# On the notion of sequentially Cohen-Macaulay modules

Giulio Caviglia  Giulio Caviglia - Department of Mathematics - Purdue
University - 150 N. University Street, West Lafayette - IN 47907-2067 - USA
<EMAIL_ADDRESS>,  Alessandro De Stefani  Alessandro De Stefani -
Dipartimento di Matematica - Università di Genova - Via Dodecaneso 35 - 16146
Genova - Italy<EMAIL_ADDRESS>,  Enrico Sbarra  Enrico Sbarra -
Dipartimento di Matematica - Università degli Studi di Pisa - Largo Bruno
Pontecorvo 5 - 56127 Pisa - Italy<EMAIL_ADDRESS>and  Francesco
Strazzanti  Francesco Strazzanti - Dipartimento di Matematica “Giuseppe Peano”
- Università degli Studi di Torino - Via Carlo Alberto 10 - 10123 Torino -
Italy<EMAIL_ADDRESS>
###### Abstract.

In this survey paper we first present the main properties of sequentially
Cohen-Macaulay modules. Some basic examples are provided to help the reader
with quickly getting acquainted with this topic. We then discuss two
generalizations of the notion of sequential Cohen-Macaulayness which are
inspired by a theorem of Jürgen Herzog and the third author.

The first author was partially supported by a grant from the Simons Foundation
(41000748, G.C.). The second author was partially supported by the PRIN 2020
project 2020355B8Y “Squarefree Gröbner degenerations, special varieties and
related topics”. The third author was partially supported by the PRIN project
2017SSNZAW004 “Moduli Theory and Birational Classification” of Italian MIUR

## Introduction

This survey note, which is dedicated to the work of Jürgen Herzog on the
topic, cannot possibly be complete: the notion of sequentially Cohen-Macaulay
module has been central in many papers in the literature from the late 90’s,
starting maybe with [20]. On the other hand, in the late 90’s Herzog’s
research activity was feverish as he counted very many visitors and
collaborators. At the same time the distribution of preprints in the form of
postscript files over the internet expedited the dissemination of mathematical
ideas. We thus must apologize in advance that our reference list is doomed to
be incomplete.

It is in 1997 that Herzog, together with who would become his top co-author,
Takayuki Hibi, published a paper on simplicial complexes [26], immediately
followed by another series of papers of the two authors together with Annetta
Aramova [5, 6, 7], where numerical problems about simplicial complexes were
addressed through the study of Hilbert functions, Gröbner bases techniques and
generic initial ideals, see also [3, 4, 19]. In 1999, another influential
paper authored by Herzog and Hibi, is published, [27]: they introduce and
study a new class of ideals, called componentwise linear. Componentwise linear
Stanley-Reisner ideals $I_{\Delta}$ are characterized as those for which the
pure $i$-th skeleton of the Alexander dual of $\Delta$ is Cohen-Macaulay for
every $i$. Thanks to [20], this means that $I_{\Delta}$ is componentwise
linear if and only its Alexander dual is sequentially Cohen-Macaulay, see also
[30]. In this way, the authors were also able to generalize a well-known
result of Eagon and Reiner, which says that the Stanley-Reisner ideal of a
simplicial complex has a linear resolution if and only if its Alexander dual
is Cohen-Macaulay. All the ideas behind the proofs of these facts led also to
another fundamental result, which is the main theorem of [31] and provides a
somewhat unexpected characterization of graded sequentially Cohen-Macaulay
modules over a polynomial ring $R$ in terms of the Hilbert function of the
local cohomology modules:

######  Theorem 1.

Let $M$ be a finitely generated graded $R$-module, and let $M\cong F/U$ a free
graded presentation of $M$. Then, $F/U$ is sequentially Cohen-Macaulay if and
only if
$\operatorname{Hilb}(H^{i}_{{\mathfrak{m}}}(F/U))=\operatorname{Hilb}(H^{i}_{{\mathfrak{m}}}(F/\operatorname{Gin}(U)))$
for all $i\geqslant 0$.

Here $\operatorname{Gin}(U)$ denotes the generic initial module of $U$ with
respect to the degree reverse lexicographic order.

This paper is organized as follows. In the first section we introduce the
definition of sequentially Cohen-Macaulay modules according to Stanley [37],
and discuss some basic properties and examples. In Section 2 we present some
of the main characterizations, or equivalent definitions, of sequential Cohen-
Macaulayness, by recalling Schenzel’s results on the dimension filtration of a
module, cf. Propositions 2.2, 2.6 and Theorem 2.7, and Peskine’s
characterization in terms of deficiency modules, cf. Theorem 2.13. In Section
3, we recollect the definition of partial sequential Cohen-Macaulayness
introduced by the third and the fourth author in [35], by requiring that only
the queue of the dimension filtration is a Cohen-Macaulay filtration, see
Definition 3.1 and some basic properties in the graded case in Lemmata 3.4 and
3.5. The next two results, Lemma 3.8 and Proposition 3.10, are dedicated to
filling a gap in the proof of a fundamental lemma in [35], and we present the
first of our generalizations of Theorem 1 in Theorem 3.11. We start Section 4
by recalling the definition of E-depth, as introduced by the first and second
author in [12], see Definition 4.1. E-depth measures how much depth the
deficiency modules of a finitely generated standard graded module $M$ have
altogether, and sequential Cohen-Macaulayness can be detected by E-depth as
observed in Remark 4.2. Some interesting homological properties of E-depth,
especially in connection with strictly filter regular elements, are shown in
Proposition 4.4. In Definition 4.9 we introduce the other crucial notion for
the following, what we might call partial generic initial ideal, making use of
a special partial revlex order. Finally, by means of Proposition 4.10, we
prove in Theorem 4.11 the main result of this section, which can be seen as
another generalization of Theorem 1.

To Jürgen Herzog, a bright example of mathematician and teacher.

##  1\. Sequentially Cohen-Macaulay modules

Throughout the paper, let $(R,\mathfrak{m},k)$ denote either a Noetherian
local ring with maximal ideal $\mathfrak{m}$ and residue field
$k=R/\mathfrak{m}$, or a standard graded $k$-algebra $R=\bigoplus_{i\geqslant
0}R_{i}$, with $R_{0}=k$ and $\mathfrak{m}=\bigoplus_{i>0}R_{i}$. In the
second case, every $R$-module we consider will be a graded $R$-module, and
homomorphisms will be graded of degree zero.

One of the features that makes the Cohen-Macaulay property significant is its
characterization in terms of the vanishing and non-vanishing of local
cohomology: for a $d$-dimensional finitely generated module $M$ with
$t=\operatorname{depth}M$, it holds that $H^{i}_{\mathfrak{m}}(M)=0$ for all
$i<t$ and $i>d$; also, $H^{t}_{\mathfrak{m}}(M)\neq 0$ and
$H^{d}_{\mathfrak{m}}(M)\neq 0$. These results are originally due to
Grothendieck, cf. [10], Theorem 3.5.8, Corollary 3.5.9, Corollary 3.5.11.a)
and b). As a consequence, $M$ is a Cohen-Macaulay module if and only if
$H^{i}_{\mathfrak{m}}(M)=0$ for all $i\neq d$.

Let $\dim(R)=n$. When $R$ is Cohen-Macaulay and has a canonical module
$\omega_{R}$, by duality, the above conditions can be checked on the Matlis
dual of the local cohomology modules, i.e., on the modules
$\operatorname{Ext}^{n-i}_{R}(M,\omega_{R})$. With the above notation, we then
have that $\operatorname{Ext}_{R}^{n-t}(M,\omega_{R})$ and
$\operatorname{Ext}_{R}^{n-d}(M,\omega_{R})$ are non-zero, and furthermore
that $\operatorname{Ext}^{i}_{R}(M,\omega_{R})=0$ for all $i<n-d$ and all
$i>n-t$. Moreover, $M$ is Cohen-Macaulay if and only if
$\operatorname{Ext}^{n-i}_{R}(M,\omega_{R})=0$ for all $i\neq d$. Often in the
literature the modules $\operatorname{Ext}^{n-i}_{R}(M,\omega_{R})$,
$i=0,\ldots,n$, are called the deficiency modules of $M$, since they also
measure how far a module is from being Cohen-Macaulay.

We now introduce the class of sequentially Cohen-Macaulay modules, following
[37], recalling their main properties, and in the next section we discuss some
of its equivalent definitions. Our main references here are [31] and [36].

######  Definition 1.1.

A finitely generated $R$-module $M$ is called sequentially Cohen-Macaulay if
it admits a filtration of submodules $0=M_{0}\subsetneq
M_{1}\subsetneq\ldots\subsetneq M_{r}=M$ such that each quotient of the
filtration $M_{i}/M_{i-1}$ is Cohen-Macaulay and
$\dim(M_{i}/M_{i-1})<\dim(M_{i+1}/M_{i})$ for all $i$. In this case, we say
that the above filtration is a sCM filtration for $M$.

######  Example 1.2.

  1. (1)

It is clear from the definition that if $M\neq 0$ is a Cohen-Macaulay module,
then it is sequentially Cohen-Macaulay. In fact $M=M_{1}\supsetneq M_{0}=0$.

  2. (2)

Any $1$-dimensional module $M$ is sequentially Cohen-Macaulay, for if $M$ is
not Cohen-Macaulay one can take $M_{1}=H^{0}_{\mathfrak{m}}(M)$ and $M_{2}=M$.

  3. (3)

A domain $R$ is sequentially Cohen-Macaulay if and only if it is Cohen-
Macaulay. In fact, any non-zero submodule of $R$ will have dimension
$d=\dim(R)$, and the only possible sCM filtration for $R$ is
$0=M_{0}\subsetneq M_{1}=R$.

  4. (4)

On the other hand, if $M$ has a Cohen-Macaulay submodule $M_{1}$ such that
$M/M_{1}$ is sequentially Cohen-Macaulay with a filtration
$0=M_{1}/M_{1}\subsetneq N_{1}/M_{1}\subsetneq\ldots\subsetneq
N_{s}/M_{1}=M/M_{1}$, and $\dim M_{1}<\dim N_{1}/M_{1}$, then $0\subsetneq
M_{1}\subsetneq N_{1}\subsetneq\ldots\subsetneq N_{s}=M$ is a sCM filtration
of $M$.

The notion of sequentially Cohen-Macaulay modules appears for the first time
in the literature in [37] in the graded setting, in connection with the theory
of Stanley-Reisner rings and simplicial complexes. Later on and independently,
in [36], the notion of Cohen-Macaulay filtered modules has been introduced in
the local case; in the same paper, it is proven that the two notions coincide.
Since then, sequentially Cohen-Macaulay modules have been extensively studied
especially in connection with shellability and graph theory, see for instance
[40] [25] [2], [22], [1], and the definition has been extended in other
directions, see [14], [33]. Section 3 and 4 are devoted to two such
extensions, due to the third and fourth author and the first two authors
respectively.

######  Example 1.3.

Let $M$ be a sequentially Cohen-Macaulay module with sCM filtration
$0=M_{0}\subsetneq M_{1}\subsetneq\dots\subsetneq M_{r}=M$ and let
$d_{i}=\dim(M_{i}/M_{i-1})$ for all $i$.

  1. (1)

Let $N$ be another sequentially Cohen-Macaulay; then $M\oplus N$ is
sequentially Cohen-Macaulay. To see this, let $0=N_{0}\subsetneq
N_{1}\subsetneq\ldots\subsetneq N_{s}=N$ be a sCM filtration of $N$ and let
$\delta_{i}=\dim(N_{i}/N_{i-1})$. For convenience, also let
$d_{0}=\delta_{0}=-1$. Then, a filtration whose terms are of the form
$M_{i}\oplus N_{j}$, where either $d_{i}=\max\\{d_{a}\mid
d_{a}\leqslant\delta_{j}\\}$ or
$\delta_{j}=\max\\{\delta_{a}\mid\delta_{a}\leqslant d_{i}\\}$, is a sCM
filtration of $M_{r}\oplus N_{s}=M\oplus N$.

  2. (2)

The completion $\widehat{M}$ of $M$ at the maximal ideal $\mathfrak{m}$ is a
sequentially Cohen-Macaulay $\widehat{R}$-module. To this end, let
$M^{\prime}_{i}=M_{i}\otimes_{R}\widehat{R}$, and by flatness of $\widehat{R}$
we have that $0=M^{\prime}_{0}\subsetneq
M^{\prime}_{1}\subsetneq\ldots\subsetneq M^{\prime}_{r}$ is a sCM filtration
of $M\otimes_{R}\widehat{R}\cong\widehat{M}$.

  3. (3)

It can be proven in general that being sequentially Cohen-Macaulay is
preserved by faithfully flat base changes; with some extra effort one can
derive that $M$ is a sequentially Cohen-Macaulay $R$-module if and only if
$M[|x|]$ is sequentially Cohen-Macaulay as an $R[|x|]$-module, cf [36, Theorem
6.2].

######  Remark 1.4.

In the above example Part (1), the converse also holds, cf. [16, Proposition
4.5], [39, Proposition 3.2] or Corollary 2.14.

In Part (2) of the same example, the vice versa does not hold in general, see
for instance [36, Example 6.1]. It is true though when $R$ is a homomorphic
image of a Cohen-Macaulay ring, see [23, Corollaries 2.8] and [23, Corollaries
2.9], where it is proved that, in such a case $M$ is a sequentially Cohen-
Macaulay $R$-module if and only if $\widehat{M}$ is a sequentially Cohen-
Macaulay $\widehat{R}$-module.

######  Proposition 1.5.

Let $M$ be a sequentially Cohen-Macaulay $R$-module with sCM filtration
$0=M_{0}\subsetneq M_{1}\subsetneq\ldots\subsetneq M_{r}=M$. Also let
$d_{i}=\dim(M_{i}/M_{i-1})$, for all $i\in\\{1,\ldots,r\\}$; then:

  1. (1)

for all $j\in{\mathbb{Z}}$, we have that $H^{j}_{\mathfrak{m}}(M)\neq 0$ if
and only if $j\in\\{d_{1},\ldots,d_{r}\\}$, and

$H^{d_{i}}_{\mathfrak{m}}(M)\cong H^{d_{i}}_{\mathfrak{m}}(M_{i})\cong
H^{d_{i}}_{\mathfrak{m}}(M_{i}/M_{i-1});$

  2. (2)

for all $i\in\\{1,\ldots,r\\}$ and $j<i$, the modules $M_{i}$ and
$M_{i}/M_{j}$ are sequentially Cohen-Macaulay.

###### Proof.

For Part (1), using the short exact sequences $0\to M_{i-1}\to M_{i}\to
M_{i}/M_{i-1}\to 0$ one inductively shows that $\dim(M_{i})=d_{i}$ for all
$i\in\\{1,\ldots,r\\}$, since $d_{i-1}<d_{i}$. The induced long exact
sequences on local cohomology

$\cdots\to H^{j-1}_{\mathfrak{m}}(M_{i}/M_{i-1})\to
H^{j}_{\mathfrak{m}}(M_{i-1})\to H^{j}_{\mathfrak{m}}(M_{i})\to
H^{j}_{\mathfrak{m}}(M_{i}/M_{i-1})\to
H^{j+1}_{\mathfrak{m}}(M_{i-1})\to\cdots$

together with the fact that $M_{i}/M_{i-1}$ is Cohen-Macaulay of dimension
$d_{i}$, imply that

$H^{d_{i}}_{\mathfrak{m}}(M_{i})\cong
H^{d_{i}}_{\mathfrak{m}}(M_{i}/M_{i-1}),\,\,\,H^{j}_{\mathfrak{m}}(M_{i})\cong
H^{j}_{\mathfrak{m}}(M_{i-1})\text{\, if }j<d_{i},\text{\, and
}H^{j}_{\mathfrak{m}}(M_{i})=0\text{\, otherwise}.$

In particular, $H^{j}_{\mathfrak{m}}(M)=0$ for all $j>d_{r}$, and
$H^{j}_{\mathfrak{m}}(M)=H^{j}_{\mathfrak{m}}(M_{i})$ for all $j\leqslant
d_{i}$. In conclusion we have shown that

$H^{j}_{\mathfrak{m}}(M)=\begin{cases}H^{d_{i}}_{\mathfrak{m}}(M_{i})=H^{d_{i}}_{\mathfrak{m}}(M_{i}/M_{i-1})&\text{
if }j=d_{i}\text{ for some }i,\\\ 0&\text{\,\,otherwise}.\end{cases}$

Part (2) is clear once we notice that $0=M_{0}\subsetneq
M_{1}\subsetneq\ldots\subsetneq M_{i}$ and $0=M_{j}/M_{j}\subsetneq
M_{j+1}/M_{j}\subsetneq\ldots\subsetneq M_{i}/M_{j}$ are sCM filtrations. ∎

######  Example 1.6.

Suppose that $x$ is an $M$-regular element and that $M/xM$ is sequentially
Cohen-Macaulay; it is not true in general that $M$ is sequentially Cohen-
Macaulay, as observed after [38, Proposition 2.2] \- the statement of [36,
Theorem 4.7] is not correct. As a counterexample one can take a
$2$-dimensional not Cohen-Macaulay local domain $R$ of depth 1, which is not
sequentially Cohen-Macaulay by Example 1.2 (3). For instance, take
$R=\mathbb{Q}[|a^{4},a^{3}b,ab^{3},b^{4}|]\cong\mathbb{Q}[|z_{1},z_{2},z_{3},z_{4}]/(z_{2}z_{3}-z_{1}z_{4},z_{3}^{2}-z_{2}z_{4}^{2},z_{1}z_{3}^{2}-z_{2}^{2}z_{4},z_{2}^{3}-z_{1}^{2}z_{3})$.
On the contrary, every $0\neq x\in R$ is regular, $R/(x)$ is a 1-dimensional
$R$-module and, therefore, always sequentially Cohen-Macaulay by Example 1.2
(2).

