# Simon Conjecture and
the v-number of monomial ideals

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.

Let $I\subset S$ be a graded ideal of a standard graded polynomial ring $S$
with coefficients in a field $K$, and let $\operatorname{v}(I)$ be the
$\operatorname{v}$-number of $I$. In previous work, we showed that for any
graded ideal $I\subset S$ generated in a single degree, then
$\operatorname{v}(I^{k})=\alpha(I)k+b$, for all $k\gg 0$, where $\alpha(I)$ is
the initial degree of $I$ and $b$ is a suitable integer. In the present paper,
using polarization, we extend Simon conjecture to any monomial ideal. As a
consequence, if Simon conjecture holds, and all powers of $I$ have linear
quotients, then $b\in\\{-1,0\\}$. This fact suggest that if $I$ is an
equigenerated monomial ideal with linear powers, then
$\operatorname{v}(I^{k})=\alpha(I)k-1$, for all $k\geq 1$. We verify this
conjecture for monomial ideals with linear powers having
$\operatorname{depth}S/I=0$, edge ideals with linear resolution, polymatroidal
ideals, and Hibi ideals.

###### Key words and phrases:

graded ideals, $\operatorname{v}$-number, asymptotic behaviour, primary
decomposition

###### 2020 Mathematics Subject Classification:

Primary 13F20; Secondary 13F55, 05C70, 05E40.

##  1\. Introduction

Let $S=K[x_{1},\dots,x_{n}]=\bigoplus_{d}S_{d}$ be the standard graded
polynomial ring with $n$ variables and coefficients in a field $K$, and let
$\mathfrak{m}=(x_{1},\dots,x_{n})$ be the graded maximal ideal. We denote the
set of associated primes of $I$ by $\operatorname{Ass}(I)$, and by
$\operatorname{Max}(I)$ the set of associated primes of $I$ that are maximal
with respect to the inclusion. The concept of $\operatorname{v}$-number was
introduced in [7], and further studied in [1, 3, 5, 17, 20, 25, 26, 29, 30,
32]. Let $I\subset S$ be a graded ideal and let
$\mathfrak{p}\in\operatorname{Ass}(I)$. Then, the $\operatorname{v}$-number of
$I$ at $\mathfrak{p}$ is defined as

$\operatorname{v}_{\mathfrak{p}}(I)\ =\ \min\\{d\ :\ \textit{there exists}\
f\in S_{d}\ \textit{such that}\ (I:f)=\mathfrak{p}\\}.$

Whereas, the $\operatorname{v}$-number of $I$ is defined as

$\operatorname{v}(I)\ =\ \min\\{d\ :\ \textit{there exists}\ f\in S_{d}\
\textit{such that}\ (I:f)\in\operatorname{Ass}(I)\\}.$

Let $I\subset S$ be a graded ideal. In [17] the asymptotic behaviour of the
function $\operatorname{v}(I^{k})$, called the $\operatorname{v}$-function of
$I$, was investigated. It is known by Brodmann [4] that
$\operatorname{Ass}(I^{k})$ stabilizes for large $k$. That is,
$\operatorname{Ass}(I^{k+1})=\operatorname{Ass}(I^{k})$ for all $k\gg 0$. A
prime ideal $\mathfrak{p}\subset S$ such that
$\mathfrak{p}\in\operatorname{Ass}(I^{k})$ for all $k\gg 0$, is called a
stable prime of $I$. The set of the stable primes of $I$ is denoted by
$\operatorname{Ass}^{\infty}(I)$. Thus, for all $k\gg 0$,

$\operatorname{v}(I^{k})=\min_{\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)}\operatorname{v}_{\mathfrak{p}}(I^{k}).$

It is expected that for any graded ideal $I\subset S$,
$\operatorname{v}(I^{k})$ becomes a linear function in $k$ for $k\gg 0$ [17,
Conjecture 4.5]. Such a conjecture has been proved when $I$ is generated in a
single degree, when $I$ does not have embedded stable primes [17, Theorem
3.1], and for all monomial ideals in two variables [17, Theorem 5.1].

To compute the $\operatorname{v}$-number, we use the following result of
Grisalde, Reyes and Villarreal [20, Theorem 3.2]. Let $M=\bigoplus_{d\geq
0}M_{d}\neq 0$ be a finitely generated graded $S$-module. We call
$\alpha(M)=\min\\{d:M_{d}\neq 0\\}$, the initial degree of $M$, and set
$\omega(M)=\max\\{d:(M/\mathfrak{m}M)_{d}\neq 0\\}$. The bar
$\overline{\phantom{p}}$ denotes the residue class modulo $I$.

######  Theorem 1.1.

Let $I\subset S$ be a graded ideal and let
$\mathfrak{p}\in\operatorname{Ass}(I)$. The following hold.

  1. (a)

If $\mathcal{G}=\\{\overline{g_{1}},\dots,\overline{g_{r}}\\}$ is a
homogeneous minimal generating set of $(I:\mathfrak{p})/I$, then

$\operatorname{v}_{\mathfrak{p}}(I)=\min\\{\deg(g_{i})\ :\ 1\leq i\leq r\
\textit{and}\ (I:g_{i})=\mathfrak{p}\\}.$

  2. (b)

$\operatorname{v}(I)=\min\\{\operatorname{v}_{\mathfrak{p}}(I):\mathfrak{p}\in\operatorname{Ass}(I)\\}$.

  3. (c)

$\operatorname{v}_{\mathfrak{p}}(I)\geq\alpha((I:\mathfrak{p})/I)$, with
equality if $\mathfrak{p}\in\operatorname{Max}(I)$.

  4. (d)

If $I$ has no embedded primes, then
$\operatorname{v}(I)=\min\\{\alpha((I:\mathfrak{p})/I):\mathfrak{p}\in\operatorname{Ass}(I)\\}$.

In this paper, we study the $\operatorname{v}$-function of monomial ideals
with linear powers.

The paper is structured as follows. In Section 2, we show in Theorem 2.3 that
for an equigenerated graded ideal $I\subset S$, we have
$\operatorname{v}(I^{k})=\alpha(I)k+b$ for some integer $b$, and all $k\gg 0$.
In general, the integer $b$ depends on $\textup{char}(K)$, as we show in
Example 2.4. However, if $I$ is a monomial ideal, then $b$ does not depend on
$\textup{char}(K)$ (Proposition 2.5) and moreover $b\geq-1$ (Theorem 2.3(c)).
Indeed, the lower bound $\alpha(I)k-1\leq\operatorname{v}(I^{k})$ (Proposition
2.2) always holds. In general, we also have the lower bound
$\alpha(I)k\leq\operatorname{reg}(I^{k})$ for the Castelnuovo–Mumford
regularity. If $I$ has linear powers, then this lower bound is achieved for
all $k$. In view of this fact, it is natural to expect that
$\operatorname{v}(I^{k})=\alpha(I)k-1$ for all $k\geq 1$, when $I$ is a
monomial ideal with linear powers (Conjecture 2.6). Next, we investigate this
conjecture.

A particular class of monomial ideals with linear powers is that of
equigenerated monomial ideals whose all powers have linear quotients. Linear
quotients is the algebraic counterpart of the concept of shellability in
simplicial complexes theory. A famous conjecture of Simon [31, Conjecture
4.2.1] can be expressed, in algebraic terms, by saying that any squarefree
monomial ideal $I\subset S$ generated in a single degree $d$ and having linear
quotients, can be extended to the squarefree Veronese ideal $I_{n,d}$ by
linear quotients (Conjecture 3.1). We extend this conjecture to all monomial
ideals (Conjecture 3.2). Surprisingly, by using polarization, it turns out
that the conjectures are equivalent (Proposition 3.4). For another instance of
such an use of polarization, see [11, Proposition 1.8]. As a consequence, if
Simon conjecture holds, then for any monomial ideal generated in a single
degree and whose all powers have linear quotients, either
$\operatorname{v}(I^{k})=\alpha(I)k-1$ or
$\operatorname{v}(I^{k})=\alpha(I)k$, for all $k\gg 0$.

