# Asymptotic behaviour of the v-number of homogeneous ideals

Antonino Ficarra, Emanuele Sgroi  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>Emanuele Sgroi, 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 $I$ be a graded ideal of a standard graded polynomial ring $S$ with
coefficients in a field $K$. The asymptotic behaviour of the
$\operatorname{v}$-number of the powers of $I$ is investigated. Natural lower
and upper bounds which are linear functions in $k$ are determined for
$\operatorname{v}(I^{k})$. We call $\operatorname{v}(I^{k})$ the
$\operatorname{v}$-function of $I$. We prove that $\operatorname{v}(I^{k})$ is
a linear function in $k$ for $k$ large enough, of the form
$\operatorname{v}(I^{k})=\alpha(I)k+b$, where $\alpha(I)$ is the initial
degree of $I$, and $b\in{\mathbb{Z}}$ is a suitable integer. For this aim, we
construct new blowup algebras associated to graded ideals. Finally, for a
monomial ideal in two variables, we compute explicitly its
$\operatorname{v}$-function.

###### Key words and phrases:

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

###### 2020 Mathematics Subject Classification:

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

## Introduction

In 1921, Emmy Noether revolutionized Commutative Algebra by establishing the
primary decomposition Theorem for Noetherian rings [30]. It says that any
ideal $I$ of a Noetherian ring $R$ can be decomposed as the irredundant
intersection of finitely many primary ideals $I=Q_{1}\cap\dots\cap Q_{t}$ and
$\operatorname{Ass}(I)=\\{\sqrt{Q_{1}},\dots,\sqrt{Q_{t}}\\}$, the set of
associated primes of $I$, is uniquely determined. This fundamental result is a
landmark in Commutative Algebra, and always inspires new exciting research
trends. A basic question in the seventies was the following. What is the
asymptotic behaviour of the set $\operatorname{Ass}(I^{k})$ for $k\gg 0$ large
enough? In 1976, it was predicted by Ratliff [31], and later proved by
Brodmann in 1979 [3], that $\operatorname{Ass}(I^{k})$ stabilizes. That is,
there exists $k_{0}>0$ such that
$\operatorname{Ass}(I^{k+1})=\operatorname{Ass}(I^{k})$ for all $k\geq k_{0}$.
Another remarkable result of Brodmann says that
$\operatorname{depth}(R/I^{k})$ is constant for $k\gg 0$ [4]. Suppose
furthermore that $I$ is a graded ideal of a standard graded polynomial ring
$S=K[x_{1},\dots,x_{n}]$ with coefficients in a field $K$. In 1999, Kodiyalam
[25], and, independently, Cutkosky, Herzog and Trung [9], showed that the
Castelnuovo–Mumford regularity of $S/I^{k}$ is a linear function in $k$ for
$k\gg 0$. The legacy of Brodmann theorem opened up the most flourished
research topic in Commutative Algebra: the asymptotic behaviour of the
homological invariants of (ordinary) powers of graded ideals, see [5].

Now, let $S=K[x_{1},\dots,x_{n}]$ be the standard graded polynomial ring with
coefficients in a field $K$, $I\subset S$ be a graded ideal and
$\mathfrak{m}=(x_{1},\dots,x_{n})$ be the maximal ideal. Note that $S$ is
Noetherian. The graded version of the primary decomposition theorem says that
for any prime $\mathfrak{p}\in\operatorname{Ass}(I)$, there exists a
homogeneous element $f\in S$ such that $(I:f)=\mathfrak{p}$. It is natural to
define the following invariants. Denote by $S_{d}$ the $d$th graded component
of $S$. 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

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

The concept of $\operatorname{v}$-number was introduced by Cooper et all in
[8], and further studied in [1, 2, 7, 13, 19, 23, 26, 27, 28, 32, 33, 34].

This invariant plays an important role in Algebraic Geometry and in the theory
of (projective) Reed-Muller-type codes [8, 10, 16, 17, 18, 20, 21]. Let
$\mathbb{X}$ be a finite set of points of the projective plane
$\mathbb{P}^{s-1}$, and let $\delta_{\mathbb{X}}(d)$ be the minimum distance
function of the projective Reed-Muller-type code $C_{\mathbb{X}}(d)$. Then
$\delta_{\mathbb{X}}(d)=1$ if and only if $\operatorname{v}(I(\mathbb{X}))\leq
d$ [8, Corollary 5.6]. In such article, for a radical complete intersection
ideal $I$, the famous Eisenbud-Green-Harris conjecture [12] is shown to be
equivalent to [8, Conjecture 6.2] (see [8, Proposition 6.8]). This latter
conjecture is related to the $\operatorname{v}$-number. Indeed, for such an
ideal $I$ we have $\operatorname{v}(I)=\operatorname{reg}(S/I)$. For a nice
summary, see [34, Section 12].

