# Dao numbers and
the asymptotic Behaviour of fullness

Antonino Ficarra  Antonino Ficarra, Department of mathematics and computer
sciences, physics and earth sciences, University of Messina, Viale Ferdinando
Stagno d’Alcontres 31, 98166 Messina, Italy<EMAIL_ADDRESS>
###### Abstract.

In the present paper, we study the Dao numbers
$\mathfrak{d}_{1}(I),\mathfrak{d}_{2}(I)$ and $\mathfrak{d}_{3}(I)$ of an
ideal $I$ of a Noetherian local ring $(R,\mathfrak{m},K)$ or a standard graded
Noetherian $K$-algebra. They are defined as the smallest $\ell\geq 0$ such
that $I\mathfrak{m}^{k}$ is $\mathfrak{m}$-full, full, weakly
$\mathfrak{m}$-full, respectively, for all $k\geq\ell$. We provide general
bounds for the Dao numbers in terms of the Castelnuovo–Mumford regularity of
certain modules over the Rees algebra $\mathcal{R}(\mathfrak{m})$. If $R$ is a
Koszul algebra, we prove that the Dao numbers are less or equal to
$\textup{reg}_{\text{gr}_{\mathfrak{m}}(R)}\text{gr}_{\mathfrak{m}}(I)$, where
$\text{gr}_{\mathfrak{m}}(I)$ is the associated graded module of $I$. Finally,
for monomial ideals, we combinatorially bound the Dao numbers in terms of
asymptotic linear quotients and bounding multidegrees.

###### Key words and phrases:

colon ideals, ${\mathfrak{m}}$-full ideals, Koszul algebras.

###### 2020 Mathematics Subject Classification:

Primary 13C15, 05E40, 05C70

## Introduction

Let $(R,{\mathfrak{m}},K)$ be either a Noetherian local ring or a standard
graded Noetherian $K$-algebra with unique graded maximal ideal
${\mathfrak{m}}$. Hereafter, we will always assume that $K$ is an infinite
field and that $\operatorname{depth}R>0$. These assumptions will be explained
in a moment. Let $I\subset R$ be an ideal, we assume that $I$ is homogeneous
if $R$ is a $K$-algebra. The ideal $I$ is called

  1. (a)

${\mathfrak{m}}$-full if $(I{\mathfrak{m}}:x)=I$ for a generic element
$x\in{\mathfrak{m}}-{\mathfrak{m}}^{2}$,

  2. (b)

full if $(I:x)=(I:{\mathfrak{m}})$ for a generic element
$x\in{\mathfrak{m}}-{\mathfrak{m}}^{2}$,

  3. (c)

weakly ${\mathfrak{m}}$-full if $(I{\mathfrak{m}}:{\mathfrak{m}})=I$.

Here, that $x\in{\mathfrak{m}}-{\mathfrak{m}}^{2}$ is generic means that $x$
belongs to some non-empty open set of the Zariski topology of $R$.

Let $P$ be one of the properties $\\{{\mathfrak{m}}\textit{-full,\ full,\
weakly}\ {\mathfrak{m}}\textit{-full}\\}$ and let $I,J\subset R$ be two
ideals. In [10, Theorem 3.1], Hailong Dao proved that $J$ has the property $P$
if and only if $J+I{\mathfrak{m}}^{k}$ has the property $P$ for all $k\gg 0$
(the result is stated in the local case, but the proof carries over in the
graded case). In particular, there exists an ideal $I\subset R$ for which
$I{\mathfrak{m}}^{k}$ is (weakly) ${\mathfrak{m}}$-full for all $k\gg 0$ if
and only if $\operatorname{depth}R>0$ [10, Corollary 3.2]. Thus, under our
assumptions, for all $k\gg 0$, $I{\mathfrak{m}}^{k}$ has any of the properties
$P$ considered above. Based on this fact, Dao introduced the following
numerical invariants [10, Definition 3.1], which we call the Dao numbers of
$I$:

$\displaystyle\mathfrak{d}_{1}(I)\ $ $\displaystyle=\ \min\\{\ell\geq 0\ :\
I{\mathfrak{m}}^{k}\ \textup{is}\ {\mathfrak{m}}\textup{-full for all}\
k\geq\ell\\},$ $\displaystyle\mathfrak{d}_{2}(I)\ $ $\displaystyle=\
\min\\{\ell\geq 0\ :\ I{\mathfrak{m}}^{k}\ \textup{is}\ \textup{full for all}\
k\geq\ell\\},$ $\displaystyle\mathfrak{d}_{3}(I)\ $ $\displaystyle=\
\min\\{\ell\geq 0\ :\ I{\mathfrak{m}}^{k}\ \textup{is weakly}\
{\mathfrak{m}}\textup{-full for all}\ k\geq\ell\\}.$

It is shown in [22, Proposition 2.2] that

$\mathfrak{d}_{2}(I)\ \leq\ \mathfrak{d}_{3}(I)\ =\ \mathfrak{d}_{1}(I)$

if $K$ is infinite and $\operatorname{depth}R>0$. This fact justifies these
recurring assumptions we imposed in the beginning. In [10], Dao raised the
problem of finding good bounds for the Dao numbers. For reduction ideals of
the maximal ideal in a local ring, this question has been answered by Miranda-
Neto and Queiroz [22].

The concept of ${\mathfrak{m}}$-full ideals was firstly introduced by David
Rees in unpublished work, and later fully developed by Junzo Watanabe [28, 29,
30]. Recently, fullness of ideals gained the attention of several reaserchers
[5, 10, 11, 15, 16, 17, 21, 24]. Surprisingly, these concepts are also related
to the classical Zariski-Lipman conjecture (about derivations and smoothness)
in the open case of surfaces, see [21, Section 4].

It is straightforward to see that a ${\mathfrak{m}}$-full ideal is weakly
${\mathfrak{m}}$-full. A reduction of ${\mathfrak{m}}$ is an ideal $I$ such
that $I{\mathfrak{m}}^{k}={\mathfrak{m}}^{k+1}$ for some $k$. If $R$ is
regular (with infinite residue field $R/{\mathfrak{m}}$) and $I$ is a
reduction of ${\mathfrak{m}}$, then $I$ is ${\mathfrak{m}}$-full [22,
Corollary 3.11]. If $R$ is a normal domain, then integrally closed ideals are
${\mathfrak{m}}$-full [14, Theorem (2.4)]. If $R=K[x_{1},\dots,x_{n}]$ is the
polynomial ring with coefficients over an infinite field $K$, then any
componentwise linear ideal is ${\mathfrak{m}}$-full [8] (see also [15,
Proposition 18]). We extend this result to any Koszul algebra (Corollary 2.3).
In [11] Burch ideals are introduced. An ideal $I$ is called Burch if
$(I{\mathfrak{m}}:{\mathfrak{m}})\neq(I:{\mathfrak{m}})$. If
$\operatorname{depth}R/I=0$ and $I$ is weakly ${\mathfrak{m}}$-full, then $I$
is Burch [11, Corollary 2.4].

The next scheme summarizes the relationship between the concepts discussed.

$\begin{matrix}{\textup{Componentwise
linear}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\[6.0pt]
\,\,\,\,\,\,\rotatebox{90.0}{$\Longleftarrow$}\,\,\begin{subarray}{c}\big{(}\\\
\phantom{ll}\\\
\phantom{..}\end{subarray}\begin{subarray}{c}\small\textup{if}\ R\ \textup{is
a}\\\ \small\textup{Koszul algebra}\\\ \phantom{aaaa}\\\
\phantom{..}\end{subarray}\begin{subarray}{c}\big{)}\\\ \phantom{ll}\\\
\phantom{..}\end{subarray}\\\\[-10.0pt] \textup{Reduction of}\
{\mathfrak{m}}\,\,\,\xRightarrow{\begin{subarray}{c}(\small R\
\textup{regular})\\\
\phantom{...}\end{subarray}}\,\,\,{\mathfrak{m}\textup{-full}}\,\,\Longrightarrow\,\,\textup{Weakly}\
\mathfrak{m}\textup{-full}\,\,\xRightarrow{\begin{subarray}{c}(\operatorname{depth}R/I=0)\\\
\phantom{aa}\end{subarray}}\,\,\textup{Burch}\\\\[7.0pt]
\,\,\,\,\\!\,\,\,\,\rotatebox{90.0}{$\Longrightarrow$}\,\,\begin{subarray}{c}\big{(}\\\
\phantom{l}\\\ \phantom{.}\end{subarray}\begin{subarray}{c}{\small\textup{if}\
R\ \textup{is a}}\\\ \small\textup{normal domain}\\\\[6.0pt]
\phantom{aaa}\end{subarray}\begin{subarray}{c}\big{)}\\\ \phantom{l}\\\
\phantom{.}\end{subarray}\\\\[-3.0pt] {\textup{Integrally
closed}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{matrix}$

In the present paper, we determine general bounds for the Dao numbers in both
the local and graded settings. Moreover, we will consider Koszul algebras and
monomial ideals, and provide more specific bounds for the Dao numbers in such
cases.

The paper proceeds as follows. In Section 1 we address Dao’s question and we
bound the Dao numbers of any ideal $I$. For this aim, we introduce the Dao
module $\mathfrak{D}_{\mathfrak{m}}(I)$ of $I$ which is defined as
$\bigoplus_{k\geq
0}(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})/(I{\mathfrak{m}}^{k})$ and has the
structure of a module over the Rees algebra
$\mathcal{R}({\mathfrak{m}})=\bigoplus_{k\geq 0}{\mathfrak{m}}^{k}$ of
${\mathfrak{m}}$. The $k$th component of this module is zero if and only if
$I{\mathfrak{m}}^{k}$ is weakly ${\mathfrak{m}}$-full. Since
$\mathfrak{D}_{\mathfrak{m}}(I)_{k}=0$ for all
$k\geq\mathfrak{d}_{3}(I)=\mathfrak{d}_{1}(I)$, it follows that
$\mathfrak{D}_{\mathfrak{m}}(I)$ has finite length, and thus is a finitely
generated $\mathcal{R}({\mathfrak{m}})$-module. In Corollary 1.2 we note that
if $I$ is not weakly ${\mathfrak{m}}$-full, then
$\mathfrak{d}_{1}(I)=\mathfrak{d}_{3}(I)=\operatorname{reg}_{\mathcal{R}_{\mathfrak{m}}(I)}\mathfrak{D}_{\mathfrak{m}}(I)+1$.
In our main Theorem 1.1 we prove that:

$\mathfrak{d}_{2}(I)\leq\mathfrak{d}_{3}(I)=\mathfrak{d}_{1}(I)\ \leq\
\max\\{\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}},I),\,\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}},I):_{\mathcal{R}(R)}{\mathfrak{m}}\\}.$

Here $\mathcal{R}({\mathfrak{m}},I)=\bigoplus_{k\geq 0}I{\mathfrak{m}}^{k}$ is
just the extension of the ideal $I\subset R$ in the ring
$\mathcal{R}({\mathfrak{m}})$, and
$\mathcal{R}({\mathfrak{m}},I):_{\mathcal{R}(R)}{\mathfrak{m}}=\bigoplus_{k\geq
0}(I{\mathfrak{m}}^{k}:{\mathfrak{m}})$. Under the more restrictive assumption
that $\operatorname{depth}\operatorname{gr}_{\mathfrak{m}}(I)>0$ (here
$\operatorname{gr}_{\mathfrak{m}}(I)=\bigoplus_{k\geq
0}(I{\mathfrak{m}}^{k}/I{\mathfrak{m}}^{k+1})$ is the associated graded module
of $I$), or when $R$ is regular (Theorems 1.3 and Corollary 1.4) then we even
have

$\mathfrak{d}_{2}(I)\leq\mathfrak{d}_{3}(I)=\mathfrak{d}_{1}(I)\ \leq\
\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}},I).$

This bounds holds because $\mathfrak{D}_{\mathfrak{m}}(I)$ is equal to the
socle
$\textup{Soc}(\mathcal{R}({\mathfrak{m}})/\mathcal{R}({\mathfrak{m}},I))$
under the above assumptions (Corollary 1.5). Due to these results, we could
expect that:

######  Conjecture 0.1.

For all ideals $I\subset R$:
$\mathfrak{d}_{2}(I)\leq\mathfrak{d}_{3}(I)=\mathfrak{d}_{1}(I)\ \leq\
\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}},I)$.