The behaviour of the $\operatorname{v}$-number of monomial ideals under
polarization was studied in [30]. The authors showed that
$\operatorname{v}(I^{\wp})\leq\operatorname{v}(I)$ and equality holds in some
special cases. In Theorem 4.1, surprisingly we show that the equality
$\operatorname{v}(I)=\operatorname{v}(I^{\wp})$ holds for any monomial ideal
$I$.

Finally, we prove that Conjecture 2.6 holds: when $\operatorname{depth}S/I=0$
(Theorem 3.6), for edge ideals with linear resolution (Theorem 5.1), for
polymatroidal ideals (Theorem 5.5) and for Hibi ideals (Theorem 5.8).

##  2\. The $\operatorname{v}$-function of monomial ideals

In this section, we summarize some basic facts about the
$\operatorname{v}$-function of monomial ideals. Hereafter, for a positive
integer $n$, we set $[n]=\\{1,2,\dots,n\\}$. If $I\subset S$ is a monomial
ideal, then each associated prime $\mathfrak{p}\in\operatorname{Ass}(I)$ is a
monomial prime ideal. That is, $\mathfrak{p}=\mathfrak{p}_{A}=(x_{i}:i\in A)$
for some nonempty subset $A$ of $[n]$.

The following result [30, Proposition 3.11], due to Saha and Sengupta, is a
special case of [17, Proposition 4.3]. It provides an useful general method to
bound $\operatorname{v}(I)$ from above, when $I$ is a monomial ideal.

######  Proposition 2.1.

Let $I\subset S$ be a monomial ideal and $f\in S\setminus I$ a monomial. Then,

$\operatorname{v}(I)\leq\operatorname{v}(I:f)+\deg(f).$

On the other hand, one always has

######  Proposition 2.2.

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

$\operatorname{v}_{\mathfrak{p}}(I)\geq\alpha(I)-1,\ \ \textit{for all}\ \
\mathfrak{p}\in\operatorname{Ass}(I).$

In particular, $\operatorname{v}(I^{k})\geq\alpha(I)k-1$, for all $k\geq 1$.

###### Proof.

Let $\mathfrak{p}\in\operatorname{Ass}(I)$ and let $u\in S$ be a monomial such
that $(I:u)=\mathfrak{p}$ and $\deg(u)=\operatorname{v}_{\mathfrak{p}}(I)$.
Then $\mathfrak{p}=\mathfrak{p}_{A}=(x_{i}:i\in A)$ for some $A\subseteq[n]$.
Thus $x_{i}u\in I$ for all $i\in A$. In particular,
$\deg(x_{i}u)\geq\alpha(I)$. Hence, $\deg(u)\geq\alpha(I)-1$, as desired.

From the work of [17], the next result follows.

######  Theorem 2.3.

Let $I\subset S$ be a graded ideal generated in a single degree.

  1. (a)

For all $\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$, there exists an
integer $b_{\mathfrak{p}}$ such that
$\operatorname{v}_{\mathfrak{p}}(I^{k})=\alpha(I)k+b_{\mathfrak{p}}$ is a
linear function in $k$, for all $k\gg 0$.

  2. (b)

$\operatorname{v}(I^{k})=\alpha(I)k+b$ for all $k\gg 0$, where
$b=\min_{\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)}b_{\mathfrak{p}}$.

  3. (c)

If $I$ is a monomial ideal, then $b_{\mathfrak{p}},b\geq-1$, for all
$\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$.

###### Proof.

The proof of [17, Theorem 3.1(b)] shows that
$\operatorname{v}_{\mathfrak{p}}(I^{k})=\alpha(I)k+b_{\mathfrak{p}}$, for all
$k\gg 0$ and all $\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$, for some
$b_{\mathfrak{p}}\in{\mathbb{Z}}$. Setting
$b=\min\limits_{\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)}b_{\mathfrak{p}}$,
then

$\operatorname{v}(I^{k})\ =\
\min_{\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)}\operatorname{v}_{\mathfrak{p}}(I^{k})\
=\
\min_{\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)}(\alpha(I)k+b_{\mathfrak{p}})\
=\ \alpha(I)k+b,$

for all $k\gg 0$, and (a) and (b) follow. Part (c) follows from Proposition
2.2.

Recall that a graded ideal $I\subset S$ has a $d$-linear resolution if it is
generated in a single degree $d$, and for all $i\geq 0$, $\beta_{i,j}(I)=0$ if
$j\neq i+d$. Equivalently, an equigenerated ideal $I$ has linear resolution,
if and only if, the Castelnuovo–Mumford regularity is as small as possible,
namely $\operatorname{reg}(I)=\alpha(I)$. We say that $I$ has linear powers if
$I^{k}$ has a linear resolution, for all $k\geq 1$. Famous examples of
monomial ideals with linear powers are given in the following list.

  1. (i)

Edge ideals with linear resolution [23, Theorem 10.2.6].

  2. (ii)

Polymatroidal ideals [23, Corollary 12.6.4].

  3. (iii)

Hibi ideals [23, Corollary 10.2.9 and Theorem 9.1.13], or [8, Corollary 4.11].

In general, the $\operatorname{v}$-function of a graded ideal $I$ depends on
the characteristic of $K$, even when $I$ has linear powers.

######  Example 2.4.

Indeed, consider the principal ideal $I=(x^{2}+y^{2})$ of
$S={\mathbb{Q}}[x,y]$. Then $I^{k}$ is principal as well, and thus $I$ has
linear powers. Since $x^{2}+y^{2}$ is an irreducible polynomial over
${\mathbb{Q}}[x,y]$, $I$ is a prime ideal, and thus $I^{k}$ is a $I$-primary
ideal and $\operatorname{Ass}(I^{k})=\\{I\\}$ for all $k$. Note that

$(I^{k}:I)/I^{k}\ =\ (((x^{2}+y^{2})^{k}):(x^{2}+y^{2}))/I^{k}\ =\
((x^{2}+y^{2})^{k-1})/I^{k}$

for all $k\geq 1$. Hence,
$\operatorname{v}(I^{k})=\operatorname{v}_{I}(I^{k})=2k-2$ for all $k\geq 1$.

On the other hand, if $I=(x^{2}+y^{2})$ is an ideal of ${\mathbb{C}}[x,y]$,
then $(x^{2}+y^{2})^{k}=(x+iy)^{k}(x-iy)^{k}$ for all $k\geq 1$. Setting
$\mathfrak{p}_{1}=(x+iy)$ and $\mathfrak{p}_{2}=(x-iy)$, then
$I^{k}=\mathfrak{p}_{1}^{k}\mathfrak{p}_{2}^{k}=\mathfrak{p}_{1}^{k}\cap\mathfrak{p}_{2}^{k}$
is the primary decomposition of $I^{k}$ and
$\operatorname{Ass}(I^{k})=\\{\mathfrak{p}_{1},\mathfrak{p}_{2}\\}$ for all
$k\geq 1$. Note that

$(I^{k}:\mathfrak{p}_{1})/I^{k}\ =\ (((x^{2}+y^{2})^{k}):(x+iy))/I^{k}\ =\
((x+iy)^{k-1}(x-iy)^{k})/I^{k}$

for all $k\geq 1$. Hence $\operatorname{v}_{\mathfrak{p}_{1}}(I^{k})=2k-1$ for
all $k\geq 1$. By symmetry, we also have
$\operatorname{v}_{\mathfrak{p}_{2}}(I^{k})=2k-1$ for all $k\geq 1$. Thus
$\operatorname{v}(I^{k})=2k-1$ for all $k\geq 1$.