In the rest of this section, let $S=k[x_{1},\ldots,x_{n}]$, with the standard
grading. Given a monomial ideal $I\subset S$, we let $G(I)$ denote the
monomial minimal system of generators of $I$ and $m(I)=\max\\{i\mid x_{i}\
{\rm divides}\ u\ {\rm for\ some}\ u\in G(I)\\}$.

An important class of sequentially Cohen-Macaulay modules is given by rings
defined by weakly stable ideals.

######  Definition 1.7.

A monomial ideal $I\subseteq S$ is said to be weakly stable if for all
monomials $u\in I$ and all integers $i,j$ with $1\leqslant j<i\leqslant n$,
there exists $t\in\mathbb{N}$ such that $x_{j}^{t}u/x_{i}^{\ell}\in I$, where
$\ell$ is the largest integer such that $x_{i}^{\ell}$ divides $u$.

Observe that the condition of the previous definition is verified if and only
if it is verified for all $u\in G(I)$. In the literature weakly stable ideals
are also called ideals of Borel type, quasi-stable ideals, or ideals of nested
type, see [11], [28], [8]. It is easy too see from the definition that stable,
strongly stable, and $p$-Borel ideals are weakly stable. In particular, no
matter what the characteristic of the field is, Borel fixed ideals are weakly
stable, see also [28, Theorem 4.2.10] and, thus, generic initial ideals are
always weakly stable.

Observe that the saturation $I^{\rm sat}$ of a weakly stable ideal $I$ is
$I:\mathfrak{m}^{\infty}=I:x_{n}^{\infty}$, see [28, Proposition 4.2.9]; thus,
$I^{\rm sat}$ is again weakly stable and $x_{n}$ does not divide any $u\in
G(I^{\rm sat})$.

######  Proposition 1.8.

Let $I$ be a weakly stable ideal of $S=k[x_{1},\dots,x_{n}]$. Then, $S/I$ is
sequentially Cohen-Macaulay.

###### Proof.

We prove the statement by induction on $m(I)$. If $m(I)=1$, then
$I=(x_{1}^{r})$ for some positive integer $r$ and $S/I$ is Cohen-Macaulay.
Assume now that $S/J$ is sequentially Cohen-Macaulay for every weakly stable
ideal $J$ for which $m(J)<m(I)$.

Let $S^{\prime}=k[x_{1},\dots,x_{m(I)}]$ and $I^{\prime}=I\cap S^{\prime}$.
Since $S/I\cong S^{\prime}/I^{\prime}\otimes_{k}k[x_{m(I)+1},\dots,x_{n}]$ and
a sCM filtration of $S^{\prime}/I^{\prime}$ is easily extended into one of
$S/I$, it is sufficient to prove that $S^{\prime}/I^{\prime}$ is sequentially
Cohen-Macaulay. Therefore, without loss of generality, we may assume that
$m(I)=n$, and that $S/I^{\rm sat}$ is sequentially Cohen-Macaulay. Now,
$I^{\rm sat}/I$ is a non-trivial Artinian module, and the first non-zero
module of a sCM filtration of $S/I^{\rm sat}$ has positive dimension, see
Proposition 1.5. We may thus conclude that $S/I$ is sequentially Cohen-
Macaulay, cf. Example 1.2 (4). ∎

######  Remark 1.9.

By [8, Proposition 3.2], see also [28, Proposition 4.2.9] and [11, Chapter 4])
a monomial ideal $I$ is weakly stable if and only if all its associated primes
are generated by initial segments of variables, i.e. are of type
$(x_{1},\ldots,x_{i})$ for some $i$. Since being sequentially Cohen-Macaulay
is independent of coordinates changes on $S$, the above proposition shows that
whenever the associated primes of a monomial ideal $I\subseteq S$ are totally
ordered by inclusion, then $S/I$ is sequentially Cohen-Macaulay.

######  Example 1.10.

Let
$I=(x_{1}^{4},x_{1}^{2}x_{2}^{2},x_{1}^{3}x_{3},x_{1}^{2}x_{2}x_{3},x_{1}^{2}x_{2}x_{4})\subseteq
S=k[x_{1},x_{2},x_{3},x_{4}]$. It is easy to verify that $I$ is a weakly
stable ideal which is not strongly stable, by checking the definition or by
computing its associated primes, by using the previous remark; thus, $S/I$ is
sequentially Cohen-Macaulay by Proposition 1.8. We can construct an explicit
sCM filtration proceeding as in its proof. We have $m(I)=4$, and $I^{\rm
sat}=I:x_{4}^{\infty}=(x_{1}^{4},x_{1}^{2}x_{2},x_{1}^{3}x_{3})$. This means
that the first non-zero module of our filtration will be
$(x_{1}^{4},x_{1}^{2}x_{2},x_{1}^{3}x_{3})/I$. Now we consider
$I^{\prime}=(x_{1}^{4},x_{1}^{2}x_{2},x_{1}^{3}x_{3})$ as an ideal of
$k[x_{1},x_{2},x_{3}]$ and compute its saturation
$I^{\prime\prime}=I:x_{3}^{\infty}=(x_{1}^{3},x_{1}^{2}x_{2})$. Hence, the
second non-zero module of the filtration is $(x_{1}^{3},x_{1}^{2}x_{2})/I$.
Proceeding in this way we obtain the filtration
$0\subseteq(x_{1}^{4},x_{1}^{2}x_{2},x_{1}^{3}x_{3})/I\subseteq(x_{1}^{3},x_{1}^{2}x_{2})/I\subseteq(x_{1}^{2})/I\subseteq(1)/I=S/I$,
and it is easily seen that it is a sCM filtration of $S/I$.

######  Example 1.11.

For the case $M=R=S/I$, when $I=I_{\Delta}$ is the Stanley-Reisner ideal of a
simplicial complex $\Delta$, there is a beautiful characterization of
sequential Cohen-Macaulayness due to Duval, [20, Theorem 3.3]. Given a
simplicial complex $\Delta$, let $\Delta(i)$ be the pure $i$-th skeleton of
$\Delta$, i.e. the pure subcomplex of $\Delta$ whose facets are the faces of
$\Delta$ of dimension $i$. Then, $S/I_{\Delta}$ is sequentially Cohen-Macaulay
if and only if $S/I_{\Delta(i)}$ is Cohen-Macaulay for all $i$. Another
important result in this context is that $I_{\Delta}$ is componentwise linear
if and only if the Stanley-Reisner ring of its Alexander dual $\Delta^{*}$ is
sequentially Cohen-Macaulay, see [27, Theorem 2.1] and [28, Theorem 8.2.20].
Moreover, it is known that if $\Delta$ is (nonpure) shellable, then
$S/I_{\Delta}$ is sequentially Cohen-Macaulay, see [28, Corollary 8.2.19].

######  Example 1.12.

Another class of examples of sequentially Cohen-Macaulay modules is given by
pretty clean modules, which have been introduced by Herzog and Popescu in [29]
in order to characterize shellability of multicomplexes. A pretty clean module
$M$ is a module that admits a pretty clean filtration, i.e. a prime filtration
of $M$ by submodules $0=M_{0}\subsetneq M_{1}\subsetneq\ldots\subsetneq
M_{s}=M$ such that each quotient $M_{i}/M_{i-1}$ is isomorphic to
$S/{\mathfrak{p}}_{i}$, for some prime ideals ${\mathfrak{p}}_{i}$, with the
following property: if ${\mathfrak{p}}_{i}\subsetneq{\mathfrak{p}}_{j}$ then
$i>j$. For example, if $M=S/I$ with $I$ weakly stable, cf. [29, Proposition
5.2], or is such that $\operatorname{Ass}(M)$ is a totally ordered set, cf.
[29, Proposition 5.1] or Remark 1.9, then $M$ is pretty clean. By [29, Theorem
4.1] pretty clean modules are sequentially Cohen-Macaulay, provided every
prime ${\mathfrak{p}}_{i}$ appearing in the pretty clean filtration is such
that $S/{\mathfrak{p}}_{i}$ is Cohen-Macaulay.

##  2\. Characterizations of sequentially Cohen-Macaulay modules

The goal of this section is to present two characterizations of sequentially
Cohen-Macaulay modules, due to Schenzel, see Theorem 2.7 and Peskine, see
Theorem 2.13. As in the previous section, we let $(R,\mathfrak{m},k)$ be
either a Noetherian local ring or a standard graded $k$-algebra with maximal
homogeneous ideal $\mathfrak{m}$; we let $\dim(R)=n$. In the second case,
modules will be graded and homomorphisms homogeneous of degree 0.

###  2.1.  Schenzel’s characterization

Our main reference here is [36]. By convention, the dimension of the zero
module is set to be $-1$. Let $M$ be a finitely generated (graded) $R$-module
of dimension $d$. Since $R$ is Noetherian, for all $i=0,\ldots,d$, we can
consider the largest (graded) submodule of $M$ of dimension $\leqslant i$, and
denote it by $\delta_{i}(M)$. By maximality, in this way one obtains a
filtration
$\mathcal{M}:0\subseteq\delta_{0}(M)\subseteq\delta_{1}(M)\subseteq\ldots\subseteq\delta_{d}(M)=M$,
called the dimension filtration of $M$. Evidently, such a filtration is
unique.

Given a set $X$ of prime ideals of $R$, we denote by
$X_{i}=\\{{\mathfrak{p}}\in X\,:\,\dim R/{\mathfrak{p}}=i\\}$. Similarly, we
define $X_{\leqslant i}$ and $X_{>i}$.

######  Remark 2.1.

Observe that $M$ has a non-zero submodule of dimension $i$ if and only if
$\operatorname{Ass}(M)_{i}\neq\emptyset$. In fact, if
${\mathfrak{p}}\in\operatorname{Ass}(M)_{i}$, then $R/{\mathfrak{p}}$ is a
non-zero submodule of $M$ of dimension $i$. Conversely, if $N$ is a non-zero
submodule of $M$ of dimension $i$, then there must exist a minimal prime
${\mathfrak{p}}$ of $N$ such that $\dim(R/{\mathfrak{p}})=\dim(N)=i$. Since
$N\subseteq M$, we must also have that
${\mathfrak{p}}\in\operatorname{Ass}(M)$, as desired. In particular, we have
that $\delta_{i}(M)=0$ if and only if $\operatorname{Ass}(M)_{\leqslant
i}=\emptyset$.

Notice that $\delta_{0}(M)=H^{0}_{\mathfrak{m}}(M)$; the other modules in
$\mathcal{M}$ can be described similarly, with the help of the minimal primary
decomposition of $0$ as a submodule of $M$. In fact, let
$\operatorname{Ass}(M)=\\{{\mathfrak{p}}_{1},\ldots,{\mathfrak{p}}_{m}\\}$
and, for all $i=0,\ldots,d$, consider the set
$\operatorname{Ass}(M)_{\leqslant i}$. We set

${\mathfrak{a}}_{i}=\begin{cases}\prod\limits_{{\mathfrak{p}}\in\operatorname{Ass}(M)_{\leqslant
i}}{\mathfrak{p}}&\text{ if }\operatorname{Ass}(M)_{\leqslant
i}\neq\emptyset,\\\ R&\text{ otherwise.}\end{cases}$

Now let us consider an irredundant primary decomposition
$0=\bigcap_{j=1}^{m}N_{j}$ inside $M$, where each $M/N_{j}$ is a
${\mathfrak{p}}_{j}$-primary module.

######  Proposition 2.2 ([36, Proposition 2.2]).

Let $M$ be a finitely generated $R$-module of dimension $d$. With the above
notation, for all $i=0,\ldots,d$ we have that

$\delta_{i}(M)=H^{0}_{{\mathfrak{a}}_{i}}(M)=\bigcap\limits_{\\{j\ \mid\
{\mathfrak{p}}_{j}\in\operatorname{Ass}(M)_{>i}\\}}N_{j},$

where we let the intersection over the empty set be equal to $M$.

###### Proof.

We start with the first equality and assume that $\delta_{i}(M)=0$. By Remark
2.1 we have that $\operatorname{Ass}(M)_{\leqslant i}=\emptyset$ and in this
case ${\mathfrak{a}}_{i}=R$. Thus,
$H^{0}_{{\mathfrak{a}}_{i}}(M)=H^{0}_{R}(M)=0=\delta_{i}(M)$.

Now assume that $\delta_{i}(M)\neq 0$ and observe that, every associated prime
of $\delta_{i}(M)$ has dimension $\leqslant i$ and, therefore,
$\operatorname{Ass}(\delta_{i}(M))\subseteq\operatorname{Ass}(M)_{\leqslant
i}$. Thus, there is a power of ${\mathfrak{a}}_{i}$ which annihilates
$\delta_{i}(M)$, and we get that $\delta_{i}(M)\subseteq
H^{0}_{{\mathfrak{a}}_{i}}(M)$. On the other hand, there is a power of
${\mathfrak{a}}_{i}$ which annihilates $H^{0}_{{\mathfrak{a}}_{i}}(M)$ and,
thus,
$\dim(H^{0}_{{\mathfrak{a}}_{i}}(M))\leqslant\dim(R/{\mathfrak{a}}_{i})\leqslant
i$. From the maximality of $\delta_{i}(M)$ it follows that
$H^{0}_{{\mathfrak{a}}_{i}}(M)\subseteq\delta_{i}(M)$, as desired.

For the second equality, observe that
$N_{j}:_{M}{\mathfrak{p}}_{j}^{\infty}=M$ and $N_{j}:_{M}x^{\infty}=N_{j}$ for
any $x\notin{\mathfrak{p}}_{j}$, with $j\in\\{1,\ldots,m\\}$, since $M/N_{j}$
is ${\mathfrak{p}}_{j}$-primary. Also notice that

$H^{0}_{{\mathfrak{a}}_{i}}(M)=0:_{M}{\mathfrak{a}}_{i}^{\infty}=\left(\bigcap_{j=1}^{m}N_{j}\right):_{M}{\mathfrak{a}}_{i}^{\infty}=\bigcap_{j=1}^{m}(N_{j}:_{M}{\mathfrak{a}}_{i}^{\infty}).$

Now, if ${\mathfrak{p}}_{j}\in\operatorname{Ass}(M)_{\leqslant i}$, then
$M\supseteq N_{j}:_{M}{\mathfrak{a}}_{i}^{\infty}\supseteq
N_{j}:_{M}{\mathfrak{p}}_{j}^{\infty}=M,$ forcing equality everywhere. Next,
assume that ${\mathfrak{p}}_{j}\notin\operatorname{Ass}(M)_{\leqslant i}$.
Then, ${\mathfrak{p}}\not\subseteq{\mathfrak{p}}_{j}$ for every
${\mathfrak{p}}\in\operatorname{Ass}(M)_{\leqslant i}$ since, otherwise, we
would have $\dim(R/{\mathfrak{p}}_{j})\leqslant\dim(R/{\mathfrak{p}})\leqslant
i$ and, thus, ${\mathfrak{p}}_{j}\in\operatorname{Ass}(M)_{\leqslant i}$. In
this case, by choosing $x\in{\mathfrak{a}}_{i}$ and
$x\notin{\mathfrak{p}}_{j}$, we have that $N_{j}\subseteq
N_{j}:_{M}{\mathfrak{a}}_{i}^{\infty}\subseteq N_{j}:_{M}x^{\infty}=N_{j},$
and equalities hold. Summing up, we conclude that

$H^{0}_{{\mathfrak{a}}_{i}}(M)=\bigcap_{\\{j\ \mid\
{\mathfrak{p}}_{j}\notin\operatorname{Ass}(M)_{\leqslant
i}\\}}N_{j}=\bigcap_{\\{j\ \mid\
{\mathfrak{p}}_{j}\in\operatorname{Ass}(M)_{>i}\\}}N_{j}.\qed$

Note that this is consistent with our convention that the intersection over
the empty set is equal to $M$; in fact, for $i=d$ we have that
$\operatorname{Ass}(M)_{\leqslant d}=\operatorname{Ass}(M)$ and
$\operatorname{Ass}(M)_{>d}=\emptyset$. This agrees with the fact that
$H^{0}_{{\mathfrak{a}}_{d}}(M)=H^{0}_{\sqrt{0}}(M)=M$.

######  Proposition 2.3 ([36], Corollary 2.3).

Let $M$ be a finitely generated $R$-module of dimension $d$; then, for all
$i=0,\ldots,d$,

  1. (1)

$\operatorname{Ass}(\delta_{i}(M))=\operatorname{Ass}(M)_{\leqslant i}$;

  2. (2)

$\operatorname{Ass}(M/\delta_{i}(M))=\operatorname{Ass}(M)_{>i}$;

  3. (3)

$\operatorname{Ass}(\delta_{i}(M)/\delta_{i-1}(M))=\operatorname{Ass}(M)_{i}$.

###### Proof.

It is well known that,
$\operatorname{Ass}(M)=\operatorname{Ass}(H^{0}_{\mathfrak{a}}(M))\sqcup\operatorname{Ass}(M/H^{0}_{\mathfrak{a}}(M))$,
for any ideal ${\mathfrak{a}}$ of $R$. Also notice that
$\operatorname{Ass}(H^{0}_{{\mathfrak{a}}}(M))=\operatorname{Ass}(M)\cap
V({\mathfrak{a}})$, for all ${\mathfrak{a}}$.

Since $\delta_{i}(M)=H^{0}_{{\mathfrak{a}}_{i}}(M)$ by Proposition 2.2, the
first equality descends from Remark 2.1 and the above observation.

For (2), consider the short exact sequence
$0\longrightarrow\delta_{i}(M)\longrightarrow M\longrightarrow
M/\delta_{i}(M)\longrightarrow 0$. By Proposition 2.2 one has that
$\operatorname{Ass}(M)=\operatorname{Ass}(\delta_{i}(M))\sqcup\operatorname{Ass}(M/\delta_{i}(M))$,
which is equal to $\operatorname{Ass}(M)_{\leqslant
i}\sqcup\operatorname{Ass}(M)_{>i}$ by (1), and the second equality follows.