The $\operatorname{v}$-number of edge ideals was studied in [27]. A graph $G$
belongs to the class $W_{2}$ if and only if $G$ is well–covered without
isolated vertices, and $G\setminus v$ is well–covered for all vertices $v\in
V(G)$. Let $I(G)\subset S=K[x_{v}:v\in V(G)]$ be the edge ideal of $G$. Then
$G$ is in $W_{2}$ if and only if $\operatorname{v}(I(G))=\dim(S/I(G))$ [27,
Theorem 4.5]. The $\operatorname{v}$-number of binomial edge ideals was
recently considered in [1] and [26].

In this article, we investigate the eventual behaviour of the function
$\operatorname{v}(I^{k})$ for $k\gg 0$, where $I\subset S$ is a graded ideal.
Such a function, for large $k$ measures the “asymptotic homogeneous growth” of
the primary decomposition of $I^{k}$. We prove that
$\operatorname{v}(I^{k})=\alpha(I)k+b$ is a linear function for all $k\gg 0$,
where $\alpha(I)$ is the initial degree of $I$ and $b\in{\mathbb{Z}}$ is a
suitable integer.

The article is structured as follows. In Section 1, we recall how to compute
the $\operatorname{v}$-number of a graded ideal $I\subset S$ (Theorem 1.1) as
shown by Grisalde, Reyes and Villarreal [23]. For a finitely generated graded
$S$-module $M=\bigoplus_{d\geq 0}M_{d}\neq 0$, we set
$\alpha(M)=\min\\{d:M_{d}\neq 0\\}$ and
$\omega(M)=\max\\{d:(M/\mathfrak{m}M)_{d}\neq 0\\}$. In the next theorem, 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
$\operatorname{v}(I)=\min\\{\alpha((I:\mathfrak{p})/I):\mathfrak{p}\in\operatorname{Ass}(I)\\}$.

Firstly, we determine the “local” numbers
$\operatorname{v}_{\mathfrak{p}}(I)$, for all
$\mathfrak{p}\in\operatorname{Ass}(I)$. After computing a basis of the
$S$-module $(I:\mathfrak{p})/I$, we select a generator
$\overline{f}\in(I:\mathfrak{p})/I$ of least degree $d$ such that
$(I:f)=\mathfrak{p}$. This latter condition is automatically satisfied if
$\mathfrak{p}\in\operatorname{Max}(I)$. Then
$\operatorname{v}_{\mathfrak{p}}(I)=d$ and
$\operatorname{v}(I)=\min\\{\operatorname{v}_{\mathfrak{p}}(I):\mathfrak{p}\in\operatorname{Ass}(I)\\}$.

Let $I\subset S$ be a graded ideal, we call $\operatorname{v}(I^{k})$ the
$\operatorname{v}$-function of $I$. In Section 2, we investigate the
asymptotic behaviour of $\operatorname{v}(I^{k})$ for $k\gg 0$. By Theorem
1.1(b), we have
$\operatorname{v}(I^{k})=\min\\{\operatorname{v}_{\mathfrak{p}}(I^{k}):\mathfrak{p}\in\operatorname{Ass}(I^{k})\\}$.
By Brodmann [3], $\operatorname{Ass}(I^{k})=\operatorname{Ass}(I^{k+1})$ for
all $k\gg 0$. We denote this common set by $\operatorname{Ass}^{\infty}(I)$,
and call each prime $\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$ a stable
prime ideal of $I$. Moreover, let $\operatorname{Max}^{\infty}(I)$ be the set
of stable prime ideals of $I$ maximal with respect to the inclusion. Thus,
$\operatorname{v}(I^{k})=\min\\{\operatorname{v}_{\mathfrak{p}}(I^{k}):\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)\\}$
for $k\gg 0$. To understand the asymptotic behaviour of the
$\operatorname{v}$-function, we consider the
$\operatorname{v}_{\mathfrak{p}}$-functions
$\operatorname{v}_{\mathfrak{p}}(I^{k})$ for each
$\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$. In the classical case, to
prove the asymptotic linearity of the Castelnuovo–Mumford regularity of the
powers of $I$, $\operatorname{reg}(I^{k})$, one introduces the Rees ring of
$I$, $\mathcal{R}(I)=\bigoplus_{k\geq 0}I^{k}$, and shows that this is a
bigraded finitely generated module over a suitable polynomial ring [9].

