$v=x_{j_{1}}x_{j_{2}}\cdots x_{j_{d}}$ monomials of $M_{n,d,t}$, with
$u\geq_{{\operatorname{slex}}}v$. Set $L=\mathcal{L}_{t}(u,v)$, $I=(L)$. We
assume that $I$ is a completely $t$–spread lexsegment ideal, _i.e._ , $I\ =\
J\cap T$, with $J=(\mathcal{L}_{t}^{i}(v))$ and $T=(\mathcal{L}_{t}^{f}(u))$.

  1. -

If $u=v$, $I=(L)=(u)$ is a principal ideal, and its standard primary
decomposition is $I=(x_{i_{1}})\cap\dots\cap(x_{i_{d}})$. Therefore, we may
assume $u>_{{\operatorname{slex}}}v$.

  2. -

If $\deg(u)=\deg(v)=d=1$, then
$I=(L)=(x_{i_{1}},x_{i_{1}+1},\dots,x_{{j_{1}}-1},x_{j_{1}})=\mathfrak{p}_{[i_{1},j_{1}]}$
is the standard primary decomposition of $I$. Thus, we may assume $d\geq 2$.

  3. -

If $I$ is initial, _i.e._ , $I=J=(\mathcal{L}_{t}^{i}(v))$, we may assume
$\min(v)=j_{1}\geq 2$, otherwise

$J=(x_{1})\cap(\mathcal{L}_{t}(x_{1+t}\cdots x_{1+(d-1)t},v/x_{1})),$

and to determine the standard primary decomposition of $J$, it suffices to
determine that of $(\mathcal{L}_{t}(x_{1+t}\cdots x_{1+(d-1)t},v/x_{j_{1}}))$
which is an initial $t$–spread lexsegment ideal in the polynomial ring
$K[x_{1+t},\dots,x_{n}]$ with fewer indeterminates than $S$.

  4. -

If $I$ is final, _i.e._ , $I=T=(\mathcal{L}_{t}^{f}(u))$, we can assume
$\min(u)=i_{1}=1$. Indeed, if $i_{1}>1$, none of the variables
$x_{1},\dots,x_{i_{1}-1}$ divides any minimal monomial generator of $T$. So
computing the standard primary decomposition of $T$ is equivalent to computing
that of $T\cap K[x_{i_{1}},\dots,x_{n}]$ in the polynomial ring
$K[x_{i_{1}},\dots,x_{n}]$ with fewer indeterminates than $S$.

Hence, from now on, we assume that $I=(\mathcal{L}_{t}(u,v))=J\cap T$, with
$u>_{{\operatorname{slex}}}v$, $\min(u)=1$, $\min(v)\geq 2$ and
$\deg(u)=\deg(v)=d\geq 2$.

To compute the standard primary decomposition of $I=J\cap T$, we can proceed
as follows. Firstly, we determine the decomposition of $J$, and if $I=J$ we
are done. Secondly, we determine that of $T$, and if $I=T$ we are done.
Finally, knowing the primary decompositions of $J$ and $T$ we take into
account the intersection $J\cap T=I$, and deleting the non minimal primes, we
obtain the standard primary decomposition of $I$ in the general case.

### 2.1. The initial $t$–spread lexsegment case

In this subsection we analyze the case of the standard primary decomposition
of an initial $t$–spread lexsegment ideal of $S=K[x_{1},\dots,x_{n}]$.

###### Theorem 2.1

Let $v=x_{j_{1}}x_{j_{2}}\dots x_{j_{d}}$ be a $t$–spread monomial of degree
$d\geq 2$ of $S$ with $j_{1}\geq 2$ and let $J=(\mathcal{L}_{t}^{i}(v))$ be an
initial $t$–spread lexsegment ideal of $S$. Then, the standard primary
decomposition of $J$ is

$J\ =\
\bigcap_{p=1}^{d}\mathfrak{p}_{F_{p}}\cap\bigcap_{F\in\mathcal{F}}\mathfrak{p}_{[n]\setminus
F},$

where

$\displaystyle F_{p}\ $
$\displaystyle=[j_{p}]\setminus{\operatorname{supp}}_{t}(x_{j_{1}}x_{j_{2}}\cdots
x_{j_{p-1}}),\ p=1,\dots,d,$ $\displaystyle\mathcal{F}\ $ $\displaystyle=\
\big{\\{}{\operatorname{supp}}_{t}(w):w\in M_{n+1-t,d-1,t},\
w>_{{\operatorname{slex}}}v/x_{\max(v)}\big{\\}}.$

Proof. Let $\Delta$ be the simplicial complex on the vertex set $[n]$
associated to $J$. From Proposition 1.5, we have to prove that the facets of
$\Delta$ are exactly the sets $[n]\setminus F_{p}$, $p=1,\dots,d$, together
with the sets of $\mathcal{F}$, _i.e._ ,

$\mathcal{F}(\Delta)=\big{\\{}[n]\setminus
F_{p}:p=1,\dots,d\big{\\}}\cup\mathcal{F}.$ (7)

Observe that if $G\in\Delta$, $G\neq\varnothing$, then $G$ is a facet of
$\Delta$ if and only if $G\cup\\{i\\}\notin\Delta$, for all $i\in[n]\setminus
G$, _i.e._ , if and only if ${\bf x}_{G\cup\\{i\\}}\in J$, for all
$i\in[n]\setminus G$.

We show that each $F\in\mathcal{F}$ is a facet of $\Delta$. Let
$F\in\mathcal{F}$, then

$\displaystyle F\ $ $\displaystyle=\
{\operatorname{supp}}_{t}(x_{\ell_{1}}\cdots x_{\ell_{d-1}})\ =\
\bigcup_{r=1}^{d-1}[\ell_{r},\ell_{r}+(t-1)]$ $\displaystyle=\
\big{\\{}\ell_{1},\ell_{1}+1,\dots,\ell_{1}+(t-1),\ \dots,\
\ell_{d-1},\ell_{d-1}+1,\dots,\ell_{d-1}+(t-1)\big{\\}},$

with $x_{\ell_{1}}\cdots x_{\ell_{d-1}}\in M_{n+1-t,d-1,t}$ and
$x_{\ell_{1}}\cdots
x_{\ell_{d-1}}>_{{\operatorname{slex}}}v/x_{\max(v)}=x_{j_{1}}\cdots
x_{j_{d-1}}$.

Clearly, ${\bf x}_{F}\notin J$, since ${\bf x}_{F}$ is not a multiple of any
$t$–spread monomial of degree $d$ of $S$. To prove that $F$ is a facet, it
suffices to show that ${\bf x}_{F\cup\\{i\\}}\in J$, for all $i\in[n]\setminus
F$.

Let $i\in[n]\setminus F$. We have:

  1. -

if $i<\ell_{1}$, then ${\bf x}_{F\cup\\{i\\}}$ is divided by the $t$–spread
monomial $x_{i}x_{\ell_{1}+(t-1)}\cdots x_{\ell_{d-1}+(t-1)}$;

  2. -

if $\ell_{j}+(t-1)<i<\ell_{j+1}$, for some $1\leq j\leq d-2$, then ${\bf
x}_{F\cup\\{i\\}}=$ ${\bf x}_{F}x_{i}$ is divided by the $t$–spread monomial
$\big{(}\prod\limits_{r=1}^{j}x_{\ell_{r}}\big{)}x_{i}\big{(}\prod\limits_{r=j+1}^{d-1}x_{\ell_{r}+(t-1)}\big{)}$;

  3. -

if $\ell_{d-1}+(t-1)<i$, then ${\bf x}_{F\cup\\{i\\}}$ is is divided by the
monomial $x_{\ell_{1}}\cdots x_{\ell_{d-1}}x_{i}$.

More in detail, in every case, ${\bf x}_{F\cup\\{i\\}}$ is a multiple of a
$t$–spread monomial of degree $d$ strictly greater than $v$, with respect to
$>_{{\operatorname{slex}}}$. Therefore ${\bf x}_{F\cup\\{i\\}}\in J$ and $F$
is a facet.

Now, we determine all the facets of $\Delta$. Let $G\in\Delta$ and define the
following integers:

  1. $\ell_{1}=\min(G)$, and

  2. $\ell_{i}=\min\big{\\{}r\in G:r\geq\ell_{i-1}+t\big{\\}}$, for $i\geq 2$.

Let $\ell_{k}$ be the last element of the sequence $\ell_{1}<\ell_{2}<\cdots$.
If $\ell_{i}\leq j_{i}$ for all $i$, then $k\leq d-1$, otherwise
$x_{\ell_{1}}x_{\ell_{2}}\cdots x_{\ell_{d}}\in J$. Against the fact that
$G\in\Delta$. Then,

$\displaystyle G\ $ $\displaystyle\subseteq\
\bigcup_{i=1}^{k}[\ell_{i},\ell_{i}+(t-1)]\cup\bigcup_{i=k+1}^{d-1}[j_{i},j_{i}+(t-1)]$
$\displaystyle=\ {\operatorname{supp}}_{t}(x_{\ell_{1}}\cdots
x_{\ell_{k}}x_{j_{k+1}}\cdots x_{j_{d-1}})=F\in\mathcal{F}.$

Assume there exists an integer $1\leq p\leq d$ such that $\ell_{p}>j_{p}$ and
let $p$ be minimal for such a property. By the meaning of $p$, for $q<p$,
$\ell_{q}\leq j_{q}$. We need to distinguish two cases.
Case 1. There exists an integer $q<p$ such that $\ell_{q}<j_{q}$.
In such a case, $x_{\ell_{1}}x_{\ell_{2}}\cdots
x_{\ell_{p-1}}>_{{\operatorname{slex}}}x_{j_{1}}x_{j_{2}}\cdots x_{j_{p-1}}$.
Moreover, $k\leq d-1$, otherwise $x_{\ell_{1}}x_{\ell_{2}}\cdots
x_{\ell_{p-1}}x_{\ell_{p}}\cdots x_{\ell_{d}}>_{{\operatorname{slex}}}v$ and
so ${\bf x}_{G}\in J$. A contradiction. Hence,

$G\subseteq\bigcup_{i=1}^{k}[\ell_{i},\ell_{i}+(t-1)]$

and we can quickly find a facet $F\in\mathcal{F}$ which contains
$\bigcup_{i=1}^{k}[\ell_{i},\ell_{i}+(t-1)]$ and consequently $G$.
Case 2. For all integers $q<p$, $\ell_{q}=j_{q}$.
In such a case, since $\ell_{p}>j_{p}$, we have $x_{j_{1}}\cdots
x_{j_{p}}>_{{\operatorname{slex}}}x_{\ell_{1}}\cdots x_{\ell_{p}}$, and

$G\subseteq\bigcup_{i=1}^{p-1}[j_{i},j_{i}+(t-1)]\cup[j_{p}+1,n]=[n]\setminus
F_{p}$

and $[n]\setminus F_{p}$ is clearly a facet. Finally, all the facets of
$\Delta$ are those described in (7).

$\Box$

###### Remark 2.2

We note that for an initial $t$–spread lexsegment ideal $J=I_{\Delta}$, all
the facets $F\in\mathcal{F}(\Delta)$ have cardinality $|F|\geq(d-1)t$. Indeed,
from Theorem 2.1, $\mathcal{F}(\Delta)=\big{\\{}[n]\setminus
F_{p}:p=1,\dots,d\big{\\}}\cup\mathcal{F}$. If $F\in\mathcal{F}$, then
$|F|=(d-1)t$. Otherwise, if $F=[n]\setminus F_{p}$ for some $p$, then
$|F|=|[n]\setminus F_{p}|=n-|F_{p}|=n-(j_{p}-(p-1)t)$. We observe that
$j_{p}\leq n-(d-p)t$. Thus

$|F|=n+(p-1)t-j_{p}\geq n+(p-1)t-n+(d-p)t=(d-1)t,$

as desired.

We illustrate the previous result with an example.

###### Example 2.3

Let $v=x_{2}x_{5}x_{7}\in S=\mathbb{Q}[x_{1},\dots,x_{7}]$ be a $2$–spread
monomial and consider the initial $2$–spread ideal
$J=(\mathcal{L}_{2}^{i}(v))$ of $S$:

$J=(x_{1}x_{3}x_{5},x_{1}x_{3}x_{6},x_{1}x_{3}x_{7},x_{1}x_{4}x_{6},x_{1}x_{4}x_{7},x_{1}x_{5}x_{7},x_{2}x_{4}x_{6},x_{2}x_{4}x_{7},x_{2}x_{5}x_{7}).$

We have $n=7,d=3,t=2$. The monomials $w\in M_{n+1-t,d-1,t}=M_{6,2,2}$ such
that $w>_{{\operatorname{slex}}}v/x_{\max(v)}=x_{2}x_{5}$ are the following

$x_{1}x_{3},\ \ x_{1}x_{4},\ \ x_{1}x_{5},\ \ x_{1}x_{6},\ \ x_{2}x_{4}.$

Therefore,

$\displaystyle\mathcal{F}\ $ $\displaystyle=\
\big{\\{}{\operatorname{supp}}_{2}(x_{1}x_{3}),{\operatorname{supp}}_{2}(x_{1}x_{4}),{\operatorname{supp}}_{2}(x_{1}x_{5}),{\operatorname{supp}}_{2}(x_{1}x_{6}),{\operatorname{supp}}_{2}(x_{2}x_{4})\big{\\}}$
$\displaystyle=\
\big{\\{}\\{1,2,3,4\\},\\{1,2,4,5\\},\\{1,2,5,6\\},\\{1,2,6,7\\},\\{2,3,4,5\\}\big{\\}}.$

Moreover, $v=x_{j_{1}}x_{j_{2}}x_{j_{3}}=x_{2}x_{5}x_{7}$, hence

$\displaystyle F_{1}$
$\displaystyle=[j_{1}]\setminus{\operatorname{supp}}_{2}(1)=[2]\setminus\varnothing=\\{1,2\\},$
$\displaystyle F_{2}$
$\displaystyle=[j_{2}]\setminus{\operatorname{supp}}_{2}(x_{j_{1}})=[5]\setminus\\{2,3\\}=\\{1,4,5\\},$
$\displaystyle F_{3}$
$\displaystyle=[j_{3}]\setminus{\operatorname{supp}}_{2}(x_{j_{1}}x_{j_{2}})=[7]\setminus\\{2,3,5,6\\}=\\{1,4,7\\}.$

Finally, the standard decomposition of $J$ is

$\displaystyle J$
$\displaystyle=(x_{1},x_{2})\cap(x_{1},x_{4},x_{5})\cap(x_{1},x_{4},x_{7})\cap(x_{1},x_{6},x_{7})\cap(x_{3},x_{4},x_{5})$
$\displaystyle\phantom{=..}\cap(x_{3},x_{4},x_{7})\cap(x_{3},x_{6},x_{7})\cap(x_{5},x_{6},x_{7}).$

Here are some corollaries of Theorem 2.1.