In Corollary 1.7 we reobtain the nice result [22, Corollary 3.11], due to
Miranda Neto and Queiroz, which says that $\mathfrak{d}_{i}(I)=0$ for all $i$
if $I$ is a reduction of the maximal ideal ${\mathfrak{m}}$ and $R$ is
regular.

In Section 2, we consider the more tame situation of a Koszul algebra. A
standard graded $K$-algebra $R$ is Koszul if $\operatorname{reg}_{R}K=0$. A
homogeneous ideal $I$ of a Koszul algebra $R$ is called componentwise linear
if $I_{\langle j\rangle}$ has a linear resolution for all $j$. It turns out
that componentwise linear ideals are full, ${\mathfrak{m}}$-full and weakly
${\mathfrak{m}}$-full (Corollary 2.3). In the polynomial ring case,
componentwise linear ideals coincide with the so-called completely
${\mathfrak{m}}$-full ideals [16]. So the above result is not surprising. To
the best of our knowledge, the concept of completely ${\mathfrak{m}}$-full
ideal has not been defined yet in a Koszul algebra. In Theorem 2.5, we notice
that for all
$k\geq\operatorname{reg}_{\operatorname{gr}_{\mathfrak{m}}(R)}\operatorname{gr}_{\mathfrak{m}}(I)$,
$I{\mathfrak{m}}^{k}$ is componentwise linear. Hence, the number
$\operatorname{reg}_{\operatorname{gr}_{\mathfrak{m}}(R)}\operatorname{gr}_{\mathfrak{m}}(I)$
is an upper bound for the Dao numbers of $I$. If $R$ is a strongly Koszul
algebra, any monomial ideal $I\subset R$ with linear quotients order
$\mathcal{O}:u_{1}<\dots<u_{m}$ with $\deg(u_{1})\leq\dots\leq\deg(u_{m})$ is
componentwise linear (Proposition 2.8). If $R$ is the polynomial ring
$K[x_{1},\dots,x_{n}]$, then by the Björner-Wachs rearrangement lemma [25,
Lemma 2.1] (see also [4]) the condition $\deg(u_{1})\leq\dots\leq\deg(u_{m})$
can be omitted. It is not clear whether this is also the case for any strongly
Koszul algebra.

In the final Section 3, we consider monomial ideals in the standard graded
polynomial ring $S=K[x_{1},\dots,x_{n}]$. Let $I\subset S$ be a monomial ideal
with minimal monomial generating set $\mathcal{G}(I)$. It was noted in [13,
Theorem 1.5] (Proposition 3.2) that $I{\mathfrak{m}}^{k}$ has linear quotients
for all $k\gg 0$, which strengthens the fact that $I{\mathfrak{m}}^{k}$ is
componentwise linear for all $k\gg 0$. Analyzing carefully the proof of this
fact, which is based on combinatorial arguments, we obtain in Theorem 3.3 the
combinatorial bound

$\mathfrak{d}_{2}(I)\leq\mathfrak{d}_{3}(I)=\mathfrak{d}_{1}(I)\ \leq\
\operatorname{reg}_{\operatorname{gr}_{\mathfrak{m}}(S)}\operatorname{gr}_{\mathfrak{m}}(I)\
\leq\ \min_{\mathcal{O}}(\max_{u\in\mathcal{G}(I)}\lambda_{I,\mathcal{O},u}).$

Here the minimum runs through all orders $\mathcal{O}$ of $\mathcal{G}(I)$. In
Corollary 3.4 we prove that
$\operatorname{reg}_{\operatorname{gr}_{\mathfrak{m}}(S)}\operatorname{gr}_{\mathfrak{m}}(I)\leq(\sum_{u\in\mathcal{G}(I)}\deg(u))+1-\mu(I)-\omega(I)$.
This bound is sharp. Indeed, equality holds for monomial complete
intersections [1, Theorem 4.1(b)]. Next, in Theorem 3.6, we show that
$\mathfrak{d}_{i}(I)\leq\max\\{|{\bf deg}(I)|-n,0\\}$ for $i=1,2,3$, where
${\bf deg}(I)$ is the bounding multidegree of $I$ [12]. In particular, if $I$
is squarefree, then all Dao numbers are zero. In Corollary 3.10 we compute the
Dao numbers of monomial complete intersections. This result shows that in
general it may be difficult to compute the Dao numbers of a general complete
intersection ideal.

##  1\. General bounds for the Dao numbers

In this first section, we answer Dao’s question. For this aim, we need to
consider Castelnuovo–Mumford regularity over general base rings.

Hereafter, we follows closely [2, Section 8.1]. Let $R=\bigoplus_{k\geq
0}R_{k}$ be a standard graded Noetherian algebra. That is, $R_{0}$ is a
commutative Noetherian ring and $R$ is generated as an $R_{0}$-algebra by
finitely many elements $f_{1},\dots,f_{n}$ of degree one. Let
${\mathfrak{m}}=R_{+}=\bigoplus_{k>0}R_{k}$ be the unique homogeneous maximal
ideal of $R$. Let $M=\bigoplus_{k\geq 0}M_{k}$ be a finitely generated graded
$R$-module. The initial degree and the final degree of $M$, denoted by
$\alpha(M)$ and $\omega(M)$, are defined as

$\displaystyle\alpha(M)\ $ $\displaystyle=\ \min\\{j\ :\
(M/{\mathfrak{m}}M)_{j}\neq 0\\}\ =\ \min\\{j\ :\ M_{j}\neq 0\\},$
$\displaystyle\omega(M)\ $ $\displaystyle=\ \max\\{j\ :\
(M/{\mathfrak{m}}M)_{j}\neq 0\\}.$

The Castelnuovo–Mumford regularity of $M$ is defined as [2, Theorem 8.1.3]:

$\displaystyle\operatorname{reg}_{R}M\ $ $\displaystyle=\ \max\\{j+i\ :\
\textup{H}^{i}_{\mathfrak{m}}(M)_{j}\neq 0\\}$ $\displaystyle=\ \max\\{j-i\ :\
H_{i}({\bf y};M)_{j}\neq 0\\}.$

Here $\textup{H}_{{\mathfrak{m}}}^{i}(M)$ is the $i$th local cohomology module
of $M$ with support on ${\mathfrak{m}}$, and $H_{i}({\bf y};M)$ is the $i$th
Koszul homology module of ${\bf y}:y_{1},\dots,y_{n}$ with respect to $M$,
where ${\bf y}$ is a minimal homogeneous system of generators of
${\mathfrak{m}}$.

Finally, we record some basic properties which we will need in a moment. These
properties are also stated in [2, Page 277, (a), (c) and (d)] and [18, page
268].

  1. (i)

Let $M(j)$ be the module $M$ whose degrees are shifted by $j$:
$M(j)_{i}=M_{i+j}$ for all $i$. Then $\omega(M(j))=\omega(M)-j$ and
$\operatorname{reg}_{R}M(j)=\operatorname{reg}_{R}M-j$.

  2. (ii)

Let $0\rightarrow M\rightarrow N\rightarrow P\rightarrow 0$ be a short exact
sequence of finitely generated graded $R$-modules. Then,

$\displaystyle\operatorname{reg}_{R}M\ $ $\displaystyle\leq\
\max\\{\operatorname{reg}_{R}N,\,\operatorname{reg}_{R}P+1\\},$
$\displaystyle\operatorname{reg}_{R}P\ $ $\displaystyle\leq\
\max\\{\operatorname{reg}_{R}N,\,\operatorname{reg}_{R}M-1\\}.$

  3. (iii)

If $M_{j}=0$ for all $j\gg 0$, then
$\operatorname{reg}_{R}M=\max\\{j:M_{j}\neq 0\\}$.

  4. (iv)

Let $I\subset R$ be a homogeneous ideal and ${\bf y}:y_{1},\dots,y_{n}$ be a
minimal homogeneous system of generators of ${\mathfrak{m}}$. Then

$\displaystyle H_{n}({\bf y};R/I)\ $ $\displaystyle\cong\
e_{1}\wedge\cdots\wedge e_{n}\ \textup{Soc}(R/I)\ =\ e_{1}\wedge\cdots\wedge
e_{n}\ (0:_{R/I}{\mathfrak{m}})\ $ $\displaystyle=\ e_{1}\wedge\cdots\wedge
e_{n}\ (I:\mathfrak{m})/I.$

Here, $\textup{Soc}(R/I)$ is called the socle of $I$ and $e_{1},\dots,e_{n}$
is the basis of a graded free $R$-module of rank $n$, with $\deg e_{i}=1$ for
all $i$. Hence, we have

$\displaystyle\max\\{j\ :\ \textup{Soc}(R/I)_{j}\neq 0\\}\ $ $\displaystyle=\
\max\\{j-n\ :\ H_{n}({\bf y};R/I)_{j}\neq 0\\}$ $\displaystyle\leq\
\max\\{j-i\ :\ H_{i}({\bf y};R/I)_{j}\neq 0\\}\ $ $\displaystyle=\
\operatorname{reg}_{R}R/I.$

We are now ready to deliver the promised bound for the Dao numbers.

Let $R$ be a commutative ring and $I$ be an ideal. The Rees algebra of $I$ is
the graded ring $\mathcal{R}(I)=\bigoplus_{k\geq 0}I^{k}$. Let $J\subset R$ be
another ideal, then $\mathcal{R}(I,J)=\bigoplus_{k\geq 0}JI^{k}$ is the
extension of $J\subset R$ in the ring $\mathcal{R}(I)$. In particular,
$\mathcal{R}(I,J)$ is a finitely generated ideal of $\mathcal{R}(I)$ if $J$ is
a finitely generated ideal of $R$.

######  Theorem 1.1.

Let $(R,{\mathfrak{m}},K)$ be either a Noetherian local ring or a standard
graded Noetherian $K$-algebra with unique homogeneous maximal ideal
${\mathfrak{m}}$. We assume that $K$ is infinite and
$\operatorname{depth}R>0$. Let $I\subset R$ be an ideal. We assume that $I$ is
homogeneous if $R$ is a $K$-algebra. Then

$\mathfrak{d}_{2}(I)\leq\mathfrak{d}_{3}(I)=\mathfrak{d}_{1}(I)\ \leq\
\max\\{\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}},I),\,\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}},I):_{\mathcal{R}(R)}{\mathfrak{m}}\\}.$

Here
$\mathcal{R}({\mathfrak{m}},I):_{\mathcal{R}(R)}{\mathfrak{m}}=\mathcal{R}({\mathfrak{m}},I):_{\mathcal{R}(R)}\mathcal{R}({\mathfrak{m}},{\mathfrak{m}})=\bigoplus_{k\geq
0}(I{\mathfrak{m}}^{k}:{\mathfrak{m}})$.