However, the primary decomposition of a monomial ideal $I\subset S$ does not
depend on the characteristic, $\textup{char}(K)$, of the field $K$ [23,
Theorem 1.3]. Next, since each $\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$
is a monomial prime ideal, it follows from [23, Proposition 1.2.2] that the
monomial $K$-basis of $(I^{k}:\mathfrak{p})/I^{k}$ is independent from
$\text{char}(K)$. This observations show that

######  Proposition 2.5.

Let $I\subset S$ be a monomial ideal. Then, the functions
$\operatorname{v}_{\mathfrak{p}}(I^{k})$ for each
$\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$ and all $k\gg 0$, and the
function $\operatorname{v}(I^{k})$ for $k\geq 1$, do not depend upon
$\textup{char}(K)$.

Now, suppose that $I\subset S$ is a monomial ideal with linear powers. Then,
the regularity of $I^{k}$ attains its minimum value possible, namely
$\alpha(I^{k})=k\alpha(I)$. Hence, $\operatorname{reg}(I^{k})=\alpha(I)k$, for
all $k\geq 1$. Thus it is natural to expect that the $\operatorname{v}$-number
of $I^{k}$ attains its minimal possible value for all $k$, namely
$\operatorname{v}(I^{k})=\alpha(I)k-1$ for all $k\geq 1$.

######  Conjecture 2.6.

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

$\operatorname{v}(I^{k})\ =\ \alpha(I)k-1,\ \ \text{for all}\ k\geq 1.$

If $I\subset S$ is a graded ideal with linear powers, but not a monomial
ideal, then the above conjecture is false, as the ideal
$(x^{2}+y^{2})\subset{\mathbb{Q}}[x,y]$ of Example 2.4 shows.

Similarly, if $I$ is a monomial ideal, but does not have linear powers,
Conjecture 2.6 is no longer valid. Next example is due to Terai [6, Remark 3].
If $\textup{char}(K)\neq 2$, the Stanley Reisner ideal
$I=(abd,abf,ace,adc,aef,bde,bcf,bce,cdf,def)$ of the minimal triangulation of
the projective plane has a linear resolution, but $I^{2}$ has not. By using
Macaulay2 [19], we have $\operatorname{v}(I)=\alpha(I)=3$ and
$\operatorname{v}(I^{k})=3k-1$ for all $k\geq 2$.

##  3\. The monomial Simon conjecture

Let $I\subset S$ be a monomial ideal and $G(I)$ be the minimal monomial
generating set of $I$. We say that $I$ has linear quotients if there exists an
order $u_{1},\dots,u_{m}$ of $G(I)$ such that $(u_{1},\dots,u_{j-1}):u_{j}$ is
generated by variables, for $j=2,\dots,m$.

The monomial ideal of $S$ generated by all squarefree monomial ideals of
degree $1\leq d\leq n$ is called the Veronese ideal of degree $d$ of $S$, and
it is denoted by $I_{n,d}$. For instance,
$I_{4,3}=(x_{1}x_{2}x_{3},x_{1}x_{2}x_{4},x_{1}x_{3}x_{4},x_{2}x_{3}x_{4})$.

Let $I\subset J$ be monomial ideals with $G(I)\subset G(J)$. We say that $I$
can be extended to $J$ by linear quotients if the set $G(J)\setminus G(I)$ can
be ordered as $v_{1},\dots,v_{r}$ such that $(I,v_{1},\dots,v_{j-1}):v_{j}$ is
generated by variables, for all $j=2,\dots,r$.

The famous Simon conjecture [31, Conjecture 4.2.1] states that the skeleton of
any simplex is extendably shellable, see [2] and the references therein for
more details on this topic. In algebraic terms, the conjecture can be
equivalently stated as follows.

######  Conjecture 3.1.

Let $I\subset S$ be a squarefree monomial ideal with linear quotients such
that $G(I)\subset G(I_{n,d})$. Then $I$ can be extended to $I_{n,d}$ by linear
quotients.

It is natural to extend the above conjecture to all monomial ideals.

######  Conjecture 3.2.

Let $I\subset S$ be a monomial ideal with linear quotients such that
$G(I)\subset G(\mathfrak{m}^{d})$. Then $I$ can be extended to
$\mathfrak{m}^{d}$ by linear quotients.

We refer to the above conjecture as the monomial Simon conjecture. Since the
squarefree part of an ideal with linear quotients has linear quotients [12,
Lemma 3.1], this latter conjecture implies the usual Simon conjecture.
Surprisingly, however, it turns out that the two statements are equivalent, as
we show next.

######  Proposition 3.3.

Simon conjecture is equivalent to the monomial Simon conjecture.

We use polarization. Let $u=x_{1}^{b_{1}}\cdots x_{n}^{b_{n}}\in S$ be a
monomial. The $x_{i}$-degree of $u$ is the integer $\deg_{x_{i}}(u)=b_{i}$ for
all $i$. The polarization of $u$ is the monomial

$u^{\wp}\ =\ \prod_{i=1}^{n}(\prod_{j=1}^{b_{i}}x_{i,j})\ =\
\prod_{\begin{subarray}{c}1\leq i\leq n\\\
b_{i}>0\end{subarray}}x_{i,1}x_{i,2}\cdots x_{i,b_{i}}$

in the polynomial ring $K[x_{i,j}:i\in[n],j\in[b_{i}]]$.

Let $I\subset S$ be a monomial ideal. Following [14], the bounding multidegree
of $I$ is the vector ${\bf deg}(I)=(\deg_{x_{1}}(I),\dots,\deg_{x_{n}}(I))$,
with $\deg_{x_{i}}(I)=\max_{u\in G(I)}\deg_{x_{i}}(u)$, for all $1\leq i\leq
n$. The polarization of $I$ is defined to be the squarefree ideal $I^{\wp}$ of
$S^{\wp}=K[x_{i,j}:i\in[n],j\in[\deg_{x_{i}}(I)]]$ such that
$G(I^{\wp})=\\{u^{\wp}:u\in G(I)\\}$.

Attached to the polarization of $I$, we have the specialization map
$\pi:S^{\wp}\rightarrow S$ defined by setting $\pi(x_{i,j})=x_{i}$ for all $i$
and $j$.

###### Proof of Proposition 3.3.

We only need to show that the Simon conjecture implies the monomial version.
Let $I\subset S$ be a monomial ideal generated in degree $d$ with linear
quotients. Say with order $u_{1},\dots,u_{m}$ such that
$(u_{1},\dots,u_{j-1}):u_{j}$ is generated by variables, for $j=2,\dots,m$.
Then, [24, Lemma 3.3] (see, also, [8, Lemma 4.10]) implies that $I^{\wp}$ also
has linear quotients with order $u_{1}^{\wp},\dots,u_{m}^{\wp}$. Denote by $J$
the Veronese ideal of degree $d$ of $S^{\wp}$. By our assumption, $I^{\wp}$
can be extended to $J$ by linear quotients. Thus, there is an order
$u_{1}^{\wp},\dots,u_{m}^{\wp},v_{1},\dots,v_{r}$ of $G(J)$ such that $J$ has
linear quotients with respect to this order. Applying the specialization map,
we obtain an order $u_{1},\dots,u_{m},\pi(v_{1}),\dots,\pi(v_{r})$ of
generators of $\mathfrak{m}^{d}$ such that $(u_{1},\dots,u_{j-1}):u_{j}$, for
$j=2,\dots,m$, and
$(u_{1},\dots,u_{m},\pi(v_{1}),\dots,\pi(v_{j-1})):\pi(v_{j})$, for
$j=2,\dots,r$, are generated by variables. Removing all minimal generators of
$\mathfrak{m}^{d}$ in this order wherever they appear again, we obtain an
order of the minimal generators of $\mathfrak{m}^{d}$ whose beginning is
$u_{1},\dots,u_{m}$ such that $\mathfrak{m}^{d}$ has linear quotients with
respect to this order. Hence, $I$ can be extended to $\mathfrak{m}^{d}$ by
linear quotients, and the monomial Simon conjecture holds.