###### Corollary 2.4

In the hypotheses of Theorem 2.1, ${\operatorname{pd}}(S/J)=n-(d-1)t$,
$0pt(S/J)=(d-1)t$ and $\dim(S/J)=n-j_{1}$.

Proof. Since $J$ is a $t$–spread strongly stable ideal, by (1),
${\operatorname{pd}}(S/J)={\operatorname{pd}}(J)+1=n-(d-1)t$. So, by the
Auslander–Buchsbaum formula,
$0pt(S/J)=0pt(S)-{\operatorname{pd}}(S/J)=n-(n-(d-1)t)=(d-1)t$.

Let $\Delta$ be the simplicial complex associated to $J$. By Theorem 2.1, the
facets of $\Delta$ are those described in (7), and so, since $j_{1}\leq
n-(d-1)t$, and $j_{1}+(i-1)t\leq j_{i}$, for $i=2,\dots,d$, we have

$\displaystyle\textup{height}(J)\ $ $\displaystyle=\
\min\big{\\{}n-(d-1)t,|F_{\ell}|\ :\ \ell=1,\dots,d\big{\\}}$ $\displaystyle=\
\min\big{\\{}n-(d-1)t,j_{1},j_{2}-t,\dots,j_{d}-(d-1)t\big{\\}}\ =\ j_{1}.$

Therefore, $\dim(K[\Delta])=\dim(S/J)=n-j_{1}$. $\Box$

###### Corollary 2.5

An initial $t$–spread lexsegment ideal generated in degree $d\geq 2$ is
Cohen–Macaulay if and only if it is the $t$–spread Veronese ideal of degree
$d$.

Proof. Let $J=(\mathcal{L}_{t}^{i}(v))$, $v=x_{j_{1}}x_{j_{2}}\cdots
x_{j_{d}}$ and let $\Delta$ be the simplicial complex associated to $J$.
Assume $J$ is Cohen–Macaulay, then $\Delta$ is pure. By Theorem 2.1, the
facets of $\Delta$ are those described in (7), and so $\Delta$ is pure if and
only if

$(d-1)t=n-j_{1}=n-(j_{2}-t)=\cdots=n-(j_{d}-(d-1)t).$

Hence $v=x_{n-(d-1)t}x_{n-(d-2)t}\cdots x_{n-t}x_{n}$ and $J=I_{n,d,t}$.
Conversely, $I_{n,d,t}$ is Cohen–Macaulay. Indeed,
$\dim(S/J)=\textup{depth}(S/J)=(d-1)t$ (Corollary 2.4). $\Box$

### 2.2. The final $t$–spread lexsegment case

In this subsection we determine the standard primary decomposition of a final
$t$–spread lexsegment ideal of $S=K[x_{1},\dots,x_{n}]$.

The next result will be crucial.

###### Proposition 2.6

Let $u$ be a $t$–spread monomial of degree $d$ of $S$ such that $\min(u)=1$
and let $T=(\mathcal{L}_{t}^{f}(u))$ be a final $t$–spread lexsegment of $S$.
Let $\Delta$ be the simplicial complex on $[n]$ associated to $T$. Then
$|F|\in\\{(d-1)t,1+(d-1)t\\}$, for all $F\in\mathcal{F}(\Delta)$.

Proof. Since we may see $T$ as an initial $t$–spread lexsegment ideal with the
order on the variables reversed $x_{n}>x_{n-1}>\cdots>x_{1}$, then each facet
of $\Delta$ has cardinality $\geq(d-1)t$ (Remark 2.2). To prove that $|F|\leq
1+(d-1)t$ for all $F\in\mathcal{F}(\Delta)$, it is enough to show that each
subset $G\subseteq[n]$ with cardinality $|G|=2+(d-1)t$ does not belong to
$\Delta$, _i.e._ , ${\bf x}_{G}\in T$.

Let $G\subseteq[n]$ with $|G|=2+(d-1)t$. Define

  1. $\ell_{1}=\min\\{g\in G:g>1\\}$, and

  2. $\ell_{j}=\min\big{\\{}k\in G:k\geq\ell_{j-1}+t\big{\\}}$, for $j\geq 2$.

Since $|G|=2+(d-1)t$, the set $\\{g\in G:g>1\\}$ has at least $1+(d-1)t$
elements. Hence, the sequence of integers $\ell_{1}<\ell_{2}<\cdots$ has at
least $d$ terms. Therefore, $H=\\{\ell_{1},\ell_{2},\dots,\ell_{d}\\}\subseteq
G$ and $u>_{{\operatorname{slex}}}{\bf x}_{H}$. Indeed, $\min(u)=1$ and
$\min({\bf x}_{H})=\ell_{1}>1$ and so ${\bf x}_{H}\in T$. It follows that
${\bf x}_{G}\in T$, as desired. $\Box$

In the next theorem, we assume that $T\neq I_{n,d,t}$, as the case of the
Veronese ideal has been covered in Theorem 2.1.

###### Theorem 2.7

Let $u=x_{i_{1}}x_{i_{2}}\dots x_{i_{d}}$ be a $t$–spread monomial of degree
$d\geq 2$ of $S$ with $i_{1}=1$ and let $T=(\mathcal{L}_{t}^{f}(u))$ be a
final $t$–spread lexsegment ideal of $S$, with $T\neq I_{n,d,t}$. Then, the
standard primary decomposition of $T$ is

$T\ =\ \bigcap_{G\in\mathcal{G}}\mathfrak{p}_{[n]\setminus
G}\cap\bigcap_{H\in\mathcal{H}}\mathfrak{p}_{[n]\setminus H},$

where

$\displaystyle\mathcal{G}\ $ $\displaystyle=\
\big{\\{}{\operatorname{cosupp}}_{t}(w)\cup\\{1\\}:x_{1}w\in M_{n,d,t},\
w>_{{\operatorname{slex}}}u/x_{1}\big{\\}},$ $\displaystyle\mathcal{H}\ $
$\displaystyle=\ \big{\\{}{\operatorname{cosupp}}_{t}(w):x_{1}w\in M_{n,d,t},\
w\leq_{{\operatorname{slex}}}u/x_{1}\big{\\}}.$

Proof. Let $\Delta$ be a simplicial complex on the vertex set $[n]$ associated
to $T$. By Proposition 1.5, we have to prove that

$\mathcal{F}(\Delta)=\mathcal{G}\cup\mathcal{H}.$ (8)

By Proposition 2.6, the facets of $\Delta$ have cardinality $(d-1)t$ or
$1+(d-1)t$.

Let $G\in\mathcal{F}(\Delta)$ such that $|G|=1+(d-1)t$. We prove that
$G\in\mathcal{G}$. First of all, $\min(G)=1$. Indeed, if $\min(G)\geq 2$, then
setting $s_{1}=\min(G)$, and

$s_{j}=\min\big{\\{}s\in G:s\geq s_{j-1}+t\big{\\}},\,\,\mbox{for $j\geq 2$},$

the sequence $s_{1}<s_{2}<\cdots$ has at least $d$ elements, otherwise
$|G|<1+(d-1)t$. Hence, $\\{s_{1},s_{2},\dots,s_{d}\\}=U\subseteq G$ and ${\bf
x}_{U}\in T$, as $\min({\bf x}_{U})=\min(G)\geq 2>1=\min(u)$. It follows that
$G\notin\Delta$. A contradiction. Therefore, $\min(G)=1$.

Consider the following integers:

  1. $\ell_{d}=\max(G)$, and

  2. $\ell_{j}=\max\big{\\{}\ell\in G:\ell\leq\ell_{j+1}-t\big{\\}}$, for $j<d$.

The sequence $\ell_{d}>\ell_{d-1}>\dots>\ell_{k}$ has at least $d$ terms,
otherwise $|G|<1+(d-1)t$. Moreover it has at most $d$ terms, otherwise
$\ell_{1}>1$, $x_{\ell_{1}}x_{\ell_{2}}\cdots x_{\ell_{d}}\in T$ and then
$G\notin\Delta$. A contradiction. Hence, $k=1$.

Finally,

$\displaystyle G\subseteq F\ $ $\displaystyle=\
\big{\\{}\ell_{d},\ell_{d}-1,\dots,\ell_{d}-(t-1),\ \ldots,\
\ell_{2},\ell_{2}-1,\dots,\ell_{2}-(t-1),\ \ell_{1}\big{\\}}$ $\displaystyle=\
\ \bigcup_{r=2}^{d}[\ell_{r}-(t-1),\ell_{r}]\cup\\{\ell_{1}\\}\ =\
{\operatorname{cosupp}}_{t}(x_{\ell_{2}}\cdots
x_{\ell_{d}})\cup\\{\ell_{1}\\}.$

Moreover $1+(d-1)t=|G|\leq|F|=1+(d-1)t$ and $\ell_{1}=1$, and so
$G=F\in\mathcal{G}$. Clearly, $G$ is a facet. Indeed ${\bf x}_{G}\notin T$.

Now, let us determine the facets of $\Delta$ with cardinality $(d-1)t$. Let
$H\in\mathcal{F}(\Delta)$ with $|H|=(d-1)t$. We prove that $H\in\mathcal{H}$.
Consider the following integers

  1. $\ell_{d}=\max(H)$, and

  2. $\ell_{j}=\max\big{\\{}\ell\in H:\ell\leq\ell_{j+1}-t\big{\\}}$, for $j<d$.

The sequence $\ell_{d}>\ell_{d-1}>\dots>\ell_{k}$ has at least $d-1$ terms,
otherwise $|H|<(d-1)t$, against our assumption. In fact, we have $k=2$.
Indeed, suppose $k=1$. If $\ell_{1}=1$, then $H\subseteq G\in\mathcal{G}$. A
contradiction since $H$ is a facet. On the other hand, if $\ell_{1}>1$, then
$x_{\ell_{1}}x_{\ell_{2}}\cdots x_{\ell_{d}}\in T$, and $H\notin\Delta$. A
contradiction.

Finally, setting $U=\\{\ell_{2},\dots,\ell_{d}\\}\subseteq H$, if $x_{1}{\bf
x}_{U}>_{{\operatorname{slex}}}u$, then $H$ is contained in a facet
$G\in\mathcal{G}$. A contradiction since $H$ is a facet. If $x_{1}{\bf
x}_{U}\leq_{{\operatorname{slex}}}u$, then $H\in\mathcal{H}$ and $H$ is
clearly a facet. The proof is complete. $\Box$

###### Example 2.8

Let $u=x_{1}x_{4}x_{6}$ be a $2$–spread monomial of degree $3$ of
$S=\mathbb{Q}[x_{1},\dots,x_{7}]$. Consider the ideal
$T=(\mathcal{L}_{2}^{f}(u))$ of $S$:

$T=(x_{1}x_{4}x_{6},x_{1}x_{4}x_{7},x_{1}x_{5}x_{7},x_{2}x_{4}x_{6},x_{2}x_{4}x_{7},x_{2}x_{5}x_{7},x_{3}x_{5}x_{7}).$

The monomials $w\in M_{7,2,2}$ such that $x_{1}w\in M_{7,3,2}$ and
$w>_{{\operatorname{slex}}}u/x_{1}=x_{4}x_{6}$ are

$x_{3}x_{5},\ \ x_{3}x_{6},\ \ x_{3}x_{7};$

whereas, the monomials $w\in M_{7,2,2}$ such that $x_{1}w\in M_{7,3,2}$ and
$w\leq_{{\operatorname{slex}}}u/x_{1}=x_{4}x_{6}$ are

$x_{4}x_{6},\ \ x_{4}x_{7},\ \ x_{5}x_{7}.$

Therefore,

$\displaystyle\mathcal{G}\ $ $\displaystyle=\
\big{\\{}{\operatorname{cosupp}}_{2}(x_{3}x_{5})\cup\\{1\\},{\operatorname{cosupp}}_{2}(x_{3}x_{6})\cup\\{1\\},{\operatorname{cosupp}}_{2}(x_{3}x_{7})\cup\\{1\\}\big{\\}}$
$\displaystyle=\
\big{\\{}\\{1,2,3,4,5\\},\\{1,2,3,5,6\\},\\{1,2,3,6,7\\}\big{\\}},\vskip 6.0pt
plus 2.0pt minus 2.0pt$ $\displaystyle\mathcal{H}\ $ $\displaystyle=\
\big{\\{}{\operatorname{cosupp}}_{2}(x_{4}x_{6}),{\operatorname{cosupp}}_{2}(x_{4}x_{7}),{\operatorname{cosupp}}_{2}(x_{5}x_{7})\big{\\}}$
$\displaystyle=\ \big{\\{}\\{3,4,5,6\\},\\{3,4,6,7\\},\\{4,5,6,7\\}\big{\\}}.$

Thus, the standard primary decomposition of $T$ is

$T=(x_{4},x_{5})\cap(x_{4},x_{7})\cap(x_{6},x_{7})\cap(x_{1},x_{2},x_{3})\cap(x_{1},x_{2},x_{5})\cap(x_{1},x_{2},x_{7}).$

Here are some corollaries of Theorem 2.7.