Finally, consider the short exact sequence
$0\to\delta_{i-1}(M)\to\delta_{i}(M)\to\delta_{i}(M)/\delta_{i-1}(M)\to 0$ and
observe that
$\operatorname{Ass}(\delta_{i}(M)/\delta_{i-1}(M))\subseteq\operatorname{Ass}(M/\delta_{i-1}(M))$.
We also have $\operatorname{Ass}(\delta_{i}(M)/\delta_{i-1}(M))$
$\subseteq\operatorname{Ass}(\delta_{i}(M))$ since
$\delta_{i-1}(M)=H^{0}_{{\mathfrak{a}}_{i-1}}(M)=H^{0}_{{\mathfrak{a}}_{i-1}}(\delta_{i}(M))$.
Thus, by (1) and (2) we have necessarily that
$\operatorname{Ass}(\delta_{i}(M)/\delta_{i-1}(M))$
$\subseteq\operatorname{Ass}(M)_{i}$. On the other hand, by (1), if
${\mathfrak{p}}\in\operatorname{Ass}(M)_{i}$ then
${\mathfrak{p}}\not\in\operatorname{Ass}(M_{i-1})$, and therefore
${\mathfrak{p}}\in\operatorname{Ass}(\delta_{i}(M)/\delta_{i-1}(M))$. ∎

As a corollary of Propositions 2.2 and 2.3, one can obtain another
characterization of the dimension filtration, cf. [29, Proposition 1.1].

######  Corollary 2.4.

Let $M$ be a $d$-dimensional $R$-module. A filtration of $M$ by submodules
$0\subseteq M_{0}\subseteq\ldots\subseteq M_{d}=M$ is the dimension filtration
of $M$ if and only if
$\operatorname{Ass}(M_{i}/M_{i-1})=\operatorname{Ass}(M)_{i}$, for all $i$.

The following definition was introduced in [36, Definition 4.1].

######  Definition 2.5.

Let $M$ be a $d$-dimensional finitely generated $R$-module, and
$\mathcal{M}:0\subseteq\delta_{0}(M)\subseteq\ldots\subseteq\delta_{d}(M)=M$
be its dimension filtration. Then, $M$ is called Cohen-Macaulay filtered if
for all $i\in\\{0,\ldots,d\\}$ the module $\delta_{i}(M)/\delta_{i-1}(M)$ is
either zero or Cohen-Macaulay.

Notice that from the definition it immediately follows that if
$\delta_{i}(M)/\delta_{i-1}(M)\neq 0$, then it has dimension $i$.

A Cohen-Macaulay filtered module $M$ is sequentially Cohen-Macaulay. In fact,
we may let $i_{1}$ denote the smallest integer such that
$\delta_{i_{1}}(M)\neq 0$ and set $M_{1}=\delta_{i_{1}}(M)$. Then, if $i_{j}$
is the smallest integer such that
$\delta_{i_{j}-1}(M)\subsetneq\delta_{i_{j}}(M)$, we let
$M_{j}=\delta_{i_{j}}(M)$. The resulting filtration $0=M_{0}\subsetneq
M_{1}\subsetneq\ldots\subsetneq M_{r}=M$ is clearly a sCM filtration of $M$.

A filtration $0=C_{-1}\subseteq C_{0}\subseteq C_{1}\subseteq\ldots\subseteq
C_{d}=M$ such that each of its quotients $C_{i}/C_{i-1}$ is either zero or
Cohen-Macaulay of dimension $i$ is called a CM filtration of $M$. To see that
the notions of sequentially Cohen-Macaulay and Cohen-Macaulay filtered modules
coincide, we need the following result, which shows that if a CM filtration of
a module exists, then it is unique and it is equal to its dimension
filtration.

######  Proposition 2.6 ([36], Proposition 4.3).

Let $M$ be a $d$-dimensional finitely generated $R$-module. If $M$ has a CM
filtration $0=C_{-1}\subseteq C_{0}\subseteq C_{1}\subseteq\ldots\subseteq
C_{d}=M$, then $C_{i}=\delta_{i}(M)$ for every $i\in\\{0,\ldots,d\\}$.

###### Proof.

We proceed by induction on $d$. If $d=0$, then $C_{0}=M=\delta_{0}(M)$ and
this case is complete. Assume henceforth that $d>0$, so that by induction we
have that $C_{i}=\delta_{i}(C_{d-1})$ for all $i<d$. Since for $i<d$ we have
that $\delta_{i}(M)=\delta_{i}(\delta_{d-1}(M))$, it suffices to show that
$\delta_{d-1}(M)=C_{d-1}$.

Observe that
$\operatorname{Ass}(C_{0})\subseteq\operatorname{Ass}(M)_{\leqslant 0}$ and,
for all ${\mathfrak{p}}\in\operatorname{Ass}(C_{i}/C_{i-1})$, we have $\dim
A/{\mathfrak{p}}=i$ if $C_{i}\neq C_{i-1}$. With this information, we can
inductively show that $\dim C_{i}\leqslant i$ and, accordingly
$C_{i}\subseteq\delta_{i}(M)$ for all $i$; in particular,
$C_{d-1}\subseteq\delta_{d-1}(M)$. If $\operatorname{Ass}(M)_{\leqslant
d-1}=\emptyset$, then $\delta_{d-1}(M)=0$ by Remark 2.1, and the desired
equality is trivial. Otherwise, we let
${\mathfrak{a}}={\mathfrak{a}}_{d-1}=\prod\limits_{{\mathfrak{p}}\in\operatorname{Ass}(M)_{\leqslant
d-1}}{\mathfrak{p}}$ and we claim that $H^{0}_{\mathfrak{a}}(M/C_{d-1})=0$.
This is obvious if $M/C_{d-1}=0$. Thus, assume $C_{d-1}\neq M$ and observe
that ${\mathfrak{a}}$ contains a regular element of $M/C_{d-1}$. To see the
latter, assume that
${\mathfrak{a}}\subseteq\bigcup\limits_{{\mathfrak{p}}\in\operatorname{Ass}(M/C_{d-1})}{\mathfrak{p}}$;
then, by prime avoidance we can find
${\mathfrak{p}}\in\operatorname{Ass}(M/C_{d-1})$ such that
${\mathfrak{a}}\subseteq{\mathfrak{p}}$ and, therefore, ${\mathfrak{p}}$
contains a prime ${\mathfrak{p}}^{\prime}\in\operatorname{Ass}(M)_{\leqslant
d-1}$. This is a contradiction, since it would imply
$d=\dim(M/C_{d-1})=\dim(R/{\mathfrak{p}})\leqslant\dim(R/{\mathfrak{p}}^{\prime})\leqslant
d-1$. Finally, consider the short exact sequence $0\longrightarrow
C_{d-1}\longrightarrow M\longrightarrow M/C_{d-1}\longrightarrow 0$. By
Proposition 2.2 we have that
$\delta_{d-1}(M)=H^{0}_{{\mathfrak{a}}}(M)=H^{0}_{{\mathfrak{a}}}(C_{d-1})\subseteq
C_{d-1}$. Thus, $C_{d-1}=\delta_{d-1}(M)$, and we are done. ∎

######  Theorem 2.7 (Schenzel).

Let $M$ be a finitely generated $R$-module; then $M$ is sequentially Cohen-
Macaulay if and only if $M$ is Cohen-Macaulay filtered. Moreover, if $M$ is
sequentially Cohen-Macaulay, then its sCM filtration is unique.

###### Proof.

We only need to show that the “only if” part. Let
$\mathcal{M}:0=M_{0}\subsetneq M_{1}\subsetneq\ldots\subsetneq M_{r}=M$ be any
sCM filtration of $M$, with $d=\dim(M)$, $d_{0}=-1$ and
$d_{i}=\dim(M_{i}/M_{i-1})$. For $j\in\\{-1,0,\ldots,d\\}$, we also let $i(j)$
be the largest integer $i$ such that $d_{i}\leqslant j$, and we set
$C_{j}=M_{i(j)}$. In this way we have constructed a filtration
$\mathcal{C}:0=C_{-1}\subseteq C_{0}\subseteq\ldots\subseteq C_{d}=M$, which
is a CM filtration of $M$. By Proposition 2.6 we may conclude that
$C_{i}=\delta_{i}(M)$, and that $\mathcal{C}$ is the unique dimension
filtration of $M$. Thus, $M$ is Cohen-Macaulay filtered.

Observe that one can reconstruct $\mathcal{M}$ from $\mathcal{C}$; it follows
that the modules $M_{i}$ only depend on $M$ as well, and $\mathcal{M}$ is also
unique. ∎

Notice that, by the above theorem and Proposition 2.3, if $M$ is sequentially
Cohen-Macaulay with sCM filtration $\mathcal{M}$, then
$\operatorname{Ass}(M)=\sqcup_{i}\operatorname{Ass}(M_{i}/M_{i-1})$. The
comparison between sCM filtrations and dimension filtrations has also the
following useful consequences.

######  Corollary 2.8.

Let $M$ be a $d$-dimensional finitely generated $R$-module. The following are
equivalent:

  1. (1)

$M$ is sequentially Cohen-Macaulay;

  2. (2)

The modules $\delta_{i}(M)$ and $M/\delta_{i}(M)$ are sequentially Cohen-
Macaulay for all $i\in\\{0,\ldots,d\\}$;

  3. (3)

There exists $i\in\\{0,\ldots,d\\}$ such that $\delta_{i}(M)$ and
$M/\delta_{i}(M)$ are sequentially Cohen-Macaulay.

In particular, $M$ is sequentially Cohen-Macaulay if and only if
$M/H^{0}_{\mathfrak{m}}(M)$ is sequentially Cohen-Macaulay.

###### Proof.

Observe that, given any $i\in\\{0,\ldots,d\\}$, we have that
$\delta_{j}(M)=\delta_{j}(\delta_{i}(M))$ for every $j\leqslant i$, and
$\delta_{j}(M/\delta_{i}(M))=\delta_{j}(M)/\delta_{i}(M)$ for every
$j\geqslant i$. It follows that
$0\subseteq\delta_{0}(M)\subseteq\ldots\subseteq\delta_{i}(M)$ and
$0=\delta_{i}(M)/\delta_{i}(M)\subseteq\delta_{i+1}(M)/\delta_{i}(M)\subseteq\ldots\subseteq\delta_{d}(M)/\delta_{i}(M)$
are the dimension filtrations of $\delta_{i}(M)$ and $M/\delta_{i}(M)$
respectively. This yields all the assertions at once. ∎

Recall now that a finitely generated $R$-module $M$ is unmixed if
$\dim(R/{\mathfrak{p}})=\dim(M)$ for all
${\mathfrak{p}}\in\operatorname{Ass}(M)$.

######  Corollary 2.9.

Let $M$ be a finitely generated unmixed $R$-module. Then, $M$ is sequentially
Cohen-Macaulay if and only if $M$ is Cohen-Macaulay.

###### Proof.

One direction is clear. Let $d=\dim(M)$; in view of Remark 2.1, the fact that
$M$ is unmixed guarantees that $\delta_{i}(M)=0$ for all $i<d$. By Theorem
2.7, $M$ is Cohen-Macaulay filtered, and thus
$\delta_{d}(M)/\delta_{d-1}(M)=M$ is Cohen-Macaulay. ∎

###  2.2. Peskine’s characterization

We will always assume that $R$ has a canonical module $\omega_{R}$. In our
setup, this assumption is not too restrictive: it is always the case when $R$
is standard graded, and it is true for instance if $R$ is complete local, cf.
Remark 1.4.

######  Proposition 2.10.

Let $R$ be an $n$-dimensional Cohen-Macaulay ring with canonical module
$\omega_{R}$ and $M$ be sequentially Cohen-Macaulay, with sCM filtration
$0=M_{0}\subsetneq M_{1}\subsetneq\ldots\subsetneq M_{r}=M$; also let
$d_{i}=\dim M_{i}/M_{i-1}$, for $i=1,\ldots,r$. Then

  1. (1)

for all $i=1,\ldots,r$, one has that
$\operatorname{Ext}_{R}^{n-d_{i}}(M,\omega_{R})\simeq\operatorname{Ext}_{R}^{n-d_{i}}(M_{i}/M_{i-1},\omega_{R})$
is Cohen-Macaulay and has dimension $d_{i}$;

  2. (2)

$\operatorname{Ext}_{R}^{n-j}(M,\omega_{R})=0$ whenever
$j\not\in\\{d_{1},\ldots,d_{r}\\}$;

  3. (3)

$\operatorname{Ext}_{R}^{n-d_{i}}(\operatorname{Ext}_{R}^{n-d_{i}}(M,\omega_{R}),\omega_{R})\simeq
M_{i}/M_{i-1}$ for $i=1,\ldots,r$.

###### Proof.

In the graded case, (1) and (2) follow immediately from Proposition 1.5,
graded local duality [10, Theorem 3.6.19] and [10, Theorem 3.3.10 (c) (i)]. In
the local case, local duality yields (1) and (2) for the completion
$\widehat{M}$ of $M$ as an $\widehat{R}$-module, see Example 1.3 (2). Since
$\omega_{\widehat{R}}\cong\widehat{\omega_{R}}\cong\omega_{R}\otimes_{R}\widehat{R}$
and
$\operatorname{Ext}^{i}_{\widehat{R}}(\widehat{N},\omega_{\widehat{R}})\cong\operatorname{Ext}^{i}_{R}(N,\omega_{R})\otimes_{R}\widehat{R}$
for any finitely generated $R$-module $N$, we conclude by faithful flatness of
$\widehat{R}$ that (1) and (2) hold also in the local case.

Finally, by [10, Theorem 3.3.10 (c) (iii)] we have that
$N\cong\operatorname{Ext}_{R}^{n-d}(\operatorname{Ext}_{R}^{n-d}(N,\omega_{R}),\omega_{R})$
for any Cohen-Macaulay module $N$ of dimension $d$ and, thus, the last
statement follows immediately from (1). ∎

######  Remark 2.11.

Let $M$ be as in the above proposition.

  1. (1)

By Parts (2) and (3) of the previous proposition, we have an isomorphism

$M_{1}\cong\operatorname{Ext}_{R}^{n-t}(\operatorname{Ext}_{R}^{n-t}(M,\omega_{R}),\omega_{R}),$

where $t=\operatorname{depth}M$ and $M_{1}$ is $t$-dimensional and Cohen-
Macaulay. We will show an extension of this fact in Lemma 2.12.

  2. (2)

Notice that if $M$ is sequentially Cohen-Macaulay with
$\operatorname{depth}(M)=0$, then in particular
$M_{1}=H^{0}_{\mathfrak{m}}(M)$. In fact, using [10, Theorem 3.3.10 (c) (iii)]
and the short exact sequence $0\longrightarrow
H^{0}_{\mathfrak{m}}(M)\longrightarrow M\longrightarrow
M/H^{0}_{\mathfrak{m}}(M)\longrightarrow 0$ we get that

$M_{1}\cong\operatorname{Ext}_{R}^{n}(\operatorname{Ext}_{R}^{n}(M,\omega_{R}),\omega_{R})\cong\operatorname{Ext}_{R}^{n}(\operatorname{Ext}_{R}^{n}(H^{0}_{\mathfrak{m}}(M),\omega_{R}),\omega_{R})\cong
H^{0}_{\mathfrak{m}}(M).$

We recall now the following crucial lemma, see [31, Lemma 1.5].

######  Lemma 2.12.

Let $R$ be a Cohen-Macaulay $n$-dimensional ring with canonical module
$\omega_{R}$ and $M$ be a finitely generated $R$-module.

Let also $\operatorname{depth}(M)=t$ and assume that
$\operatorname{Ext}_{R}^{n-t}(M,\omega_{R})$ is Cohen-Macaulay of dimension
$t$. Then, there is a natural monomorphism
$\alpha\,:\,\operatorname{Ext}_{R}^{n-t}(\operatorname{Ext}_{R}^{n-t}(M,\omega_{R}),\omega_{R})\longrightarrow
M$ such that

$\operatorname{Ext}^{n-t}(\alpha)\,:\,\operatorname{Ext}_{R}^{n-t}(M,\omega_{R})\longrightarrow\operatorname{Ext}_{R}^{n-t}(\operatorname{Ext}_{R}^{n-t}(\operatorname{Ext}_{R}^{n-t}(M,\omega_{R}),\omega_{R})\omega_{R})$

is an isomorphism.

###### Proof.

We only give a proof in the graded case; the local one is handled similarly.
Without loss of generality we may assume that $R$ is a polynomial ring, as we
show next. We write $R=S/I$, where $S$ is a standard graded polynomial ring of
dimension $m$ over a field $k$ and $I\subseteq S$ is homogeneous. Let
${\mathfrak{m}}$ and ${\mathfrak{n}}$ denote the graded maximal ideals of $R$
and $S$ respectively. By graded Local Duality [10, Theorem 3.6.19] we have
that
$\operatorname{Ext}^{m-i}_{S}(M,\omega_{S})\cong\operatorname{Hom}_{S}(H^{i}_{\mathfrak{n}}(M),E_{S}(k))$
and
$\operatorname{Ext}^{n-i}_{R}(M,\omega_{R})\cong\operatorname{Hom}_{R}(H^{i}_{\mathfrak{m}}(M),E_{R}(k))$.
By Base Independence $H^{i}_{{\mathfrak{n}}}(M)\cong H^{i}_{\mathfrak{m}}(M)$
for all $i$. Thus, by $\operatorname{Hom}-\otimes$ adjointness and [10, Lemma
3.1.6] we obtain

$\operatorname{Hom}_{S}(H^{i}_{\mathfrak{n}}(M),E_{S}(k))\cong\operatorname{Hom}_{R}(H^{i}_{\mathfrak{m}}(M),\operatorname{Hom}_{S}(R,E_{S}(k)))\cong\operatorname{Hom}_{R}(H^{i}_{\mathfrak{m}}(M),E_{R}(k)).$

In particular, we have shown that
$\operatorname{Ext}^{n-t}_{R}(M,\omega_{R})\cong\operatorname{Ext}^{m-t}_{S}(M,\omega_{S})$
and that we may assume that $R$ is a polynomial ring.