As a consequence, we have the following result.

######  Proposition 3.4.

Assume the monomial Simon conjecture. Let $I\subset S$ be a monomial ideal
generated in a single degree and having linear quotients. Then

$\alpha(I)-1\leq\operatorname{v}(I)\leq\alpha(I).$

###### Proof.

Let $\alpha(I)=d$. If $I=\mathfrak{m}^{d}$, then
$\operatorname{Ass}(I)=\\{\mathfrak{m}\\}$,
$(I:\mathfrak{m})/I=\mathfrak{m}^{d-1}/\mathfrak{m}^{d}$ and so
$\operatorname{v}(I)=\operatorname{v}_{\mathfrak{m}}(I)=d-1=\alpha(I)-1$.
Suppose now that $I\neq\mathfrak{m}^{d}$. Then, $I$ can be extended to
$\mathfrak{m}^{d}$ by linear quotients. Hence there exists a monomial
$u\in\mathfrak{m}^{d}\setminus I$ such that $(I:u)$ is generated by variables.
Applying Proposition 2.1, we get

$\operatorname{v}(I)\ \leq\ \operatorname{v}(I:u)+\deg(u)=0+d=\alpha(I).$

This inequality together with Proposition 2.2 yields the conclusion.

The next result supports Conjecture 2.6.

######  Corollary 3.5.

Assume the monomial Simon conjecture. Let $I\subset S$ be a monomial ideal
generated in single degree whose all powers have linear quotients. Then,
either $\operatorname{v}(I^{k})=\alpha(I)k-1$ or
$\operatorname{v}(I^{k})=\alpha(I)k$, for all $k\gg 0$.

###### Proof.

By Theorem 2.3(c), $\operatorname{v}(I^{k})=\alpha(I)k+b$ for all $k\gg 0$,
with $b\geq-1$. On the other hand, Proposition 3.4 implies that $b\leq 0$.
Thus either $b=-1$ or $b=0$.

We end the section, with the next general result which supports Conjecture
2.6.

######  Theorem 3.6.

Let $I\subset S$ be a monomial ideal with linear powers, such that
$\operatorname{depth}S/I=0$. Then,

$\operatorname{v}(I^{k})=\alpha(I)k-1,\ \ \textit{for all}\ k\geq 1.$

###### Proof.

Let $\alpha(I)=d$. By [23, Proposition 10.3.4],
$\operatorname{depth}(S/I^{k})$ is a nonincreasing function. Hence,
$\operatorname{depth}S/I^{k}=0$ for all $k\geq 1$. Therefore,
$\mathfrak{m}\in\operatorname{Ass}(I^{k})$ for all $k\geq 1$.

On the other hand, by [14, Lemma 3.1], the condition
$\operatorname{depth}S/I^{k}=0$ is equivalent to
$G(I^{k}:\mathfrak{m})_{dk-1}\neq 0$, for all $k\geq 1$. Here $G(J)_{j}$
denotes the set of all monomials of degree $j$ belonging to $G(J)$. Since
$I^{k}$ is generated in degree $dk$, it follows that
$\alpha((I^{k}:\mathfrak{m})/I^{k})=dk-1$ for all $k\geq 1$. Hence, Theorem
1.1(c) implies that
$\operatorname{v}_{\mathfrak{m}}(I^{k})=\alpha((I^{k}:\mathfrak{m})/I^{k})=dk-1$
for all $k\geq 1$. Thus
$\operatorname{v}(I^{k})\leq\operatorname{v}_{\mathfrak{m}}(I^{k})=\alpha(I)k-1$
for all $k\geq 1$. Equality follows from Proposition 2.2.

##  4\. Reduction to the squarefree case

In [30], the authors studied the behaviour of the $\operatorname{v}$-number of
monomial ideals under polarization. They showed that
$\operatorname{v}(I^{\wp})\leq\operatorname{v}(I)$ and equality holds in some
special cases. In this section, surprisingly we show that the equality
$\operatorname{v}(I)=\operatorname{v}(I^{\wp})$ holds for all monomial ideals.

As in the previous section, let $I\subset S$ be a monomial ideal, and let
$I^{\wp}$ be the polarization of $I$ in the polynomial ring
$S^{\wp}=K[x_{i,j}:i\in[n],j\in[\deg_{x_{i}}(I)]]$.

######  Theorem 4.1.

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

  1. (a)

$\mathfrak{p}\in\operatorname{Ass}(I)$ if and only if there exists
$\mathfrak{q}\in\operatorname{Ass}(I^{\wp})$ with
$\pi(\mathfrak{q})=\mathfrak{p}$.

  2. (b)

For all $\mathfrak{q}\in\operatorname{Ass}(I^{\wp})$, we have
$\operatorname{v}_{\mathfrak{p}}(I)\leq\operatorname{v}_{\mathfrak{q}}(I^{\wp})$,
where $\mathfrak{p}=\pi(\mathfrak{q})$.

  3. (c)

For all $\mathfrak{p}\in\operatorname{Ass}(I)$, we have

$\operatorname{v}_{\mathfrak{p}}(I)=\min\\{\operatorname{v}_{q}(I^{\wp})\ :\
\mathfrak{q}\in\operatorname{Ass}(I^{\wp}),\pi(\mathfrak{q})=\mathfrak{p}\\}.$

  4. (d)

$\operatorname{v}(I)=\operatorname{v}(I^{\wp})$.

###### Proof.

(a) is proved in [13, Corollary 3.7].

(b) Let $\mathfrak{q}\in\operatorname{Ass}(I^{\wp})$ and let
$\mathfrak{p}=\pi(\mathfrak{q})$ be, thanks to part (a), the corresponding
prime of $\operatorname{Ass}(I)$. Suppose that $f\in S^{\wp}$ is a monomial
such that $\operatorname{v}_{\mathfrak{q}}(I^{\wp})=\deg(f)$ and
$(I^{\wp}:f)=\mathfrak{q}$. In view of [23, Proposition 1.2.2] it is clear
that $\mathfrak{p}=\pi(\mathfrak{q})=\pi(I^{\wp}:f)=(I:\pi(f))$. Thus,
$\operatorname{v}_{\mathfrak{p}}(I)\leq\deg(\pi(f))=\deg(f)=\operatorname{v}_{\mathfrak{q}}(I^{\wp})$.