###### Corollary 2.9

Let $T=(\mathcal{L}_{t}^{f}(u))$ be a final $t$–spread lexsegment ideal of
$S$. Then ${\operatorname{pd}}(S/T)=n-(d-1)t$, $0pt(S/T)=(d-1)t$ and

$\dim(S/T)=\begin{cases}(d-1)t&\textup{if}\ u=x_{1}x_{1+t}\cdots
x_{1+(d-1)t},\\\ 1+(d-1)t&\textup{otherwise}.\end{cases}$

Proof. The ideal $T$ is $t$–spread strongly stable but with the order on the
variables reversed, $x_{n}>x_{n-1}>\dots>x_{1}$. Hence, by (2), as $\min(u)=1$
and $u\in T$,
${\operatorname{pd}}(S/T)={\operatorname{pd}}(T)+1=n-1-(d-1)t+1=n-(d-1)t$. By
the Auslander–Buchsbaum formula, $0pt(S/T)=(d-1)t$.

Suppose $T\neq I_{n,d,t}$. Let $\Delta$ the simplicial complex associated to
$T$. By Theorem 2.7, the facets of $\Delta$ are those in (8). Now,
$\mathcal{H}$ is always non empty, as $u\in\mathcal{L}_{t}^{f}(u)$. Moreover,
$\mathcal{G}$ is non empty too, as $u\neq\min(M_{n,d,t})$, otherwise
$T=I_{n,d,t}$. Thus

$\displaystyle\textup{height}(T)\ $ $\displaystyle=\
\min\big{\\{}\textup{height}(\mathfrak{p}_{[n]\setminus
G}),\textup{height}(\mathfrak{p}_{[n]\setminus
H}):G\in\mathcal{G},H\in\mathcal{H}\big{\\}}$ $\displaystyle=\
\min\big{\\{}n-(d-1)t,n-1-(d-1)t\big{\\}}\ =\ n-1-(d-1)t.$

Thus, $\dim(K[\Delta])=1+(d-1)t$.
Otherwise, if $T=I_{n,d,t}$, _i.e._ , $u=\max(M_{n,d,t})=x_{1}x_{1+t}\cdots
x_{1+(d-1)t}$, we have $\dim(K[\Delta])$ $=(d-1)t$ (Corollary 2.4). $\Box$

###### Corollary 2.10

A final $t$–spread lexsegment ideal of $S$ generated in degree $d\geq 2$ is
Cohen–Macaulay if and only if it is the $t$–spread Veronese ideal of degree
$d$.

Proof. Let $T=(\mathcal{L}_{t}^{f}(u))$ and let $\Delta$ be the simplicial
complex associated to $T$. Assume $T$ is Cohen–Macaulay, then $\Delta$ is
pure. By contradiction, suppose $T\neq I_{n,d,t}$. By Theorem 2.7, the facets
of $\Delta$ are those described in (8). The families $\mathcal{G}$ and
$\mathcal{H}$ are both non empty, the first as $u\in\mathcal{L}_{t}^{f}(u)$,
the second as $u\neq\max(M_{n,d,t})$. But, any $G\in\mathcal{G}$ has
cardinality $|G|=1+(d-1)t$ and any $H\in\mathcal{H}$ has cardinality
$|H|=(d-1)t$. So $\Delta$ should not be pure. A contradiction. Thus
$T=I_{n,d,t}$. The converse easily follows. $\Box$

###### Remark 2.11

A simplicial complex $\Delta$ is pure if and only if the Stanley–Reisner ideal
$I_{\Delta}$ is unmixed, in the sense that all associated prime ideals have
the same height (see, for instance, [18]). One can observe that the proofs of
Corollary 2.5 and Corollary 2.10 have pointed out that the classification for
initial Cohen–Macaulay $t$–spread lexsegment ideal and for final
Cohen–Macaulay $t$–spread lexsegment ideal is equivalent to the classification
for initial unmixed $t$–spread lexsegment ideal and for final unmixed
$t$–spread lexsegment ideal.

### 2.3. The case of a completely $t$–spread lexsegment ideal

The aim of this subsection is to determine the minimal primary decomposition
of a completely $t$–spread lexsegment ideal $I=(\mathcal{L}_{t}(u,v))$ of
$S=K[x_{1},\dots,x_{n}]$, $u,v\in M_{n,d,t}$, $u>_{{\operatorname{slex}}}v$
with $\min(u)=1$ and $\min(v)\geq 2$.

###### Theorem 2.12

Let $u=x_{i_{1}}x_{i_{2}}\dots x_{i_{d}},v=x_{j_{1}}x_{j_{2}}\cdots x_{j_{d}}$
be $t$–spread monomials of degree $d\geq 2$ of $S$ with $i_{1}=1$ and
$j_{1}\geq 2$. Let $I=(\mathcal{L}_{t}(u,v))$ be a completely $t$–spread
lexsegment ideal of $S$, $I\neq I_{n,d,t}$. Then, the standard primary
decomposition of $I$ is

$I\ =\ \bigcap_{G\in\mathcal{G}}\mathfrak{p}_{[n]\setminus
G}\cap\bigcap_{p\in\mathcal{I}}\mathfrak{p}_{F_{p}}\cap\bigcap_{F\in\widetilde{\mathcal{F}}}\mathfrak{p}_{[n]\setminus
F},$

where $F_{p}$ $(p=1,\dots,d)$ and $\mathcal{G}$ are the families described in
Theorems 2.1 and 2.7,

  1. -

$\mathcal{I}=[d]\setminus\big{\\{}p\in[d]:\big{|}[n]\setminus
F_{p}\big{|}=(d-1)t\,\,\mbox{and
$v/x_{j_{p}}>_{{\operatorname{slex}}}u/x_{1}$}\big{\\}}$,

  2. -

$\widetilde{\mathcal{F}}$ is the family consisting of the sets
$F={\operatorname{supp}}_{t}(w)\in\mathcal{F}$ with $w=x_{\ell_{1}}\cdots
x_{\ell_{d-1}}$ such that

    1. -

$1\notin F$ and $x_{\ell_{1}+(t-1)}\cdots
x_{\ell_{d-1}+(t-1)}\leq_{{\operatorname{slex}}}u/x_{1}$, or

    2. -

$1\in F$ and $F\cup\\{j\\}\notin\mathcal{G}$, for all $j\notin F$,

where $\mathcal{F}$ is the family described in Theorem 2.1.

Proof. By hypothesis, $I=J\cap T$ with $J=(\mathcal{L}^{i}_{t}(v))$ and
$T=(\mathcal{L}_{t}^{f}(u))$. By Theorems 2.1 and 2.7 we have

$I=T\cap J=\bigcap_{G\in\mathcal{G}}\mathfrak{p}_{[n]\setminus
G}\cap\bigcap_{H\in\mathcal{H}}\mathfrak{p}_{[n]\setminus
H}\cap\bigcap_{p=1}^{d}\mathfrak{p}_{F_{p}}\cap\bigcap_{F\in\mathcal{F}}\mathfrak{p}_{[n]\setminus
F}.$ (9)

To find the standard primary decomposition of $I$, it suffices to delete from
this presentation those primes that are not minimal. Equivalently, if $\Delta$
is the simplicial complex on vertex set $[n]$ associated to $I$, we must
determine the facets of $\Delta$. By (9) we have that

$\mathcal{F}(\Delta)\subseteq\mathcal{G}\cup\mathcal{H}\cup\big{\\{}[n]\setminus
F_{p}:p=1,\dots,d\big{\\}}\cup\mathcal{F}.$ (10)

where $\mathcal{G}$, $\mathcal{H}$, $F_{p}$ ($p=1,\dots,d$), $\mathcal{F}$ are
the families described in Theorems 2.1 and 2.7.

Claim. $\mathcal{F}(\Delta)=\mathcal{G}\cup\big{\\{}[n]\setminus
F_{\ell}:\ell\in\mathcal{I}\big{\\}}\cup\widetilde{\mathcal{F}}$.
Proof of the Claim. We analyze the four families in (10), and in so doing we
determine the facets of $\Delta$.

  1. -

Let $G\in\mathcal{G}$. Since $I\subseteq T$ and ${\bf x}_{G}\notin T$, it
follows that ${\bf x}_{G}\notin I$, _i.e._ , $G\in\Delta$. Let us prove that
$G$ is a facet. Indeed, $G\not\subseteq F,H$, for all $F\in\mathcal{F}$ and
all $H\in\mathcal{H}$. Indeed, $|F|=|H|=(d-1)t<1+(d-1)t=|G|$. Moreover,
$G\not\subseteq[n]\setminus F_{p}$, as $1\in G$ and $1\in F_{p}$. In fact,
$\min(v)=j_{1}\geq 2$.

  2. -

Let $H\in\mathcal{H}$. Arguing as before, $H\in\Delta$. We show that we can
“eliminate” the family $\mathcal{H}$. In fact, we can write
$H={\operatorname{cosupp}}_{t}(x_{\ell_{2}}\cdots x_{\ell_{d}})$, with
$x_{1}x_{\ell_{2}}\cdots x_{\ell_{d}}\in M_{n,d,t}$ and $x_{\ell_{2}}\cdots
x_{\ell_{d}}\leq_{{\operatorname{slex}}}u/x_{1}$. Let

$\widetilde{H}=\\{\ell_{2}-(t-1),\ell_{3}-(t-1),\ldots,\ell_{d}-(t-1)\\}\subseteq
H.$

Setting $h_{i}=\ell_{i+1}-(t-1)$, for $i=1,\ldots,d-1$, then
$\widetilde{H}=\\{h_{1}<h_{2}<\dots<h_{d-1}\\}$. We distinguish three cases.

    1. -

Case 1. Let ${\bf x}_{\widetilde{H}}>_{{\operatorname{slex}}}v/x_{\max(v)}$.
Since ${\operatorname{supp}}_{t}({\bf
x}_{\widetilde{H}})={\operatorname{cosupp}}_{t}(x_{\ell_{2}}\cdots
x_{\ell_{d}})=H$, then $H\in\mathcal{F}$ (we will determine when
$H\in\mathcal{F}$ is a facet of $\Delta$ later).

    2. -

Case 2. Let ${\bf x}_{\widetilde{H}}<_{{\operatorname{slex}}}v/x_{\max(v)}$.
Then there exists an integer $s\in[d-1]$ such that
$j_{1}=h_{1},\ldots,j_{s-1}=h_{s-1}$ and $j_{s}<h_{s}$. Hence

$\displaystyle H\ $ $\displaystyle\subseteq\
\big{\\{}j_{1},\dots,j_{1}+(t-1),\ldots,j_{s-1},\dots,j_{s-1}+(t-1),\
j_{s}+1,\dots,n\big{\\}}$ $\displaystyle=\ [n]\setminus F_{s}=F.$

Moreover, ${\bf x}_{F}\notin I$, as $I\subseteq J$ and ${\bf x}_{F}\notin J$.
So $H$ is included in $F\in\Delta$.

    3. -

Case 3. Let ${\bf x}_{\widetilde{H}}=v/x_{\max(v)}$. Then we have
$H\subseteq[n]\setminus F_{d}=F$, and also $F\in\Delta$. Thus $H$ is included
in the face $F\in\mathcal{F}$.

  3. -

Let $F=[n]\setminus F_{p}$, some some $1\leq p\leq d$. Since $I\subseteq J$
and ${\bf x}_{F}\notin J$, then ${\bf x}_{F}\notin I$ and so $F\in\Delta$. If
$|F|\geq 1+(d-1)t$, then $F$ is a facet as $F$ is not contained in any other
set of the families $\mathcal{F},\mathcal{H}$. Moreover, $F\neq G$ for all
$G\in\mathcal{G}$, as $1\in G\setminus F$.
Suppose $|F|=(d-1)t$. Then $F$ is not a facet if and only if $F\subsetneq G$,
for some $G\in\mathcal{G}$. Therefore, if and only if $F=G\setminus\\{1\\}$,
for some $G\in\mathcal{G}$. On the other hand,
$G=\big{\\{}\ell_{d},\ell_{d}-1,\dots,\ell_{d}-(t-1),\ \ldots,\
\ell_{2},\ell_{2}-1,\dots,\ell_{2}-(t-1),\ 1\big{\\}}$ is characterized by the
conditions $x_{1}x_{\ell_{2}}\cdots x_{\ell_{d}}\in M_{n,d,t}$ and
$x_{1}x_{\ell_{2}}\cdots x_{\ell_{d}}>_{{\operatorname{slex}}}u$. We observe
that ${\operatorname{supp}}(v/x_{j_{p}})\subseteq F=[n]\setminus F_{p}$.
Hence, if $F\subseteq G$, then

$v/x_{j_{p}}\geq_{{\operatorname{slex}}}x_{\ell_{2}}\cdots
x_{\ell_{d}}>_{{\operatorname{slex}}}u/x_{1}.$

Thus, if $F\subseteq G$, then $v/x_{j_{p}}>_{{\operatorname{slex}}}u/x_{1}$.
On the other hand, if $v/x_{j_{p}}\leq_{{\operatorname{slex}}}u/x_{1}$, then
$x_{\ell_{2}}\cdots
x_{\ell_{d}}\leq_{{\operatorname{slex}}}v/x_{j_{p}}\leq_{{\operatorname{slex}}}u/x_{1}$
and $G$ is not a facet. Finally, we have verified that $F=[n]\setminus F_{p}$
is a facet if and only if $p\in\mathcal{I}$.