###### Proof.

Notice that for all $k\geq 0$ we have
$I{\mathfrak{m}}^{k}\subseteq(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})$.
Therefore, we obtain the short exact sequence

$0\rightarrow
I{\mathfrak{m}}^{k}\rightarrow(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})\rightarrow\frac{(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})}{I{\mathfrak{m}}^{k}}\rightarrow
0.$ (1)

The Rees algebra $\mathcal{R}({\mathfrak{m}})$ of ${\mathfrak{m}}$ is a
standard graded Noetherian $R$-algebra. Since
$\mathcal{R}({\mathfrak{m}},I)=\bigoplus_{k\geq 0}I{\mathfrak{m}}^{k}$ is just
the extension of the ideal $I\subset R$ in the ring
$\mathcal{R}({\mathfrak{m}})$, it is a finitely generated graded
$\mathcal{R}({\mathfrak{m}})$-module. Taking the direct sum of the exact
sequences (1) for all $k\geq 0$, we obtain the short exact sequence:

$0\rightarrow\bigoplus_{k\geq 0}I{\mathfrak{m}}^{k}\rightarrow\bigoplus_{k\geq
0}(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})\rightarrow\bigoplus_{k\geq
0}\frac{(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})}{I{\mathfrak{m}}^{k}}\rightarrow
0.$ (2)

Notice that

$\displaystyle\mathcal{R}({\mathfrak{m}},I)(1):_{\mathcal{R}(R)}{\mathfrak{m}}$
$\displaystyle=\ \ (\bigoplus_{k\geq
0}I{\mathfrak{m}}^{k+1}):_{\mathcal{R}(R)}{\mathfrak{m}}\ =\ \bigoplus_{k\geq
0}(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})$ (3) $\displaystyle=\ \
(\mathcal{R}({\mathfrak{m}},I):_{\mathcal{R}(R)}{\mathfrak{m}})(1).$

We set
$\mathfrak{D}_{\mathfrak{m}}(I)=(\mathcal{R}({\mathfrak{m}},I):_{\mathcal{R}(R)}{\mathfrak{m}})(1)/\mathcal{R}({\mathfrak{m}},I)$
and notice that for all $k$,

$\mathfrak{D}_{\mathfrak{m}}(I)_{k}\ =\
\frac{(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})}{I{\mathfrak{m}}^{k}}.$

By [10, Theorem 3.1 and Corollary 3.2], we have that $I{\mathfrak{m}}^{k}$ is
weakly ${\mathfrak{m}}$-full for all $k\geq\mathfrak{d}_{3}(I)$, that is
$(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})=I{\mathfrak{m}}^{k}$. Thus, we have
$\mathfrak{D}_{\mathfrak{m}}(I)_{k}=0$ for all $k\geq\mathfrak{d}_{3}(I)$, and
if $\mathfrak{d}_{3}(I)-1\geq 0$ then
$\mathfrak{D}_{\mathfrak{m}}(I)_{\mathfrak{d}_{3}(I)-1}\neq 0$. It follows
that $\mathfrak{D}_{\mathfrak{m}}(I)$ is a finitely generated
$\mathcal{R}({\mathfrak{m}})$-module, and from (iv)

$\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathfrak{D}_{\mathfrak{m}}(I)\
=\ \max\\{\mathfrak{d}_{3}(I)-1,\,0\\}.$

If $\mathfrak{d}_{3}(I)=0$ there is nothing to prove. Thus, we suppose that
$\mathfrak{d}_{3}(I)>0$. Then
$\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathfrak{D}_{\mathfrak{m}}(I)=\mathfrak{d}_{3}(I)-1$.
From equations (2) and (3) we obtain the following short exact sequence of
graded $\mathcal{R}({\mathfrak{m}})$-modules:

$0\rightarrow\mathcal{R}({\mathfrak{m}},I)\rightarrow(\mathcal{R}({\mathfrak{m}},I):_{\mathcal{R}(R)}{\mathfrak{m}})(1)\rightarrow\mathfrak{D}_{\mathfrak{m}}(I)\rightarrow
0.$ (4)

Since $\mathcal{R}({\mathfrak{m}},I)$ and $\mathfrak{D}_{\mathfrak{m}}(I)$ are
finitely generated $\mathcal{R}({\mathfrak{m}})$-modules, it follows that
$(\mathcal{R}({\mathfrak{m}},I):_{\mathcal{R}(R)}{\mathfrak{m}})(1)$ is a
finitely generated graded $\mathcal{R}({\mathfrak{m}})$-module, as well.
Therefore, applying the rules (ii) and (i) to the sequence (4) we obtain that

$\displaystyle\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathfrak{D}_{\mathfrak{m}}(I)\
$ $\displaystyle\leq\
\max\\{\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}(\mathcal{R}({\mathfrak{m}},I):_{\mathcal{R}(R)}{\mathfrak{m}})(1),\,\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}},I)-1\\}$
$\displaystyle\leq\
\max\\{\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}},I):_{\mathcal{R}(R)}{\mathfrak{m}},\,\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}},I)\\}-1.$

Taking into account that
$\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathfrak{D}_{\mathfrak{m}}(I)=\mathfrak{d}_{3}(I)-1$,
we obtain the asserted inequality for $\mathfrak{d}_{3}(I)$. Finally, under
the assumptions that $K$ is infinite and $\operatorname{depth}R>0$, it follows
from [22, Proposition 2.2] that
$\mathfrak{d}_{2}(I)\leq\mathfrak{d}_{1}(I)=\mathfrak{d}_{3}(I)$.

We call the $\mathcal{R}({\mathfrak{m}})$-module, considered in the above
proof,

$\mathfrak{D}_{\mathfrak{m}}(I)\ =\ \bigoplus_{k\geq
0}\frac{(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})}{I{\mathfrak{m}}^{k}}$

the Dao module of $I$. As shown in the proof, we have

######  Corollary 1.2.

Under the same assumptions and notation of Theorem 1.1, we have

$\mathfrak{d}_{2}(I)\ \leq\ \mathfrak{d}_{1}(I)=\mathfrak{d}_{3}(I)\ =\
\begin{cases}\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathfrak{D}_{\mathfrak{m}}(I)+1&\textit{if}\
\mathfrak{D}_{\mathfrak{m}}(I)\neq 0,\\\ \hfill
0&\hfill\textit{otherwise.}\end{cases}$

Recall that the associated graded ring of $R$ is
$\operatorname{gr}_{\mathfrak{m}}(R)=\bigoplus_{k\geq
0}({\mathfrak{m}}^{k}/{\mathfrak{m}}^{k+1})$. If $R$ is Noetherian, as we
assume in our case, then $\operatorname{gr}_{\mathfrak{m}}(R)$ is a graded
Noetherian ring. We can strengthen Theorem 1.1 as follows.

######  Theorem 1.3.

Under the same assumptions and notation of Theorem 1.1, if we suppose in
addition that $\operatorname{depth}\operatorname{gr}_{\mathfrak{m}}(R)>0$,
then

$\mathfrak{d}_{2}(I)\leq\mathfrak{d}_{3}(I)=\mathfrak{d}_{1}(I)\ \leq\
\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}},I).$

###### Proof.

If $\mathfrak{d}_{1}(I)=\mathfrak{d}_{3}(I)=0$, there is nothing to prove.
Thus, we suppose that $\mathfrak{d}_{1}(I)=\mathfrak{d}_{3}(I)>0$. In
particular,
$\mathfrak{d}_{1}(I)-1=\mathfrak{d}_{3}(I)-1=\max\\{k:\mathfrak{D}_{\mathfrak{m}}(I)_{k}\neq
0\\}$. Let $\mathcal{R}({\mathfrak{m}})_{+}=\bigoplus_{k>0}{\mathfrak{m}}^{k}$
be the graded maximal ideal of $\mathcal{R}({\mathfrak{m}})$. Notice that

$\displaystyle(\mathcal{R}({\mathfrak{m}},I):_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}})_{+})_{k}$
$\displaystyle=\ \\{f\in\mathcal{R}({\mathfrak{m}})_{k}={\mathfrak{m}}^{k}\ :\
f{\mathfrak{m}}\subseteq I{\mathfrak{m}}^{k+1}\\}$ (5) $\displaystyle=\
{\mathfrak{m}}^{k}\cap\\{f\in R\ :\ f{\mathfrak{m}}\subseteq
I{\mathfrak{m}}^{k+1}\\}$ $\displaystyle=\
{\mathfrak{m}}^{k}\cap(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}}).$

We claim that for all $k\geq 0$,

${\mathfrak{m}}^{k}\cap(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})\ =\
(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}}).$ (6)

The assumptions that
$\operatorname{depth}\operatorname{gr}_{\mathfrak{m}}(R)>0$ and that $K$ is
infinite guarantee the existence of a non-zero divisor
$x_{0}\in\operatorname{gr}_{\mathfrak{m}}(R)$ of degree one. Suppose for a
contradiction that there exists $k\geq 0$ and an element
$f\in(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})$ which does not belong to
${\mathfrak{m}}^{k}$. Then
$f\in{\mathfrak{m}}^{i}\setminus{\mathfrak{m}}^{i+1}$ for a unique $i<k$. Thus
$\overline{f}=f+{\mathfrak{m}}^{i+1}\in\operatorname{gr}_{\mathfrak{m}}(R)_{i}$
is a non-zero element. Then
$\overline{x_{0}}\overline{f}=x_{0}f+{\mathfrak{m}}^{i+2}\in\operatorname{gr}_{\mathfrak{m}}(I)_{i+1}$
is non-zero as well. Thus $x_{0}f\notin{\mathfrak{m}}^{i+2}$ and since $i<k$,
then $x_{0}f\notin{\mathfrak{m}}^{k+1}$ as well. This is a contradiction,
because $x_{0}\in{\mathfrak{m}}$ and
$f\in(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})$, thus $x_{0}f\in
I{\mathfrak{m}}^{k+1}\subseteq{\mathfrak{m}}^{k+1}$. It follows that
$(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})\subseteq{\mathfrak{m}}^{k}$.

By our claim (6) and the computation (5) we obtain

$\textup{Soc}(\mathcal{R}({\mathfrak{m}})/\mathcal{R}({\mathfrak{m}},I))\ =\
\bigoplus_{k\geq
0}\frac{{\mathfrak{m}}^{k}\cap(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})}{I{\mathfrak{m}}^{k}}\
=\ \bigoplus_{k\geq
0}\frac{(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})}{I{\mathfrak{m}}^{k}}\ =\
\mathfrak{D}_{\mathfrak{m}}(I).$

Now, let ${\bf y}:y_{1},\dots,y_{n}$ be a minimal homogeneous system of
generators of $\mathcal{R}({\mathfrak{m}})_{+}$. By fact (iv) and the previous
computation, we obtain that

$H_{n}({\bf y};\mathcal{R}({\mathfrak{m}})/\mathcal{R}({\mathfrak{m}},I))\
\cong\ e_{1}\wedge\cdots\wedge e_{n}\ \mathfrak{D}_{\mathfrak{m}}(I),$

and
$\mathfrak{d}_{1}(I)-1=\mathfrak{d}_{3}(I)-1=\max\\{k:\mathfrak{D}_{\mathfrak{m}}(I)_{k}\neq
0\\}\leq\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}})/\mathcal{R}({\mathfrak{m}},I)$.
Finally, since
$\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}})/\mathcal{R}({\mathfrak{m}},I)=\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}},I)-1$,
the assertion follows.