(c) Let $f\in S$ be a monomial such that $(I:f)=\mathfrak{p}$ and
$\deg(f)=\operatorname{v}_{\mathfrak{p}}(I)$. Then, by [24, Lemma 3.3] we have
that $(I^{\wp}:f^{\wp})$ is prime and
$\mathfrak{p}=(I:f)=\pi((I^{\wp}:f^{\wp}))$. Thus
$(I^{\wp}:f^{\wp})=\mathfrak{q}\in\operatorname{Ass}(I^{\wp})$. This shows
$\operatorname{v}_{\mathfrak{q}}(I^{\wp})\leq\operatorname{v}_{\mathfrak{p}}(I)$
and together with part (b) we have
$\operatorname{v}_{\mathfrak{p}}(I)=\operatorname{v}_{\mathfrak{q}}(I^{\wp})$.
Since
$\operatorname{v}_{\mathfrak{p}}(I)\leq\operatorname{v}_{\mathfrak{r}}(I^{\wp})$
for all $\mathfrak{r}\in\operatorname{Ass}(I^{\wp})$ such that
$\pi(\mathfrak{r})=\mathfrak{p}$, it follows that
$\operatorname{v}_{\mathfrak{q}}(I)$ and so, also
$\operatorname{v}_{\mathfrak{p}}(I)$, is equal to
$\min\\{\operatorname{v}_{\mathfrak{r}}(I):\mathfrak{r}\in\operatorname{Ass}(I^{\wp}),\pi(\mathfrak{r})=\mathfrak{p}\\}$.

(d) By part (c) we obtain that

$\operatorname{v}(I)\ =\
\min_{\mathfrak{p}\in\operatorname{Ass}(I)}\operatorname{v}_{\mathfrak{p}}(I)\
=\
\min_{\mathfrak{p}\in\operatorname{Ass}(I)}(\min_{\begin{subarray}{c}\mathfrak{q}\in\operatorname{Ass}(I^{\wp})\\\
\pi(\mathfrak{q})=\mathfrak{p}\end{subarray}}\operatorname{v}_{\mathfrak{q}}(I^{\wp})).$

By part (a), we conclude that the above minimum is
$\operatorname{v}(I^{\wp})$, as desired.

##  5\. The $\operatorname{v}$-number of monomial ideals with linear powers

In this section, we study equigenerated monomial ideals $I$ arising from
combinatorial contexts. In particular, we show that for any ideal $I$ in the
list (i)-(ii)-(iii) at page 2, Conjecture 2.6 holds true.

###  5.1. Edge ideals with linear resolution

Let $G$ be a finite simple graph with vertex set $V(G)=[n]$ and edge set
$E(G)$. The edge ideal of $G$ is the monomial ideal $I(G)$ of $S$ generated by
the monomials $x_{i}x_{j}$ such that $\\{i,j\\}\in E(G)$. A graph $G$ is
complete if every $\\{i,j\\}$ with $i,j\in[n]$, $i\neq j$, is an edge of $G$.
The open neighbourhood of $i\in V(G)$ is the set

$N_{G}(i)=\\{j\in V(G):\\{i,j\\}\in E(G)\\}.$

Whereas, the closed neighbourhood of $i\in V(G)$ is defined as
$N_{G}[i]=N_{G}(i)\cup\\{i\\}$.

A graph $G$ is called chordal if it has no induced cycles of length bigger
than three. Recall that a perfect elimination order of $G$ is an ordering
$v_{1},\dots,v_{n}$ of its vertex set $V(G)$ such that $N_{G_{i}}(v_{i})$
induces a complete subgraph on $G_{i}$, where $G_{i}$ is the induced subgraph
of $G$ on the vertex set $\\{i,i+1,\dots,n\\}$. Hereafter, if $1,2,\dots,n$ is
a perfect elimination order of $G$, we denote it by $x_{1}>x_{2}>\dots>x_{n}$.

A famous theorem of Dirac guarantees that a finite simple graph $G$ is chordal
if and only if $G$ admits a perfect elimination order [10].

The complementary graph $G^{c}$ of $G$ is the graph with vertex set
$V(G^{c})=V(G)$ and where $\\{i,j\\}$ is an edge of $G^{c}$ if and only if
$\\{i,j\\}\notin E(G)$. A graph $G$ is called cochordal if and only if $G^{c}$
is chordal. In [18], Fröberg proved that $I(G)$ has a linear resolution if and
only if $G$ is cochordal.

By Theorem 2.3 and Proposition 2.5, for any graph $G$ and all $k\gg 0$, we
have

$\operatorname{v}(I(G)^{k})\ =\ 2k+b(G),$

where $b(G)\geq-1$ is a constant independent from $\textup{char}(K)$. In
particular, $\operatorname{v}(I(G)^{k})\geq 2k-1$ for all $k\geq 1$. As a
consequence of Dirac and Fröberg theorems, we show that for any edge ideal
with linear resolution this lower bound is achieved.

######  Theorem 5.1.

Let $I(G)$ be the edge ideal of a graph $G$. Suppose that $I(G)$ has a linear
resolution. Then,

$\operatorname{v}(I(G)^{k})\ =\ 2k-1,\ \ \textit{for all}\ \ k\geq 1.$

The proof is based upon the next lemma. If $A$ is a subset of $V(G)$, the
induced subgraph of $G$ on $A$, is the graph on vertex set $A$ and edges
$\\{i,j\\}\in E(G)$ such that $i,j\in A$.

######  Lemma 5.2.

Let $I(G)$ be an edge ideal with linear resolution, and let
$x_{1}>x_{2}>\cdots>x_{n}$ be a perfect elimination order of $G^{c}$. Then,

$(I(G):x_{1})\ =\ (x_{j}:j\in N_{G}(1)).$ (1)

###### Proof.

It is clear that

$(x_{j}:j\in N_{G}(1))\ \subseteq\ (I(G):x_{1}).$

To end the proof, we show the opposite inclusion. Let $x_{k}x_{\ell}\in I(G)$
and suppose that both $k$ and $\ell$ are not in $N_{G}(1)$. Then
$\\{1,k\\},\\{1,\ell\\}\in E(G^{c})$, that is $k,\ell\in N_{G^{c}}(1)$. Since,
$x_{1}>x_{2}>\dots>x_{n}$ is a perfect elimination order of $G^{c}$, it
follows that that $N_{G^{c}}(1)$ induces a complete subgraph of $G_{2}^{c}$,
where $G_{2}^{c}$ is the induced subgraph of $G^{c}$ on the vertex set
$\\{2,\dots,n\\}$. Since $k,\ell>1$, it follows that $\\{k,\ell\\}\in
E(G^{c})$, in contradiction with $\\{k,\ell\\}\in E(G)$. Thus either $k\in
N_{G}(1)$ or $\ell\in N_{G}(1)$ and formula (1) follows.

As a consequence, we obtain by a different method the next result already
proved in [26, Proposition 3.19].

######  Corollary 5.3.

Let $I(G)$ be the edge ideal of a graph $G$. Suppose that $I(G)$ has a linear
resolution. Then, $\operatorname{v}(I(G))=1$.

###### Proof.

We proceed by induction on $|V(G)|\geq 2$ with the base case being trivial.
Let $|V(G)|>2$. Let $x_{1}>x_{2}>\cdots>x_{n}$ be a perfect elimination order
of $G^{c}$. Then, by Lemma 5.2, equation (1) holds. Thus, by Proposition 2.1,
$\operatorname{v}(I(G))\leq\operatorname{v}(I(G):x_{1})+\deg(x_{1})=0+1=1$. On
the other hand, by Proposition 2.2,
$\operatorname{v}(I(G))\geq\alpha(I(G))-1=1$. The assertion follows.

######  Remark 5.4.

Let $I\subset S$ be a graded ideal. Suppose that $(I^{k+1}:I)=I^{k}$ and
$\mathfrak{p}\in\operatorname{Ass}(I^{k})$ for all $k\geq 1$. Then, the proof
of [17, Proposition 3.1] shows that
$\operatorname{v}_{\mathfrak{p}}(I^{k+1})\leq\operatorname{v}_{\mathfrak{p}}(I^{k})+\omega(I)$
for all $k\geq 1$.

Now, we are in the position to prove Theorem 5.1.