Let now $F_{\bullet}\colon 0\longrightarrow
F_{n-t}\longrightarrow\ldots\longrightarrow F_{1}\longrightarrow
F_{0}\longrightarrow 0$ be the minimal graded free resolution of $M$; applying
the functor $\operatorname{Hom}_{R}(-,\omega_{R})$ we obtain the dual complex

$F^{*}_{\bullet}\,:\,\,\,\,\,\,\,0\longrightarrow
F^{*}_{n-t}\longrightarrow\ldots\longrightarrow F^{*}_{1}\longrightarrow
F^{*}_{0}\longrightarrow 0,$

with $H_{0}(F^{*})=\operatorname{Ext}_{R}^{n-t}(M,\omega_{R})$. Observe that
$F^{*}_{\bullet}$ is a complex of free modules, since $\omega_{R}\cong R(-n)$.
If we also let $G_{\bullet}$ denote the minimal graded free resolution of
$\operatorname{Ext}_{R}^{n-t}(M,\omega_{R})$, then there is a map of complexes
$\phi_{\bullet}\,:\,F^{*}_{\bullet}\longrightarrow G_{\bullet}$ which lifts
the identity map of $\operatorname{Ext}_{R}^{n-t}(M,\omega_{R})$. Since the
last one is Cohen-Macaulay of dimension $t$, the length of $G_{\bullet}$ is
the same as that of $F^{*}_{\bullet}$, namely $n-t$. Applying the functor
$\operatorname{Hom}_{R}(-,\omega_{R})$ to $\phi_{\bullet}$, we obtain a map of
complexes $\phi^{*}_{\bullet}\colon G^{*}_{\bullet}\longrightarrow
F_{\bullet}^{**}$ which, in turn, gives a map on the zero-th cohomology:

$\alpha=H_{0}(\phi^{*}_{\bullet})\,:\,\operatorname{Ext}_{R}^{n-t}(\operatorname{Ext}_{R}^{n-t}(M,\omega_{R}),\omega_{R})=H_{0}(G^{*}_{\bullet})\to
H_{0}(F_{\bullet}^{**})\cong H_{0}(F_{\bullet})=M.$

Applying again the functor $\operatorname{Hom}_{R}(-,\omega_{R})$, this time
to $\phi^{*}_{\bullet}$, we obtain a map of complexes
$\phi^{**}_{\bullet}:F_{\bullet}^{***}\to G^{**}_{\bullet}$ and, thus, a map

$\begin{split}H_{0}(\phi^{**}_{\bullet}):H_{0}(F^{***}_{\bullet})\cong
H^{0}(F^{*}_{\bullet})&\cong\operatorname{Ext}^{n-t}_{R}(M,\omega_{R})\\\
&\longrightarrow\operatorname{Ext}^{n-t}_{R}(\operatorname{Ext}^{n-t}_{R}(\operatorname{Ext}^{n-t}_{R}(M,\omega_{R}),\omega_{R}),\omega_{R})=H_{0}(G^{**}_{\bullet}),\end{split}$

which coincides with the map $\operatorname{Ext}^{n-t}(\alpha)$. The canonical
isomorphism $\psi:G_{\bullet}^{**}{\longrightarrow}G_{\bullet}$, together with
the fact that $\operatorname{Ext}^{n-t}_{R}(M,\omega_{R})$ is Cohen-Macaulay
of dimension $t$, gives an isomorphism
$H_{0}(\psi):\operatorname{Ext}^{n-t}_{R}(\operatorname{Ext}^{n-t}_{R}(\operatorname{Ext}^{n-t}_{R}(M,\omega_{R}),\omega_{R}),\omega_{R})\longrightarrow\operatorname{Ext}^{n-t}_{R}(M,\omega_{R})$,
and one can verify that

$H_{0}(\psi)\circ H_{0}(\phi_{\bullet}^{**})=H_{0}(\phi_{\bullet})={\rm
id}_{\operatorname{Ext}^{n-t}_{R}(M,\omega_{R})}.$

Note that $H_{0}(\psi)$ is the inverse of the isomorphism of [10, Theorem
3.3.10 (c) (iii)], and this implies that
$H_{0}(\phi^{**}_{\bullet})=\operatorname{Ext}^{n-t}(\alpha)$ is the natural
isomorphism of the same theorem.

To conclude the proof, it remains to be shown that $\alpha$ is a monomorphism.

Let
$N=\operatorname{Ext}^{n-t}_{R}(\operatorname{Ext}^{n-t}_{R}(M,\omega_{R}),\omega_{R})$.
We show that $\alpha$ is injective once we localize at every associated prime
of $N$ and then we are done, since, if $\ker(\alpha)\neq 0$, its associated
primes would be contained in those of $N$.

Let ${\mathfrak{p}}\in\operatorname{Ass}(N)$; since $N$ is Cohen-Macaulay of
dimension $t$, we have that $\dim(R/{\mathfrak{p}})=t$ and
$\dim(R_{\mathfrak{p}})=n-t$. By replacing $R$ with $R_{\mathfrak{p}}$, $M$
with $M_{\mathfrak{p}}$, and $\omega_{R}$ with
$(\omega_{R})_{\mathfrak{p}}\cong\omega_{R_{\mathfrak{p}}}$, the proof will be
complete once we show that $\alpha$ is injective in the case $t=0$. To this
end observe that, as in Remark 2.11, the short exact sequence $0\to
H^{0}_{\mathfrak{m}}(M)\to M\to M/H^{0}_{\mathfrak{m}}(M)\to 0$ and [10,
Theorem 3.3.10 (c) (iii)] yield

$\operatorname{Ext}^{n}_{R}(\operatorname{Ext}^{n}_{R}(M,\omega_{R}),\omega_{R})\cong\operatorname{Ext}^{n}_{R}(\operatorname{Ext}^{n}_{R}(H^{0}_{\mathfrak{m}}(M),\omega_{R}),\omega_{R})\cong
H^{0}_{\mathfrak{m}}(M),$

and $\alpha$ composed with this isomorphism becomes just the inclusion of
$H^{0}_{\mathfrak{m}}(M)$ inside $M$. ∎

The equivalence between the first two conditions in the next theorem was
announced in [37] without a proof, but citing a spectral sequence argument due
to Peskine. Here we give another proof of this fact, see also [36, Theorem
5.5] where the equivalence with the third condition is proved.

######  Theorem 2.13 (Peskine).

Let $R$ be a Cohen-Macaulay ring of dimension $n$ with canonical module
$\omega_{R}$, and $M$ be a finitely generated $d$-dimensional $R$-module.
Then, the following are equivalent:

  1. (1)

$M$ is sequentially Cohen-Macaulay;

  2. (2)

$\operatorname{Ext}_{R}^{n-i}(M,\omega_{R})$ is either $0$ or Cohen-Macaulay
of dimension $i$ for all $i\in\\{0,\ldots,d\\}$;

  3. (3)

$\operatorname{Ext}_{R}^{n-i}(M,\omega_{R})$ is either $0$ or Cohen-Macaulay
of dimension $i$ for all $i\in\\{1,\ldots,d-1\\}$.

###### Proof.

The implication (1) $\Rightarrow$ (2) follows at once by Proposition 2.10, and
clearly (2) implies (3).

Now assume (3); we proceed by induction on $d-t$, where
$t=\operatorname{depth}(M)$. If $t=d$, then $M$ is Cohen-Macaulay, and hence
sequentially Cohen-Macaulay. Assume that $t<d$. If $t=0$, then
$0\neq\operatorname{Ext}^{n}_{R}(M,\omega_{R})$ has finite length, and hence
it is Cohen-Macaulay. Either way, thanks to our assumption we have that
$0\neq\operatorname{Ext}^{n-t}_{R}(M,\omega_{R})$ is $t$-dimensional and
Cohen-Macaulay. By Lemma 2.12, there is an injective homomorphism
$\alpha\,:\,\operatorname{Ext}_{R}^{n-t}(\operatorname{Ext}_{R}^{n-t}(M,\omega_{R}),\omega_{R})\longrightarrow
M$. Since the first module is $t$-dimensional Cohen-Macaulay by [10, Theorem
3.3.10 (c)], then so is its image, say $M_{1}$, which is a submodule of $M$.
It follows that
$\operatorname{depth}(M_{1})=\dim(M_{1})=t=\operatorname{depth}(M)<d=\dim(M)$.

Consider the short exact sequence $0\longrightarrow M_{1}\longrightarrow
M\longrightarrow M/M_{1}\longrightarrow 0$ and the induced sequence in
cohomology obtained by applying the functor
$\operatorname{Hom}_{R}(-,\omega_{R})$. We then have isomorphisms
$\operatorname{Ext}_{R}^{j}(M/M_{1},\omega_{R})\cong\operatorname{Ext}_{R}^{j}(M,\omega_{R})$
for all $j\neq n-t,n-t+1$ and the exact sequence

$0\to\operatorname{Ext}_{R}^{n-t}(M/M_{1},\omega_{R})\to\operatorname{Ext}_{R}^{n-t}(M,\omega_{R})\stackrel{{\scriptstyle\beta}}{{\to}}\operatorname{Ext}_{R}^{n-t}(M_{1},\omega_{R})\to\operatorname{Ext}_{R}^{n-t+1}(M/M_{1},\omega_{R})\to
0.$

By Lemma 2.12 we know that the map $\operatorname{Ext}^{n-t}(\alpha)$ is an
isomorphism, and therefore $\beta$ is an isomorphism as well. It follows that
$\operatorname{Ext}^{j}_{R}(M,\omega_{R})\cong\operatorname{Ext}^{j}_{R}(M/M_{1},\omega_{R})$
for every $j\neq n-t$, while
$\operatorname{Ext}^{n-t}_{R}(M/M_{1},\omega_{R})=0$.

This shows in particular that $\operatorname{depth}(M/M_{1})>t$; since
$\dim(M/M_{1})=d$, we may apply induction and obtain that $M/M_{1}$ is
sequentially Cohen-Macaulay. Let $0=M_{1}/M_{1}\subsetneq
M_{2}/M_{1}\subsetneq\ldots\subsetneq M_{r}/M_{1}=M/M_{1}$ be a sCM
filtration. Since $M_{1}$ is Cohen-Macaulay of dimension $t$ and
$\operatorname{depth}(M_{2}/M_{1})=\operatorname{depth}(M/M_{1})>t$ by Example
1.2 (4) , we deduce that $0=M_{0}\subsetneq M_{1}\subsetneq
M_{2}\subsetneq\ldots\subsetneq M_{r}=M$ is a sCM filtration and, thus $M$ is
sequentially Cohen-Macaulay. ∎

######  Corollary 2.14.

Let $R$ be Cohen-Macaulay with canonical module $\omega_{R}$, and $M$, $N$ be
finitely generated $R$-modules. Then, $M$ and $N$ are sequentially Cohen-
Macaulay if and only if $M\oplus N$ is sequentially Cohen-Macaulay.

###### Proof.

Let $n=\dim R$. We have already showed in Example 1.3 (1) that, if $M$ and $N$
are sequentially Cohen-Macaulay, then so is $M\oplus N$. This also follows
immediately from Theorem 2.13, since if
$\operatorname{Ext}^{n-i}_{R}(M,\omega_{R})$ and
$\operatorname{Ext}^{n-i}_{R}(N,\omega_{R})$ is either zero or Cohen-Macaulay
of dimension $i$, then so is
$\operatorname{Ext}^{n-i}_{R}(M,\omega_{R})\oplus\operatorname{Ext}^{n-i}_{R}(N,\omega_{R})\cong\operatorname{Ext}^{n-i}_{R}(M\oplus
N,\omega_{R})$. For the converse, it suffices to observe that if the direct
sum of two modules is zero or Cohen-Macaulay of a given dimension, then so is
each of its summand. ∎

######  Remark 2.15.

We remark that sequential Cohen-Macaulayness behaves well with respect to
localization; see for instance [16, Proposition 4.7] or [15, Proposition 2.6].
For any sequentially Cohen-Macaulay $R$-module $M$ and
${\mathfrak{p}}\in\operatorname{Supp}(M)$ one has that $M_{\mathfrak{p}}$ is a
sequentially Cohen-Macaulay $R_{\mathfrak{p}}$-module and, in fact, one can
recover its dimension filtration from that of $M$. Let
$s=\dim(R/{\mathfrak{p}})$ and consider the quotients
$\delta_{i}(M)/\delta_{i-1}(M)$ of the dimension filtration of $M$. If not
zero, they are $i$-dimensional Cohen-Macaulay, and their localization is
either zero or Cohen-Macaulay of dimension $i-s$. Now let
$N_{i}=(\delta_{i+s}(M))_{\mathfrak{p}}$ for all $i\geqslant 0$ such that
$i+s\leqslant d=\dim(M)$, i.e., for $i=0,\ldots,d-s$, and observe that
$0\subseteq N_{0}\subseteq\ldots\subseteq N_{d-s}=M_{\mathfrak{p}}$ is a CM
filtration of $M_{\mathfrak{p}}$, with quotients
$N_{i}/N_{i-1}\simeq(\delta_{i+s}(M)/\delta_{i+s-1}(M))_{\mathfrak{p}}$. If
$R$ is Cohen-Macaulay, the fact that $M_{\mathfrak{p}}$ is sequentially Cohen-
Macaulay for all ${\mathfrak{p}}\in\operatorname{Supp}(M)$ is also a
consequence of Theorem 2.13. See [39] for other results about localization and
sequentially Cohen-Macaulay modules.

We now recall some definitions needed to state the next result, and that we
will use frequently in the next sections. Given an $R$-module $N$, we let
$\operatorname{Ass}^{\circ}(N)=\operatorname{Ass}(N)\smallsetminus\\{\mathfrak{m}\\}$.

######  Definition 2.16.

Let $M\neq 0$ be a finitely generated graded $R$-module.

  1. (1)

A homogeneous element $0\neq y\in\mathfrak{m}$ is filter regular for $M$ if
$y\notin\bigcup_{{\mathfrak{p}}\in\operatorname{Ass}^{\circ}(M)}{\mathfrak{p}}$.

A sequence of homogeneous elements $y_{1},\ldots,y_{t}\in\mathfrak{m}$ is a
filter regular sequence for $M$ if $y_{i+1}$ is a filter regular element for
$M/(y_{1},\ldots,y_{i})M$ for all $i\in\\{0,\ldots,t-1\\}$.

  2. (2)

A homogeneous element $0\neq y\in\mathfrak{m}$ is strictly filter regular for
$M$ if
$y\notin\bigcup_{{\mathfrak{p}}\in\operatorname{Ass}^{\circ}(X(M))}{\mathfrak{p}}$,
where $X(M)=\bigoplus_{i\in{\mathbb{N}}}\operatorname{Ext}^{i}_{R}(M,R)$.

A sequence of homogeneous elements $y_{1},\ldots,y_{t}\in\mathfrak{m}$ is a
strictly filter regular sequence for $M$ if $y_{i+1}$ is a strictly filter
regular element for $M/(y_{1},\ldots,y_{i})M$ for all
$i\in\\{0,\ldots,t-1\\}$.

######  Remark 2.17.

Regular sequences are clearly filter regular sequences. Moreover, strictly
filter regular sequences are filter regular by the graded version of [9,
Corollary 11.3.3]. When the field $k$ is infinite, by Prime Avoidance any
sequence of general forms is strictly filter regular.

We conclude this section with the following result, which clarifies how
sequential Cohen-Macaulayness behaves with respect to quotients, cf. Example
1.6; it will also be useful later on.

######  Proposition 2.18.

Let $R$ be a Cohen-Macaulay ring of dimension $n$ with canonical module
$\omega_{R}$; let $M$ be a $d$-dimensional finitely generated $R$-module, and
$x\in R$ a strictly filter regular element for $M$. Then,

  1. (1)

If $M$ is sequentially Cohen-Macaulay, then $M/xM$ is sequentially Cohen-
Macaulay.

  2. (2)

The converse holds if $x$ is regular for all non-zero
$\operatorname{Ext}^{n-i}_{R}(M,\omega_{R})$ with $i>0$.

###### Proof.

Assume that $M$ is sequentially Cohen-Macaulay. Since $x$ is strictly filter
regular, it is filter regular by Remark 2.17, $L=0:_{M}x$ has finite length
and
$\operatorname{Ext}^{n-i}_{R}(\overline{M},\omega_{R})\cong\operatorname{Ext}^{n-i}_{R}(M,\omega_{R})$
for all $i>0$, where $\overline{M}=M/L$. Then, the long exact sequence
obtained by applying the functor $\operatorname{Hom}_{R}(-,\omega_{R})$ to the
short exact sequence $0\longrightarrow\overline{M}\stackrel{{\scriptstyle\cdot
x}}{{\longrightarrow}}M\longrightarrow M/xM\longrightarrow 0$ gives a long
exact sequence

$\begin{split}\ldots&\stackrel{{\scriptstyle\cdot
x}}{{\longrightarrow}}\operatorname{Ext}^{n-i}_{R}(M,\omega_{R})\longrightarrow\operatorname{Ext}^{n-(i-1)}_{R}(M/xM,\omega_{R})\longrightarrow\operatorname{Ext}^{n-(i-1)}_{R}(M,\omega_{R})\to\ldots\\\
\ldots&\stackrel{{\scriptstyle\cdot
x}}{{\longrightarrow}}\operatorname{Ext}^{n-1}_{R}(M,\omega_{R})\longrightarrow\operatorname{Ext}^{n}_{R}(M/xM,\omega_{R})\longrightarrow\operatorname{Ext}^{n}_{R}(M,\omega_{R})\longrightarrow
0.\end{split}$

By Theorem 2.13 we have that each non-zero
$\operatorname{Ext}^{n-i}_{R}(M,\omega_{R})$ is Cohen-Macaulay of dimension
$i$, and in this case $x$ is
$\operatorname{Ext}^{n-i}_{R}(M,\omega_{R})$-regular when $i>0$. For $i>1$ we
then have short exact sequences

$0\longrightarrow\operatorname{Ext}^{n-i}_{R}(M,\omega_{R})\stackrel{{\scriptstyle\cdot
x}}{{\longrightarrow}}\operatorname{Ext}^{n-i}_{R}(M,\omega_{R})\longrightarrow\operatorname{Ext}^{n-(i-1)}_{R}(M/xM,\omega_{R})\longrightarrow
0,$

and it follows that
$\operatorname{Ext}^{n-(i-1)}_{R}(M/xM,\omega_{R})\cong\operatorname{Ext}^{n-i}_{R}(M,\omega_{R})\otimes_{R}R/(x)$
is Cohen-Macaulay of dimension $i-1$ for all $i>1$. We conclude that $M/xM$ is
sequentially Cohen-Macaulay using the implication (3) $\Rightarrow$ (1) of
Theorem 2.13.