######  Corollary 1.4.

Under the same assumptions and notation of Theorem 1.1, if we suppose in
addition that $R$ is regular, then

$\mathfrak{d}_{2}(I)\leq\mathfrak{d}_{3}(I)=\mathfrak{d}_{1}(I)\ \leq\
\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}},I).$

###### Proof.

Since $R$ is regular, we have $\operatorname{gr}_{\mathfrak{m}}(R)\cong
K[x_{1},\dots,x_{n}]$, where $n=\dim R$ (see [3, Proposition 2.2.5] and [3,
2.2.25(c)]). Hence
$\operatorname{depth}\operatorname{gr}_{\mathfrak{m}}(R)=n\geq\operatorname{depth}R>0$
and the assertion follows from Theorem 1.3.

In particular, the proof of Theorem 1.3 and Corollary 1.4 show that

######  Corollary 1.5.

Under the same assumptions and notation of Theorem 1.1, if we suppose in
addition that $\operatorname{depth}\operatorname{gr}_{\mathfrak{m}}(R)>0$ or
that $R$ is regular, then

$\textup{Soc}(\mathcal{R}({\mathfrak{m}})/\mathcal{R}({\mathfrak{m}},I))\ =\
\mathfrak{D}_{\mathfrak{m}}(I).$

An ideal $J$ is called a reduction of $I$ if $JI^{k}=I^{k+1}$ for some $k\geq
0$. Let $J$ be a reduction of $I$. The reduction number of $I$ with respect to
$J$ is defined as the integer

$\textup{r}_{J}(I)\ =\ \min\\{k\geq 0\ :\ JI^{k}=I^{k+1}\\}.$

######  Proposition 1.6.

Let $(R,{\mathfrak{m}},K)$ be either a Noetherian local ring or a standard
graded Noetherian $K$-algebra with unique homogeneous maximal ideal
${\mathfrak{m}}$. Let $I\subset R$ be a reduction of ${\mathfrak{m}}$, which
we assume to be homogeneous if $R$ is a $K$-algebra. Then,

$\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}},I)\
\leq\
\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}}).$

###### Proof.

For all $k\geq 0$, we have the short exact sequence

$0\rightarrow
I{\mathfrak{m}}^{k}\rightarrow{\mathfrak{m}}^{k+1}\rightarrow\frac{{\mathfrak{m}}^{k+1}}{I{\mathfrak{m}}^{k}}\rightarrow
0.$

Taking the direct sum of these short exact sequences for all $k\geq 0$, we
obtain the short exact sequence
$0\rightarrow\mathcal{R}({\mathfrak{m}},I)\rightarrow\mathcal{R}({\mathfrak{m}})(1)\rightarrow\mathcal{R}({\mathfrak{m}})(1)/\mathcal{R}({\mathfrak{m}},I)\rightarrow
0$ of finitely generated $\mathcal{R}({\mathfrak{m}})$-modules. Notice that

$\frac{\mathcal{R}({\mathfrak{m}})(1)}{\mathcal{R}({\mathfrak{m}},I)}\ \ =\ \
\bigoplus_{k\geq 0}\frac{{\mathfrak{m}}^{k+1}}{I{\mathfrak{m}}^{k}}\ =\
\bigoplus_{k=0}^{\textup{r}_{I}({\mathfrak{m}})-1}\frac{{\mathfrak{m}}^{k+1}}{I{\mathfrak{m}}^{k}}.$

Hence, from fact (iii) we have
$\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}})(1)/\mathcal{R}({\mathfrak{m}},I)=\max\\{\textup{r}_{I}({\mathfrak{m}})-1,0\\}$.
Now, if $\textup{r}_{I}({\mathfrak{m}})=0$, then
$\mathcal{R}({\mathfrak{m}},I)\cong\mathcal{R}({\mathfrak{m}})(1)$. Thus, by
rule (i) we have that

$\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}},I)\
=\
\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}})(1)\
=\
\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}})-1\
<\
\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}}).$

Suppose now that $\textup{r}_{I}({\mathfrak{m}})>0$. Then
$\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}})(1)/\mathcal{R}({\mathfrak{m}},I)=\textup{r}_{I}({\mathfrak{m}})-1$.
By [26, Theorem 4.8] we have
$\textup{r}_{I}({\mathfrak{m}})\leq\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}})$.
Applying rule (ii) to the above sequence we get

$\displaystyle\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}},I)\
$ $\displaystyle\leq\
\max\\{\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}})(1),\,\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}})(1)/\mathcal{R}({\mathfrak{m}},I)+1\\}$
$\displaystyle=\
\max\\{\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}})-1,\,\textup{r}_{I}({\mathfrak{m}})\\}$
$\displaystyle\leq\
\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}}),$

as desired.

Using the theory developed thus far we can reprove [22, Corollary 3.11].

######  Corollary 1.7.

Under the same assumptions and notation of Theorem 1.1, if we suppose in
addition that $R$ is a regular ring, and that $I$ is a reduction of
${\mathfrak{m}}$, then

$\mathfrak{d}_{1}(I)=\mathfrak{d}_{2}(I)=\mathfrak{d}_{3}(I)=0.$

###### Proof.

Under the assumption that $R$ is regular, ${\mathfrak{m}}$ is generated by a
regular sequence. Therefore [26, Corollary 5.2] implies that
$\operatorname{reg}_{\mathcal{R}({\mathfrak{m}})}\mathcal{R}({\mathfrak{m}})=0$.
By Proposition 1.6 and Corollary 1.4 we conclude that
$\mathfrak{d}_{1}(I)=\mathfrak{d}_{2}(I)=\mathfrak{d}_{3}(I)=0$.

We conclude this section by providing a lower bound for
$\mathfrak{d}_{1}(I)=\mathfrak{d}_{3}(I)$.

######  Proposition 1.8.

Under the same assumptions and notation of Theorem 1.1, we have

$\omega(\mathcal{R}({\mathfrak{m}},I):_{\mathcal{R}(R)}{\mathfrak{m}})-1\
\leq\ \mathfrak{d}_{3}(I)=\mathfrak{d}_{1}(I)$

###### Proof.

Let $M=\bigoplus_{k\geq 0}M_{k}$ be a finitely generated module over a
standard graded Noetherian algebra $A=\bigoplus_{k\geq 0}A_{k}$ with
homogeneous maximal ideal ${\mathfrak{n}}=(y_{1},\dots,y_{n})$ and $\deg
y_{i}=1$ for all $i$. Then
$M_{k}={\mathfrak{n}}_{1}M_{k-1}=\\{\sum_{i}y_{i}f_{i}:f_{i}\in M_{k-1}\\}$
for all $k\gg 0$.

Now, let $k\geq\mathfrak{d}_{3}(I)+1$, then $I{\mathfrak{m}}^{k-1}$ and
$I{\mathfrak{m}}^{k}$ are weakly ${\mathfrak{m}}$-full. Hence

$(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})\ =\ I{\mathfrak{m}}^{k}\ =\
{\mathfrak{m}}(I{\mathfrak{m}}^{k-1})\ =\
{\mathfrak{m}}(I{\mathfrak{m}}^{k}:{\mathfrak{m}}).$

Thus

$(\mathcal{R}({\mathfrak{m}},I):_{\mathcal{R}({\mathfrak{m}})}{\mathfrak{m}})_{k+1}\
=\
\mathcal{R}({\mathfrak{m}})_{1}(\mathcal{R}({\mathfrak{m}},I):_{\mathcal{R}({\mathfrak{m}})}{\mathfrak{m}})_{k}$

for all $k\geq\mathfrak{d}_{3}(I)+1$. This shows that
$\omega(\mathcal{R}({\mathfrak{m}},I):_{\mathcal{R}({\mathfrak{m}})}{\mathfrak{m}})\leq\mathfrak{d}_{3}(I)+1$.

##  2\. Fullness in Koszul algebras

In this section, we consider the more tame situation of a Koszul algebra and
determine bounds for the Dao numbers of an ideal in such a ring. Let $K$ be a
field and let $R=\bigoplus_{k\geq 0}$ be a standard graded $K$-algebra. That
is,

  1. (i)

$R$ admits a decomposition $R=\bigoplus_{k\geq 0}R_{k}$ as an abelian group

  2. (ii)

$R$ is generated as an $R_{0}$-algebra by the finite dimensional $K$-vector
space $R_{1}$,

  3. (iii)

and $R_{0}=K$.

In this case, we have the surjective homogeneous ring map

$\varphi\ :\ S=K[x_{1},\dots,x_{n}]\rightarrow R$ (7)

such that $R_{1}$ has a $K$-basis given by elements $f_{1},\dots,f_{n}$ of
degree one, $\varphi(x_{i})=f_{i}$ for all $i$ and $\varphi(r)=r$ for all
$r\in R_{0}=K$. We call (7) the canonical presentation of $R$. Hence $R\cong
S/\ker\varphi$. We denote by ${\mathfrak{m}}=R_{+}=\bigoplus_{k>0}R_{k}$ the
unique homogeneous maximal ideal of $R$.

Notice that we may view the field $K$ as a finitely generated graded
$R$-module by identifying it with $R/{\mathfrak{m}}$. Following [7, Definition
4], we say that the algebra $R$ is Koszul if $\operatorname{reg}_{R}K=0$. In
this case, $\operatorname{reg}_{R}{\mathfrak{m}}=1$.

Let $M=\bigoplus_{k\geq 0}M_{k}$ be a finitely generated graded $R$-module. We
denote by $M_{\langle k\rangle}$ the submodule of $M$ generated by the
$K$-vector space $M_{k}$.

Now, let $I\subset R$ be a homogeneous ideal. We say that $I$ has linear
resolution if $\alpha(I)=\omega(I)$ (that is, $I$ is generated in a single
degree) and $\operatorname{reg}_{R}I=\alpha(I)$. We say that $I$ is
componentwise linear if $I_{\langle j\rangle}$ has linear resolution, for all
$j$.

######  Lemma 2.1.

$(I{\mathfrak{m}})_{\langle k\rangle}=I_{\langle k-1\rangle}{\mathfrak{m}}$
for all $k\geq 1$.

###### Proof.

It is clear that $I_{\langle k-1\rangle}{\mathfrak{m}}$ is contained in
$(I{\mathfrak{m}})_{\langle k\rangle}$. For the opposite inclusion, observe
that the $K$-vector space $(I{\mathfrak{m}})_{k}$ is generated by the non-zero
elements of the form $ug$ where $u\in{\mathfrak{m}}$ is a monomial and $g$ is
homogeneous element of $I$ such that $\deg(ug)=k$. Therefore, it is enough to
show that if $f=ug$ is as before, then $f\in I_{\langle
k-1\rangle}{\mathfrak{m}}$. We may assume that $x_{j}$ divides $u$. Then, we
can write $f=x_{j}((u/x_{j})g)$, where $(u/x_{j})g\in I$ has degree $k-1$.
Thus, $f\in I_{\langle k-1\rangle}{\mathfrak{m}}$.

######  Proposition 2.2.

Let $R$ be a Koszul algebra with homogeneous maximal ideal ${\mathfrak{m}}$,
and let $I\subset R$ be a componentwise linear ideal. We assume that
$\operatorname{depth}R>0$ and $K$ is infinite. Then the following properties
hold:

  1. (a)

$I{\mathfrak{m}}^{\ell}$ is componentwise linear for any $\ell$.

  2. (b)

$(I{\mathfrak{m}}:{\mathfrak{m}})=I$.

  3. (c)

$\mathfrak{d}_{3}(I)=0$.

###### Proof.

We first note that $I{\mathfrak{m}}$ is non-zero. Indeed, since
$\operatorname{depth}R>0$ there exists a non-zero divisor
$x\in{\mathfrak{m}}$.