###### Proof of Theorem 5.1.

By the previous result, $\operatorname{v}(I(G))=1$. Therefore, for some
$\mathfrak{p}\in\operatorname{Ass}(I(G))$, we have
$\operatorname{v}_{\mathfrak{p}}(I(G))=1$. By [27, Theorem 2.15], we have

$\operatorname{Ass}(I(G))\subseteq\operatorname{Ass}(I(G)^{2})\subseteq\dots\subseteq\operatorname{Ass}(I(G)^{k})\subseteq\cdots.$

Hence, $\mathfrak{p}\in\operatorname{Ass}(I(G)^{k})$ for all $k\geq 1$. By
[27, Lemma 2.12], $(I(G)^{k+1}:I(G))=I(G)^{k}$ for all $k\geq 1$. Since
$\alpha(I(G)^{k+1})=2(k+1)$ and $\omega(I(G))=2$, by Remark 5.4 and
Proposition 2.2, we have

$2(k+1)-1\leq\operatorname{v}_{\mathfrak{p}}(I(G)^{k+1})\leq\operatorname{v}_{\mathfrak{p}}(I(G)^{k})+2$

for all $k\geq 1$. By induction on $k\geq 1$, we may assume that
$\operatorname{v}_{\mathfrak{p}}(I(G)^{k})=2k-1$. The above chain of
inequalities gives
$\operatorname{v}_{\mathfrak{p}}(I(G)^{k+1})=2(k+1)-1=\alpha(I(G)^{k+1})-1$.
By Proposition 2.2 it follows that $\operatorname{v}(I(G)^{k})=2k-1$ for all
$k\geq 1$, as well.

###  5.2. Polymatroidal ideals

Let $I\subset S$ be a monomial ideal generated in a single degree. Then $I$ is
called polymatroidal if the following exchange property holds: for all $u,v\in
G(I)$ and all $i$ such that $\deg_{x_{i}}(u)>\deg_{x_{i}}(v)$ there exists $j$
such that $\deg_{x_{j}}(u)<\deg_{x_{j}}(v)$ and $x_{j}(u/x_{i})\in G(I)$.

Such a name is justified by the fact that the set of the multidegrees of the
minimal generators of $I$ is the set of the bases of a discrete polymatroid
[23, Chapter 12].

This class of monomial ideals is very rich. Indeed, it includes

  1. (i)

Graphic matroids. They are the ideals generated by the monomials $\prod_{i\in
F}x_{i}$, for all spanning forests $F$ of a finite simple graph $G$ on $n$
vertices.

  2. (ii)

Transversal polymatroidal ideals. They are of the form
$I=\mathfrak{p}_{A_{1}}\mathfrak{p}_{A_{2}}\cdots\mathfrak{p}_{A_{r}}$, for
some finite collection $A_{1},\dots,A_{r}$ of arbitrary nonempty subsets of
$[n]$.

  3. (iii)

Ideals of Veronese type. Let ${\bf c}=(c_{1},\dots,c_{n})\in{\mathbb{Z}}^{n}$
be a vector with non negative entries. Then, the ideal of Veronese type
$(d,{\bf c})$ is defined as

$I_{n,d,{\bf c}}\ =\ (x_{1}^{a_{1}}\cdots x_{n}^{a_{n}}\in S\ :\
\sum_{i=1}^{n}a_{i}=d,\ \ a_{i}\leq c_{i},\ \text{for}\ i\in[n]).$

Polymatroidal ideals also satisfies a dual exchange property [22, Lemma 2.1],
namely: for all $u,v\in G(I)$ and all $i$ such that
$\deg_{x_{i}}(u)<\deg_{x_{i}}(v)$ there exists $j$ such that
$\deg_{x_{j}}(u)>\deg_{x_{j}}(v)$ and $x_{i}(u/x_{j})\in G(I)$.

######  Theorem 5.5.

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

$\operatorname{v}(I^{k})=\alpha(I)k-1,\ \ \textit{for all}\ \ k\geq 1.$

The proof is based upon the next lemma.

######  Lemma 5.6.

Let $I\subset S$ be a polymatroidal ideal generated in degree $\alpha(I)\geq
2$. Then $(I:x_{i})$ is again a polymatroidal ideal generated in degree
$\alpha(I)-1$, for all $i\in[n]$.

###### Proof.

We may assume that $i=1$. We can write $I=x_{1}I_{1}+I_{2}$, where $I_{1}$ and
$I_{2}$ are the unique monomial ideals of $S$ such that

$\displaystyle G(x_{1}I_{1})\ $ $\displaystyle=\ \\{u\in G(I):x_{1}\
\text{divides}\ u\\},$ $\displaystyle G(I_{2})\ $ $\displaystyle=\ \\{u\in
G(I):x_{1}\ \text{does not divide}\ u\\}.$

We claim that $I_{2}\subset I_{1}$. It is enough to show that
$G(I_{2})\subseteq I_{1}$. Let $u\in G(I_{2})$ and let $v\in G(x_{1}I_{1})$.
Then $\deg_{x_{1}}(u)=0<\deg_{x_{1}}(v)$. Thus, by the dual exchange property,
we can find $j$ such that $\deg_{x_{j}}(u)>\deg_{x_{j}}(v)$ and
$x_{1}(u/x_{j})\in G(I)$. Hence $x_{1}(u/x_{j})\in x_{1}I_{1}$, and so
$u/x_{j}\in I_{1}$. Consequently, $u\in I_{1}$ too, and thus $I_{2}\subset
I_{1}$.

By the previous paragraph, we have $(I:x_{1})=I_{1}+I_{2}=I_{1}$. It is clear
that $I_{1}$ is equigenerated in degree $\alpha(I)-1$. It remains to prove
that $I_{1}$ is polymatroidal. Let $u_{1},v_{1}\in G(I_{1})$ and $i$ such that
$\deg_{x_{i}}(u_{1})>\deg_{x_{i}}(v_{1})$. Our job is to find $j$ such that
$\deg_{x_{j}}(u_{1})<\deg_{x_{j}}(v_{1})$ and $x_{j}(u_{1}/x_{i})\in
G(I_{1})$. Set $u=x_{1}u_{1}$ and $v=x_{1}v_{1}$. Then $u,v\in
G(x_{1}I_{1})\subset G(I)$ and $\deg_{x_{i}}(u)>\deg_{x_{i}}(v)$. Since $I$ is
polymatroidal, there exists $j$ such that $\deg_{x_{j}}(u)<\deg_{x_{j}}(v)$
and $x_{j}(u/x_{i})\in G(I)$. We claim that $x_{1}$ divides $x_{j}(u/x_{i})$.
Indeed $x_{1}$ divides $u$. If $i\neq 1$, then $x_{1}$ divides
$x_{j}(u/x_{i})$ as well. If $i=1$, since $x_{1}$ divides $v$ and
$\deg_{x_{i}}(u)>\deg_{x_{i}}(v)>0$, it follows that $x_{1}^{2}$ divides $u$
and so $x_{1}$ divides $x_{j}(u/x_{i})=x_{j}(u/x_{1})$. Therefore, in any case
$x_{1}$ divides $x_{j}(u/x_{i})$. Hence,
$(x_{j}(u/x_{i}))/x_{1}=x_{j}(u_{1}/x_{i})\in G(I_{1})$ and the proof is
complete.

We are ready for the proof of the theorem.