For the converse, the fact that $x$ is regular for all non-zero
$\operatorname{Ext}^{n-i}_{R}(M,\omega_{R})$ with $i>0$ shows that the above
long exact sequence of $\operatorname{Ext}$ modules breaks into short exact
sequences

$0\longrightarrow\operatorname{Ext}^{n-i}_{R}(M,\omega_{R})\stackrel{{\scriptstyle\cdot
x}}{{\longrightarrow}}\operatorname{Ext}^{n-i}_{R}(M,\omega_{R})\longrightarrow\operatorname{Ext}^{n-(i-1)}_{R}(M/xM,\omega_{R})\longrightarrow
0.$

for all $i>1$. By Theorem 2.13 we have that
$\operatorname{Ext}^{n-(i-1)}_{R}(M/xM,\omega_{R})$ is either zero or Cohen-
Macaulay of dimension $i-1$, and thus
$\operatorname{Ext}^{n-i}_{R}(M,\omega_{R})$ is either zero or Cohen-Macaulay
of dimension $i$ for all $i>1$. Since $x$ is assumed to be regular on
$\operatorname{Ext}^{n-1}_{R}(M,\omega_{R})$, and
$\dim(\operatorname{Ext}^{n-1}_{R}(M,\omega_{R}))\leqslant 1$, we have that
$\operatorname{Ext}^{n-1}_{R}(M,\omega_{R})$ is either zero, or Cohen-Macaulay
of dimension $1$. It follows again from the implication (3) $\Rightarrow$ (1)
of Theorem 2.13 that $M$ is sequentially Cohen-Macaulay. ∎

######  Remark 2.19.

There are many other interesting results about sequentially Cohen-Macaulay
modules and their characterizations which do not find space in this note. For
instance, in [14, Theorem 5.1] it is proven that a module is sequentially
Cohen-Macaulay if and only if each module of its dimension filtration is
pseudo Cohen-Macaulay. In [18, Theorem 1.1], the sequential Cohen-Macaulayness
of $M$ is characterized in terms of the existence of one good system of
parameters of $M$ which has the property of parametric decomposition; see also
Theorems 3.9 and 4.2 in [14] for other characterizations which involve good
systems of parameters and $dd$-sequences.

Moreover, in [15] it is investigated how the sequential Cohen-Macaulay
property behaves in relation to taking associated graded rings and Rees
algebras, see also [38] for more results of this type.

##  3\. Partially sequentially Cohen-Macaulay modules

We are going to study next the notion of partially sequentially Cohen-Macaulay
module, as introduced in [35], which naturally generalizes that of
sequentially Cohen-Macaulay modules. Thanks to Schenzel’s Theorem 2.7, the
definition can be given in terms of the dimension filtration of the module.
Throughout this section we let $R=k[x_{1},\dots,x_{n}]$ be a standard graded
polynomial ring over an infinite field $k$ with homogeneous maximal ideal
${\mathfrak{m}}=(x_{1},\dots,x_{n})$. Recall that, in this case, $R$ has a
graded canonical module $\omega_{R}\cong R(-n)$. We consider finitely
generated graded $R$-modules $M$; when $M=0$, we set
$\operatorname{depth}(M)=+\infty$ and $\dim(M)=-1$, as usual. We let
$d=\dim(M)$.

######  Definition 3.1.

Let $i\in\\{0,\ldots,d\\}$ and let $\\{\delta_{j}(M)\\}_{j}$ be the dimension
filtration of $M$; $M$ is called $i$-partially sequentially Cohen-Macaulay,
$i$-sCM for short, if $\delta_{j}(M)/\delta_{j-1}(M)$ is either zero or Cohen-
Macaulay for all $i\leqslant j\leqslant d$.

By definition and Corollary 2.8, a module $M$ is $0$-sCM if and only if $M$
sequentially Cohen-Macaulay if and only if $M$ is $1$-sCM.

######  Example 3.2.

  1. (1)

Let $M$ be a sequentially Cohen-Macaulay module. The simplest way of
constructing an $i$-sCM module which is not sequentially Cohen-Macaulay, is
perhaps taking a non-sequentially Cohen-Macaulay module $N$ of dimension
strictly smaller than $i$, and condider their direct sum $M\oplus N$, cf.
Example 1.3 (1).

  2. (2)

Let $M=R/I$, where
$I=(x_{1})\cap(x_{2},x_{3})\cap(x_{1}^{2},x_{4},x_{5})\subset
R=k[x_{1},x_{2},x_{3},x_{4},x_{5}]$. With the help of Proposition 2.2, we can
construct the dimension filtration
$0=\delta_{-1}=\delta_{0}=\delta_{1}\subseteq\delta_{2}=((x_{1})\cap(x_{2},x_{3}))/I\subseteq\delta_{3}=(x_{1})/I\subseteq\delta_{4}=R/I$
of $M$. Then, $\delta_{4}/\delta_{3}$ and $\delta_{3}/\delta_{2}$ are Cohen-
Macaulay of dimension $4$ and $3$ respectively, but
$0\neq\delta_{2}/\delta_{1}$ is not Cohen-Macaulay. Hence, $M$ is an example
of a $3$-sCM which is not $2$-sCM.

######  Remark 3.3.

Observe that $M$ is $i$-sCM if and only if $M/\delta_{i-1}(M)$ is sequentially
Cohen-Macaulay. This follows at once recalling that the dimension filtration
$\\{\gamma_{j}\\}_{j}$ of $M/\delta_{i-1}(M)$ is such that
$\gamma_{j}=\delta_{j}(M)/\delta_{i-1}(M)$ for $j\geqslant i$ and
$\gamma_{j}=0$ otherwise. Notice that, since $\gamma_{j}=0$ for all
$j\leqslant i-1$, if $M$ is $i$-sCM then
$H^{j}_{\mathfrak{m}}(M/\delta_{i-1}(M))=0$ for all $j\leqslant i-1$, by
Proposition 1.5.

Given a graded free presentation of $M\cong F/U$, we denote by
$\\{e_{1},\ldots,e_{r}\\}$ a graded basis of $F$. We consider $R$ together
with the pure reverse lexicographic ordering $>$ such that
$x_{1}>\ldots>x_{n}$; recall that $>$ is not a monomial order on $R$, but by
definition it agrees with the reverse lexicographic order that refines it on
monomials of the same degree.

We extend $>$ to $F$ in the following way: given monomials $ue_{i}$ and
$ve_{j}$ of $F$, set

$\begin{split}ue_{i}>ve_{j}\text{ if
}&\big{(}\deg(ue_{i})>\deg(ve_{j})\big{)},\text{ or
}\big{(}\deg(ue_{i})=\deg(ve_{j})\text{ and }u>v\big{)}\\\ \text{ or
}&\big{(}\deg(ue_{i})=\deg(ve_{j}),\,\,u=v\text{ and }i<j\big{)}.\end{split}$

We shall consider this order until the end of the section, and denote by
$\operatorname{Gin}(U)$ the generic initial module of $U$ with respect to $>$.
Since the action of ${\rm GL}_{n}(k)$ on $R$ as change of coordinates can be
extended in an obvious way to $F$, $\operatorname{Gin}(U)$ simply results to
be the initial submodule ${\rm in}_{>}(gU)$ where $g$ is a general change of
coordinates.

We prove next some preliminary facts which are needed later on. Given a graded
submodule $V\subseteq F$ with $\dim(F/V)=d$, for all $j\in\\{-1,\ldots,d\\}$
we denote by $V^{\langle j\rangle}$ the $R$-module such that $V^{\langle
j\rangle}/V=\delta_{j}(F/V)$. Several results contained in the next two
lemmata can be found in [22], where they are proved in the ideal case.

######  Lemma 3.4.

With the above notation,

  1. (1)

$\operatorname{Gin}(U^{\langle
j\rangle})\subseteq\operatorname{Gin}(U)^{\langle j\rangle}$;

  2. (2)

$U^{\langle j\rangle}=(U^{\langle j\rangle})^{\langle j\rangle}$;

  3. (3)

if $V$ is a graded submodule of $F$ such that $U\subseteq V$, then $U^{\langle
j\rangle}\subseteq V^{\langle j\rangle}$;

  4. (4)

$\operatorname{Gin}(U^{\langle j\rangle})^{\langle
j\rangle}=\operatorname{Gin}(U)^{\langle j\rangle}$.

###### Proof.

(1) Notice that $\operatorname{Gin}(U^{\langle
j\rangle})/\operatorname{Gin}(U)$ and $U^{\langle j\rangle}/U$ have the same
Hilbert series, hence the same dimension, which is less than or equal to $j$.
Since $\operatorname{Gin}(U)^{\langle
j\rangle}/\operatorname{Gin}(U)=\delta_{j}(F/\operatorname{Gin}(U))$, we have
the desired inclusion.

(2) Since $U\subseteq U^{\langle j\rangle}$, one inclusion is clear. Now
consider the short exact sequence

$0\longrightarrow U^{\langle j\rangle}/U\longrightarrow(U^{\langle
j\rangle})^{\langle j\rangle}/U\longrightarrow(U^{\langle j\rangle})^{\langle
j\rangle}/U^{\langle j\rangle}\longrightarrow 0.$

Since the dimensions of $U^{\langle j\rangle}/U$ and $(U^{\langle
j\rangle})^{\langle j\rangle}/U^{\langle j\rangle}$ are less than or equal to
$j$, it follows that also $\dim((U^{\langle j\rangle})^{\langle
j\rangle}/U)\leqslant j$ and, hence, $(U^{\langle j\rangle})^{\langle
j\rangle}\subseteq U^{\langle j\rangle}$.

(3) Since $U\subseteq U^{\langle j\rangle}\cap V$, we have that
$\dim((U^{\langle j\rangle}+V)/V)\leqslant\dim(U^{\langle
j\rangle}/U)\leqslant j$, which implies $U^{\langle j\rangle}\subseteq
U^{\langle j\rangle}+V\subseteq V^{\langle j\rangle}$.

(4) Since $U\subseteq U^{\langle j\rangle}$, it immediately follows from (1)
that $\operatorname{Gin}(U)\subseteq\operatorname{Gin}(U^{\langle
j\rangle})\subseteq\operatorname{Gin}(U)^{\langle j\rangle}$ and, accordingly,
$\operatorname{Gin}(U)^{\langle
j\rangle}\subseteq\operatorname{Gin}(U^{\langle j\rangle})^{\langle
j\rangle}$. On the other hand, by Parts (1) and (2), the latter is contained
in $(\operatorname{Gin}(U)^{\langle j\rangle})^{\langle
j\rangle}=\operatorname{Gin}(U)^{\langle j\rangle}$. ∎

We denote the Hilbert series of a graded $R$-module $N$ by
$\operatorname{Hilb}(N)=\operatorname{Hilb}(N,z)$. We also let
$h^{j}(N)=\operatorname{Hilb}(H^{j}_{\mathfrak{m}}(N))$.

######  Lemma 3.5.

Let $M\cong F/U$, with dimension filtration $\\{\delta_{j}\\}_{j}$; then, the
following holds:

  1. (1)

$M$ is $i$-sCM if and only if $M/H^{0}_{\mathfrak{m}}(M)$ is $i$-sCM.

  2. (2)

If $M$ is $i$-sCM, then $H^{j}_{\mathfrak{m}}(M)\cong
H^{j}_{\mathfrak{m}}(\delta_{j})\cong
H^{j}_{\mathfrak{m}}(\delta_{j}/\delta_{j-1})$ for all $j\geqslant i$.

  3. (3)

If $M$ is $i$-sCM, then
$(z-1)^{j}h^{j}(M)=(1-z)^{j}\operatorname{Hilb}(\delta_{j}/\delta_{j-1})$ for
all $j\geqslant i$.

  4. (4)

Let $\\{\gamma_{j}\\}_{j}$ be the dimension filtration of
$F/\operatorname{Gin}(U)$; if $M$ is $i$-sCM, then
$\operatorname{Hilb}(\delta_{j}/\delta_{j-1})=\operatorname{Hilb}(\gamma_{j}/\gamma_{j-1})$
for all $j\geqslant i$.

###### Proof.

Since $H^{0}_{\mathfrak{m}}(R)=\delta_{0}$, the dimension filtration of
$M/H^{0}_{\mathfrak{m}}(R)$ is $\\{\delta_{j}/\delta_{0}\\}_{j}$, which shows
the first part. The long exact sequence in cohomology induced by
$0\to\delta_{j-1}\to\delta_{j}\to\delta_{j}/\delta_{j-1}\to 0$ and the Cohen-
Macaulayness of $\delta_{j}/\delta_{j-1}$ for all $j\geqslant i$ easily imply
(2).

Let us fix $j\geqslant i$ and prove (3). If $j=0$ the assertion is clear; thus
we may assume $j>0$ and by way of Part (2) that $\delta_{j}/\delta_{j-1}$ is
$j$-dimensional Cohen-Macaulay. Let $x$ be a
$(\delta_{j}/\delta_{j-1})$-regular element of degree one; then, the short
exact sequence given by multiplication by $x$ induces a short exact sequence
in cohomology

$0\longrightarrow
H_{\mathfrak{m}}^{j-1}\left(\frac{\delta_{j}/\delta_{j-1}}{x(\delta_{j}/\delta_{j-1})}\right)\longrightarrow
H^{j}_{\mathfrak{m}}(\delta_{j}/\delta_{j-1})(-1)\longrightarrow
H^{j}_{\mathfrak{m}}(\delta_{j}/\delta_{j-1})\longrightarrow 0$

together with (2) imply that
$h^{j-1}((\delta_{j}/\delta_{j-1})/(x(\delta_{j}/\delta_{j-1}))=(z-1)\,h^{j}(\delta_{j}/\delta_{j-1})=(z-1)\,h^{j}(M)$.

Thus, one can easily prove that $h^{0}((\delta_{j}/\delta_{j-1})/({\bf
x}(\delta_{j}/\delta_{j-1})))=(z-1)^{j}\,h^{j}(M)$, where ${\bf x}$ is a
$(\delta_{j}/\delta_{j-1})$-maximal regular sequence. Since
$\dim((\delta_{j}/\delta_{j-1})/({\bf x}\delta_{j}/\delta_{j-1}))=0$, we also
have $h^{0}((\delta_{j}/\delta_{j-1})/({\bf
x}\delta_{j}/\delta_{j-1}))=\operatorname{Hilb}((\delta_{j}/\delta_{j-1})/({\bf
x}\delta_{j}/\delta_{j-1}))=(1-z)^{j}\operatorname{Hilb}(\delta_{j}/\delta_{j-1})$,
and the proof of (3) is complete.

Finally, to prove (4) we show that $\operatorname{Hilb}(U^{\langle
j\rangle}/U^{\langle
j-1\rangle})=\operatorname{Hilb}(\operatorname{Gin}(U)^{\langle
j\rangle}/(\operatorname{Gin}(U)^{\langle j-1\rangle}))$ holds for all
$j\geqslant i$. Actually, we prove more, i.e. that
$\operatorname{Hilb}(U^{\langle
j\rangle})=\operatorname{Hilb}(\operatorname{Gin}(U)^{\langle j\rangle})$ for
all $j\geqslant i$; since $\operatorname{Gin}(U^{\langle
j\rangle})\subseteq\operatorname{Gin}(U)^{\langle j\rangle}$ by Lemma 3.4 (1)
for all $j$, the last equality is equivalent to proving that
$\operatorname{Gin}(U^{\langle j\rangle})=\operatorname{Gin}(U)^{\langle
j\rangle}$ for all $j\geqslant i$, and this is what we do.

Consider now, for all $j$, the short exact sequences $0\to U^{\langle
j\rangle}/U^{\langle j-1\rangle}\to F/U^{\langle j-1\rangle}\to F/U^{\langle
j\rangle}\to 0;$ we see inductively that $\operatorname{depth}(F/U^{\langle
j\rangle})\geqslant j+1$ for all $j\geqslant i$. For $j=d$ and if $U^{\langle
j\rangle}/U^{\langle j-1\rangle}=0$ this is obvious; otherwise, since $M$ is
$i$-sCM, $U^{\langle j\rangle}/U^{\langle j-1\rangle}$ is $j$-dimensional
Cohen-Macaulay for all $j\geqslant i$ and by [10, Proposition 1.2.9] we get
that $\operatorname{depth}F/U^{\langle j-1\rangle}\geqslant\min\\{j,j+1\\}=j$.