(a) This fact is well-known. We sketch the proof for completeness. It is
enough to prove that $I{\mathfrak{m}}$ is componentwise linear. We must show
that for each $j$ such that $(I{\mathfrak{m}})_{\langle j\rangle}=I_{\langle
j-1\rangle}{\mathfrak{m}}$ is non-zero, this ideal has linear resolution.
Notice that $\operatorname{reg}_{R}(I{\mathfrak{m}})_{\langle j\rangle}\geq
j$. On the other hand, by [7, Lemma 4] and since
$\operatorname{reg}_{R}I_{\langle j-1\rangle}=j-1$, we obtain that
$\operatorname{reg}_{R}(I{\mathfrak{m}})_{\langle
j\rangle}\leq\operatorname{reg}_{R}I_{\langle
j-1\rangle}+\operatorname{reg}_{R}{\mathfrak{m}}=j$.

(b) Firstly, we assume that $I$ has linear resolution, say $d$-linear. It is
clear that $I\subseteq(I{\mathfrak{m}}:{\mathfrak{m}})$. Suppose for a
contradiction that there exists $f\in(I{\mathfrak{m}}:{\mathfrak{m}})\setminus
I$. Then $\overline{f}=f+I\in(I:{\mathfrak{m}})/I$, because
$f{\mathfrak{m}}\subseteq I{\mathfrak{m}}\subseteq I$ and $f\notin I$. By fact
(iv) recalled at page (iv),
$\operatorname{Tor}^{R}_{n-1}(R/{\mathfrak{m}},I)\cong\operatorname{Tor}^{R}_{n}(R/{\mathfrak{m}},R/I)\cong
e_{1}\wedge\dots\wedge e_{n}\ \textup{Soc}(R/I)$. Since $I$ has a $d$-linear
resolution, $\operatorname{Tor}^{R}_{n-1}(R/{\mathfrak{m}},I)$ is concentrated
in degree $n+d-1$. Thus, the $K$-vector space $\textup{Soc}(R/I)$ is generated
in a single degree $d-1$, and so $\deg(f)=d-1$. By part (a), $I{\mathfrak{m}}$
has a $(d+1)$-linear resolution. Hence, the above argument shows that any non-
zero homogeneous element of $(I{\mathfrak{m}}:{\mathfrak{m}})\setminus
I{\mathfrak{m}}$ has degree $d$. Since $I{\mathfrak{m}}$ is generated in
degree $d+1$, $(I{\mathfrak{m}}:{\mathfrak{m}})$ is generated in degrees $\geq
d$. This contradicts the fact that $\deg(f)=d-1$ and shows that
$(I{\mathfrak{m}}:{\mathfrak{m}})=I$.

Now, suppose that $I$ is generated in more than one degree. Notice that the
ideal $(I{\mathfrak{m}}:{\mathfrak{m}})$ is homogeneous as well. Thus, by the
first part we have

$\displaystyle(I{\mathfrak{m}}:{\mathfrak{m}})\ $ $\displaystyle=\
\bigoplus_{j}(I{\mathfrak{m}}:{\mathfrak{m}})_{j}\ =\
\bigoplus_{j}((I{\mathfrak{m}})_{\langle j+1\rangle}:{\mathfrak{m}})_{j}$
$\displaystyle=\ \bigoplus_{j}(I_{\langle
j\rangle}{\mathfrak{m}}:{\mathfrak{m}})_{j}\ =\ \bigoplus_{j}(I_{\langle
j\rangle})_{j}$ $\displaystyle=\ \bigoplus_{j}I_{j}\ =\ I.$

(c) It follows by combining (a) with (b).

######  Corollary 2.3.

Let $R$ be a Koszul $K$-algebra and let $I\subset R$ be a componentwise linear
ideal. Assume that $K$ is infinite and $\operatorname{depth}R>0$. Then
$\mathfrak{d}_{1}(I)=\mathfrak{d}_{2}(I)=\mathfrak{d}_{3}(I)=0$. In
particular, $I$ is ${\mathfrak{m}}$-full, full and weakly
${\mathfrak{m}}$-full.

###### Proof.

By Proposition 2.2(c), $\mathfrak{d}_{3}(I)=0$. Thus [22, Proposition 2.2]
implies that $\mathfrak{d}_{1}(I)=\mathfrak{d}_{2}(I)=0$. The assertion
follows.

######  Proposition 2.4.

Let $R$ be a Koszul algebra with homogeneous maximal ideal ${\mathfrak{m}}$,
and let $I\subset R$ be a homogeneous ideal generated in a single degree.
Suppose that $K$ is infinite and $\operatorname{depth}R>0$. Then

$\mathfrak{d}_{2}(I)\leq\mathfrak{d}_{3}(I)=\mathfrak{d}_{1}(I)\ \leq\
\operatorname{reg}_{R}I-\alpha(I).$

###### Proof.

Since $I$ is generated in a single degree, we have
$I{\mathfrak{m}}^{k}=I_{\langle\alpha(I)+k\rangle}$ for all $k$. From [7,
Proposition 2.8] it follows that for all
$k\geq\operatorname{reg}_{R}I-\alpha(I)$, the ideal $I_{\langle
k\rangle}=I{\mathfrak{m}}^{k-\alpha(I)}$ has linear resolution. By Proposition
2.2, it follows that $I{\mathfrak{m}}^{k}$ is weakly ${\mathfrak{m}}$-full,
for all $k\geq\operatorname{reg}_{R}I-\alpha(I)$. Hence
$\mathfrak{d}_{3}(I)\leq\operatorname{reg}_{R}I-\alpha(I)$. Since $K$ is
infinite and $\operatorname{depth}R>0$, [22, Proposition 2.2] implies that
$\mathfrak{d}_{2}(I)\leq\mathfrak{d}_{1}(I)=\mathfrak{d}_{3}(I)\leq\operatorname{reg}_{R}I-\alpha(I)$.

The following result generalizes [13, Theorem 1.1] and the above proposition
when $I$ is generated in more than one degree.

######  Theorem 2.5.

Let $R$ be a Koszul algebra with maximal ideal ${\mathfrak{m}}$, and let
$I\subset R$ be a homogeneous ideal. Suppose that $K$ is infinite and
$\operatorname{depth}R>0$. Then

$\mathfrak{d}_{2}(I)\leq\mathfrak{d}_{3}(I)=\mathfrak{d}_{1}(I)\ \leq\
\operatorname{reg}_{\operatorname{gr}_{\mathfrak{m}}(R)}\operatorname{gr}_{\mathfrak{m}}(I).$

###### Proof.

Recall that the associated graded module of $I$

$\operatorname{gr}_{\mathfrak{m}}(I)\ =\ \bigoplus_{k\geq
0}(I{\mathfrak{m}}^{k}/I{\mathfrak{m}}^{k+1})\ =\
(I/{\mathfrak{m}}I)\oplus(I{\mathfrak{m}}/I{\mathfrak{m}}^{2})\oplus(I{\mathfrak{m}}^{2}/I{\mathfrak{m}}^{3})\oplus\cdots$

is a finitely generated graded module over the associated graded ring
$\operatorname{gr}_{\mathfrak{m}}(R)$. It follows from [19, 1.5 Proposition
(1)-(3)] (see also [23, Theorem 3.2.8]) that a finitely generated graded
$R$-module $M$ is componentwise linear if and only if
$\operatorname{gr}_{\mathfrak{m}}(M)$ has a linear resolution. By [7,
Proposition 8], $(\operatorname{gr}_{\mathfrak{m}}(I))_{\langle k\rangle}$ has
a linear resolution for all
$k\geq\operatorname{reg}_{\operatorname{gr}_{\mathfrak{m}}(R)}\operatorname{gr}_{\mathfrak{m}}(I)$.
Notice that $(\operatorname{gr}_{\mathfrak{m}}(I))_{\langle\ell\rangle}\ =\
\operatorname{gr}_{\mathfrak{m}}(I{\mathfrak{m}}^{\ell})$ for any integer
$\ell$. Thus, $I{\mathfrak{m}}^{k}$ is componentwise linear for all
$k\geq\operatorname{reg}_{\operatorname{gr}_{\mathfrak{m}}(R)}\operatorname{gr}_{\mathfrak{m}}(I)$.
By Proposition 2.2, it follows that $I{\mathfrak{m}}^{k}$ is weakly
${\mathfrak{m}}$-full for all
$k\geq\operatorname{reg}_{\operatorname{gr}_{\mathfrak{m}}(R)}\operatorname{gr}_{\mathfrak{m}}(I)$.
Hence
$\mathfrak{d}_{3}(I)\leq\operatorname{reg}_{\operatorname{gr}_{\mathfrak{m}}(R)}\operatorname{gr}_{\mathfrak{m}}(I)$.
Since $K$ is infinite and $\operatorname{depth}R>0$, by [22, Proposition 2.2]
we conclude that
$\mathfrak{d}_{2}(I)\leq\mathfrak{d}_{1}(I)=\mathfrak{d}_{3}(I)\leq\operatorname{reg}_{\operatorname{gr}_{\mathfrak{m}}(R)}\operatorname{gr}_{\mathfrak{m}}(I)$.

######  Question 2.6.

Let $a\geq b\geq c\geq 0$ be non-negative integers. Can we find a graded ideal
$I\subset R$ such that
$\operatorname{reg}_{\operatorname{gr}_{\mathfrak{m}}(R)}\operatorname{gr}_{\mathfrak{m}}(I)=a$,
$\mathfrak{d}_{1}(I)=\mathfrak{d}_{3}(I)=b$ and $\mathfrak{d}_{2}(I)=c$?

Next, we present a large class of componentwise linear ideals.

Let $u=x_{1}^{a_{1}}\cdots x_{n}^{a_{n}}\in S$ be a monomial where $S$ is the
polynomial ring in the canonical presentation (7). Since $R\cong
S/\ker\varphi$, if the residue class $\overline{u}=u+\ker\varphi$ is non-zero,
then we call $\overline{u}$ a monomial of $R$. To simplify the notation, we
denote $\overline{u}$ again by $u$. Notice that $u$ may have different
representations in $R$. For example, in the Koszul algebra
$K[x_{1},x_{2},x_{3},x_{4}]/(x_{1}x_{2}-x_{3}x_{4})$ we have
$x_{1}x_{2}=x_{3}x_{4}$.

Again let $R$ be a Koszul algebra and $I\subset R$ be an ideal. We say that
$I$ is a monomial ideal of $R$ if $I$ can be generated by monomials of $R$. In
such case, we denote by $\mathcal{G}(I)$ any minimal monomial generating set
of $I$.

We say that $I$ has linear quotients if there exists a minimal monomial
generating set $\mathcal{G}(I)$ of $I$ and an order
$\mathcal{O}:u_{1}<\dots<u_{m}$ of $\mathcal{G}(I)$ such that
$(u_{1},\dots,u_{j-1}):_{R}(u_{j})$ is generated by variables, for
$j=2,\dots,m$.

######  Example 2.7.

[6, Example 1.20] Consider the Koszul algebra

$R\ =\ \frac{K[a,b,c,d]}{(ac,ad,ab-bd,a^{2}+bc,b^{2})}.$

In this algebra, the ideal $I=(b)$ does not have linear resolution. Notice
that in the usual polynomial ring $S=K[a,b,c,d]$ the ideal $I=(b)$ has linear
quotients, and thus a linear resolution [18, Proposition 8.2.1].

To guarantee that monomial ideals $I\subset R$ with linear quotients are
componentwise linear we need to impose further conditions on the algebra $R$.
As shown in [27], we must assume that $R$ is strongly Koszul. A Koszul algebra
$R$ is called strongly Koszul if there exists a basis $X$ of $R_{1}$ such that
for every proper subset $Y\subset X$ and every $x\in X\setminus Y$, there
exists a subset $Z\subset X$ such that $(Y):_{R}(x)=(Z)$.