###### Proof of Theorem 5.5.

Firstly, we show that $\operatorname{v}(I)=\alpha(I)-1$. We proceed by strong
induction on $\alpha(I)\geq 1$. If $\alpha(I)=1$, then $I=\mathfrak{p}_{A}$
for some $A\subseteq[n]$, $\alpha(I)=1$ and
$\operatorname{v}(I)=\operatorname{v}_{\mathfrak{p}_{A}}(I)=0$. Suppose
$\alpha(I)>1$. By the previous proposition, $(I:x_{1})$ is a polymatroidal
ideal and $\alpha(I:x_{1})=\alpha(I)-1$. By induction hypothesis,
$\operatorname{v}(I:x_{1})=\alpha(I:x_{1})-1=\alpha(I)-2$. Hence, by
Proposition 2.1,

$\operatorname{v}(I)\leq\operatorname{v}(I:x_{1})+\deg(x_{1})=\alpha(I)-1.$

By Proposition 2.2, $\operatorname{v}(I)\geq\alpha(I)-1$. Equality follows.

Let $k>1$. It is well–known that the product of polymatroidal ideals is
polymatroidal [23, Theorem 12.6.3]. Hence, $I^{k}$ is a polymatroidal ideal
generated in degree $\alpha(I)k$. By what shown above,
$\operatorname{v}(I^{k})=\alpha(I^{k})-1=\alpha(I)k-1$.

###  5.3. Hibi ideals

Let $(P,\succeq)$ be a finite partially ordered set (a poset, for short) with
$P=\\{p_{1},\ldots,p_{n}\\}$. A poset ideal of $P$ is a subset $\mathcal{I}$
of $P$ such that if $p_{i}\in P$, $p_{j}\in\mathcal{I}$ and $p_{i}\preceq
p_{j}$, then $p_{i}\in\mathcal{I}$. To any poset ideal $\mathcal{I}$ of $P$,
we associate the monomial
$u_{\mathcal{I}}=(\prod_{p_{i}\in\mathcal{I}}x_{i})(\prod_{p_{i}\in
P\setminus\mathcal{I}}y_{i})\in S=K[x_{1},\ldots,x_{n},y_{1},\ldots,y_{n}]$.
The set of all poset ideals of $P$ is denoted by $\mathcal{J}(P)$. Then the
Hibi ideal of $P$ is the monomial ideal of $S$ defined as

$H_{P}\ =\ (u_{\mathcal{I}}\ :\ \mathcal{I}\in\mathcal{J}(P)).$

######  Example 5.7.

Consider the poset $(P,\succeq)$ with $P=\\{p_{1},p_{2},p_{3}\\}$, $p_{1}\prec
x_{2}$ and $p_{1}\prec p_{3}$. The poset $(P,\succeq)$ and the distributive
lattice $\mathcal{J}(P)$ are depicted below:

$x_{1}$$x_{2}$$x_{3}$$\\{x_{1},x_{2}\\}$$\\{x_{1}\\}$$\\{x_{1},x_{2},x_{3}\\}$$\\{x_{1},x_{3}\\}$$\emptyset$$(P,\succeq)$$\mathcal{J}(P)$

Then,
$H_{P}=(x_{1}x_{2}x_{3},\,x_{1}x_{2}y_{3},\,x_{1}y_{2}x_{3},\,x_{1}y_{2}y_{3},\,y_{1}y_{2}y_{3})$.

The Hibi ideal $H_{P}$ is equigenerated in degree $|P|$. It is well known that
Hibi ideals have linear powers. Next, we calculate the
$\operatorname{v}$-function of $H_{P}$.

######  Theorem 5.8.

Let $H_{P}$ be a Hibi ideal. Then,

$\operatorname{v}(H_{P}^{k})=k|P|-1,\ \ \textit{for all}\ \ k\geq 1.$

In order to prove the theorem, we need some preliminary lemmata.

######  Lemma 5.9.

Let $H_{P}$ be a Hibi ideal. Then,

$\operatorname{Ass}(H_{P}^{k})\ =\ \\{(x_{i},y_{j}):p_{i}\preceq p_{j}\\},\ \
\textit{for all}\ k\geq 1.$

###### Proof.

By [21, Corollary 1.2] the Rees algebra $\mathcal{R}(H_{P})$ is a normal
domain. Hence, $H_{P}$ is a normal ideal (see, also, [9, Corollary 3.5]).
Thus, by [28, Proposition (4.7)], $(H_{P}^{k+1}:H_{P})=H_{P}^{k}$, for all
$k\geq 1$. By [23, Theorem 9.1.13] $H_{P}$ is the cover ideal $J(G)$ of a
Cohen–Macaulay bipartite graph $G$. Next, by [32, Theorem 6.10] we have that
$J(G)^{k}=J(G)^{(k)}$ for all $k\geq 1$, that is, ordinary and symbolic powers
of $J(G)=H_{P}$ coincide. Since $H_{P}$ is a squarefree monomial ideal, by
[23, Proposition 1.4.4 and Corollary 1.3.6] we have

$H_{P}^{k}\ =\ H_{P}^{(k)}\ =\ \bigcap_{p_{i}\preceq p_{j}}(x_{i},y_{j})^{k}.$

Hence, $\operatorname{Ass}(H_{P}^{k})=\operatorname{Ass}(H_{P})$ for all
$k\geq 1$.

######  Lemma 5.10.

Let $H_{P}$ be a Hibi ideal. Then

  1. (a)

$\operatorname{v}_{(x_{i},y_{i})}(H_{P})=|P|-1$, for all $p_{i}\in P$.

  2. (b)

$\operatorname{v}_{(x_{i},y_{j})}(H_{P})=|P|+|\\{p_{\ell}\in P:p_{i}\prec
p_{\ell}\prec p_{j}\\}|$, if $p_{i}\neq p_{j}\in P$ with $p_{i}\prec p_{j}$.

###### Proof.

(a) Let $\mathcal{I}$ be the smallest poset ideal containing $p_{i}$. Hence,
up to a suitable relabeling we may assume that
$\mathcal{I}=\\{p_{1},\dots,p_{i-1},p_{i}\\}$. It is clear that
$\mathcal{I}\setminus\\{p_{i}\\}$ is again a poset ideal. Thus $x_{1}\cdots
x_{i}y_{i+1}\cdots y_{n},x_{1}\cdots x_{i-1}y_{i}\cdots y_{n}\in H_{P}$. This
shows that $(H_{P}:x_{1}\cdots x_{i-1}y_{i+1}\cdots y_{n})=(x_{i},y_{i})$.
Hence $\operatorname{v}_{(x_{i},y_{i})}(H_{P})\leq|P|-1$. By Proposition 2.2,
we conclude that $\operatorname{v}_{(x_{i},y_{i})}(H_{P})=|P|-1$, as wanted.

(b) Let $\mathcal{I}_{1},\mathcal{I}_{2}$ be the two smallest poset ideals of
$P$ containing $p_{i}$, $p_{j}$, respectively. Since $p_{i}\prec p_{j}$, we
have $\mathcal{I}_{1}\subset\mathcal{I}_{2}$. After a suitable relabeling, we
may assume that $\mathcal{I}_{1}=\\{p_{1},\dots,p_{i-1},p_{i}\\}$ and
$\mathcal{I}_{2}=\\{p_{1},\dots,p_{i},\dots,p_{j-1},p_{j}\\}$. Consider the
monomial

$u\ =\ x_{1}\cdots
x_{i-1}y_{i}(\\!\prod_{\ell=i+1}^{j-1}x_{\ell}y_{\ell})x_{j}y_{j+1}\cdots
y_{n}.$

Then, it is easily seen that $(H_{P}:u)=(x_{i},y_{j})$. Therefore,

$\operatorname{v}_{(x_{i},y_{j})}(H_{P})\ \leq\ \deg(u)=|P|+|\\{p_{\ell}\in
P:p_{i}\prec p_{\ell}\prec p_{j}\\}|.$