For all graded submodules $V\subseteq F$ it is well-known that
$\operatorname{depth}(F/\operatorname{Gin}(V))=\operatorname{depth}(F/V)$ and
that $F/\operatorname{Gin}(V)$ is sequentially Cohen-Macaulay. Thus, $j+1$
$\leqslant$ $\operatorname{depth}(F/\operatorname{Gin}(U^{\langle j\rangle}))$
by what we proved above, and Proposition 1.5 together with Theorem 2.7 imply
that the latter is also equal to the smallest integer $t$ such that
$\operatorname{Gin}(U^{\langle
j\rangle})\subsetneq\operatorname{Gin}(U^{\langle j\rangle})^{\langle
t\rangle}$. Therefore, we have shown that $\operatorname{Gin}(U^{\langle
j\rangle})=\operatorname{Gin}(U^{\langle j\rangle})^{\langle j\rangle}$ for
all $j\geqslant i$. Now the conclusion follows from Lemma 3.4 (4). ∎

######  Definition 3.6.

Let $M$ be a finitely generated graded $R$-module. We let $\delta(M)=0$ if
$M=0$, and we let $\delta(M)$ be the largest graded $R$-submodule of $M$ of
dimension at most $\dim(M)-1$ otherwise. Given $j\geqslant 0$, we also define
the module $\delta^{j}(M)$ inductively by letting

$\delta^{0}(M)=M,\,\,\,\delta^{1}(M)=\delta(M),\,\,\,\text{and}\,\,\,\delta^{j}(M)=\delta(\delta^{j-1}(M)).$

Since $\delta_{i}(M)$ is the largest submodule of $M$ of dimension at most
$i$, it is easy to see that for all $i\in\\{0,\ldots,d\\}$ there exists
$j=j(i)\geqslant 0$ such that $\delta_{i}(M)=\delta^{j}(M)$. In particular
$\delta_{d-1}(M)=\delta(M)=\delta^{1}(M)$.

In the following remark we collect two known facts which are useful in the
following.

######  Remark 3.7.

Recall that in our setting $R\cong\omega_{R}(n)$. It is a well-known fact that
for a $d$-dimensional graded $R$-module $M$ it holds that
$\dim\operatorname{Ext}^{n-i}_{R}(M,\omega_{R})\leqslant i$ for all $i$.

  1. (1)

Given a graded submodule $N\subseteq M$ such that $\dim(N)<\dim(M)=d$, we have
that $N=\delta(M)$ if and only if $M/N$ is unmixed of dimension $d$, i.e.,
$\dim(R/{\mathfrak{p}})=\dim(M/N)=d$ for all
${\mathfrak{p}}\in\operatorname{Ass}(M/N)$. This is a straightforward
application of Proposition 2.2.

  2. (2)

A $d$-dimensional finitely generated graded $R$-module $M$ is unmixed if and
only if $\dim(\operatorname{Ext}^{n-j}_{R}(M,R))<j$ for all
$j\in\\{0,\ldots,d-1\\}$. In fact, if ${\mathfrak{p}}\in\operatorname{Ass}(M)$
were a prime of height $n-j$ for some $j<d$, we would have that
$\operatorname{Ext}^{n-j}_{R}(M,R)_{\mathfrak{p}}\cong\operatorname{Ext}^{n-j}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}},R_{\mathfrak{p}})\neq
0$, since the latter is, up to shift, the Matlis dual of
$H^{0}_{{\mathfrak{p}}R_{\mathfrak{p}}}(M_{\mathfrak{p}})$ which is not zero
because $\operatorname{depth}(M_{\mathfrak{p}})=0$ by [10, Proposition
1.2.13]. It follows that
${\mathfrak{p}}\in\operatorname{Supp}(\operatorname{Ext}^{n-j}_{R}(M,R))$ and,
thus, $\dim(\operatorname{Ext}^{n-j}_{R}(M,R))$ $\geqslant j$, contradiction.
The converse is analogous, observing that if
$\operatorname{Ext}^{n-j}_{R}(M,R)$ has dimension at least $j$ and, hence,
necessarily equal to $j$, then it must have a prime of height $n-j$ in its
support.

The following lemma can be regarded as an enhanced graded version of [17,
Proposition 4.16].

######  Lemma 3.8.

Let $M$ be a finitely generated graded $R$-module. For a sufficiently general
homogeneous element $x\in\mathfrak{m}$ and for all $j\geqslant 0$ there is a
short exact sequence
$0\to\delta^{j}(\delta(M)/x\delta(M))\to\delta^{j+1}(M/xM)\to L_{j}\to 0$,
where $L_{j}$ is a module of finite length.

###### Proof.

Let $d=\dim(M)$. The case $d\leqslant 1$ is trivial, therefore we will assume
that $d\geqslant 2$ and proceed by induction on $j\geqslant 0$.

First assume that $j=0$; let $\overline{M}=M/\delta(M)$, and observe that
$\overline{M}$ is unmixed of positive depth and dimension $d$. In particular,
$\operatorname{Ext}^{n}_{R}(\overline{M},R)=0$ and, by Remark 3.7, we have
$\dim(\operatorname{Ext}^{n-\ell}_{R}(\overline{M},R))<\ell$ for all
$0<\ell<d$.

For $x$ sufficiently general, we have that $x$ is $\overline{M}$-regular and,
thus, $\delta(M)\cap xM=x(\delta(M):_{M}x)=x\delta(M)$; moreover, either
$\dim(\delta(M)/x\delta(M))\leqslant 0$ or
$\dim(\delta(M)/x\delta(M))=\dim(\delta(M))-1<d-1=\dim(\overline{M}/x\overline{M})=\dim(M/xM)$.
If we let $T=\delta(M)+xM$, then $\delta(M)/x\delta(M)\cong
T/xM\subseteq\delta(M/xM)$, and therefore we have an exact sequence
$0\to\delta(M)/x\delta(M)\stackrel{{\scriptstyle\varphi}}{{\to}}\delta(M/xM)$.
Moreover, we have that ${\rm
coker}(\varphi)\cong\delta((M/xM)/(T/xM))\cong\delta(M/T)$. Since
$M/T\cong\overline{M}/x\overline{M}$, we then have a short exact sequence

$0\longrightarrow\delta(M)/x\delta(M)\stackrel{{\scriptstyle\varphi}}{{\longrightarrow}}\delta(M/xM)\longrightarrow\delta(\overline{M}/x\overline{M})\longrightarrow
0.$

Since $x$ is sufficiently general, by Remark 2.17 we may assume that $x$ is
also strictly filter regular for $\overline{M}$. We then have that either
$\dim(\operatorname{Ext}^{n-\ell}_{R}(\overline{M},R)\otimes_{R}R/(x))=\dim(\operatorname{Ext}^{n-\ell}_{R}(\overline{M},R))\leqslant
0$ or
$\dim(\operatorname{Ext}^{n-\ell}_{R}(\overline{M},R)\otimes_{R}R/(x))=\dim(\operatorname{Ext}^{n-\ell}_{R}(\overline{M},R))-1<\ell-1$
for all $0<\ell<d$. Since $x$ is strictly filter regular for $\overline{M}$,
from the short exact sequences

$0\longrightarrow\operatorname{Ext}^{n-\ell}_{R}(\overline{M},R)\otimes_{R}R/(x)\longrightarrow\operatorname{Ext}^{n-(\ell-1)}_{R}(\overline{M}/x\overline{M},R)\longrightarrow
0:_{\operatorname{Ext}^{n-(\ell-1)}_{R}(\overline{M},R)}x\longrightarrow 0$

it also follows that
$\dim(\operatorname{Ext}^{n-(\ell-1)}_{R}(\overline{M}/x\overline{M},R))=\dim(\operatorname{Ext}^{n-\ell}_{R}(\overline{M},R)\otimes_{R}R/(x))<\ell-1$
for all $0\leqslant\ell-1<d-1$. Thus, by Remark 3.7 it follows that
$\overline{M}/x\overline{M}$ is unmixed of dimension $d-1$ and that
$L_{0}=\delta(\overline{M}/x\overline{M})$ is necessarily
$\delta_{0}(\overline{M}/x\overline{M})=H^{0}_{\mathfrak{m}}(\overline{M}/x\overline{M})$,
which has finite length.

Now suppose that the statement of the lemma is proved for $j-1$, so that we
have a short exact sequence
$0\to\delta^{j-1}(\delta(M)/x\delta(M))\to\delta^{j}(M/xM)\to L_{j-1}\to 0$,
with $L_{j-1}$ of finite length.

It is clear from the definition of $\delta$ that there is an exact sequence
$0\longrightarrow\delta^{j}(\delta(M)/x\delta(M))\stackrel{{\scriptstyle\varphi_{j}}}{{\longrightarrow}}U\longrightarrow
L_{j}\longrightarrow 0$, where we let $U=\delta^{j+1}(M/xM)$ and $L_{j}={\rm
coker}(\varphi_{j})$. If $U$ has finite length we are done, so let us assume
that $\dim(U)>0$. In this case we necessarily have that
$h=\dim(\delta^{j}(M/xM))>0$, and since $L_{j-1}$ has finite length we
conclude that $\dim(\delta(M)/x\delta(M))=h$. Again because $L_{j-1}$ has
finite length, we can find $p\gg 0$ such that
$\mathfrak{m}^{p}U\subseteq\delta^{j-1}(\delta(M)/x\delta(M))$. Moreover,
$\dim(\mathfrak{m}^{p}U)\leqslant\dim(U)<h$, and therefore $\mathfrak{m}^{p}U$
is contained in $\delta^{j}(\delta(M)/x\delta(M))$. This shows that
$\mathfrak{m}^{p}L_{j}=0$, and thus $L_{j}$ has finite length. ∎

In [35] it is claimed that, if $x\in R$ is $M$-regular, it is possible to
prove that $M$ is $i$-sCM if and only if $M/xM$ is $(i-1)$-sCM following the
same lines of the proof [36, Theorem 4.7], which is not utterly correct, as we
already pointed out in Example 1.6. The claim is indeed false: if $x$ is
$M$-regular and $M/xM$ is $(i-1)$-sCM, then $M$ is not necessarily $i$-sCM, as
the following example shows.

######  Example 3.9.

Let $R$ be a $2$-dimensional domain which is not Cohen-Macaulay, cf. Example
1.6. For all $0\neq x\in R$ we have that $R/(x)$ is $0$-sCM, but $R$ is not
even $2$-sCM.

In the following proposition we show that the claimed result holds true under
some additional assumption.

######  Proposition 3.10.

Let $i$ be a positive integer, and $M$ be a finitely generated graded
$R$-module with $\operatorname{depth}(M)>0$. Also assume that
$\operatorname{depth}(\operatorname{Ext}^{n-\ell}_{R}(M,R))>0$ for all
$\ell\geqslant i-1$. For a sufficiently general $x\in R$, if $M/xM$ is
$(i-1)$-sCM, then $M$ is $i$-sCM.

###### Proof.

Since a module is $0$-sCM if and only if it is $1$-sCM, cf. Definition 3.1 and
Corollary 2.8, when $i=1,2$ the statement follows immediately from Proposition
2.18; therefore we may let $3\leqslant i\leqslant d=\dim(M)$. Let $j$ be the
smallest integer such that $\delta_{i-1}(M)=\delta^{j}(M)$, and let
$N=M/\delta^{j}(M)$. Observe that $\operatorname{depth}(N)>0$ by Proposition
2.3. By Remark 3.3 it is enough to prove that $N$ is sequentially Cohen-
Macaulay, and this is what we do.

Observe that, since $\dim(\delta^{j}(M))\leqslant i-1$, we have that
$\operatorname{Ext}^{n-\ell}_{R}(M,R)\cong\operatorname{Ext}^{n-\ell}_{R}(N,R)$
for all $\ell\geqslant i$ and there is an injection
$0\to\operatorname{Ext}^{n-(i-1)}_{R}(N,R)\to\operatorname{Ext}^{n-(i-1)}_{R}(M,R)$.
In particular, from our assumptions we get that

$\operatorname{depth}(\operatorname{Ext}^{n-\ell}_{R}(N,R))>0\text{\,\, for
all \,\,}\ell\geqslant i-1.$

By a repeated application of Lemma 3.8 there exists a short exact sequence

$0\longrightarrow\delta^{j}(M)/x\delta^{j}(M)\stackrel{{\scriptstyle\varphi}}{{\longrightarrow}}\delta^{j}(M/xM)\longrightarrow
L\longrightarrow 0,$

where $L$ is a module of finite length. Now, either
$\dim(\delta^{j}(M/xM))=\dim(\delta^{j}(M)/x\delta^{j}(M))=\dim(\delta^{j}(M))-1$,
or $\dim(\delta^{j}(M/xM))\leqslant 0$, and in both cases we have that
$\dim(\delta^{j}(M/xM))$ $\leqslant$ $i-2$. By minimality of $j$, we have that
$\dim(\delta^{j-1}(M))>i-1$ and applying iteratively Lemma 3.8 as we did
above, we also obtain $\dim(\delta^{j-1}(M/xM))=\dim(\delta^{j-1}(M))-1>i-2$;
thus, we may conclude that

$\delta^{j}(M/xM)=\delta_{i-2}(M/xM).$

Since $L={\rm coker}(\varphi)$ and $x$ is regular for $N=M/\delta^{j}(M)$, we
have $(M/xM)/(\delta^{j}(M)/x\delta^{j}(M))$ $\cong$
$M/(\delta^{j}(M)+xM)\cong N/xN$, and the above yields a short exact sequence
$0\to L\to N/xN\to(M/xM)/\delta_{i-2}(M/xM)\to 0.$

Since $L$ has finite length, the associated long exact sequence of
$\operatorname{Ext}$-modules yields that

$\operatorname{Ext}^{n-\ell}_{R}(N/xN,R)\cong\operatorname{Ext}^{n-\ell}_{R}((M/xM)/\delta_{i-2}(M/xM),R)\text{\,\,for
all\,\,}\ell\neq 0.$

In particular, being $M/xM$ a $(i-1)$-sCM module by assumption, Remark 3.3 and
Peskine Theorem 2.13 imply that $\operatorname{Ext}^{n-\ell}_{R}(N/xN,R)$ is
either zero or Cohen-Macaulay of dimension $\ell$ for all $\ell\neq 0$. Remark
3.3 together with Local Duality also imply that

$\operatorname{Ext}^{n-(\ell-1)}_{R}(N/xN,R)=0\text{ for all
}1\leqslant\ell-1\leqslant i-2.$

We may assume that $x$, which is $N$-regular, is also
$\operatorname{Ext}^{n-\ell}_{R}(N,R)$-regular for all $\ell\geqslant i-1$,
for we proved above that all these modules have positive depth. For
$\ell\geqslant i$ we thus have short exact sequences
$0\to\operatorname{Ext}^{n-\ell}_{R}(N,R)\stackrel{{\scriptstyle\cdot
x}}{{\longrightarrow}}\operatorname{Ext}^{n-\ell}_{R}(N,R)\to\operatorname{Ext}^{n-(\ell-1)}_{R}(N/xN,R)\to
0$, from which it follows that $\operatorname{Ext}^{n-\ell}_{R}(N,R)$ is
either zero or Cohen-Macaulay of dimension $\ell$ for all $\ell\geqslant i$.

From the above, we also have that for all $2\leqslant\ell\leqslant i-1$ the
maps $\operatorname{Ext}^{n-\ell}_{R}(N,R)\stackrel{{\scriptstyle\cdot
x}}{{\longrightarrow}}\operatorname{Ext}^{n-\ell}_{R}(N,R)$ are isomorphisms
and, by graded Nakayama’s Lemma, that $\operatorname{Ext}^{n-\ell}_{R}(N,R)=0$
for all $2\leqslant\ell\leqslant i-1$. Finally, we also have an injection
$0\to\operatorname{Ext}^{n-1}_{R}(N,R)\stackrel{{\scriptstyle\cdot
x}}{{\longrightarrow}}\operatorname{Ext}^{n-1}_{R}(N,R)$, which implies that
$\operatorname{Ext}^{n-1}_{R}(N,R)$ is Cohen-Macaulay of dimension one by
Remark 3.7.

Applying Peskine’s Theorem 2.9, we have thus showed that $N$ is sequentially
Cohen-Macaulay, that is, $M$ is $i$-sCM.

∎

The next theorem provides a characterization of partially sequentially Cohen-
Macaulay modules. It was proved for the first time in [35, Theorem 3.5] in the
ideal case. Here, we generalize the result to finitely generated modules and
fix the gap in the original proof thanks to Proposition 3.10. We let
$R_{[n-1]}=k[x_{1},\dots,x_{n-1}]\cong R/x_{n}R$ and denote by $N_{[n-1]}$ the
$R_{[n-1]}$-module $N/x_{n}N\simeq N\otimes_{R}R/x_{n}R$ by restriction of
scalars. We let $x\in R$ be a general linear form which, without loss of
generality, we may write as $l=a_{1}x_{1}+\dots+a_{n-1}x_{n-1}-x_{n}$ and
consider the map $g_{n}\colon R\to R_{[n-1]}$, defined by $x_{i}\mapsto x_{i}$
for $i=1,\ldots,n-1$ and $x_{n}\mapsto a_{1}x_{1}+\dots+a_{n-1}x_{n-1}$. Then,
the surjective homomorphism $F/U\rightarrow F_{[n-1]}/g_{n}(U)$ has kernel
$(U+xF)/U$ and induces the isomorphism

(3.1) $\frac{F}{U+xF}\cong\frac{F_{[n-1]}}{g_{n}(U)}.$

Moreover, the image of $\operatorname{Gin}(U)$ in $F_{[n-1]}$ via the mapping
$x_{n}\mapsto 0$ is $\operatorname{Gin}(U)_{[n-1]}$. With this notation, the
module version of [24, Corollary 2.15] states that

(3.2) $\operatorname{Gin}(g_{n}(U))=\operatorname{Gin}(U)_{[n-1]}.$

######  Theorem 3.11.

Let $M$ be a finitely generated graded $R$-module of dimension $d$, and let
$M\cong F/U$ be a free graded presentation of $M$. The following conditions
are equivalent:

  1. (1)

$F/U$ is $i$-sCM;

  2. (2)

$h^{j}(F/U)=h^{j}(F/\operatorname{Gin}(U))$ for all $i\leqslant j\leqslant d$.

###### Proof.

(1) $\Rightarrow$ (2) is a direct consequence of Lemma 3.5 (3) and (4), since
also $F/\operatorname{Gin}(U)$ is $i$-sCM.