A pivotal property of a strongly Koszul algebra $R$ is that any ideal of $R$
generated by linear forms has linear resolution [27, Lemma 3.3]. Hence, this
fact together with [20, Theorem 3.1] implies that

######  Proposition 2.8.

Let $R$ be a strongly Koszul algebra and $I\subset R$ be a monomial ideal with
linear quotients order $\mathcal{O}:u_{1}<\dots<u_{m}$ satisfying
$\deg(u_{1})\leq\dots\leq\deg(u_{m})$. Then $I$ is componentwise linear. In
particular, $\mathfrak{d}_{3}(I)=0$.

See also [9] for the exterior algebra case.

If $I$ is a monomial ideal in the standard graded polynomial ring
$S=K[x_{1},\dots,x_{n}]$ having linear quotients order
$\mathcal{O}:u_{1}<\dots<u_{m}$, then by the usual Björner-Wachs rearrangement
lemma [25, Lemma 2.1] (see also [4]) we can always also find an order
$\mathcal{O}^{\prime}:u_{i_{1}}<\dots<u_{i_{m}}$ such that
$\deg(u_{i_{1}})\leq\dots\leq\deg(u_{i_{m}})$. Thus, at least in the case of a
polynomial ring, the condition on the degrees
$\deg(u_{1})\leq\dots\leq\deg(u_{m})$ given in Proposition 2.8 can be removed.
It is not clear whether this is also the case for monomial ideals with linear
quotients in a strongly Koszul algebra

##  3\. The Dao numbers of monomial ideals

While Theorem 2.5 shows that the number
$\operatorname{reg}_{\operatorname{gr}_{\mathfrak{m}}(R)}\operatorname{gr}_{\mathfrak{m}}(I)$
is an upper bound for the Dao numbers, it may be difficult to compute it. In
this section, we consider the more specific class of monomial ideals in a
standard graded polynomial ring, and we provide combinatorial bounds for the
Dao numbers.

Let $S=K[x_{1},\dots,x_{n}]$ be the standard graded polynomial ring over a
infinite field $K$ and let ${\mathfrak{m}}=(x_{1},\dots,x_{n})$ be the unique
homogeneous maximal ideal.

Let ${\bf a}=(a_{1},\dots,a_{n})\in{\mathbb{Z}}_{\geq 0}^{n}$. We set ${\bf
a}[i]=a_{i}$ for all $i$, ${\bf x^{a}}=\prod_{i}x_{i}^{{\bf a}[i]}$. In
particular, ${\bf x^{0}}=1$ for ${\bf 0}=(0,0,\dots,0)$. The monomial ${\bf
x^{a}}$ is called squarefree if ${\bf a}[i]\in\\{0,1\\}$ for all $i$. A
monomial ideal $I\subset S$ is called squarefree if $I$ is generated by
squarefree monomials. The next easy observation will be used several times.

######  Remark 3.1.

Let $I\subset S$ be a monomial ideal. Notice that $\operatorname{depth}S=n\geq
1$. Since $K$ is infinite, by [22, Proposition 2.2] we have
$\mathfrak{d}_{2}(I)\leq\mathfrak{d}_{1}(I)=\mathfrak{d}_{3}(I)$.

We first combinatorially bound
$\operatorname{reg}_{\operatorname{gr}_{\mathfrak{m}}(S)}\operatorname{gr}_{\mathfrak{m}}(I)$
for a monomial ideal $I\subset S$. Then, by Theorem 2.5 we obtain some bounds
for the Dao numbers. For such aim, we recall the next result [13, Theorem
1.5].

######  Proposition 3.2.

Let $S=K[x_{1},\dots,x_{n}]$ be the standard graded polynomial ring and let
$I\subset S$ be a monomial ideal. Then $I{\mathfrak{m}}^{k}$ has linear
quotients for all $k\gg 0$.

Analyzing carefully the proof of [13, Theorem 1.5] we will obtain the desired
combinatorial bounds for the Dao numbers.

Let $I\subset S$ be a monomial ideal. In this case, $I$ has a unique minimal
monomial generating set, denoted as usual by $\mathcal{G}(I)$. Fix any order
$\mathcal{O}:u_{1}<\cdots<u_{m}$ of $\mathcal{G}(I)$. For each $2\leq i\leq
m$, let $(u_{1},\ldots,u_{i-1}):_{S}(u_{i})=(w_{i,1},\ldots,w_{i,\ell_{i}})$.
We set

$\lambda_{I,\mathcal{O},u_{i}}=\sum_{j=1}^{\ell_{i}}(\deg(w_{i,j})-1),$

for $2\leq i\leq m$, and $\lambda_{I,\mathcal{O},u_{1}}=0$. It is clear that
$I$ has linear quotients with respect to the order $\mathcal{O}$ if and only
if $\lambda_{I,\mathcal{O},u_{i}}=0$ for all $2\leq i\leq r$.

Notice that $I{\mathfrak{m}}$ is generated by the set $\\{x_{j}u_{i}\ :\
j=1,\dots,n,\ i=1,\dots,m\\}$. After removing the non-minimal generators of
$I{\mathfrak{m}}$ and any repeated element whenever it appears again, we
obtain the minimal generating set $\mathcal{G}(I{\mathfrak{m}})$. Let
$x_{j_{1}}u_{i_{1}},x_{j_{2}}u_{i_{2}}$ be two minimal generators of
$I{\mathfrak{m}}$. We set $x_{j_{1}}u_{i_{1}}<x_{j_{2}}u_{i_{2}}$ if
$i_{1}=i_{2}$ and $j_{1}<j_{2}$ or $i_{1}<i_{2}$. We denote by
$\mathcal{O}_{1}$ this order. It is shown in the proof of [13, Theorem 1.5]
that for all $i$ and $j$ such that $x_{j}u_{i}\in\mathcal{G}(I{\mathfrak{m}})$
we have that
$\lambda_{{\mathfrak{m}}I,\mathcal{O}_{1},x_{j}u_{i}}\leq\lambda_{I,\mathcal{O},u_{i}}$
and if $\lambda_{I,\mathcal{O},u_{i}}>0$, then
$\lambda_{{\mathfrak{m}}I,\mathcal{O}_{1},x_{j}u_{i}}<\lambda_{I,\mathcal{O},u_{i}}$.
Hence, iterating this process for all $k\geq 1$, and calling $\mathcal{O}_{k}$
the order of $\mathcal{G}(I{\mathfrak{m}}^{k})$ obtained as explained above,
we see that for all

$k\ \geq\ \max\\{\lambda_{I,\mathcal{O},u_{i}}\ :\ i=1,\dots,m\\}$

we have $\lambda_{I,\mathcal{O}_{k},v}=0$ for all
$v\in\mathcal{G}(I{\mathfrak{m}}^{k})$. This means, as we observed above, that
$I{\mathfrak{m}}^{k}$ indeed has linear quotients with linear quotients order
$\mathcal{O}_{k}$.

Our discussion together with Theorem 2.5 shows that

######  Theorem 3.3.

Let $S=K[x_{1},\dots,x_{n}]$ be the standard graded polynomial ring over an
infinite field $K$ and let $I\subset S$ be a monomial ideal. Then

$\mathfrak{d}_{2}(I)\leq\mathfrak{d}_{3}(I)=\mathfrak{d}_{1}(I)\ \leq\
\operatorname{reg}_{\operatorname{gr}_{\mathfrak{m}}(S)}\operatorname{gr}_{\mathfrak{m}}(I)\
\leq\ \min_{\mathcal{O}}(\max_{u\in\mathcal{G}(I)}\lambda_{I,\mathcal{O},u}),$

where the minimum is taken over all possible orders $\mathcal{O}$ of
$\mathcal{G}(I)$.

For two monomials $u$ and $v$, we set $u:v=u/\gcd(u,v)$. Let
$u_{1},\dots,u_{j}\in S$ be monomials. Then
$(u_{1},\dots,u_{j-1}):_{S}(u_{j})$ is generated by the monomials
$u_{i}:u_{j}$ for $i=1,\dots,j-1$ [18, Proposition 1.2.2]. We denote by
$\mu(I)=\dim_{K}(I/{\mathfrak{m}}I)$ the minimal number of generators of $I$.

######  Corollary 3.4.

Let $S=K[x_{1},\dots,x_{n}]$ be the standard graded polynomial ring over an
infinite field $K$ and let $I\subset S$ be a monomial ideal. Then

$\mathfrak{d}_{2}(I)\leq\mathfrak{d}_{3}(I)=\mathfrak{d}_{1}(I)\ \leq\
\operatorname{reg}_{\operatorname{gr}_{\mathfrak{m}}(S)}\operatorname{gr}_{\mathfrak{m}}(I)\
\leq\ (\\!\\!\,\\!\sum_{u\in\mathcal{G}(I)}\deg(u))+1-\mu(I)-\omega(I).$

In particular, if $I$ is generated in a single degree, then

$\operatorname{reg}_{S}I\ \leq\
(\\!\\!\,\\!\\!\sum_{u\in\mathcal{G}(I)}\deg(u))+1-\mu(I).$

###### Proof.

Let $\mathcal{O}:u_{1}<\dots<u_{m}$ be an order of $\mathcal{G}(I)$ with
$\deg(u_{1})\leq\dots\leq\deg(u_{m})$. For all $2\leq j\leq m$, we have that
$(u_{1},\dots,u_{j-1}):_{S}(u_{j})=(u_{i}:u_{j},i=1,\dots,j-1)$. Hence, for
all $2\leq j\leq m$, we have

$\displaystyle\lambda_{I,\mathcal{O},u_{j}}\ $ $\displaystyle=\
\sum_{i=1}^{j-1}(\deg(u_{i}:u_{j})-1)\ \leq\ \sum_{i=1}^{m-1}(\deg(u_{i})-1).$

Since
$\sum_{i=1}^{m-1}\deg(u_{i})=\sum_{u\in\mathcal{G}(I)}\\!\deg(u)-\deg(u_{m})=\sum_{u\in\mathcal{G}(I)}\deg(u)-\omega(I)$
and $m=\mu(I)$, we conclude that
$\lambda_{I,\mathcal{O},u_{j}}\leq(\sum_{u\in\mathcal{G}(I)}\deg(u))+1-\mu(I)-\omega(I)$,
for all $j=1,\dots,m$. The assertion follows from Theorem 3.3 and Proposition
2.4.

Let $I\subset S$ be a graded ideal. Even if $I$ does not have a unique minimal
generating set, all minimal generating sets $f_{1},\dots,f_{m}$ of $I$ have
the same number $\mu(I)=\dim_{K}(I/{\mathfrak{m}}I)$ of generators and the
number $\sum_{i=1}^{m}\deg(f_{i})$ does not depend on the particular minimal
generating set. This follows because $I/{\mathfrak{m}}I$ is a graded
$K$-vector space. Therefore, in light of the above corollary, it makes sense
to ask

######  Question 3.5.

Let $I\subset S$ be a graded ideal. It is true that

$\mathfrak{d}_{2}(I)\leq\mathfrak{d}_{3}(I)=\mathfrak{d}_{1}(I)\ \leq\
(\sum_{f\in\mathcal{B}}\deg(f))+1-\mu(I)-\omega(I),$

where $\mathcal{B}$ is any graded $K$-basis of $I/{\mathfrak{m}}I$?

Now, we propose a different bound for the Dao numbers of monomial ideals.