  4. -

Let $F\in\mathcal{F}$. Clearly ${\bf x}_{F}\notin I$ and so $F\in\Delta$. We
have that

$F=\big{\\{}\ell_{1},\ell_{1}+1,\ \ldots,\
\ell_{1}+(t-1),\dots,\ell_{d-1},\ell_{d-1}+1,\dots,\ell_{d-1}+(t-1)\big{\\}},$

with $x_{\ell_{1}}\cdots x_{\ell_{d-1}}\in M_{n+1-t,d-1,t}$ and
$x_{\ell_{1}}\cdots x_{\ell_{d-1}}>_{{\operatorname{slex}}}v/x_{\max(v)}$. If
$F$ is not a facet, then $F\subsetneq G$, for some $G\in\mathcal{G}$.

    1. -

Case 1. Let $1\notin F$. Then we have $F=G\setminus\\{1\\}$. $G$ is
characterized by the condition $x_{\ell_{1}+(t-1)}\cdots
x_{\ell_{d-1}+(t-1)}>_{{\operatorname{slex}}}u/x_{1}$. Hence, in such a case,
$F$ is a facet if and only if $x_{\ell_{1}+(t-1)}\cdots
x_{\ell_{d-1}+(t-1)}\leq_{{\operatorname{slex}}}u/x_{1}$.

    2. -

Case 2. Let $1\in F$. Set $\ell=\ell_{j}+t$, if
$j=\min\big{\\{}j\in[d-2]:\ell_{j}+(t-1)<\ell_{j+1}\big{\\}}$ does exist,
otherwise set $\ell=\ell_{d-1}+t$. Then $F\subseteq F\cup\\{\ell\\}$.

Finally $F$ is a facet if and only if $F\cup\\{\ell\\}\notin\mathcal{G}$.
Thus, if and only if

$\displaystyle x_{\ell_{1}+t}\cdots x_{\ell_{j-1}+t}x_{\ell_{j}+t}\cdots
x_{\ell_{d-1}+(t-1)}\leq_{{\operatorname{slex}}}u/x_{1},$ $\displaystyle\ \
\text{if}\ \ell=\ell_{j}+t,$ $\displaystyle x_{\ell_{1}+t}\cdots
x_{\ell_{j-1}+t}x_{\ell_{j}+t}\cdots
x_{\ell_{d-1}+t}\leq_{{\operatorname{slex}}}u/x_{1},$ $\displaystyle\ \
\text{if}\ \ell=\ell_{d-1}+t.$

The claim follows. $\Box$

###### Example 2.13

Let $u=x_{1}x_{4}x_{6}$ and $v=x_{2}x_{5}x_{7}$ be $2$–spread monomials of
degree $3$ of $S=\mathbb{Q}[x_{1},\dots,x_{7}]$. Consider the $t$–spread
lexsegment ideal

$I=(\mathcal{L}_{2}(u,v))=(x_{1}x_{4}x_{6},x_{1}x_{4}x_{7},x_{1}x_{5}x_{7},x_{2}x_{4}x_{6},x_{2}x_{4}x_{7},x_{2}x_{5}x_{7})$

of $S$. Firstly, we verify that $I$ is a completely $2$–spread lexsegment
ideal, _i.e._ , $I=J\cap T$, where $J$ and $T$ are the $2$–spread lexsegment
ideals in Examples 2.3 and 2.8, respectively. By [15, Theorem 3.7. (b)], $I$
is a completely $2$–spread lexsegment ideal if and only if for all $w\in
M_{7,3,2}$ with $w<_{{\operatorname{slex}}}v=x_{2}x_{5}x_{7}$, there exists
$s\in{\operatorname{supp}}(w)$ such that
$x_{\min(u)}w/x_{s}=x_{1}w/x_{s}\leq_{{\operatorname{slex}}}u=x_{1}x_{4}x_{6}$.
We have $w=x_{3}x_{5}x_{7}$ and we may choose $s=\min(w)$. Thus, $I$ is a
completely $2$–spread lexsegment ideal.
Setting, $u=x_{i_{1}}x_{i_{2}}x_{i_{3}}=x_{1}x_{i_{2}}x_{i_{3}}$ and
$v=x_{j_{1}}x_{j_{2}}x_{j_{3}}=x_{2}x_{j_{2}}x_{j_{3}}$, we have

$\mathcal{I}=[3]\setminus\big{\\{}p\in[3]:\big{|}[7]\setminus
F_{p}\big{|}=2\cdot 2\ \text{and}\
u/x_{1}=x_{4}x_{6}<_{{\operatorname{slex}}}v/x_{j_{p}}=x_{2}x_{5}x_{7}/x_{j_{\ell}}\big{\\}}.$

The only sets $F_{p}$ with $\big{|}[7]\setminus F_{p}\big{|}=2\cdot 2=4$ are
$F_{2}=\\{1,4,5\\}$ and $F_{3}=\\{1,4,7\\}$ (Example 2.3). Moreover,
$v/x_{j_{2}}=x_{2}x_{7}>_{{\operatorname{slex}}}x_{4}x_{6}$ and
$v/x_{j_{3}}=x_{2}x_{5}>_{{\operatorname{slex}}}x_{4}x_{6}$. Thus,
$\mathcal{I}=[3]\setminus\\{2,3\\}=\\{1\\}$.

From Examples 2.3 and 2.8, we have:

$\mathcal{G}=\big{\\{}\\{1,2,3,4,5\\},\\{1,2,3,5,6\\},\\{1,2,3,6,7\\}\big{\\}}$

and

$\mathcal{F}=\big{\\{}\\{1,2,3,4\\},\\{1,2,4,5\\},\\{1,2,5,6\\},\\{1,2,6,7\\},\\{2,3,4,5\\}\big{\\}}.$

Let $F\in\mathcal{F}$ such that $1\notin F$. Then
$F=\\{2,3,4,5\\}={\operatorname{supp}}_{2}(x_{2}x_{4})$. We have that
$x_{3}x_{5}>_{{\operatorname{slex}}}u/x_{1}=x_{4}x_{6}$, so
$F\notin\widetilde{\mathcal{F}}$.
Let $F\in\mathcal{F}$ such that $1\in F$. Then $F\in\widetilde{\mathcal{F}}$
if and only if $F\not\subseteq G$, for all $G\in\mathcal{G}$. Since such sets
do not exist, $\widetilde{\mathcal{F}}=\varnothing$.

Finally, the standard primary decomposition of $I$ is

$I=(x_{1},x_{2})\cap(x_{4},x_{5})\cap(x_{4},x_{7})\cap(x_{6},x_{7}).$

## 3\. Cohen–Macaulay $t$–spread lexsegment ideals

In Section 2, we have seen that if $I$ is an initial or a final $t$–spread
lexsegment ideal of $S=K[x_{1},\dots,x_{n}]$, $t\geq 1$, then $I$ is
Cohen–Macaulay if and only if $I=I_{n,d,t}$ is the Veronese $t$–spread ideal
of degree $d$ of $S$ (Corollaries 2.5, 2.10). In this section, we consider the
more general problem of classifying all Cohen–Macaulay $t$–spread lexsegment
ideals $I=(\mathcal{L}_{t}(u,v))$ of $S$, with $t\geq 1$. The case $t=1$ has
been examined also in [9] by using Serre’s condition $(S_{2})$.

As in Section 2, we may assume that $I=(\mathcal{L}_{t}(u,v))$ with
$u>_{{\operatorname{slex}}}v$, $\min(u)=1$, $\min(v)\geq 2$ and
$\deg(u)=\deg(v)=d\geq 2$. In order to achieve our purpose, we will
distinguish two cases: $\min(v)=2$ and $\min(v)>2$.

First, we note that we can always suppose $n>2+(d-1)t$. Indeed, for
$n=1+(d-1)t$, there is only one $t$–spread lexsegment ideal, namely
$I=(x_{1}x_{1+t}\cdots x_{1+(d-1)t})=I_{1+(d-1)t,d,t}$ which is always a
Cohen–Macaulay ideal; whereas, for $n=2+(d-1)t$, $I=(\mathcal{L}_{t}(u,v))$
with $v=x_{2}x_{2+t}\cdots x_{2+(d-1)t}$, necessarily. In fact, in such a case
there does exist only one $t$–spread monomial whose minimum is equal to $2$.
Hence, $I$ is final and therefore Cohen–Macaulay if and only if
$I=I_{2+(d-1)t,d,t}$ (Corollary 2.10).

In what follows we denote the squarefree lex order
$\geq_{{\operatorname{slex}}}$ simply by $\geq$.

### 3.1. The height two case

In this subsection we study the Cohen–Macaulayness of a $t$–spread lexsegment
ideal $I=(\mathcal{L}_{t}(u,v))$ of $S$ with $\min(v)=2$. We will distinguish
two cases: $3+(d-1)t\leq n\leq 3+(2d-3)t$ and $n\geq 4+(2d-3)t$.

We start with the case $3+(d-1)t\leq n\leq 3+(2d-3)t$. The notion of Betti
splitting [12] (see also [16] and the reference therein) will be crucial.

Let $I$, $P$, $Q$ be monomial ideals of $S=K[x_{1},\dots,x_{n}]$ such that
$G(I)$ is the disjoint union of $G(P)$ and $G(Q)$. We say that $I=P+Q$ is a
Betti splitting if

$\beta_{i,j}(I)=\beta_{i,j}(P)+\beta_{i,j}(Q)+\beta_{i-1,j}(P\cap Q),\ \ \
\textup{for all}\ i,j.$

If $I=P+Q$ is a Betti splitting [16, Corollary 2.2], then

$\displaystyle{\operatorname{pd}}(I)\ $ $\displaystyle=\
\max\\{{\operatorname{pd}}(P),{\operatorname{pd}}(Q),{\operatorname{pd}}(P\cap
Q)+1\\}.$ (11)

###### Lemma 3.1

[16, Corollary 2.7] Let $I$ be a monomial ideal of $S$. Let $Q$ be the ideal
generated by all elements of $G(I)$ divisible by $x_{i}$ and let $P$ be the
ideal generated by all other elements of $G(I)$. If the ideal $Q$ has a linear
resolution, then $I=P+Q$ is a Betti splitting.

For $I$ a monomial ideal of $S$, we set $\gcd(I)=\gcd(u:u\in G(I))$.

###### Theorem 3.2

Let $u=x_{i_{1}}x_{i_{2}}\dots x_{i_{d}}$, $v=x_{j_{1}}x_{j_{2}}\cdots
x_{j_{d}}$ be $t$–spread monomials of degree $d\geq 2$ of $S$ with $i_{1}=1$,
$j_{1}=2$ and let $I$ be the $t$–spread lexsegment ideal
$I=(\mathcal{L}_{t}(u,v))$ of $S$. Assume $3+(d-1)t\leq n\leq 3+(2d-3)t$ and
set

$P=(\mathcal{L}_{t}(u,x_{1}x_{n-(d-2)t}\cdots x_{n-t}x_{n})),\ \ \ \
Q=(\mathcal{L}_{t}(x_{2}x_{2+t}\cdots x_{2+(d-1)t},v)).$

Then $I$ is Cohen–Macaulay if and only if $\gcd(I)=1$ and $P\cap Q$ is a
principal ideal.

Proof. Let $I$ be Cohen–Macaulay and let $\Delta$ be the simplicial complex on
the vertex set $[n]$ associated to $I$. Then $\Delta$ is pure. Since each
monomial in $G(I)$ has minimum equal to 1 or 2, the set
$[n]\setminus\\{1,2\\}$ is a facet of $\Delta$. Thus, $\Delta$ must be pure of
dimension $\dim(\Delta)=n-3$.