Let $\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$. For the
$\operatorname{v}_{\mathfrak{p}}$-function
$\operatorname{v}_{\mathfrak{p}}(I^{k})$, we consider a similar approach as
above. It should be noted, however, that the $S$-module
$(I^{k}:\mathfrak{p})/I^{k}$ has a more subtle module structure than the
ordinary power $I^{k}$. We introduce the module

$\textup{Soc}_{\mathfrak{p}}(I)\ =\ \bigoplus_{k\geq
0}(I^{k}:\mathfrak{p})/I^{k}.$

over the ring $\mathcal{F}_{\mathfrak{p}}(I)\ =\ \bigoplus_{k\geq
0}(I^{k}/\mathfrak{p}I^{k})$.

A priori, it is not clear that $\textup{Soc}_{\mathfrak{p}}(I)$ is a finitely
generated bigraded $\mathcal{F}_{\mathfrak{p}}(I)$-module. This is shown in
Theorem 2.2 by carefully analyzing the module structure of
$\operatorname{Soc}_{\mathfrak{p}}(I)$. We prove that
$\operatorname{Soc}_{\mathfrak{p}}(I)$ is equal to a truncation of the ideal
$(0:_{\operatorname{gr}_{I}(S)}\mathfrak{p})$, and this ideal of
$\operatorname{gr}_{I}(S)$ is finitely generated as a
$\mathcal{F}_{\mathfrak{p}}(I)$-module. Here
$\operatorname{gr}_{I}(S)=\bigoplus_{k\geq 0}(I^{k}/I^{k+1})$ denotes the
associated graded ring of $I$. The proof relies essentially on a property
showed by Ratliff [31, Corollary 4.2], namely that $(I^{k+1}:I)=I^{k}$ for all
$k\gg 0$, and on the fact that $\operatorname{gr}_{I}(S)$ is Noetherian ring
[29, Proposition (10.D)].

The first main result in Section 2 is
Theorem 2.1. Let $I\subset S=K[x_{1},\dots,x_{n}]$ be a graded ideal, and let
$\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$. Then, the following holds.

  1. (a)

For all $k\geq 1$, we have

$\alpha((I^{k}:\mathfrak{p})/I^{k})\ \leq\
\operatorname{v}_{\mathfrak{p}}(I^{k})\ \leq\
\omega((I^{k}:\mathfrak{p})/I^{k}).$

  2. (b)

The functions $\alpha((I^{k}:\mathfrak{p})/I^{k})$,
$\omega((I^{k}:\mathfrak{p})/I^{k})$ are linear in $k$ for $k\gg 0$.

  3. (c)

There exist eventually linear functions $f(k)$ and $g(k)$ such that

$f(k)\leq\operatorname{v}(I^{k})\leq g(k),\ \ \textit{for all}\ \ k\gg 0.$

Theorem 2.1(b) follows by a careful analysis of the bigraded structure of
$\textup{Soc}_{\mathfrak{p}}(I)$ and in the end boils down to a linear
programming argument (Proposition 2.5).

As another remarkable consequence of Theorem 2.2, if $I\subset S$ is a graded
ideal and $\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$, we show in
Proposition 2.7 that
$(I^{k}:\mathfrak{p})/I^{k}=(I(I^{k-1}:\mathfrak{p}))/I^{k}$ for all $k\gg 0$.
In the more general case that $I$ is an ideal of a Noetherian commutative
domain $R$ and $\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$, with an
analogous proof, one can also show that for all $k\gg 0$ we have
$(I^{k}:\mathfrak{p})=I(I^{k-1}:\mathfrak{p})$ (Corollary 2.9). This result
complements the property $(I^{k}:I)=I^{k-1}$, $k\gg 0$, firstly proved by
Ratliff [31, Corollary 4.2]. To the best of our knowledge, it was previously
unknown.

Our main result in the paper shows that the functions
$\operatorname{v}_{\mathfrak{p}}(I^{k})$,
$\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$, and $\operatorname{v}(I^{k})$
are indeed linear functions in $k$ for all large $k$.
Theorem 3.1. Let $I\subset S=K[x_{1},\dots,x_{n}]$ be a graded ideal. Then,
$\operatorname{v}_{\mathfrak{p}}(I^{k})$, for all for
$\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$, and $\operatorname{v}(I^{k})$
are linear functions in $k$, for $k\gg 0$.

Let $I\subset S$ be a graded ideal, and let
$\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$. To prove this theorem, we
construct another ${\mathcal{F}}_{\mathfrak{p}}(I)$-module, which we denote by
$\operatorname{Soc}_{\mathfrak{p}}^{*}(I)$, and is defined as a suitable
quotient of $\operatorname{Soc}_{\mathfrak{p}}(I)$. In the course of the proof
of Theorem 3.1, we show that
$\operatorname{v}_{\mathfrak{p}}(I^{k})=\alpha(\operatorname{Soc}_{\mathfrak{p}}^{*}(I)_{(*,k)})$.
Using again the linear programming argument given in Proposition 2.5, we then
conclude that this is a linear function in $k$ for $k\gg 0$.