For this aim, following [12], we define the bounding multidegree of a monomial
ideal $I\subset S$ to be the vector ${\bf
deg}(I)=(\deg_{x_{1}}(I),\dots,\deg_{x_{n}}(I))$, with

$\deg_{x_{i}}(I)\ =\ \max_{u\in\mathcal{G}(I)}\deg_{x_{i}}(u),\ \ \textup{for
all}\ \ \ 1\leq i\leq n.$

Here for a monomial $u={\bf x^{a}}$, we set $\deg_{x_{i}}(u)={\bf
a}[i]=\max\\{j\ :\ x_{i}^{j}\ \textup{divides}\ u\\}$. Furthermore, for ${\bf
a}\in{\mathbb{Z}}_{\geq 0}^{n}$, we set $|{\bf a}|=\sum_{i=1}^{n}{\bf a}[i]$.
Thus $\deg({\bf x^{a}})=|{\bf a}|$.

######  Theorem 3.6.

Let $I\subset S$ be a monomial ideal. Then

$\mathfrak{d}_{2}(I)\leq\mathfrak{d}_{3}(I)=\mathfrak{d}_{1}(I)\ \leq\
\max\\{|{\bf deg}(I)|-n,0\\}.$

In particular, if $I$ is squarefree then
$\mathfrak{d}_{1}(I)=\mathfrak{d}_{2}(I)=\mathfrak{d}_{3}(I)=0.$

In order to prove the theorem, we shall need the following lemma. For an
integer $n\geq 1$, we let $[n]=\\{1,2,\dots,n\\}$.

######  Lemma 3.7.

For all $t\in[n]$ and all $k\geq 0$, we have

$(I{\mathfrak{m}}^{k+1}:x_{t})\ =\ I{\mathfrak{m}}^{k}+({\bf x^{a}}/x_{t}\ :\
{\bf x^{a}}\in\mathcal{G}(I),\ {\bf a}[t]>0){\mathfrak{m}}^{k+1}.$

###### Proof.

Notice that $\\{{\bf x^{a}}u\ :\ {\bf x^{a}}\in\mathcal{G}(I),\
u\in\mathcal{G}({\mathfrak{m}}^{k+1})\\}$ is a generating set of
$I{\mathfrak{m}}^{k+1}$. Hence $\\{{\bf x^{a}}u:x_{t}\ ,\ {\bf
x^{a}}\in\mathcal{G}(I),\ u\in\mathcal{G}({\mathfrak{m}}^{k+1})\\}$ is a
generating set of $(I{\mathfrak{m}}^{k+1}:x_{t})$. Let ${\bf
x^{a}}\in\mathcal{G}(I)$ and $u\in\mathcal{G}({\mathfrak{m}}^{k+1})$. Then,

${\bf x^{a}}u:x_{t}\ =\ \begin{cases}\ ({\bf x^{a}}/x_{t})u&\text{if}\ {\bf
a}[t]>0,\\\ \ {\bf x^{a}}(u/x_{t})&\text{if}\ {\bf a}[t]=0,\ x_{t}\
\textup{divides}\ u,\\\ \ {\bf x^{a}}u&\textup{if}\ {\bf a}[t]=0,\ x_{t}\
\textup{does not divide}\ u.\end{cases}$

Since $u\in\mathcal{G}({\mathfrak{m}}^{k+1})$ is arbitrary, from the above
computations we see that

$(I{\mathfrak{m}}^{k+1}:x_{t})\ =\ I{\mathfrak{m}}^{k}\ +\ ({\bf x^{a}}/x_{t}\
:\ {\bf x^{a}}\in\mathcal{G}(I),\ {\bf a}[t]>0){\mathfrak{m}}^{k+1}\ +\
I{\mathfrak{m}}^{k+1}.$

Finally, from the inclusion $I{\mathfrak{m}}^{k+1}\subseteq
I{\mathfrak{m}}^{k}$, the asserted formula follows.

We are now ready for the proof of the main theorem.

###### Proof of Theorem 3.6.

By Remark 3.1 we have
$\mathfrak{d}_{2}(I)\leq\mathfrak{d}_{1}(I)=\mathfrak{d}_{3}(I)$ for any
monomial ideal $I\subset S$. Suppose that for some $i\in[n]$ the variable
$x_{i}$ does not divide any monomial generator of $I$, then by Lemma 3.7 we
have that $(I{\mathfrak{m}}^{k+1}:x_{i})=I{\mathfrak{m}}^{k}$ for all $k$.
Then, for all $k\geq 0$,

$I{\mathfrak{m}}^{k}\ \subseteq\ (I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})\ =\
\bigcap_{t=1}^{n}(I{\mathfrak{m}}^{k+1}:x_{t})\ \subseteq\
(I{\mathfrak{m}}^{k+1}:x_{i})\ =\ I{\mathfrak{m}}^{k}.$

Equality follows, and the Dao numbers of $I$ are zero in this case.

Suppose now that for all $i\in[n]$ the variable $x_{i}$ divides some monomial
generator ${\bf x^{a}}\in\mathcal{G}(I)$. By Lemma 3.7, for all $k\geq 0$ we
have

$\displaystyle(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})\ $ $\displaystyle=\
\bigcap_{t=1}^{n}\,(I{\mathfrak{m}}^{k+1}:x_{t})$ $\displaystyle=\
\bigcap_{t=1}^{n}\big{[}I{\mathfrak{m}}^{k}+({\bf x^{a}}/x_{t}\ :\ {\bf
x^{a}}\in\mathcal{G}(I),\ {\bf a}[t]>0){\mathfrak{m}}^{k+1}\big{]}.$

Now, let $k\geq|{\bf deg}(I)|-n$. It is clear that $I{\mathfrak{m}}^{k}$ is
contained in $(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})$. Suppose for a
contradiction that there exists a monomial
$v\in(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})$ which does not belong to
$I{\mathfrak{m}}^{k}$. Then, by the above formula, we see that

$v\ \in\ \bigcap_{t=1}^{n}\big{[}({\bf x^{a}}/x_{t}\ :\ {\bf
x^{a}}\in\mathcal{G}(I),\ {\bf a}[t]>0){\mathfrak{m}}^{k+1}\big{]}.$ (8)

From this equation, we see that for all $t\in[n]$, there exist a monomial
$u_{t}\in\mathcal{G}({\mathfrak{m}}^{k+1})$ and ${\bf x^{a}}\in\mathcal{G}(I)$
with ${\bf a}[t]>0$ such that $({\bf x^{a}}/x_{t})u_{t}$ divides $v$. Let
$p\in[n]$ such that $x_{p}$ divides $v$. Since the ideal $J=({\bf
x^{a}}/x_{p}:{\bf x^{a}}\in\mathcal{G}(I),{\bf a}[p]>0){\mathfrak{m}}^{k+1}$
is non-zero by our assumption and $v\in J$, there exist ${\bf
x^{a}}\in\mathcal{G}(I)$ with ${\bf a}[p]>0$ and a monomial
$u\in\mathcal{G}({\mathfrak{m}}^{k+1})$ such that $({\bf x^{a}}/x_{p})u$
divides $v$. From this we see that

$\deg_{x_{t}}(v)\geq\deg_{x_{t}}(({\bf x^{a}}/x_{p})u)\geq\deg_{x_{t}}({\bf
x^{a}}/x_{p})=\deg_{x_{t}}({\bf x^{a}})-1.$

We claim that $\deg_{x_{t}}(v)=\deg_{x_{t}}({\bf x^{a}})-1$. Suppose that
$\deg_{x_{t}}(v)>\deg_{x_{t}}({\bf x^{a}})-1$. Then $x_{t}^{{\bf a}[t]}$
divides $v$. Now, if $x_{t}$ divides $u$, then $({\bf
x^{a}}/x_{t})(x_{t}(u/x_{t}))={\bf x^{a}}(u/x_{t})\in I{\mathfrak{m}}^{k}$
divides $v$, which would imply that $v\in I{\mathfrak{m}}^{k}$, against our
assumption. Consequently $x_{t}$ does not divide $u$. Then $x_{t}({\bf
x^{a}}/x_{t})u={\bf x^{a}}u\in I{\mathfrak{m}}^{k+1}\subset
I{\mathfrak{m}}^{k}$ divides $v$ which is again a contradiction. Hence, we see
that $\deg_{x_{t}}(v)=\deg_{x_{t}}({\bf x^{a}})-1\leq\deg_{x_{t}}(I)-1$. Since
$t\in[n]$ is arbitrary, we obtain that

$\deg(v)\ =\ \sum_{t\in[n]}\deg_{x_{t}}(v)\ \leq\
\sum_{t\in[n]}(\deg_{x_{t}}(I)-1)\ =\ |{\bf deg}(I)|-n.$

However, by equation (8), $v$ must be of degree at least $k+1\geq|{\bf
deg}(I)|-n+1$, which is a contradiction. Hence
$(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})=I{\mathfrak{m}}^{k}$ for all
$k\geq|{\bf deg}(I)|-n$, and so $\mathfrak{d}_{3}(I)\leq|{\bf deg}(I)|-n$.
Finally, if $I$ is squarefree, then $|{\bf deg}(I)|\leq n$. Hence
$\max\\{|{\bf deg}(I)|-n,0\\}=0$ and so
$\mathfrak{d}_{1}(I)=\mathfrak{d}_{2}(I)=\mathfrak{d}_{3}(I)=0$.

######  Examples 3.8.

(a) Let $I=(x^{a},y^{a})\subset S=K[x,y]$ with $a\geq 1$. Then
$\mathfrak{d}_{i}(I)=a-1$ for all $i$ [10, Example 4.5]. The bound
$(\sum_{u\in\mathcal{G}(I)}\deg(u))+1-\mu(I)-\omega(I)=a-1$ given in Corollary
3.4 is optimal in this case. The bound $|{\bf deg}(I)|-n=2a-2$ provided in
Theorem 3.6 is, however, far from being optimal when $a>1$.

(b) Let
$I=(x_{1}x_{2}x_{3},x_{1}x_{2}x_{4},x_{1}x_{2}x_{5},x_{1}x_{3}x_{4},x_{1}x_{3}x_{5},x_{1}x_{4}x_{5})\subset
S=K[x_{1},\dots,x_{5}]$. Since $I$ is squarefree, $\mathfrak{d}_{i}(I)=0$ for
$i=1,2,3$. The bound $|{\bf deg}(I)|-n=0$ given in Theorem 3.6 is optimal,
while the bound $(\sum_{u\in\mathcal{G}(I)}\deg(u))+1-\mu(I)-\omega(I)=10$
given in Corollary 3.4 is not.

(c) For any $d\geq 1$, ${\mathfrak{m}}^{d}$ is componentwise linear and so
$\mathfrak{d}_{i}({\mathfrak{m}}^{d})=0$ for all $i$. Notice that
$|\mathcal{G}({\mathfrak{m}}^{d})|=\binom{n+d-1}{d}$ and ${\bf
deg}({\mathfrak{m}}^{d})=(d,d,\dots,d)$. Thus, the bound given in Corollary
3.4 is $(d-1)\binom{n+d-1}{d}+1$, and the one provided in Theorem 3.6 is
$|{\bf deg}({\mathfrak{m}}^{d})|-n=(d-1)n$. If $d>1$, then both bounds are not
optimal.

We have the next nice consequence.

######  Corollary 3.9.

Let $I\subset S$ be a monomial ideal. For all $k\geq\max\\{|{\bf
deg}(I)|-n,0\\}$,

$\textup{Soc}(S/(I{\mathfrak{m}}^{k+1}))\ =\
\frac{(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})}{I{\mathfrak{m}}^{k+1}}\ =\
\frac{I{\mathfrak{m}}^{k}}{I{\mathfrak{m}}^{k+1}}.$

In particular, $\beta_{n-1}(I{\mathfrak{m}}^{k+1})\ =\
\mu(I{\mathfrak{m}}^{k})$.