Now, we prove that $v\in\\{v_{1},v_{2}\\}$, where

$\displaystyle v_{1}$
$\displaystyle=\textstyle\phantom{\big{(}}\prod_{s=0}^{d-1}x_{2+st},$ (12)
$\displaystyle v_{2}$
$\displaystyle=\textstyle\big{(}\prod_{s=0}^{d-2}x_{2+st}\big{)}x_{3+(d-1)t}.$

First, note that $v_{1}>v_{2}$ are the greatest monomials of $M_{n,d,t}$ with
minimum 2. So $v_{1}\in I$. Let
$E=\big{\\{}\max(w):w\in\mathcal{L}_{t}(v_{1},v)\big{\\}}$ and set
$k=\max(E)$. Then $k\geq 2+(d-1)t$. Indeed $v_{1}\in I$. If $v=v_{1}$, there
is nothing to prove. Assume $v>v_{1}$, then $|E|\geq 2$. We show that
$v=v_{2}$. The monomial prime ideal $\mathfrak{p}_{\\{1\\}\cup E}$ contains
$I$, and $\textup{height}(\mathfrak{p}_{\\{1\\}\cup E})=1+|E|\geq 3$. Since
$I$ is Cohen–Macaulay, thus height–unmixed of height two, there must be a
minimal prime $\mathfrak{p}\in{\operatorname{Min}}(I)$ of height $2$ that
contains properly $\mathfrak{p}_{\\{1\\}\cup E}$. For all $j\in E$,
$\mathfrak{p}_{\\{1\\}\cup(E\setminus\\{j\\})}$ does not contain $I$. Indeed,
the monomial
$\big{(}\prod_{s=0}^{d-2}x_{2+st}\big{)}x_{j}\in\mathcal{L}_{t}(v_{1},v)$ is
not in $\mathfrak{p}_{\\{1\\}\cup(E\setminus\\{j\\})}$. Hence, $x_{1}$ can be
omitted and $\mathfrak{p}_{E}$ must contain $I$. One can quickly observe that
no variable $x_{j}$, $j\in E$, can be omitted again. Hence, $\mathfrak{p}_{E}$
is a minimal prime with $\textup{height}(\mathfrak{p}_{E})=|E|=2$, _i.e._ ,
$k=2+(d-1)t$. So $v\in\\{v_{1},v_{2}\\}$ as desired.

For $\ell=0,\dots,d-1$, let us define the following monomials:

$\displaystyle u_{\ell}$ $\displaystyle=\
x_{1}\big{(}\textstyle\prod_{s=\ell}^{d-2}x_{n-st-1}\big{)}\big{(}\prod_{s=0}^{\ell-1}x_{n-st}\big{)}$
(13) $\displaystyle=\ x_{1}x_{n-(d-2)t-1}\cdots x_{n-\ell
t-1}x_{n-(\ell-1)t}\cdots x_{n-t}x_{n}.$

One can observe that
$\mathcal{L}_{t}(u_{0},u_{d-1})=\\{u_{0}>u_{1}>\dots>u_{d-1}\\}$ and
furthermore $u_{d-1}$ is the smallest monomial of $M_{n,d,t}$ whose minimum is
equal to $1$. Thus $u>u_{d-1}>v$ and $u_{d-1}\in I$. We prove that
$u=u_{\ell}$, for some $\ell\in\\{0,\dots,d-1\\}$.
Indeed, writing $u=x_{1}x_{i_{2}}\cdots x_{i_{d}}$, we note that $i_{2}\leq
n-(d-2)t$. The condition $u\leq u_{0}$ is equivalent to $i_{2}\geq
n-(d-2)t-1$. As before, we consider a suitable monomial prime ideal that
contains $I$, namely $\mathfrak{p}_{[i_{2},n-(d-2)t]\cup\\{2\\}}$. None of
$x_{j}$ with $j\in[i_{2},n-(d-2)t]$ can be omitted from
$\mathfrak{p}_{[i_{2},n-(d-2)t]\cup\\{2\\}}$. Thus,
$\mathfrak{p}_{[i_{2},n-(d-2)t]}$ must be a minimal prime of height two.
Hence, $i_{2}=n-(d-2)t-1$, as desired.

By what shown so far, if $I$ is Cohen–Macaulay, then
$I=(\mathcal{L}_{t}(u_{\ell},v_{k}))$, for some $\ell\in\\{0,\dots,d-1\\}$ and
$k\in\\{1,2\\}$. On the other hand, $I$ is Cohen–Macaulay if and only if
$\dim(S/I)=\textup{depth}(S/I)$. We note that $\dim(S/I)=n-2$. Indeed, we have
shown previously that $\textup{height}(I)=2$. By the Auslander–Buchsbaum
formula,
$\textup{depth}(S/I)=n-{\operatorname{pd}}(S/I)=n-1-{\operatorname{pd}}(I)$.
Hence, $I$ is Cohen–Macaulay if and only if $\textup{height}(I)=2$ and
${\operatorname{pd}}(I)=1$.

Hence, in order to establish the theorem we need to prove the next facts.
Claim 1. $0pt(I)=2$ if and only if $\gcd(I)=\gcd(w:w\in G(I))=1$.
Proof of Claim 1. Indeed, $\mathfrak{p}_{[2]}=(x_{1},x_{2})$ is a minimal
prime of $I$ having height two, so $0pt(I)\leq 0pt(\mathfrak{p}_{[2]})=2$.
Moreover, $0pt(I)=1$ if and only if there is a minimal prime $\mathfrak{q}$ of
$I$ whose height is $1$, _i.e._ , there exists a variable $x_{s}$ dividing all
generators of $I$. The claimed statement follows.
Claim 2. ${\operatorname{pd}}(I)=1$ if and only if the ideal
$(\mathcal{L}_{t}(u,u_{d-1}))\cap(\mathcal{L}_{t}(v_{1},v))=P\cap Q$ is a
principal ideal.
Proof of Claim 2. We can observe that

$P=(w\in G(I):\min(w)=1),\ \ \ Q=(w\in G(I):\min(w)=2).$

Clearly $I=P+Q$. We claim that $I=P+Q$ is a Betti splitting of $I$. Indeed
using Lemma 3.1, it sufficies to note that $Q=(w\in G(I):x_{2}\
\textup{divides}\ w)$ and that $Q$ has a linear resolution. For all monomials
$w\in G(I)$ with $\min(w)=1$, $\min(w/x_{1})\geq n-(d-2)t-1>2$, as by
hypothesis $n\geq 3+(d-1)t>3+(d-2)t$. Moreover $Q$ is an equigenerated initial
$t$–spread lexsegment ideal in $K[x_{2},\dots,x_{n}]$, so it has a linear
resolution. Hence, by (11),

${\operatorname{pd}}(I)\ =\
\max\\{{\operatorname{pd}}(P),{\operatorname{pd}}(Q),{\operatorname{pd}}(P\cap
Q)+1\\}.$

We have ${\operatorname{pd}}(P)\leq 1$. Indeed, $P$ is isomorphic to the final
$t$–spread lexsegment ideal $(\mathcal{L}_{t}(u/x_{1},x_{n-(d-2)t}\cdots
x_{n-t}x_{n}))$ generated in degree $d-1$, and formula (2) gives

${\operatorname{pd}}(I)=\begin{cases}n-(n-(d-2)t-1)-(d-2)t=1&\textup{if}\
u>u_{d-1},\\\ n-(n-(d-2)t)-(d-2)t=0&\textup{if}\ u=u_{d-1}.\end{cases}$

Analogously, as $Q$ is an initial $t$–spread lexsegment ideal in
$K[x_{2},\dots,x_{n}]$, ${\operatorname{pd}}(Q)=0$ if $v=v_{1}$, or
${\operatorname{pd}}(Q)=1$ if $v=v_{2}$. Thus, (11) implies that
${\operatorname{pd}}(I)=1$ if and only if ${\operatorname{pd}}(P\cap Q)=0$,
_i.e._ , if and only if $P\cap Q$ is a principal ideal. $\Box$

The next result points out that for $n\gg 0$, precisely $n\geq 4+(2d-3)t$, the
Cohen–Macaulay $t$–spread lexsegment ideals with $j_{1}=2$ are just complete
intersection ideals.

###### Theorem 3.3

Let $u=x_{i_{1}}x_{i_{2}}\dots x_{i_{d}}$ and $v=x_{j_{1}}x_{j_{2}}\cdots
x_{j_{d}}$ be $t$–spread monomials of degree $d\geq 2$ of $S$ with $i_{1}=1$,
$j_{1}=2$, and let $I$ be the $t$–spread lexsegment ideal
$I=(\mathcal{L}_{t}(u,v))$ of $S$. Assume $n\geq 4+(2d-3)t$, then the
following conditions are equivalent:

  1. (a)

$I$ is Cohen–Macaulay;

  2. (b)

$u=x_{1}x_{n-(d-2)t}\cdots x_{n-t}x_{n}$, $v=x_{2}x_{2+t}\cdots x_{2+(d-1)t}$.

Proof. Let $\Delta$ be the simplicial complex on vertex set $[n]$ associated
to $I$.
(a)$\Longrightarrow$(b). If $I=I_{\Delta}$ is Cohen–Macaulay, then $\Delta$ is
pure. The set $[n]\setminus\\{1,2\\}$ is a facet of $\Delta$. Indeed, any
monomial $w\in I$ has $1\in{\operatorname{supp}}(w)$ or
$2\in{\operatorname{supp}}(w)$. Therefore, $\Delta$ pure implies
$\dim(\Delta)=n-3$. Consider the set
$G=[n]\setminus\\{2,i_{2},i_{2}+1,\dots,n-(d-2)t\\}$. Observe that
$G\in\Delta$, indeed $\mathfrak{p}_{[n]\setminus
G}=(x_{2},x_{i_{2}},x_{i_{2}+1},\dots,x_{n-(d-2)t})$ is a monomial prime ideal
that contains $I$. Now, the condition $\Delta$ is pure implies that there
exists a minimal prime $\mathfrak{p}\in{\operatorname{Min}}(I)$ of height two
such that $\mathfrak{p}\subseteq\mathfrak{p}_{[n]\setminus G}$. The $t$–spread
monomial $w=x_{2}x_{2+t}\cdots x_{2+(d-1)t}\in I$ as $u>w\geq v$. We have
$x_{2}\in\mathfrak{p}$. In fact, if $x_{2}\notin\mathfrak{p}$, then
$\mathfrak{p}\subseteq\mathfrak{p}_{[i_{2},n-(d-2)t]}$. We must have
$\mathfrak{p}=\mathfrak{p}_{[i_{2},n-(d-2)t]}$. Indeed, if we omit some
$x_{j}$, $j\in[i_{2},n-(d-2)t]$, then we can find a monomial $z\in G(I)$, with
$\min(z)=1$ and $\min(z/x_{1})=j$, and this monomial does not belong to
$\mathfrak{p}$, a contradiction. $\Delta$ pure implies
$\textup{height}(\mathfrak{p})=2$, thus $i_{2}=n-(d-2)t-1$. But
$[n-(d-2)t-1,n-(d-2)t]\cap{\operatorname{supp}}(w)=\varnothing$. Indeed,
$n\geq 4+(2d-3)t$, so

$\displaystyle n-(d-2)t-2$ $\displaystyle\geq\ 4+(2d-3)t-(d-2)t-2=2+(d-1)t$
(14) $\displaystyle=\ \max(w).$

Thus $\max(w)<n-(d-2)t-1$. This implies that $w\in I\setminus\mathfrak{p}$, a
contradiction. Hence, $x_{2}\in\mathfrak{p}$. Therefore,
$\textup{height}(\mathfrak{p})=2$ and $\mathfrak{p}=(x_{2},x_{i_{2}})$. The
monomial $z=x_{1}x_{n-(d-2)t}\cdots x_{n-t}x_{n}$ belongs to $I$. We have
$z\in\mathfrak{p}$ if and only if $x_{i_{2}}$ divides $z$. Thus, as $1+t\leq
i_{2}\leq n-(d-2)t$, we have $i_{2}=n-(d-2)t$ and $u=x_{1}x_{n-(d-2)t}\cdots
x_{n-t}x_{n}$.
Let $r=\max\\{\max(w):w\in G(I)\,\,\mbox{and $\min(w)=2$}\\}$. Observe that
$r\geq 2+(d-1)t$, as $x_{2}x_{2+t}\cdots x_{2+(d-1)t}\in I$. The ideal
$\mathfrak{q}=(x_{1},x_{2+(d-1)t},x_{3+(d-1)t},\dots,x_{r})$ is a minimal
prime of $I$. The condition $\Delta$ pure implies
$\textup{height}(\mathfrak{q})=2$ and $r=2+(d-1)t$. Thus, $\max(v)=2+(d-1)t$
and $v=x_{2}x_{2+t}\cdots x_{2+(d-1)t}$.
(b)$\Longrightarrow$(a). Let $u=x_{1}x_{n-(d-2)t}\cdots x_{n-t}x_{n}$ and
$v=x_{2}x_{2+t}\cdots x_{2+(d-1)t}$. By (14), we have
${\operatorname{supp}}(u)\cap{\operatorname{supp}}(v)=\varnothing$. Moreover
$I=(u,v)$. Thus, $I$ is a complete intersection and so it is Cohen–Macaulay.
$\Box$

###### Remark 3.4

One can observe that under the hypotheses of Theorem 3.3, the classification
for Cohen–Macaulay $t$–spread lexsegment ideal is equivalent to the
classification for unmixed $t$–spread lexsegment ideal (see also, Remark
2.11).

### 3.2. The general case

In this subsection, we classify all Cohen–Macaulay $t$–spread lexsegment
ideals $I=(\mathcal{L}_{t}(u,v))$ of $S=K[x_{1},\ldots,x_{n}]$ with
$\min(v)>2$.