We prove the converse by induction on $d$. If $d=0$, $F/U$ is Cohen-Macaulay
and sequentially Cohen-Macaulay. Therefore, without loss of generality we may
assume that $F/U$ and $F/\operatorname{Gin}(U)$ have positive dimension and,
by Lemma 3.5 (1), also positive depth. Since $F/\operatorname{Gin}(U)$ is
sequentially Cohen-Macaulay, Peskine’s Theorem 2.13 implies that there exists
a linear form $l\in R$ which is $F/\operatorname{Gin}(U)$-regular and also
regular for all non-zero
$\operatorname{Ext}^{n-j}_{R}(F/\operatorname{Gin}(U),\omega_{R})$ with $j>0$.
Starting with the exact sequence $0\rightarrow
F/\operatorname{Gin}(U)(-1)\stackrel{{\scriptstyle\cdot
l}}{{\rightarrow}}F/\operatorname{Gin}(U)\rightarrow
F/(\operatorname{Gin}(U)+lF)\rightarrow 0$, by the above and Local Duality we
obtain the short exact sequences

$0\longrightarrow
H^{j-1}_{{\mathfrak{m}}}(F/(\operatorname{Gin}(U)+lF))\longrightarrow
H^{j}_{{\mathfrak{m}}}(F/\operatorname{Gin}(U))(-1)\stackrel{{\scriptstyle\cdot
l}}{{\longrightarrow}}H^{j}_{{\mathfrak{m}}}(F/\operatorname{Gin}(U))\longrightarrow
0,$

from which it follows that
$h^{j-1}(F/(\operatorname{Gin}(U)+lF))=(z-1)h^{j}(F/\operatorname{Gin}(U))$
for all $j$.

Consider now a sufficiently general linear form $x\in R$. For all $j$, there
are exact sequences

$0\longrightarrow B^{(j)}\longrightarrow
H^{j-1}_{{\mathfrak{m}}}(F/(U+xF))\longrightarrow
H^{j}_{{\mathfrak{m}}}(F/U)(-1)\longrightarrow
H^{j}_{{\mathfrak{m}}}(F/U)\longrightarrow C^{(j)}\longrightarrow 0$

for some $R$-modules $B^{(j)}$ and $C^{(j)}$, and these imply that
$h^{j-1}(F/(U+xF))=(z-1)h^{j}(F/U)+\operatorname{Hilb}(B^{(j)})+\operatorname{Hilb}(C^{(j)})$
for all $j$. By (3.1) and (3.2), we obtain

$\begin{split}(z-1)h^{j}(F/U)&\leqslant(z-1)h^{j}(F/U)+\operatorname{Hilb}(B^{(j)})+\operatorname{Hilb}(C^{(j)})\\\
&=h^{j-1}(F/(U+xF))=h^{j-1}(F_{[n-1]}/g_{n}(U))\\\ &\leqslant
h^{j-1}(F_{[n-1]}/\operatorname{Gin}(g_{n}(U)))=h^{j-1}(F_{[n-1]}/\operatorname{Gin}(U)_{[n-1]})\\\
&=h^{j-1}(F/(\operatorname{Gin}(U)+lF))=(z-1)h^{j}(F/\operatorname{Gin}(U)).\end{split}$

Thus, from our hypothesis it follows that the above inequalities are
equalities for all $j\geqslant i$, that means that
$\operatorname{Hilb}(B^{(j)})=\operatorname{Hilb}(C^{(j)})=0$, i.e.
$B^{(j)}=C^{(j)}=0$ for $j\geqslant i$. Moreover, since $C^{(i-1)}=B^{(i)}=0$,
it follows that $x$ is regular for all non-zero
$\operatorname{Ext}_{R}^{n-j}(F/U,\omega_{R})$ with $j\geqslant i-1$, which
thus have positive depth.

From the above equalities we also get that
$h^{j}(F_{[n-1]}/g_{n}(U))=h^{j}(F_{[n-1]}/\operatorname{Gin}(g_{n}(U)))$ for
all $j\geqslant i-1$; by induction, this implies that $F/(U+xF)\cong
F_{[n-1]}/g_{n}(U)$ is $(i-1)$-sCM.

The conclusion follows now by a straightforward application of Proposition
3.10. ∎

As a corollary, we immediately obtain Theorem 1.

######  Theorem 3.12.

Let $M$ be a finitely generated graded $R$-module, and let $M\cong F/U$ a free
graded presentation of $M$. Then, $F/U$ is sequentially Cohen-Macaulay if and
only if $h^{i}(F/U)=h^{i}(F/\operatorname{Gin}(U))$ for all $j\geqslant 0$.

One can wonder whether the equality
$h^{i}(F/U)=h^{i}(F/\operatorname{Gin}(U))$ is enough to imply that $F/U$ is
$i$-sCM; however, this is not the case.

######  Example 3.13.

  1. (1)

Consider a graded Cohen-Macaulay $k[x_{1},\dots,x_{n}]$-module $M_{1}$ of
dimension $i$ and a graded non-sequentially Cohen-Macaulay
$k[x_{i+2},\dots,x_{n}]$-module $N$. Let
$M_{2}=N\otimes_{k}k[x_{1},\ldots,x_{i+1}]$ and take $M=M_{1}\oplus M_{2}$.
Then $x_{1},\ldots,x_{i+1}$ is a strictly filter-regular sequence for $M_{2}$,
and it follows from Proposition 2.18 (1) that $M_{2}$ is not sequentially
Cohen-Macaulay. By Corollary 2.14 also $M$ is not sequentially Cohen-Macaulay.
On the other hand, since $\operatorname{depth}(M_{2})>i$ we have that
$H^{i}_{\mathfrak{m}}(M)\cong H^{i}_{\mathfrak{m}}(M_{1})$. If we write
$M_{1}\cong F_{1}/U_{1}$ and $M_{2}\cong F_{2}/U_{2}$, where $F_{1}$ and
$F_{2}$ are graded free $R$-modules and $U_{1},U_{2}$ are graded submodules,
then $M_{1}\oplus M_{2}\cong F/U$ where $F=F_{1}\oplus F_{2}$ and
$U=U_{1}\oplus U_{2}$, and it follows that
$\operatorname{Gin}(U)=\operatorname{Gin}(U_{1})\oplus\operatorname{Gin}(U_{2})$.
Since $M_{1}$ is Cohen-Macaulay, hence sequentially Cohen-Macaulay, and
$\operatorname{depth}(F_{2}/\operatorname{Gin}(U_{2}))=\operatorname{depth}(M_{2})>i$
we therefore conclude by Theorem 3.12 that
$h^{i}(F/U)=h^{i}(F_{1}/U_{1})=h^{i}(F_{1}/\operatorname{Gin}(U_{1}))=h^{i}(F/\operatorname{Gin}(U))$.

  2. (2)

The following is another explicit example of such instance in the ideal case.

Consider the polynomial ring $R=k[x_{1},x_{2},x_{3},x_{4},x_{5},x_{6},x_{7}]$
and the monomial ideal
$I=(x_{1}^{3},x_{1}^{2}x_{2}x_{4},x_{1}x_{5},x_{1}x_{6},x_{2}x_{5},x_{2}x_{6},x_{2}^{2}x_{7}^{2},x_{3}x_{5},x_{3}x_{6},x_{3}x_{7},x_{4}x_{5},x_{4}x_{6},x_{4}x_{7},x_{7}^{3})$.
Then, one can check that $\operatorname{depth}(R/I)=0$ and $\dim(R/I)=3$;
moreover, $h^{j}(R/I)=h^{j}(R/\operatorname{Gin}(I))$ for $j=0,3$, and
$h^{j}(R/I)\neq h^{j}(R/\operatorname{Gin}(I))$ for $j=1,2$. By Theorem 3.11,
this means that $R/I$ is $3$-sCM but not $2$-sCM and, a fortiori, not $1$-sCM.

On the other hand, if $I\operatorname{{}^{lex}}$ is the lexicographic ideal
associated with $I$, then the equality
$h^{i}(R/I)=h^{i}(R/I\operatorname{{}^{lex}})$ ensures the $i$-partial
sequential Cohen-Macaulayness of $R/I$. Notice that this is a stronger
condition, though, since $h^{i}(R/I)\leqslant
h^{i}(R/\operatorname{Gin}(I))\leqslant h^{i}(R/I\operatorname{{}^{lex}})$,
coefficientwise, see [34, Theorems 2.4 and 5.4]. Actually, in [35, Theorem
4.4] the following result is proved.

######  Theorem 3.14.

Let $i$ be a positive integer and $I$ a homogeneous ideal of $R$; then, the
following conditions are equivalent:

  1. (1)

$h^{i}(R/I)=h^{i}(R/I\operatorname{{}^{lex}})$;

  2. (2)

$h^{i}(R/\operatorname{Gin}(I))=h^{i}(R/I\operatorname{{}^{lex}})$;

  3. (3)

$h^{j}(R/I)=h^{j}(R/I\operatorname{{}^{lex}})$ for all $j\geqslant i$.

If any of the above holds, then $I$ is $i$-sCM.

We conclude this section by observing that the conditions in the previous
theorem are still equivalent if we replace $\operatorname{Gin}(I)$ with
$\operatorname{Gin}_{0}(I)$, the zero-generic initial ideal of $I$ introduced
in [13]. We also remark that the equivalence between conditions (2) and (3) is
not true if we replace the Hilbert series of local cohomology modules with
graded Betti numbers, see [32, Theorem 3.1].

##  4\. E-depth

As in the previous section, $k$ will denote an infinite field,
$R=k[x_{1},\ldots,x_{n}]$ a standard graded polynomial ring and
$\mathfrak{m}=(x_{1},\ldots,x_{n})$ its homogeneous maximal ideal. As before,
when $M=0$ we let $\dim(M)=-1$ and $\operatorname{depth}(M)=\infty$. We start
with the main definition of this section.

######  Definition 4.1.

Given a non-zero finitely generated graded ${\mathbb{Z}}$-module $M$ and a
non-negative integer $r$, we say that $M$ satisfies condition $(E_{r})$ if
$\operatorname{depth}(\operatorname{Ext}^{i}_{R}(M,R))\geqslant\min\\{r,n-i\\}$
for all $i\in{\mathbb{Z}}$. We let

$\operatorname{E-depth}(M)=\min\left\\{n,\sup\\{r\in{\mathbb{N}}\mid M\text{
satisfies }(E_{r})\\}\right\\}.$

######  Remark 4.2.

Observe that a non-zero module $M$ is sequentially Cohen-Macaulay if and only
if it satisfies condition $(E_{r})$ for all $r\geqslant 0$, see Theorem 2.13.
It is therefore clear that a sequentially Cohen-Macaulay $R$-module has
$\operatorname{E-depth}$ equal to $n$. The converse is also true, since if
$\operatorname{E-depth}(M)=n$, then
$\operatorname{depth}(\operatorname{Ext}^{i}_{R}(M,R))\geqslant n-i$; as
$\dim(\operatorname{Ext}^{i}_{R}(M,R))\leqslant n-i$ always holds, cf. Remark
3.7, this implies the claim.

######  Lemma 4.3.

Let $M\neq 0$ be a finitely generated ${\mathbb{Z}}$-graded $R$-module with
positive depth and E-depth, and let $\ell$ be a linear form which is a
strictly filter regular for $M$; then, the graded short exact sequence
$0\longrightarrow
M(-1)\stackrel{{\scriptstyle\cdot\ell}}{{\longrightarrow}}M\longrightarrow
M/\ell M\longrightarrow 0$ induces graded short exact sequences

$0\longrightarrow\operatorname{Ext}^{i}_{R}(M,R)\stackrel{{\scriptstyle\cdot\ell}}{{\longrightarrow}}\operatorname{Ext}^{i}_{R}(M,R)(1)\longrightarrow\operatorname{Ext}^{i+1}_{R}(M/\ell
M,R)\longrightarrow 0,\,\,\,\text{ for all }i<n,$

and

$0\longrightarrow H^{i-1}_{\mathfrak{m}}(M/\ell M)\longrightarrow
H^{i}_{\mathfrak{m}}(M)(-1)\stackrel{{\scriptstyle\cdot\ell}}{{\longrightarrow}}H^{i}_{\mathfrak{m}}(M)\longrightarrow
0,\,\,\,\text{ for all }i>0.$

###### Proof.

Consider the induced long exact sequence of $\operatorname{Ext}$ modules

$\textstyle{\ldots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\operatorname{Ext}^{i}_{R}(M,R)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cdot\ell}$$\textstyle{\operatorname{Ext}^{i}_{R}(M(-1),R)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\operatorname{Ext}^{i+1}_{R}(M/\ell
M,R)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\ldots\,\,,}$

and observe that
$\operatorname{Ext}^{i}_{R}(M(-1),R)\cong\operatorname{Ext}^{i}_{R}(M,R)(1)$.
Our assumption that $\operatorname{E-depth}(M)>0$ guarantees that
$\operatorname{depth}(\operatorname{Ext}^{i}_{R}(M,R))>0$ for all $i<n$. Since
$\ell$ is strictly filter regular for $M$, the multiplication by $\ell$ is
injective on $\operatorname{Ext}^{i}_{R}(M,R)$ for all $i<n$, and the long
exact sequence breaks into short exact sequences, as claimed, and the graded
short exact sequences of local cohomology modules are obtained by graded Local
Duality. ∎

######  Proposition 4.4.

Let $M$ $\neq 0$ be a finitely generated ${\mathbb{Z}}$-graded $R$-module.

  1. (1)

$\operatorname{E-depth}(M)=\operatorname{E-depth}(M/H^{0}_{\mathfrak{m}}(M))$.

  2. (2)

Assume that $\operatorname{E-depth}(M)>0$ and $\ell$ is homogeneous strictly
filter regular for $M$; then, either
$\operatorname{E-depth}(M/(H^{0}_{\mathfrak{m}}(M)+\ell
M))=\operatorname{E-depth}(M)=n$, or
$\operatorname{E-depth}(M/(H^{0}_{\mathfrak{m}}(M)+\ell
M))=\operatorname{E-depth}(M)-1$.

###### Proof.

We first prove (1); clearly, we may assume that $H^{0}_{\mathfrak{m}}(M)\neq
0$. Applying the functor $\operatorname{Hom}_{R}(-,R)$ to the short exact
sequence $0\longrightarrow H^{0}_{\mathfrak{m}}(M)\longrightarrow
M\longrightarrow M/H^{0}_{\mathfrak{m}}(M)\longrightarrow 0$ we get a long
exact sequence of $\operatorname{Ext}$ modules, from which we obtain that
$\operatorname{Ext}^{n-i}_{R}(M,R)\cong\operatorname{Ext}^{n-i}_{R}(M/H^{0}_{\mathfrak{m}}(M),R)$
for all $i\neq 0$ because $H^{0}_{\mathfrak{m}}(M)$ has finite length; the
first statement is now clear, since
$\operatorname{depth}(\operatorname{Ext}^{n}_{R}(H^{0}_{\mathfrak{m}}(M),R))=0$
.

For the proof of (2), let $N=M/H^{0}_{\mathfrak{m}}(M)$; by (1), $M$ and $N$
have same E-depth and
$\operatorname{Ext}^{i}_{R}(M,R)\cong\operatorname{Ext}^{i}_{R}(N,R)$ for all
$i\neq n$. First assume that $M$ is sequentially Cohen-Macaulay, i.e.,
$\operatorname{E-depth}(M)=n$. Then, $N$ is sequentially Cohen-Macaulay by
Remark 4.2 or by Corollary 2.8, and since $\ell$ is filter regular for $M$ by
Remark 2.17, $\ell$ is $N$-regular. It follows from Proposition 2.18 that
$M/(H^{0}_{\mathfrak{m}}(M)+\ell M)$ is sequentially Cohen-Macaulay and,
therefore, has E-depth equal to $n$.

Now assume that $0<\operatorname{E-depth}(N)=\operatorname{E-depth}(M)=r<n$;
from an application of Lemma 4.3 we obtain short exact sequences
$0\to\operatorname{Ext}^{i}_{R}(N,R)\stackrel{{\scriptstyle\cdot\ell}}{{\to}}\operatorname{Ext}^{i}_{R}(N,R)(1)\to\operatorname{Ext}^{i+1}_{R}(N/\ell
N,R)\to 0$ and, therefore, $\operatorname{Ext}^{i+1}_{R}(N/\ell
N,R)\cong\operatorname{Ext}^{i}_{R}(M,R)(1)/\ell\operatorname{Ext}^{i}_{R}(M,R)$
for all $i<n$. It follows that
$\operatorname{depth}(\operatorname{Ext}^{i+1}_{R}(N/\ell
N,R))=\operatorname{depth}(\operatorname{Ext}^{i}_{R}(M,R))-1$ for all $i<n$,
which clearly implies $\operatorname{E-depth}(N/\ell
N)=\operatorname{E-depth}(M)-1$, as desired. ∎

We now introduce a special grading on a polynomial ring which refines the
standard grading, and that can be further refined to the monomial
${\mathbb{Z}}^{n}$-grading.