Let $u={\bf x^{a}}\in S$ be a monomial. Its support is the set defined as

$\operatorname{supp}(u)\ =\ \\{i\ :\ x_{i}\ \textup{divides}\ u\\}\ =\ \\{i\
:\ {\bf a}[i]>0\\}.$

Next, we compute the Dao numbers of monomial complete intersections.

We call an ideal $I\subset S$ a complete intersection if $I$ is generated by a
regular sequence. In particular, it is easy to see that a monomial ideal
$I\subset S$ with minimal monomial generating set
$\mathcal{G}(I)=\\{u_{1},\dots,u_{m}\\}$ is a complete intersection if and
only if $\operatorname{supp}(u_{i})\cap\operatorname{supp}(u_{j})\ =\
\emptyset$ for all $1\leq i<j\leq n$.

######  Corollary 3.10.

Let $I\subset S$ be a monomial complete intersection with minimal monomial
generating set $\mathcal{G}(I)=\\{u_{1},\dots,u_{m}\\}$.

  1. (a)

If $\bigcup_{i=1}^{m}\operatorname{supp}(u_{i})\neq[n]$, then
$\mathfrak{d}_{1}(I)=\mathfrak{d}_{2}(I)=\mathfrak{d}_{3}(I)=0.$

  2. (b)

If $\bigcup_{i=1}^{m}\operatorname{supp}(u_{i})=[n]$ and $\mathcal{G}(I)$
contains only pure powers, then $n=m$, up to relabeling $u_{i}=x_{i}^{a_{i}}$
for all $i\in[n]$ with $1\leq a_{1}\leq\dots\leq a_{n}$ and

$\mathfrak{d}_{1}(I)=\mathfrak{d}_{3}(I)=\sum_{i=1}^{n-1}a_{i}-(n-1).$

  3. (c)

If $\bigcup_{i=1}^{m}\operatorname{supp}(u_{i})=[n]$, and $\mathcal{G}(I)$
contains a non-pure power, then

$\mathfrak{d}_{1}(I)=\mathfrak{d}_{2}(I)=\mathfrak{d}_{3}(I)=0.$

###### Proof.

If $\bigcup_{i=1}^{m}\operatorname{supp}(u_{i})\neq[n]$, then the Dao numbers
are zero as shown in the first part of the proof of Theorem 3.6. Statement (a)
follows.

Now, assume that $\bigcup_{i=1}^{m}\operatorname{supp}(u_{i})=[n]$, and
$\mathcal{G}(I)$ contains only pure powers. Thus $m=n$ and up to relabeling we
can assume that $u_{i}=x_{i}^{a_{i}}$ for all $i\in[n]$ with $1\leq
a_{1}\leq\dots\leq a_{n}$. By Corollary 3.4, we have

$\mathfrak{d}_{2}(I)\leq\mathfrak{d}_{3}(I)=\mathfrak{d}_{1}(I)\leq(\\!\\!\,\\!\\!\sum_{u\in\mathcal{G}(I)}\deg(u))+1-\mu(I)-\omega(I)=\sum_{i=1}^{n-1}a_{i}-(n-1).$

Thus, it suffices to show that $I\mathfrak{m}^{k}$ is not weakly
${\mathfrak{m}}$-full for $k=\sum_{i=1}^{n-1}a_{i}-n$. For this aim, consider
the monomial $v=\prod_{i\in[n]}x_{i}^{a_{i}-1}$. Let $j\in[n]$. Notice that

$x_{j}v=u_{j}(\prod_{i\in[n]\setminus\\{j\\}}x_{i}^{a_{i}-1})\ \in\
I{\mathfrak{m}}^{\sum_{i\in[n]\setminus\\{j\\}}a_{i}-(n-1)}.$

Now, since for all $j\in[n]$ we have

$\sum_{i\in[n]\setminus\\{j\\}}\\!\\!\\!a_{i}-(n-1)\ \geq\
\sum_{i=1}^{n-1}a_{i}-n+1\ =\ k+1$

we see that $x_{j}v\in I{\mathfrak{m}}^{k+1}$ for all $j\in[n]$. Hence
$v\in(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})$. On the other hand $v\notin
I{\mathfrak{m}}^{k}$, because for each monomial $w\in
I{\mathfrak{m}}^{k}\subset I$ there exists $i\in[n]$ such that
$\deg_{x_{i}}(w)\geq a_{i}$. This shows that
$\mathfrak{d}_{1}(I)=\mathfrak{d}_{3}(I)=\sum_{i=1}^{n-1}a_{i}-(n-1)$ and (b)
follows.

Finally, assume that $\bigcup_{i=1}^{m}\operatorname{supp}(u_{i})=[n]$ and
that $\mathcal{G}(I)$ contains a non-pure power. Up to relabeling, we may
assume that $x_{1}x_{2}$ divides $u_{1}\in\mathcal{G}(I)$. Since $I$ is a
complete intersection, $x_{1}$ and $x_{2}$ do not divide any other generator
$u\in\mathcal{G}(I)\setminus\\{u_{1}\\}$. Hence, by Lemma 3.7, for all $k\geq
0$, we have

$\displaystyle(I{\mathfrak{m}}^{k+1}:x_{1})\ $ $\displaystyle=\
I{\mathfrak{m}}^{k-1}+(u_{1}/x_{1}){\mathfrak{m}}^{k},$
$\displaystyle(I{\mathfrak{m}}^{k+1}:x_{2})\ $ $\displaystyle=\
I{\mathfrak{m}}^{k-1}+(u_{1}/x_{2}){\mathfrak{m}}^{k}.$

Assume for a contradiction that $(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})\neq
I{\mathfrak{m}}^{k}$ and let $v\in(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})$ be a
monomial not belonging to $I{\mathfrak{m}}^{k}$. Arguing as in the proof of
Theorem 3.6, we see that
$v\in(I{\mathfrak{m}}^{k+1}:x_{1})\cap(I{\mathfrak{m}}^{k+1}:x_{2})$. Thus
$v=(u_{1}/x_{1})w_{1}=(u_{1}/x_{2})w_{2}$ where
$w_{1},w_{2}\in{\mathfrak{m}}^{k}$ are monomials. The previous equation
implies that $\deg_{x_{1}}(v)\geq a_{1}$. Thus $x_{1}$ divides $w_{1}$ and so
$v=(u_{1}/x_{1})(x_{1}(w/x_{1}))=u_{1}(w/x_{1})\in I{\mathfrak{m}}^{k-1}$, a
contradiction. Hence,
$(I{\mathfrak{m}}^{k+1}:{\mathfrak{m}})=I{\mathfrak{m}}^{k}$ for all $k\geq
0$, and so $\mathfrak{d}_{1}(I)=\mathfrak{d}_{2}(I)=\mathfrak{d}_{3}(I)=0$.

Acknowledgment. The author would like to thank Cleto B. Miranda-Neto and
Douglas Queiroz for their careful reading of the manuscript and several
valuable comments.

## References

  * [1] L. Amata, M. Crupi, A. Ficarra. Projective dimension and Castelnuovo–Mumford regularity of $t$-spread ideals, International Journal of Algebra and Computation, 2022, 32(4), pp. 837–858.
  * [2] W. Bruns, A. Conca, C. Raicu, M. Varbaro, Determinants, Gröbner bases and cohomology. Springer, 2022.
  * [3] W. Bruns and J. Herzog, Cohen–Macaulay rings, Cambridge University Press, 1998.
  * [4] A. Björner, M.L. Wachs, Shellable nonpure complexes and posets, I, Trans. Amer. Math. Soc. 348 (4) (1996) 1299–1327.
  * [5] O. Celikbas, T. Kobayashi, On a class of Burch ideals and a conjecture of Huneke and Wiegand, Collect. Math. 73 (2021), 221-236.
  * [6] A. Conca, Koszul algebras and their syzygies, Combinatorial algebraic geometry, Springer, 2014, pp. 1–31.
  * [7] A. Conca, E. De Negri, M.E. Rossi. Koszul algebras and regularity. In Commutative algebra, pages 285–315. Springer, New York, 2013.
  * [8] A. Conca, E. De Negri, M.E. Rossi. Integrally closed and componentwise linear ideals, Math. Z. 265 (2010), 715–734.
  * [9] M. Crupi, A. Ficarra, E. Lax, Mapping cones of monomial ideals over exterior algebras, Communications in Algebra, 2023, https://doi.org/10.1080/00927872.2023.2278665, available at arxiv.org/abs/2303.06953.
  * [10] H. Dao, On colon operations and special types of ideals, Palestine J. Math. 10 (2021), 383-388.
  * [11] H. Dao, T. Kobayashi, R. Takahashi, Burch ideals and Burch rings, Algebra Number Theory 10 (2020), 2121-2150.
  * [12] A. Ficarra, Homological shifts of polymatroidal ideals, 2022, arxiv.org/abs/2205.04163.
  * [13] A. Ficarra, J. Herzog, S. Moradi, The eventual shape of the Betti table of ${\mathfrak{m}}^{k}M$, 2023, preprint arxiv.org/abs/2308.00639.
  * [14] S. Goto, Integral closedness of complete intersection ideals, J. Algebra 108, 151–160 (1987).
  * [15] T. Harima, J. Watanabe, The weak Lefschetz property for ${\mathfrak{m}}$-full ideals and componentwise linear ideals, Illinois J. Math. 156 (2012), 957-966.
  * [16] T. Harima, J. Watanabe, Completely ${\mathfrak{m}}$-full ideals and componentwise linear ideals, Math. Proc. Camb. Phil. Soc. 158 (2015), 239-248.
  * [17] W. Heinzer, L. J. Ratliff, D. E. Rush, Basically full ideals in local rings, J. Algebra 250 (2002), 371-396.
  * [18] J. Herzog, T. Hibi. _Monomial ideals_ , Graduate texts in Mathematics 260, Springer, 2011.
  * [19] J. Herzog, S. Iyengar, Koszul modules, J. Pure and Appl. Alg. 201 (2005), 154-188.
  * [20] D. Lu, D. Zhou, Componentwise linear modules over a Koszul algebra, Taiwanese J. Math. 17 (2013), no. 6, 2135–2147. MR 3141878
  * [21] C. B. Miranda-Neto, Free logarithmic derivation modules over factorial domains, Math. Res. Lett. 24 (2017), 153-172.
  * [22] C.B. Miranda-Neto, D.S. Queiroz, Dao’s question on the asymptotic behaviour of fullness, 2023, preprint arxiv.org/abs/2308.03997.
  * [23] T. Römer, On minimal graded free resolutions, Dissertation, Universität Duisburg-Essen, 2001.
  * [24] D. E. Rush, Contracted, ${\mathfrak{m}}$-full and related classes of ideals in local rings, Glasgow Math. J. 55 (2013), 669-675.
  * [25] A. Soleyman Jahan, X. Zheng, Ideals with linear quotients, J. Combin. Theory Ser. A 117 (2010), 104–110. MR2557882 (2011d:13030).
  * [26] N. V. Trung, The Castelnuovo regularity of the Rees algebra and the associated graded ring, Trans. Amer. Math. Soc. 350 (1998), 2813-2832.
  * [27] K. Vandebogert, Iterated Mapping Cones for Strongly Koszul Algebras, 2022, preprint arxiv.org/abs/2104.00037.
  * [28] J. Watanabe, ${\mathfrak{m}}$-full ideals, Nagoya Math. J. 106 (1987), 101–111.
  * [29] J. Watanabe, The syzygies of m-full ideals, Math. Proc. Cambridge Philos. Soc. 109 (1991), 7–13.
  * [30] J. Watanabe, m-full ideals II, Math. Proc. Cambridge Philos. Soc. 111 (1991), 231–240.