####  3.2.1 The underlying idea behind the classification

Let $I=(\mathcal{L}_{t}(u,v))$ be a Cohen–Macaulay $t$–spread lexsegment ideal
of $S$ with $u=x_{i_{1}}x_{i_{2}}\dots x_{i_{d}}$, $v=x_{j_{1}}x_{j_{2}}\cdots
x_{j_{d}}\in M_{n,d,t}$ such that $\min(u)=i_{1}=1$, $\min(v)=j_{2}>2$, $d\geq
2$. If $\Delta$ is the simplicial complex on the vertex set $[n]$ associated
to $I$, then $\Delta$ is pure. This is a key property in order to get the
classification. First, note that since $\min(u)=1$, then

$u\in\mathcal{L}_{t}(x_{1}x_{1+t}\cdots
x_{1+(d-1)t},x_{1}x_{n-(d-2)t}x_{n-(d-3)t}\cdots x_{n-t}x_{n}).$

Assume $u=x_{1}x_{1+t}\cdots x_{1+(d-1)t}=\max(M_{n,d,t})$, then
$I=(\mathcal{L}_{t}(u,v))=(\mathcal{L}_{t}^{i}(v))$ is an initial $t$–spread
lexsegment ideal. Hence, $\Delta$ is Cohen–Macaulay if and only if
$I=I_{n,d,t}$ (Corollary 2.5).

Now, let $u<x_{1}x_{1+t}\cdots x_{1+(d-1)t}$. Then, $F=[1+(d-1)t]\in\Delta$.
Moreover, $[1+(d-1)t]\cup\\{k\\}\notin\Delta$, for all $k\notin[1+(d-1)t]$. In
fact, $j_{1}>2$ implies that $w=x_{2}x_{2+t}\cdots x_{2+(d-2)t}x_{k}\in
I=I_{\Delta}$, thus $G={\operatorname{supp}}(w)\notin\Delta$ and $G\subseteq
F\cup\\{k\\}$. Therefore, $F$ is a facet of $\Delta$ and by the purity of
$\Delta$ we have that

$\dim(\Delta)=|F|-1=(d-1)t.$ (15)

We show that $v\leq x_{n-1-(d-1)t}x_{n-1-(d-2)t}\cdots x_{n-1-t}x_{n-1}$.
Suppose on the contrary that $v>x_{n-1-(d-1)t}x_{n-1-(d-2)t}\cdots
x_{n-1-t}x_{n-1}$, then $A=[n-1-(d-1)t,n]$ is a face of $\Delta$ and
$\dim(\Delta)\geq\dim A=|A|-1=(d-1)t+1>(d-1)t=\dim(\Delta)$, a contradiction.
Thus, $v\leq x_{n-1-(d-1)t}x_{n-1-(d-2)t}\cdots x_{n-1-t}x_{n-1}$. Hence,

$v\in\\{v_{0}>v_{1}>\dots>v_{d-1}>v_{d}=\min(M_{n,d,t})\\},$

where

$v_{\ell}=\Big{(}\prod_{s=\ell}^{d-1}x_{n-st-1}\Big{)}\Big{(}\prod_{s=0}^{\ell-1}x_{n-st}\Big{)},\quad\ell\in\\{0,\dots,d\\}.$
(16)

We can observe that $v\neq v_{d}$. Otherwise,
$I=(\mathcal{L}_{t}(u,v))=(\mathcal{L}_{t}^{f}(u))$ is a final $t$–spread
lexsegment ideal and so $I$ is Cohen–Macaulay if and only if $I$ is the
$t$–spread Veronese ideal $I_{n,d,t}$ (Corollary 2.10). Thus
$u=\max(M_{n,d,t})$, against the assumption that $u<\max(M_{n,d,t})$.
Finally, for $v$ there do exist three admissible choices:

  1. -

$v=v_{d-1}$,

  2. -

$t=1$ and $v=v_{\ell}$, for $\ell\in\\{0,\dots,d-2\\}$,

  3. -

$t>1$ and $v=v_{\ell}$, for $\ell\in\\{0,\dots,d-2\\}$.

In particular, if
$v=v_{\ell}=\big{(}\textstyle\prod_{s=\ell}^{d-1}x_{n-st-1}\big{)}\big{(}\prod_{s=0}^{\ell-1}x_{n-st}\big{)}$
($\ell\in\\{0,\dots,d-2\\}$), then

$H=[n-(d-1)t-1,n]\setminus\\{n-\ell t-1\\}\in\Delta.$ (17)

Moreover, $|H|=1+(d-1)t$ and $\Delta$ pure implies that
$H\cup\\{1\\}\notin\Delta$, _i.e._ , ${\bf x}_{H\cup\\{1\\}}\in I=I_{\Delta}$.

####  3.2.2 The classification

Now we are in position to prove the main result in the article.

###### Theorem 3.5

Let $u=x_{i_{1}}x_{i_{2}}\dots x_{i_{d}}$, $v=x_{j_{1}}x_{j_{2}}\cdots
x_{j_{d}}$ be $t$–spread monomials of degree $d\geq 2$ of $S$ with $i_{1}=1$,
$j_{1}>2$ and let $I$ be the $t$–spread lexsegment ideal
$I=(\mathcal{L}_{t}(u,v))$ of $S$. Assume $I\neq I_{n,d,t}$. Then $I$ is
Cohen–Macaulay if and only if one of the following conditions holds true

  1. (a)

$u=x_{1}x_{n-(d-2)t}\cdots x_{n-t}x_{n}$ and
$v=x_{n-(d-1)t-1}\big{(}\textstyle\prod_{s=0}^{d-2}x_{n-st}\big{)}$;

  2. (b)

$t=1$ and for some integer $\ell\in\\{0,\dots,d-2\\}$,

$\displaystyle u$
$\displaystyle=x_{1}\big{(}\textstyle\prod_{s=\ell+1}^{d-2}x_{n-s-1}\big{)}\big{(}\prod_{s=0}^{\ell}x_{n-s}\big{)},\
\ \ \ $ $\displaystyle
v=\big{(}\textstyle\prod_{s=\ell}^{d-1}x_{n-s-1}\big{)}\big{(}\prod_{s=0}^{\ell-1}x_{n-s}\big{)};$

  3. (c)

$t>1$ and for some integer $\ell\in\\{0,\dots,d-2\\}$,

$\displaystyle u$
$\displaystyle\in\mathcal{L}_{t}\big{(}x_{1}\big{(}\textstyle\prod_{s=\ell}^{d-2}x_{n-st-1}\big{)}\big{(}\prod_{s=0}^{\ell-1}x_{n-st}\big{)},x_{1}\big{(}\prod_{s=0}^{d-2}x_{n-st}\big{)}\big{)},$
$\displaystyle v$
$\displaystyle=\big{(}\textstyle\prod_{s=\ell}^{d-1}x_{n-st-1}\big{)}\big{(}\prod_{s=0}^{\ell-1}x_{n-st}\big{)}.$

Proof. (a) Let
$v=v_{d-1}=x_{n-(d-1)t-1}\big{(}\prod_{s=0}^{d-2}x_{n-st}\big{)}$. The only
monomial $w\in M_{n,d,t}$ with $w<v$ is
$w=\min(M_{n,d,t})=\big{(}\textstyle\prod_{s=0}^{d-1}x_{n-st}\big{)}$. Since
$x_{1}(w/x_{n-(d-1)t})\leq u$, then $I$ is a completely $t$–spread lexsegment
ideal (Theorem 1.4(2)). Moreover, $I$ has a linear resolution (Theorem
1.4(2)). Indeed, setting $u=x_{1}x_{i_{2}}\cdots x_{i_{d}}$, we have
$i_{2}>1+t$ and

$x_{1}x_{i_{2}-t}x_{i_{3}-t}\cdots x_{i_{d}-t}\geq
x_{1}x_{n-(d-1)t}x_{n-(d-2)t}\cdots x_{n-t}=x_{1}(w/x_{\max(w)}).$

Note that the Cohen–Macaulayness of $\Delta$ implies $\dim(S/I)=0pt(S/I)$.
From (15), $\dim(S/I)=n-0pt(I)=1+(d-1)t=0pt(S/I)$.

On the other hand, by the Auslander–Buchsbaum formula,
$0pt(S/I)=n-{\operatorname{pd}}(S/I)=n-1-{\operatorname{pd}}(I)$. Thus, $S/I$
is Cohen–Macaulay if and only if ${\operatorname{pd}}(I)=n-(d-1)t-2$. Setting
$a=n-1-(d-1)t$ and $b=a+d$, we can note that ${\operatorname{pd}}(I)\leq
n-(d-1)t-1$ if and only if (Theorem 1.4(3))

$\displaystyle\beta_{a}(I)=\beta_{a,b}(I)$
$\displaystyle=\sum_{w\in\mathcal{L}_{t}^{f}(u)}\binom{n-\min(w)-(d-1)t}{n-1-(d-1)t}-\sum_{\begin{subarray}{c}w\in\mathcal{L}_{t}^{f}(v)\\\
w\neq v\end{subarray}}\binom{\max(w)-(d-1)t-1}{n-1-(d-1)t}$
$\displaystyle=\big{|}\big{\\{}w\in\mathcal{L}_{t}^{f}(u):\min(w)=1\big{\\}}\big{|}-\big{|}\big{\\{}w\in\mathcal{L}_{t}^{f}(v)\setminus\\{v\\}:\max(w)=n\big{\\}}\big{|}$
$\displaystyle=\big{|}\big{\\{}w\in\mathcal{L}_{t}^{f}(u):\min(w)=1\big{\\}}\big{|}-1=0,$

_i.e._ , $u=x_{1}\big{(}\prod_{s=0}^{d-2}x_{n-st}\big{)}$. Condition (a)
follows.

Conversely, let $u=x_{1}x_{n-(d-2)t}\cdots x_{n-t}x_{n}$ and
$v=x_{n-(d-1)t-1}\big{(}\textstyle\prod_{s=0}^{d-2}x_{n-st}\big{)}$. We prove
that $\dim(S/I)=0pt(S/I)$.

First, using the same arguments as before, we can verify that $I$ is a
completely $t$–spread lexsegment ideal which has a linear resolution.

Setting $a=n-(d-1)t-1$ and $c=a-1=n-(d-1)t-2$, we have
$m=\big{|}\\{w\in\mathcal{L}_{t}^{f}(u):\min(w)=2\\}\big{|}>1$ and

$\displaystyle\beta_{c}(I)$
$\displaystyle=\sum_{w\in\mathcal{L}_{t}^{f}(u)}\binom{n-\min(w)-(d-1)t}{n-2-(d-1)t}-\sum_{\begin{subarray}{c}w\in\mathcal{L}_{t}^{f}(v)\\\
w\neq v\end{subarray}}\binom{\max(w)-(d-1)t-1}{n-2-(d-1)t}$
$\displaystyle=\binom{a}{a-1}+m\binom{a-1}{a-1}-\binom{a}{a-1}=m>0.$

by Theorem 1.4(3). Thus, ${\operatorname{pd}}(I)=c=n-(d-1)t-2$. Hence, from
the Auslander–Buchsbaum formula, $0pt(S/I)=1+(d-1)t$.

Let us prove that $\dim(S/I)=1+(d-1)t$. It is sufficient to verify that
$\dim(\Delta)=(d-1)t$. Using the notation of Theorem 2.12, since $I$ is a
completely $t$–spread lexsegment ideal, we have
$\mathcal{F}(\Delta)=\mathcal{G}\cup\\{[n]\setminus
F_{p}:p\in\mathcal{I}\\}\cup\widetilde{\mathcal{F}}$. If $F\in\mathcal{G}$,
then $|F|=1+(d-1)t$. If $F\in\widetilde{\mathcal{F}}$, then
$|F|=(d-1)t<1+(d-1)t$. Finally, we check that $\mathcal{I}=\\{1\\}$ and
$|[n]\setminus F_{1}|=1+(d-1)t$. Indeed, for $p=1$, $[n]\setminus
F_{1}=[j_{1}+1,n]=[n-(d-1)t,n]$ has cardinality $1+(d-1)t$, so
$1\in\mathcal{I}$. Let $2\leq p\leq d$, then

$\displaystyle[n]\setminus F_{p}$
$\displaystyle=\bigcup_{i=1}^{p-1}[j_{i},j_{i}+(t-1)]\cup[j_{p}+1,n]$
$\displaystyle=\bigcup_{i=1}^{p-1}[j_{i},j_{i}+(t-1)]\cup[n-(d-p)t+1,n]$

has cardinality $(p-1)t+n+1-(n-(d-p)t+1)=(d-1)t$ and moreover,

$\min(v/x_{j_{p}})=\min(v)=n-(d-2)t-1<n-(d-2)t-1\leq\min(u/x_{1}).$

Thus $v/x_{j_{p}}>u/x_{1}$ and $p\notin\mathcal{I}$ and so
$\mathcal{I}=\\{1\\}$ and $|[n]\setminus F_{1}|=1+(d-1)t$. Hence,
$\dim(\Delta)=(d-1)t$ and consequently $\dim(S/I)=n-0pt(I)=1+(d-1)t=0pt(S/I)$
and $I$ is Cohen–Macaulay.