######  Definition 4.5.

Let $r$ be a positive integer, $A$ be a ${\mathbb{Z}}$-graded ring and
$S=A[y_{1},\ldots,y_{r}]$ a polynomial ring over $A$. For
$i\in\\{0,\ldots,r\\}$ let $\eta_{i}\in{\mathbb{Z}}^{r+1}$ be the vector whose
$(i+1)$-st entry is $1$, and all other entries equal $0$. We consider $S$ as a
${\mathbb{Z}}\times{\mathbb{Z}}^{r}$-graded ring by letting

$\deg_{S}(a)=\deg_{A}(a)\cdot\eta_{0},\,\,\text{for all }\,a\in A\,\,\,\text{
and }\,\deg_{S}(y_{i})=\eta_{i},\,\,\text{ for }i\in\\{1,\ldots,r\\}.$

By means of the previous definition, we may consider $R=k[x_{1},\ldots,x_{n}]$
as a ${\mathbb{Z}}\times{\mathbb{Z}}^{r}$-graded ring for any $0\leqslant
r\leqslant n-1$ by letting $\deg_{R}(x_{i})=\eta_{0}$ for all $1\leqslant
i\leqslant n-r$ and $\deg_{R}(x_{i})=\eta_{i}$ for all $n-r+1\leqslant
i\leqslant n$. Observe that an element $f\in R$ is graded with respect to such
grading if and only if $f$ can be written as $f=\overline{f}\cdot u$, where
$\overline{f}\in k[x_{1},\ldots,x_{n-r}]$ is homogeneous with respect to the
standard grading, and $u$ is a monomial in $k[x_{n-r+1},\ldots,x_{n}]$. In
particular, when $r=0$ this is just the standard grading on $R$, while for
$r=n-1$ it coincides with the monomial ${\mathbb{Z}}^{n}$-grading.

######  Remark 4.6.

We can extend in a natural way such a grading to free $R$-modules $F$ with a
basis by assigning degrees to the elements of the given basis. Accordingly,
any $R$-module $M$ is ${\mathbb{Z}}\times{\mathbb{Z}}^{r}$-graded if and only
if $M\cong F/U$, where $F$ is a free
${\mathbb{Z}}\times{\mathbb{Z}}^{r}$-graded $R$-module and $U$ is a
${\mathbb{Z}}\times{\mathbb{Z}}^{r}$-graded submodule of $F$.

We see next that the grading just introduced is very relevant to the purpose
of estimating the E-depth of a module; the following can be seen as a
refinement of Proposition 1.8.

######  Proposition 4.7.

Let $R=k[x_{1},\ldots,x_{n}]$, and $M$ be a finitely generated
${\mathbb{Z}}\times{\mathbb{Z}}^{r}$-graded $R$-module such that
$x_{n},\ldots,x_{n-r+1}$ is a filter regular sequence for $M$. Then,
$x_{n},\ldots,x_{n-r+1}$ is a strictly filter regular sequence for $M$, and
$\operatorname{E-depth}(M)\geqslant r$.

###### Proof.

Write $M$ as $M=F/U$, where $F$ is a finitely generated
${\mathbb{Z}}\times{\mathbb{Z}}^{r}$-graded free $R$-module, and $U$ is a
${\mathbb{Z}}\times{\mathbb{Z}}^{r}$-graded submodule of $F$. Since $U$ is
${\mathbb{Z}}\times{\mathbb{Z}}^{r}$-graded, $x_{n}$ is
${\mathbb{Z}}\times{\mathbb{Z}}^{r}$-homogeneous and filter regular,
$U\operatorname{{}^{\rm sat}}=U:_{F}x_{n}^{\infty}$ is also
${\mathbb{Z}}\times{\mathbb{Z}}^{r}$ graded and, accordingly
$M/H^{0}_{\mathfrak{m}}(M)\cong F/U\operatorname{{}^{\rm sat}}$ is too. If
$F/U\operatorname{{}^{\rm sat}}=0$, then $M$ has dimension zero and, thus, is
sequentially Cohen-Macaulay; then $\operatorname{E-depth}(M)=n\geqslant t$,
and every sequence of non-zero elements of $(x_{1},\ldots,x_{n})$ is a
strictly filter regular sequence for $M$. In particular,
$x_{n},\ldots,x_{n-r+1}$ is a strictly filter regular sequence for $M$.

Now suppose that $F/U\operatorname{{}^{\rm sat}}\neq 0$. By assumption,
$x_{n}$ is filter regular for $M$, and thus regular for
$F/U\operatorname{{}^{\rm sat}}$. Since the latter is
${\mathbb{Z}}\times{\mathbb{Z}}^{r}$-graded, we have that
$F/U\operatorname{{}^{\rm
sat}}\cong\overline{F}/\overline{U}\otimes_{k}k[x_{n}]$ for some
${\mathbb{Z}}\times{\mathbb{Z}}^{r-1}$-graded
$\overline{R}=k[x_{1},\ldots,x_{n-1}]$-module $\overline{F}$, and some
${\mathbb{Z}}\times{\mathbb{Z}}^{r-1}$-graded submodule $\overline{U}$ of
$\overline{F}$ such that $\overline{F}/\overline{U}$ can be identified with
the hyperplane section $F/U\operatorname{{}^{\rm sat}}\otimes_{R}R/(x_{n})$.
Thus, for $i<n$ we have

$\operatorname{Ext}^{i}_{R}(M,R)\cong\operatorname{Ext}^{i}_{R}(M/H^{0}_{\mathfrak{m}}(M),R)\cong\operatorname{Ext}^{i}_{R}(F/U\operatorname{{}^{\rm
sat}},R)\cong\operatorname{Ext}_{\overline{R}}^{i}(\overline{F}/\overline{U},\overline{R})\otimes_{k}k[x_{n}].$

It follows that $x_{n}$ is a non-zero divisor on
$\operatorname{Ext}^{i}_{R}(M,R)$ for all $i<n$, and thus
$\operatorname{E-depth}(M)>0$. Since $\operatorname{Ext}^{n}_{R}(F/U,R)$ has
finite length it also follows that $x_{n}$ is a strictly filter regular
element for $M$. Now we can consider $\overline{F}/\overline{U}$, and an
iteration of this argument will imply the desired conclusion. ∎

We now introduce a weight order on $R$. Given integers $0\leqslant r\leqslant
n$, consider the following $r\times n$ matrix

$\Omega_{r,n}=\begin{bmatrix}0&0&\ldots&0&0&0&0&\ldots&0&-1\\\
0&0&\ldots&0&0&0&0&\ldots&-1&0\\\
\vdots&\vdots&&\vdots&\vdots&\vdots&\vdots&\iddots&\vdots&\vdots\\\
0&0&\ldots&0&0&0&-1&\ldots&0&0\\\ 0&0&\ldots&0&0&-1&0&\ldots&0&0\\\
0&0&\ldots&0&-1&0&0&\ldots&0&0\end{bmatrix},$

and let $\omega_{i}$ be its $i$-th row; then, this induces a “partial” revlex
order $\operatorname{rev}_{r}$ on $R$ by declaring that a monomial
$x^{\underline{a}}=x_{1}^{a_{1}}\cdots x_{n}^{a_{n}}$ is greater than a
monomial $x^{\underline{b}}=x_{1}^{b_{1}}\cdots x_{n}^{b_{n}}$ if and only if
there exists $1\leqslant j\leqslant r$ such that
$\omega_{i}\cdot\underline{a}=\omega_{i}\cdot\underline{b}$ for all
$i\leqslant j$ and
$\omega_{j+1}\cdot\underline{a}>\omega_{j+1}\cdot\underline{b}$.

######  Remark 4.8.

Observe that $\operatorname{rev}_{n}$ coincides with the usual revlex order on
$R=k[x_{1},\ldots,x_{n}]$.

Give a graded free $R$-module $F$ with basis $\\{e_{1},\ldots,e_{s}\\}$, we
can write any $f\in F$ uniquely as a finite sum $f=\sum u_{j}e_{i_{j}}$, where
$u_{j}$ are monomials and we assume the sum has minimal support. We let the
initial form ${\rm in}_{\operatorname{rev}_{r}}(f)$ of $f$ written as above be
the sum of those $u_{j}e_{i_{j}}$ for which $u_{j}$ is maximal with respect to
the order $\operatorname{rev}_{r}$ introduced above.

######  Definition 4.9.

Let $F$ be a finitely generated graded free $R$-module, and $U$ be a graded
submodule; let also $M=F/U$. We say that the $r$-partial general initial
submodule of $U$ satisfies a given property $\mathcal{P}$ if there exists a
non-empty Zariski open set $\mathcal{L}$ of $r$-uples of linear forms such
that $F/{\rm in}_{\operatorname{rev}_{r}}(g_{\ell}(U))$ satisfies
$\mathcal{P}$ for any $\ell=(\ell_{n-r+1},\ldots,\ell_{n})\in\mathcal{L}$,
where $g_{\ell}$ is the automorphism on $F$ induced by the change of
coordinates of $R$ which sends $\ell_{i}\mapsto x_{i}$ and fixes the other
variables.

With some abuse of notation, we denote any partial initial submodule ${\rm
in}_{\operatorname{rev}_{r}}(g_{\ell}(U))$ which satisfies $\mathcal{P}$ the
$r$-partial general submodule of $U$, and denote it by
$\operatorname{gin}_{r}(U)$.

######  Proposition 4.10.

Let $F$ be a finitely generated graded free $R$-module, and $U\subseteq F$ be
a graded submodule. For every $0\leqslant r\leqslant n$ we have that
$\operatorname{E-depth}(F/\operatorname{gin}_{r}(U))\geqslant r$.

###### Proof.

For a sufficiently general change of coordinates we have that
$x_{n},\ldots,x_{n-r+1}$ form a filter regular sequence for $F/(g_{\ell}(U))$.
Since ${\rm in}_{\operatorname{rev}_{r}}(U):_{F}x_{n}={\rm
in}_{\operatorname{rev}_{r}}(U:_{F}x_{n})$, see [21, 15.7], we have that
$x_{n},\ldots,x_{n-r+1}$ also form a filter regular sequence for $F/{\rm
in}_{\operatorname{rev}_{r}}(g_{\ell}(U))$, and hence for
$F/\operatorname{gin}_{r}(U)$. By construction, the module
$F/\operatorname{gin}_{r}(U)$ is ${\mathbb{Z}}\times{\mathbb{Z}}^{r}$-graded,
and the claim now follows from Proposition 4.7. ∎

######  Theorem 4.11.

Let $F$ be a finitely generated graded free $R$-module, and $U\subseteq F$ be
a graded submodule. For every $0\leqslant r\leqslant n$ we have that
$\operatorname{E-depth}(F/U)\geqslant r$ if and only if
$h^{i}(F/U)=h^{i}(F/\operatorname{gin}_{r}(U))$ for all $i\in{\mathbb{N}}$.

###### Proof.

After performing a sufficiently general change of coordinates, we may assume
that $V={\rm in}_{\operatorname{rev}_{r}}(U)$ has the same properties as
$\operatorname{gin}_{\operatorname{rev}_{r}}(U)$, and that
$x_{n},\ldots,x_{n-r+1}$ form a strictly filter regular sequence for $F/U$. By
using again that ${\rm in}_{\operatorname{rev}_{r}}(U:_{F}x_{n})={\rm
in}_{\operatorname{rev}_{r}}(U):_{F}x_{n}$, and because strictly filter
regular sequences are filter regular by Remark 2.17, we have that
$x_{n},\ldots,x_{n-r+1}$ form a filter regular sequence for $F/V$. Since $V$
is ${\mathbb{Z}}\times{\mathbb{Z}}^{r}$-graded, it follows from Proposition
4.7 that $x_{n},\ldots,x_{n-r+1}$ is a strictly filter regular sequence for
$F/V$. By [21, 15.7] we also have that

$V\operatorname{{}^{\rm sat}}={\rm
in}_{\operatorname{rev}_{r}}(U):_{F}x_{n}^{\infty}={\rm
in}_{\operatorname{rev}_{r}}(U:_{F}x_{n}^{\infty})={\rm
in}_{\operatorname{rev}_{r}}(U\operatorname{{}^{\rm sat}}),$

and

$V\operatorname{{}^{\rm sat}}+x_{n}F={\rm
in}_{\operatorname{rev}_{r}}(U\operatorname{{}^{\rm sat}})+x_{n}F={\rm
in}_{\operatorname{rev}_{r}}(U\operatorname{{}^{\rm sat}}+x_{n}F).$

Viewing $F/(U\operatorname{{}^{\rm sat}}+x_{n}F)$ as a quotient of a free
$\overline{S}=k[x_{1},\ldots,x_{n-1}]$-module $\overline{F}$ by a graded
submodule $\overline{U}$, we see that $V\operatorname{{}^{\rm sat}}+x_{n}F$
can be identified with a submodule $\overline{V}\subseteq\overline{F}$, with
$\overline{V}={\rm in}_{\operatorname{rev}_{r-1}}(\overline{U})$ and
$\overline{V}$ has the same properties of
$\operatorname{gin}_{\operatorname{rev}_{t-1}}(\overline{U})$, see [12, Lemma
3.4] and the proof of [12, Theorem 3.6] for more details.

First assume that $\operatorname{E-depth}(F/U)\geqslant r$, and we prove
equality for the Hilbert series of $H^{i}_{\mathfrak{m}}(F/U)$ and
$H^{i}_{\mathfrak{m}}(F/V)$ for all $i\in{\mathbb{N}}$ by induction on
$r\geqslant 0$. The base case is trivial since ${\rm
in}_{\operatorname{rev}_{0}}(U)=U$. By Proposition 4.4 we have that
$\operatorname{E-depth}(F/U\operatorname{{}^{\rm sat}})\geqslant r>0$ and,
accordingly, $\operatorname{E-depth}(F/(U\operatorname{{}^{\rm
sat}}+x_{n}F))\geqslant r-1$. By induction, we have that
$H^{i}_{\mathfrak{m}}(F/(U\operatorname{{}^{\rm
sat}}+x_{n}F))=H^{i}_{\mathfrak{m}}(F/(V\operatorname{{}^{\rm sat}}+x_{n}F))$.
Since $x_{n}$ is strictly filter regular for both $F/U\operatorname{{}^{\rm
sat}}$ and $F/V\operatorname{{}^{\rm sat}}$, and these have positive
$\operatorname{E-depth}$, by Lemma 4.3 we have graded short exact sequences
for all $i>0$:

(4.1)
$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H^{i-1}_{\mathfrak{m}}(F/(U\operatorname{{}^{\rm
sat}}+x_{n}F))\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H^{i}_{\mathfrak{m}}(F/U\operatorname{{}^{\rm
sat}})(-1)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cdot
x_{n}}$$\textstyle{H^{i}_{\mathfrak{m}}(F/U\operatorname{{}^{\rm
sat}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0,}$

and

(4.2)
$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H^{i-1}_{\mathfrak{m}}(F/(V\operatorname{{}^{\rm
sat}}+x_{n}F))\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H^{i}_{\mathfrak{m}}(F/V\operatorname{{}^{\rm
sat}})(-1)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cdot
x_{n}}$$\textstyle{H^{i}_{\mathfrak{m}}(F/V\operatorname{{}^{\rm
sat}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

Since the first modules in both sequences have the same Hilbert series, a
straightforward computation shows that $h^{i}(F/U\operatorname{{}^{\rm
sat}})=h^{i}(F/V\operatorname{{}^{\rm sat}})$. Since
$H^{i}_{\mathfrak{m}}(F/U\operatorname{{}^{\rm sat}})\cong
H^{i}_{\mathfrak{m}}(F/U)$ and $H^{i}_{\mathfrak{m}}(F/V\operatorname{{}^{\rm
sat}})\cong H^{i}_{\mathfrak{m}}(F/V)$ for all $i>0$, the equality between
Hilbert series is proved for $i>0$. Finally, since
$\operatorname{Hilb}(F/U)=\operatorname{Hilb}(F/V)$ and
$\operatorname{Hilb}(F/U\operatorname{{}^{\rm
sat}})=\operatorname{Hilb}(F/V\operatorname{{}^{\rm sat}})$, we have that

$h^{0}(F/U)=\operatorname{Hilb}(U\operatorname{{}^{\rm
sat}}/U)=\operatorname{Hilb}(V\operatorname{{}^{\rm sat}}/V)=h^{0}(F/V).$

Conversely, assume that the local cohomology modules of $F/U$ and $F/V$ have
the same Hilbert series. If $r=0$ there is nothing to show, otherwise it is
enough to prove that $\operatorname{E-depth}(F/U\operatorname{{}^{\rm
sat}})\geqslant r$.

Since $x_{n}$ is a filter regular element for $F/V$, it is strictly filter
regular for $F/V$, being the latter
${\mathbb{Z}}\times{\mathbb{Z}}^{r}$-graded, and hence it is filter regular
for $F/V\operatorname{{}^{\rm sat}}$. Thus, the sequence (4.2) is exact for
$i>0$. On the other hand, suppose by way of contradiction that the sequence
(4.1) is not exact for some $i\in{\mathbb{Z}}$, and let $i$ be the smallest
such integer, so that we still have an exact sequence

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H^{i-1}_{\mathfrak{m}}(F/(U\operatorname{{}^{\rm
sat}}+x_{n}F))\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H^{i}_{\mathfrak{m}}(F/U\operatorname{{}^{\rm
sat}})(-1)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cdot
x_{n}}$$\textstyle{H^{i}_{\mathfrak{m}}(F/U\operatorname{{}^{\rm sat}}).}$

Counting dimensions in such a sequence, and comparing them to those obtained
from (4.2) we obtain that $h^{i-1}(F/(U\operatorname{{}^{\rm
sat}}+x_{n}F))_{j}>h^{i-1}(F/(V\operatorname{{}^{\rm sat}}+x_{n}F))_{j}$.
However, by upper semi-continuity we know that the reverse inequality always
holds, which gives a contradiction. Thus, the sequence (4.1) is exact for all
$i\in{\mathbb{N}}$. By induction we have that
$\operatorname{E-depth}(F/(U\operatorname{{}^{\rm sat}}+x_{n}F))\geqslant
r-1$. If $\operatorname{E-depth}(F/(U\operatorname{{}^{\rm sat}}+x_{n}F))=n$,
by Proposition 4.4 we see that $\operatorname{E-depth}(F/U)=n\geqslant r$, as
desired. Otherwise, again by Proposition 4.4 we have that
$\operatorname{E-depth}(F/U)=\operatorname{E-depth}(F/(U\operatorname{{}^{\rm
sat}}+x_{n}F))+1\geqslant r$. ∎

Recalling Remark 4.8, Theorem 4.11 can be viewed as another extension of
Theorem 1.

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