To show the converse inequality, let $v$ be a monomial such that
$(H_{P}:v)=(x_{i},y_{j})$ and
$\deg(v)=\operatorname{v}_{(x_{i},y_{j})}(H_{P})$. Then $v\notin H_{P}$, but
$x_{i}v\in H_{P}$ and $y_{j}v\in H_{P}$. The definition of $H_{P}$ implies
that $x_{i}v=w_{1}u_{\mathcal{J}_{1}}$ and $y_{j}v=w_{2}u_{\mathcal{J}_{2}}$,
where $x_{i}$ divides $u_{\mathcal{J}_{1}}\in G(H_{P})$ but not $v$, $y_{j}$
divides $u_{\mathcal{J}_{2}}\in G(H_{P})$ but not $v$, and $w_{1}$ and $w_{2}$
are suitable monomials. Since $x_{i}$ does not divide $v$ but $y_{j}v\in
H_{P}$, it follows that $y_{i}$ divides $v$. Hence $y_{i+1}\cdots y_{j-1}$
divides $v$, because if for some $\ell$, $y_{\ell}$ does not divide $v$, then
$x_{\ell}$ should divide $y_{j}v$ and since $p_{i}\prec p_{\ell}$ this would
imply that $x_{i}$ divides $v$, which is impossible. On the other hand, since
$y_{j}$ does not divide $v$, but $x_{i}v\in H_{P}$, it follows that $x_{j}$
divides $v$, and in particular divides $u_{\mathcal{J}_{1}}$, hence
$x_{1}\cdots x_{i}\cdots x_{j-1}$ divides $u_{\mathcal{J}_{1}}$, and so
$x_{1}\cdots x_{i-1}x_{i+1}\cdots x_{j-1}$ divides $v$.

Summarizing our reasoning, we have shown that $x_{1}\cdots
x_{i-1}y_{i}(\prod_{\ell=i+1}^{j-1}x_{\ell}y_{\ell})x_{j}$ divides $v$. Since
$x_{i}v\in H_{P}$, a monomial $z$ of the type $z_{j+1}\cdots z_{n}$, with
$z_{\ell}\in\\{x_{\ell},y_{\ell}\\}$ for $\ell=j+1,\dots,n$, divides $v$.
Hence $x_{1}\cdots x_{i-1}y_{i}(\prod_{\ell=i+1}^{j-1}x_{\ell}y_{\ell})x_{j}z$
divides $v$ and so

$\displaystyle\operatorname{v}_{(x_{i},y_{j})}(H_{P})=\deg(v)\ $
$\displaystyle\geq\ \deg(x_{1}\cdots
x_{i-1}y_{i}(\\!\prod_{\ell=i+1}^{j-1}x_{\ell}y_{\ell})x_{j}z)\ =\ n+(j-i-1)$
$\displaystyle=\ |P|+|\\{p_{\ell}\in P:p_{i}\prec p_{\ell}\prec p_{j}\\}|,$

as desired.

Let $(P,\succeq)$ be a finite poset, with $P=\\{p_{1},\dots,p_{n}\\}$.
Following [8], for any integer $k\geq 1$, we construct a new poset
$(P(k),\succeq_{k})$ defined as follows:

  1. -

$P(k)\ =\ \\{p_{i,1},p_{i,2},\dots,p_{i,k}\ :\ i=1,\dots,n\\}$,

  2. -

$p_{i,r}\succeq_{k}p_{j,s}$ if and only if $p_{i}\succeq p_{j}$ and $r\geq s$.

In order to preserve the structure of Hibi ideals, we innocuously modify
polarization. Let $1\leq\ell\leq k$ and $i\in[n]$, then we set

$\displaystyle(x_{i}^{\ell})^{\wp}\ $ $\displaystyle=\ x_{i,1}x_{i,2}\cdots
x_{i,\ell},$ $\displaystyle(y_{i}^{\ell})^{\wp}\ $ $\displaystyle=\
y_{i,\ell}y_{i,\ell-1}\cdots y_{i,k+1-\ell},$

and extend the polarization of an arbitrary monomial in the obvious way. In
other words, with respect to the usual polarization, we are just applying the
relabeling of the variables $y_{i,\ell}\mapsto y_{i,k+1-\ell}$ for
$i=1,\dots,n$ and $\ell=1,\dots,k$.

The next result was proved in [8, Theorem 4.9].

######  Lemma 5.11.

Let $(P,\succeq)$ be a finite poset. Then $(H_{P}^{k})^{\wp}=H_{P(k)}$ for all
$k\geq 1$.

######  Example 5.12.

Consider the poset $(P,\succeq)$ of Example 5.7. The poset
$(P(2),\succeq_{2})$ and the distributive lattice $\mathcal{J}(P(2))$ are
depicted below.

$x_{1,1}$$x_{2,1}$$x_{3,1}$$x_{1,2}$$x_{2,2}$$x_{3,2}$$(P(2),\succeq_{2})$$\\{x_{1,1},x_{1,2},x_{2,1},x_{2,2},x_{3,1}\\}$$\\{x_{1,1},x_{1,2},x_{2,1},x_{3,1},x_{3,2}\\}$$\\{x_{1,1},x_{1,2},x_{2,1},x_{2,2},x_{3,1},x_{3,2}\\}$$\\{x_{1,1},x_{1,2},x_{2,1},x_{2,2}\\}$$\\{x_{1,1},x_{1,2},x_{3,1},x_{3,2}\\}$$\\{x_{1,1},x_{1,2},x_{2,1},x_{3,1}\\}$$\\{x_{1,1},x_{1,2},x_{2,1}\\}$$\\{x_{1,1},x_{1,2},x_{3,1}\\}$$\\{x_{1,1},x_{2,1},x_{3,1}\\}$$\\{x_{1,1},x_{2,1}\\}$$\\{x_{1,1},x_{1,2}\\}$$\\{x_{1,1},x_{3,1}\\}$$\\{x_{1,1}\\}$$\emptyset$$\mathcal{J}(P(2))$

One can easily verify that $(H_{P}^{2})^{\wp}=H_{P(2)}$, with respect to our
modified polarization.

Finally, Theorem 5.8 follows immediately from the next result.

######  Corollary 5.13.

Let $H_{P}$ be a Hibi ideal and $k\geq 1$ any integer. Then

  1. (a)

$\operatorname{v}_{(x_{i},y_{i})}(H_{P}^{k})=|P|k-1$, for all $p_{i}\in P$.

  2. (b)

$\operatorname{v}_{(x_{i},y_{j})}(H_{P}^{k})=|P|k+|\\{p_{\ell}\in P:p_{i}\prec
p_{\ell}\prec p_{j}\\}|$, if $p_{i}\neq p_{j}\in P$ with $p_{i}\prec p_{j}$.

###### Proof.

By Lemma 5.9, $\operatorname{Ass}(H_{P}^{k})=\\{(x_{i},y_{j}):p_{i}\preceq
p_{j}\\}$. Whereas, by Lemma 5.11, we have $(H_{P}^{k})^{\wp}=H_{P(k)}$.
Hence, by Theorem 4.1(c) and Lemma 5.9,

$\displaystyle\operatorname{v}_{(x_{i},y_{i})}(H_{P}^{k})\ $ $\displaystyle=\
\min_{1\leq\ell\leq h\leq
k}\operatorname{v}_{(x_{i,\ell},y_{i,h})}(H_{P(k)}),\ \ \text{for all}\
p_{i}\in P,\ \text{and},$
$\displaystyle\operatorname{v}_{(x_{i},y_{j})}(H_{P}^{k})\ $ $\displaystyle=\
\min_{1\leq\ell\leq h\leq
k}\operatorname{v}_{(x_{i,\ell},y_{j,h})}(H_{P(k)}),\ \ \text{for all}\
p_{i}\prec p_{j}.$

A quick calculation shows that (a) and (b) indeed hold.

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