(b) Let $t=1$. Let us consider the set $H$ defined in (17). We note that the
smallest $1$–spread (squarefree) monomial of degree $d$ that we can “extract”
from $H\cup\\{1\\}=\\{1,n-d,n-d+1,\dots,n-\ell-2,n-\ell,\dots,n-1,n\\}$ is

$\displaystyle z$ $\displaystyle=x_{1}x_{n-d+1}\cdots x_{n-\ell-2}\cdot
x_{n-\ell}\cdots x_{n-1}x_{n}$
$\displaystyle=x_{1}\big{(}\textstyle\prod_{s=\ell+1}^{d-2}x_{n-s-1}\big{)}\big{(}\prod_{s=0}^{\ell}x_{n-s}\big{)}.$

Thus, we must have $z\in I$, and so $u\geq z$, _i.e._ ,

$u\in\mathcal{L}_{t}\big{(}x_{1}\big{(}\textstyle\prod_{s=\ell}^{d-2}x_{n-s-2}\big{)}\big{(}\prod_{s=0}^{\ell-1}x_{n-s-1}\big{)},x_{1}\big{(}\prod_{s=\ell}^{d-2}x_{n-s-1}\big{)}\big{(}\prod_{s=0}^{\ell+1}x_{n-s}\big{)}\big{)}.$

Let $v=v_{\ell}$, for some $\ell\in\\{0,\ldots,d-2\\}$. The monomials $w\in
M_{n,d,1}$ with $w<v$ are $v_{\ell+1}>\dots>v_{d}$. Since, for
$r=\ell+1,\dots,d$,

$x_{1}(v_{r}/x_{n-(d-1)})\leq
x_{1}\big{(}\textstyle\prod_{s=\ell+1}^{d-2}x_{n-s-1}\big{)}\big{(}\prod_{s=0}^{\ell}x_{n-s}\big{)}\leq
u,$

$I$ is a completely $1$–spread lexsegment ideal. Moreover, $I$ has a linear
resolution. Indeed, setting $u=x_{1}x_{i_{2}}\cdots x_{i_{d}}$, we have
$i_{2}>1+t=2$ and

$x_{1}x_{i_{2}-1}x_{i_{3}-1}\cdots x_{i_{d}-1}\geq
x_{1}\big{(}\textstyle\prod_{s=\ell+1}^{d-2}x_{n-s-2}\big{)}\big{(}\prod_{s=0}^{\ell}x_{n-s-1}\big{)}>x_{1}(v_{\ell+1}/x_{\max(w)}).$

From (15), $\dim(S/I)=1+(d-1)t=d$ and, since we are assuming $I$
Cohen–Macaulay, then $0pt(I)=d$. On the other hand from the
Auslander–Buchsbaum formula, $S/I$ is Cohen–Macaulay if and only if
${\operatorname{pd}}(I)=n-d-1$. Setting $a=n-d$ and $b=a+d$,
${\operatorname{pd}}(I)\leq n-d$, $S/I$ is Cohen–Macaulay if and only if

$\displaystyle\beta_{a}(I)=\beta_{a,b}(I)$
$\displaystyle=\big{|}\big{\\{}w\in\mathcal{L}_{t}^{f}(u):\min(u)=1\big{\\}}\big{|}-\big{|}\big{\\{}w\in\mathcal{L}_{t}^{f}(v)\setminus\\{v\\}:\max(w)=n\big{\\}}\big{|}$
$\displaystyle=\big{|}\big{\\{}w\in\mathcal{L}_{t}^{f}(u):\min(w)=1\big{\\}}\big{|}-\big{|}\big{\\{}v_{\ell+1}>\dots>v_{d}\big{\\}}\big{|}$
$\displaystyle=\big{|}\big{\\{}w\in\mathcal{L}_{t}^{f}(u):\min(w)=1\big{\\}}\big{|}-(d-\ell)=0,$

_i.e._ ,
$u=x_{1}\big{(}\textstyle\prod_{s=\ell+1}^{d-2}x_{n-s-1}\big{)}\big{(}\prod_{s=0}^{\ell}x_{n-s}\big{)}$.
The assertion (b) follows.

Conversely, let
$u=x_{1}\big{(}\textstyle\prod_{s=\ell+1}^{d-2}x_{n-s-1}\big{)}\big{(}\prod_{s=0}^{\ell}x_{n-s}\big{)}$
and $v=v_{\ell}$, for some $\ell$ $\in\\{0,\dots,d-2\\}$. We compute
$\dim(S/I)$ and $0pt(S/I)$.

Using the notation of Theorem 2.12,
$\mathcal{F}(\Delta)=\mathcal{G}\cup\\{[n]\setminus
F_{p}:p\in\mathcal{I}\\}\cup\widetilde{\mathcal{F}}$. We show that
$\dim(\Delta)=d$. If $F\in\mathcal{G}$, then $|F|=d$. Arguing as in the proof
of condition (b), one can observe that $\mathcal{I}=[d-\ell]$ and for
$p\in\mathcal{I}$, $|[n]\setminus F_{p}|=d$. Furthermore, for any
$F\in{\mathcal{F}}$, $|F|=d-1<d$. Hence $\dim(\Delta)=d$ and so $\dim(S/I)=d$.

Setting $a=n-d$ and $c=a-1=n-d-1$, we have
$\big{|}\\{w\in\mathcal{L}_{t}^{f}(u):\min(w)=1\\}\big{|}=d-\ell$,
$m=\big{|}\\{w\in\mathcal{L}_{t}^{f}(u):\min(w)=2\\}\big{|}>1$ and

$\displaystyle\beta_{c}(I)$
$\displaystyle=\sum_{w\in\mathcal{L}_{t}^{f}(u)}\binom{n-\min(w)-d+1}{n-d-1}-\sum_{\begin{subarray}{c}w\in\mathcal{L}_{t}^{f}(v)\\\
w\neq v\end{subarray}}\binom{\max(w)-d}{n-d-1}$
$\displaystyle=(d-\ell)\binom{a}{a-1}+m\binom{a-1}{a-1}-(d-\ell)\binom{a}{a-1}=m>0.$

Thus, ${\operatorname{pd}}(I)=n-d-1$ and from the Auslander–Buchsbaum formula
$0pt(I)=d$ and $I$ is Cohen–Macaulay. $\Box$

For the proof of the last case (c), we need some more preparations.

First, recall that if $0\rightarrow L\rightarrow M\rightarrow N\rightarrow 0$
is a short exact sequence of finitely generated graded $S$–modules, then
${\operatorname{pd}}(M)\leq\max\\{{\operatorname{pd}}(L),{\operatorname{pd}}(N)\\}$.
Moreover, given a monomial ideal $I$ of $S$ and a monomial $u$ of $S$, there
exists the short exact sequence

$0\rightarrow S/(I:u)\rightarrow S/I\rightarrow S/(I,u)\rightarrow 0.$

Therefore
${\operatorname{pd}}(S/I)\leq\max\\{{\operatorname{pd}}(S/(I:u)),{\operatorname{pd}}(S/(I,u))\\}$.

###### Lemma 3.6

Let $I,J$ equigenerated monomial ideals of $S$ having initial degree $d$.
Suppose $J\subseteq I$, then

$\beta_{i,i+d}(J)\leq\beta_{i,i+d}(I),$

for all $i\geq 0$. Moreover, if $J$ has a linear resolution,
${\operatorname{pd}}(I)\geq{\operatorname{pd}}(J)$.

Proof. The first assertion follows from [18, Lemma 10.3.5]. For the second
assertion, assume $J$ has a linear resolution, and let
${\operatorname{pd}}(J)=s$. Then
$\beta_{s}(J)=\beta_{s,s+d}(J)\leq\beta_{s,s+d}(I)$. Finally,
${\operatorname{pd}}(I)\geq s={\operatorname{pd}}(J)$. $\Box$

Now we are in position to prove the last statement of Theorem 3.5.

(c) Let $t\geq 2$. From (17), the smallest $t$–spread monomial we can
“extract” from $H\cup\\{1\\}$ is

$z=x_{1}x_{n-(d-2)t}x_{n-(d-3)t}\cdots x_{n-t}x_{n}.$

Since $z$ is the smallest $t$–spread monomial of $S$ with minimum equal to
$1$, then $z\in I$, and there are no further conditions to impose on the
monomial $u$. Thus

$u\in\mathcal{L}_{t}\big{(}x_{1}\big{(}\textstyle\prod_{s=\ell}^{d-2}x_{n-st-2}\big{)}\big{(}\prod_{s=0}^{\ell-1}x_{n-st-1}\big{)},x_{1}\big{(}\prod_{s=0}^{d-2}x_{n-st}\big{)}\big{)}.$
(18)

Setting

$u_{p}=x_{1}\Big{(}\prod_{s=p}^{d-2}x_{n-st-1}\Big{)}\Big{(}\prod_{s=0}^{p-1}x_{n-st}\Big{)},\,\,\,p=\ell,\dots,d-1,$

we have that

$\mathcal{L}_{t}\big{(}x_{1}\big{(}\textstyle\prod_{s=\ell}^{d-2}x_{n-st-1}\big{)}\big{(}\prod_{s=0}^{\ell-1}x_{n-st}\big{)},x_{1}\big{(}\prod_{s=0}^{d-2}x_{n-st}\big{)}\big{)}=\big{\\{}u_{\ell}>u_{\ell+1}>\dots>u_{d-1}\big{\\}}.$

We prove that if

$u>u_{\ell}=x_{1}\big{(}\textstyle\prod_{s=\ell}^{d-2}x_{n-st-1}\big{)}\big{(}\prod_{s=0}^{\ell-1}x_{n-st}\big{)},$
(19)

then $I$ is not Cohen–Macaulay.

By hypothesis $v=v_{\ell}$ ($\ell=0,\ldots,d-2$) and from (19)

$u\geq
x_{1}\big{(}\textstyle\prod_{s=\ell-1}^{d-2}x_{n-st-1}\big{)}\big{(}\prod_{s=0}^{\ell-2}x_{n-st}\big{)},$

where
$x_{1}\big{(}\textstyle\prod_{s=\ell-1}^{d-2}x_{n-st-1}\big{)}\big{(}\prod_{s=0}^{\ell-2}x_{n-st}\big{)}$
is the greatest monomial with minimum equal to $1$ preceding $u_{\ell}$, with
respect to the squarefree lex order. The monomials $w\in M_{n,d,t}$ smaller
than $v$ are $v_{\ell+1}>\dots>v_{d}$ and for each $v_{q}$,
$q=\ell+1,\ldots,d$, we have

$u\geq
x_{1}\big{(}\textstyle\prod_{s=\ell-1}^{d-2}x_{n-st-1}\big{)}\big{(}\prod_{s=0}^{\ell-2}x_{n-st}\big{)}=x_{1}(v_{\ell+1}/x_{\min(v_{\ell+1})})>x_{1}(v_{q}/x_{\min(v_{q})}).$

Hence, setting $u=x_{1}x_{i_{2}}\cdots x_{i_{d}}$, we have $i_{2}>1+t$ and

$\displaystyle x_{1}x_{i_{2}-t}\cdots x_{i_{d}-t}$ $\displaystyle\geq
x_{1}x_{n-(d-1)t-1}\cdots x_{n-\ell t-1}\cdot x_{n-(\ell-1)t}\cdots x_{n-t}$
$\displaystyle=x_{1}(v_{\ell}/x_{\max(v_{\ell})})\geq
x_{1}(v_{\ell+1}/x_{\max(v_{\ell+1})}).$

Therefore, $I$ is a completely $t$–spread lexsegment ideal with a linear
resolution. Setting $a=n-1-(d-1)t$ and $b=a+d$, we have

$\displaystyle\beta_{a}(I)=\beta_{a,b}(I)$
$\displaystyle=\big{|}\big{\\{}w\in\mathcal{L}_{t}^{f}(u):\min(u)=1\big{\\}}\big{|}-\big{|}\big{\\{}w\in\mathcal{L}_{t}^{f}(v)\setminus\\{v\\}:\max(w)=n\big{\\}}\big{|}$
$\displaystyle\geq\big{|}\big{\\{}u>u_{\ell}>\dots>u_{d-1}\\}\big{|}-(d-\ell)=d+1-\ell-(d-\ell)=1.$

It follows that ${\operatorname{pd}}(I)=n-1-(d-1)t$ and consequently $I$ is
not Cohen–Macaulay. Condition (c) follows.

Conversely, assume (c) holds. We prove that $\dim(S/I)=1+(d-1)t$. This is
equivalent to verify that $\dim(\Delta)=(d-1)t$. For this aim it sufficies to
show the following facts:

  1. (i)

there exists $F\in\mathcal{F}(\Delta)$ with cardinality $|F|=1+(d-1)t$;

  2. (ii)

for any $A\subseteq[n]$ with $|A|=2+(d-1)t$, then $A\notin\Delta$.

(i) Let us consider $F=[n-(d-1)t,n]$. By the structure of $v$, it follows that
$F\in\Delta$ and furthermore $|F|=1+(d-1)t$, as desired.
(ii) Let $A\subseteq[n]$ with $|A|=2+(d-1)t$. We have $1\leq\min(A)\leq
n-(d-1)t-1$.

If $\min(A)=n-(d-1)t-1$, then $A=[n-(d-1)t-1,n]\notin\Delta$, as

$x_{n-(d-1)t-1}x_{n-(d-2)t-1}\cdots x_{n-t-1}x_{n-1}\in I.$

If $1<\min(A)<n-(d-1)t-1$, setting $\ell_{1}=\min(A)$ and
$\ell_{r}=\min\\{\ell\in A:\ell\geq\ell_{r-1}+t\\}$, for $r\geq 2$, the
sequence $L:\ell_{1}>\ell_{2}>\cdots$ has at least $d$ terms, and ${\bf
x}_{L}\in I$. Thus, $A\notin\Delta$.

If $\min(A)=1$, we consider the sequence $L$ above defined. If ${\bf
x}_{L}\leq u$, then $A\notin\Delta$. Otherwise, set $B=A\setminus\\{1\\}$.

If $\min(B)=n-(d-1)t$, then $A=\\{1\\}\cup
B=\\{1\\}\cup[n-(d-1)t,n]\notin\Delta$, as $x_{1}x_{n-(d-2)t}\cdots
x_{n-t}x_{n}\in I$. If $\min(B)<n-(d-1)t-1$, setting $\nu_{1}=\min(B)$, and
$\nu_{r}=\min\\{\nu\in B:\nu\geq\nu_{r-1}+t\\}$, for $r\geq 2$, the sequence
$N:\nu_{1}>\nu_{2}>\cdots$ has at least $d$ terms, and ${\bf x}_{N}\in I$,
implying that $A\notin\Delta$.

Let $\min(B)=n-(d-1)t-1$. Then $A=\\{1\\}\cup
B=\\{1\\}\cup[n-(d-1)t-1,n]\setminus\\{m\\}$, for some $m>n-(d-1)t-1$. We can
write $m=n-pt-q$, with $0\leq p\leq d-3$ and $0\leq q\leq t-1$. If $q\geq 2$,
then ${\operatorname{supp}}(v)\subseteq B\subseteq A$, and $A\notin\Delta$. If
$q=1$, then $\\{1,n-(d-2)t,\dots,n-t,n\\}\subseteq A$, and
$x_{1}x_{n-(d-2)t}\cdots x_{n-t}x_{n}\in I$, thus $A\notin\Delta$. If $q=0$,
we consider $p$. If $p<\ell-1$, the structure of $v$ implies that
${\operatorname{supp}}(v)\subseteq A$, and once again $A\notin\Delta$. If
$p\geq\ell-1$, then $C=\\{n-kt-1:k=0,\dots,d-1\\}\subseteq A$, and ${\bf
x}_{C}>v$, so ${\bf x}_{C}\in I$, implying $A\notin\Delta$. Finally, in all
possible cases, $A\notin\Delta$, and (ii) follows.

Finally, $\dim(S/I)=1+(d-1)t$.

Now, set $I_{u_{p}}=(\mathcal{L}_{t}(u_{p},v))$, $p=\ell,\dots,d-1$. We prove
that $I_{u_{p}}$ is Cohen–Macaulay, _i.e._ ,
${\operatorname{pd}}(I_{u_{p}})=n-2-(d-1)t$.

First, assume $p=\ell$. We have

$u=u_{\ell}=x_{1}(v_{\ell}/x_{\min(v_{\ell})})>x_{1}(v_{q}/x_{\min(v_{q})}),\,\,\,q=\ell+1,\dots,d$

and setting $u=x_{1}x_{i_{2}}\cdots x_{i_{d}}$, it follows that

$x_{1}x_{i_{2}-t}\cdots x_{i_{d}-t}=x_{1}(v_{\ell}/x_{\max(v_{\ell})}).$

Thus, $I_{u_{\ell}}$ is a completely $t$–spread lexsegment ideal with a linear
resolution. Set $a=n-1-(d-1)t$, $b=a+d$, we have

$\displaystyle\beta_{a}(I_{u_{\ell}})=\beta_{a,b}(I_{u_{\ell}})$
$\displaystyle=\big{|}\big{\\{}w\in\mathcal{L}_{t}^{f}(u):\min(u)=1\big{\\}}\big{|}-\big{|}\big{\\{}w\in\mathcal{L}_{t}^{f}(v)\setminus\\{v\\}:\max(w)=n\big{\\}}\big{|}$
$\displaystyle=\big{|}\big{\\{}u_{\ell}>u_{\ell+1}>\dots>u_{d-1}\\}\big{|}-|\\{v_{\ell+1}>\dots>v_{d}\\}|=0,$

and, as $m=\big{|}\\{w\in\mathcal{L}_{t}^{f}(u):\min(w)=2\\}\big{|}>1$, then

$\displaystyle\beta_{a-1}(I_{u_{\ell}})$
$\displaystyle=\sum_{w\in\mathcal{L}_{t}^{f}(u)}\binom{n-\min(w)-(d-1)t}{n-2-(d-1)t}-\sum_{\begin{subarray}{c}w\in\mathcal{L}_{t}^{f}(v)\\\
w\neq v\end{subarray}}\binom{\max(w)-(d-1)t-1}{n-2-(d-1)t}$
$\displaystyle=(d-\ell)\binom{a}{a-1}+m\binom{a-1}{a-1}-(d-\ell)\binom{a}{a-1}=m>0.$

Thus, ${\operatorname{pd}}(I_{u_{\ell}})=a-1=n-(d-1)t-2$, and $I_{u_{\ell}}$
is Cohen–Macaulay, as desired.

Now let $p=\ell+1$. Firstly, observe that
$J=(\mathcal{L}_{t}(x_{2}x_{2+t}\cdots x_{2+(d-1)t},v))\subseteq
I_{u_{\ell+1}}$. Moreover, $J$ is an initial $t$–spread lexsegment ideal in
$K[x_{2},\dots,x_{n}]$ with ${\operatorname{pd}}(J)=n-2-(d-1)t$. Thus $J$ has
a linear resolution, and Lemma 3.6 implies
${\operatorname{pd}}(I_{u_{\ell+1}})\geq n-2-(d-1)t$. For the other
inequality, consider the short exact sequence:

$0\rightarrow S/(I_{u_{\ell+1}}:u_{\ell})\rightarrow
S/I_{u_{\ell+1}}\rightarrow S/(I_{u_{\ell+1}},u_{\ell})\rightarrow 0.$

Observe that $S/(I_{u_{\ell+1}},u_{\ell})=S/I_{u_{\ell}}$, and
${\operatorname{pd}}(S/I_{u_{\ell}})={\operatorname{pd}}(I_{u_{\ell}})+1=n-1-(d-1)t$.
Let us verify that
$I_{u_{\ell+1}}:u_{\ell}=\mathfrak{p}_{[2,n-(d-1)t-1]\cup\\{n-\ell t\\}}$ is a
complete intersection and
${\operatorname{pd}}(S/(I_{u_{\ell+1}}:u_{\ell}))=n-1-(d-1)t$.

By [18, Proposition 1.2.2], a set of generators for $I_{u_{\ell+1}}:u_{\ell}$
is given by

$\big{\\{}w/\gcd(w,u_{\ell}):w\in G(I_{u_{\ell+1}})\big{\\}}.$

If $w=u_{p}$, $p=\ell+1,\dots,d-1$, we have
$u_{p}/\gcd(u_{p},u_{\ell})=x_{n-(p-1)t}\cdots x_{n-\ell t}$ and for
$p=\ell+1$, $u_{\ell+1}/\gcd(u_{\ell+1},u_{\ell})=x_{n-\ell t}$, thus
$x_{n-\ell t}$ divides all these generators.

If $w\in\mathcal{L}_{t}(x_{2}x_{2+t}\cdots x_{2+(d-1)t},v)$, then
$\min(w)\in[2,n-(d-1)t-1]$. Note that
$[2,n-(d-1)t-1]\cap{\operatorname{supp}}(u_{\ell})=\varnothing$, thus
$x_{\min(w)}$ divides $w/\gcd(w,u_{\ell})$ and for
$z=x_{\min(w)}(u_{\ell}/x_{1})\in G(I_{u_{\ell+1}})$, we have that
$z/\gcd(z,u_{\ell})=x_{\min(w)}\in I_{u_{\ell+1}}:u_{\ell}$.

Finally, we have verified that
$I_{u_{\ell+1}}:u_{\ell}=\mathfrak{p}_{[2,n-(d-1)t-1]\cup\\{n-\ell t\\}}$ and
consequently ${\operatorname{pd}}(S/(I_{u_{\ell+1}}:u_{\ell}))=n-1-(d-1)t$.

Hence,

${\operatorname{pd}}(S/I_{u_{\ell+1}})\leq\max\big{\\{}{\operatorname{pd}}(S/I_{u_{\ell}}),{\operatorname{pd}}(S/(I_{u_{\ell+1}}:u_{\ell}))\big{\\}}=n-1-(d-1)t,$

so ${\operatorname{pd}}(I_{u_{\ell+1}})\leq n-2-(d-1)t$. Thus
${\operatorname{pd}}(I_{u_{\ell+1}})=n-2-(d-1)t$ and $I_{u_{\ell+1}}$ is
Cohen–Macaulay. For $p\in\\{\ell+2,\dots,d\\}$ the same arguments work and the
proof of Theorem 3.5 is complete.

## 4\. Conclusions and Perspectives

In this article, we have investigated the Cohen–Macaulayness of $t$–spread
lexsegment ideals using the theory of simplicial complexes. The completely
$t$–spread lexsegment ideals have played an essential role. Non–completely
$t$–spread lexsegment ideals are much less understood. Here are some possible
questions to investigate.

###### Question 4.1

Determine the standard primary decomposition of non–completely $t$–spread
lexsegment ideals.

###### Question 4.2

Classify the pure simplicial complexes associated to $t$–spread lexsegment
ideals.

In [19], Olteanu classified all sequentially Cohen–Macaulay squarefree
completely lexsegment ideals. To the best of our knowledge, a classification
for squarefree non–completely lexsegment ideals is unknown. More generally, we
ask the following

###### Question 4.3

Classify all sequentially Cohen–Macaulay $t$–spread lexsegment ideals.

Finally, due to many examples that we performed and our results on completely
$t$–spread lexsegment ideals, we are led to conjecture the following:

###### Conjecture 4.4

Let $I$ be a $t$–spread lexsegment ideal generated in degree $d\geq 2$. Then
$\dim(S/I)\geq(d-1)t$.

Note that Conjecture 4.4 is trivially true when $t=1$. Indeed, if $\Delta$ is
the simplicial complex associated to $I$, then each set $A\subseteq[n]$, with
$|A|=d-1$, is in $\Delta$. Thus $\dim(S/I)\geq d-1$. Conjecture 4.4 is also
true for completely $t$–spread lexsegment ideals by virtue of Theorems 2.1,
2.7, 2.12.

_Acknowledgement_. We thank the the referee for his/her helpful suggestions
that allowed us to improve the quality of the paper.

## References

  * [1] L. Amata. Computational methods for $t$–spread monomial ideals, available at arXiv preprint https://arxiv.org/abs/2110.11801 [math.AC].
  * [2] L. Amata, M. Crupi, A. Ficarra. Upper bounds for extremal Betti numbers of $t$–spread strongly stable ideals, Bull. Math. Soc. Sci. Math. Roumanie, 65 (113), No. 1 (2022), 13–34.
  * [3] L. Amata, M. Crupi, A. Ficarra. Projective dimension and Castelnuovo–Mumford regularity of $t$–spread ideals, International Journal of Algebra and Computation 32, No. 4 (2022), 837–858.
  * [4] L. Amata, A. Ficarra, M. Crupi. A numerical characterization of extremal Betti numbers of $t$–spread strongly stable ideals, J. Algebraic Combin. 32(4) (2022), 837–858.
  * [5] C. Andrei, V. Ene, B. Lajmiri. Powers of t–spread principal Borel ideals, Archiv der Mathematik 112 (6) (2019), 587–597.
  * [6] C. Andrei–Ciobanu, Kruskal–Katona Theorem for $t$–spread strongly stable ideals, Bull.Math. Soc. Sci. Math. Roumanie (N.S.) 62(110)(2) (2019), 107–122.
  * [7] A. Aramova, E. De Negri, J. Herzog, Lexsegment ideals with a linear resolution, Illinois J. Math. 42(3) (1998), 509–52.
  * [8] A. Aramova, J. Herzog, T. Hibi. Squarefree lexsegment ideals. Math.Z. 228 (1998), 353–378.
  * [9] V. Bonanzinga, L. Sorrenti, N. Terai, Pure and Cohen–Macaulay simplicial complexes associated with squarefree lexsegment Ideals, Comm. Alg. 40(11) (2012), 4195–4214
  * [10] E. De Negri, J. Herzog, Completely lexsegment ideals, Proc. Amer. Math. Soc., 126 (12) (1998), 3467–3473.
  * [11] R. Dinu, Gorenstein $T$–spread Veronese algebras, Osaka J. Math. 57(4) (2020), 935–947.
  * [12] S. Eliahou, M. Kervaire, Minimal resolutions of some monomial ideals, J. Algebra 129 (1990), 1–25.
  * [13] V. Ene, J. Herzog, A. A. Qureshi, t–spread strongly stable monomial ideals, Comm. Algebra 47(12), 5303–5316 (2019).
  * [14] V. Ene, A. Olteanu, L. Sorrenti, Properties of lexsegment ideals, Osaka J. Math. 47 (1) (2010) 67–87.
  * [15] A. Ficarra, M. Crupi, Shifting operations and completely $t$–spread lexsegment ideals, Comm. Alg. 50(8) (2022), 3320–3337.
  * [16] C. A. Francisco, H. T. Ha, A. Van Tuyl, Splittings of monomial ideals, Proc. Amer. Math. Soc., 137 (10) (2009), 3271-3282.
  * [17] D. R. Grayson, M. E. Stillman, Macaulay2, a software system for research in algebraic geometry. Available at http://www.math.uiuc.edu/Macaulay2.
  * [18] J. Herzog, T. Hibi, _Monomial ideals_ , Graduate texts in Mathematics 260, Springer–Verlag, 2011.
  * [19] O. Olteanu, Classes of sequentially Cohen–Macaulay squarefree monomial ideals, Algebra Colloq. 21 (04) (2011), 575–590.