In the last part of the section, we prove some other structural properties of
the functions $\operatorname{v}_{\mathfrak{p}}(I^{k})$. In Proposition 3.2, we
show that

$\operatorname{v}_{\mathfrak{p}}(I^{k-1})+\alpha(I)\leq\operatorname{v}_{\mathfrak{p}}(I^{k})\leq\operatorname{v}_{\mathfrak{p}}(I^{k-1})+\omega(I),$

for all $k\gg 0$. As nice consequences of this fact, we obtain that the
functions $\operatorname{v}_{\mathfrak{p}}(I^{k})$,
($\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$), and
$\operatorname{v}(I^{k})$ are strictly increasing, for $k\gg 0$ (Corollary
3.3).

Due to Theorems 3.1, we have that $\operatorname{v}(I^{k})=ak+b$ for all $k\gg
0$, where $a$ and $b$ are suitable integers. Hence, the limit
$\lim_{k\rightarrow\infty}\operatorname{v}(I^{k})/k$ exists and it is a finite
integer equal to $a$. A natural question arises, what is the number $a$? In
Section 4 we address this question, and we obtain the following quite
surprising answer:
Theorem 4.1 Let $I\subset S$ be a graded ideal. Then

$\lim_{k\rightarrow\infty}\frac{\operatorname{v}(I^{k})}{k}\ =\ \alpha(I).$

The proof of this theorem is based upon Lemma 3.4(d). In order to apply it, we
find lower and upper bounds for $\operatorname{v}(I^{k})$, $k\gg 0$, of the
form $\alpha(I)k+c$, where $c$ is a suitable constant. The lower bound is
easily obtained in Lemma 4.4. To obtain the upper bound, we follow an argument
given in [2, Theorem 4.7] and generalize a result due to Saha and Sengupta
[33, Proposition 3.11], who proved it only for monomial ideals (Proposition
4.3). In view of Theorem 3.1 and [9, 25], we could expect that
$\operatorname{v}(I^{k})\leq\operatorname{reg}(I^{k})$ for all $k\gg 0$.

Note that the local version of Theorem 4.1, namely
$\operatorname{v}_{\mathfrak{p}}(I^{k})=\alpha(I)k+b$ for all $k\gg 0$ and
some integer $b$, is false in general, as we show with an example.

Section 5 concerns monomial ideals $I\subset S=K[x,y]$ in two variables.
Denote by $G(I)=\\{u_{1},\dots,u_{m}\\}$ the unique minimal monomial
generating set of $I$, with $u_{i}=x^{a_{i}}y^{b_{i}}$ for all $i$. Then $I$
determines the sequences ${\bf a}:a_{1}>a_{2}>\dots>a_{m}$ and ${\bf
b}:b_{1}<b_{2}<\dots<b_{m}$, and we set $I=I_{\bf a,b}$. Conversely, given any
two such sequences ${\bf a}$ and ${\bf b}$,
$\\{x^{a_{1}}y^{b_{1}},\dots,x^{a_{m}}y^{b_{m}}\\}$ is the minimal generating
set of a monomial ideal of $S$. In terms of ${\bf a}$ and ${\bf b}$ we
determine $\operatorname{Ass}^{\infty}(I_{\bf a,b})$ (Corollary 5.3) and the
$\operatorname{v}$-number $\operatorname{v}(I_{\bf a,b})$ (Theorem 5.6). Let
$\operatorname{v}(I_{\bf a,b}^{k})=ak+b$ for $k\gg 0$, where $a$ and $b$ are
suitable integers. By [13, Theorem 2.3(c)] we have $b\geq-1$. On the other
hand, for any integers $a\geq 1$ and $b\geq-1$, we are able to construct a
monomial ideal of $S$ such that $\operatorname{v}(I^{k})=ak+b$ for all $k\geq
1$ (Theorem 5.7).

##  1\. How to compute the $\operatorname{v}$-number of a graded ideal?

Let $I$ be an ideal of a Noetherian domain $R$. 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. It is clear that $I$ has no embedded primes if
and only if $\operatorname{Ass}(I)=\operatorname{Max}(I)$.

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. The concept of
$\operatorname{v}$-number was introduced by Cooper et all in [8]. 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

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

Note that if $I=\mathfrak{m}=(x_{1},\dots,x_{n})$, then
$\operatorname{v}_{\mathfrak{m}}(I)=\operatorname{v}(I)=0$.

The following result due to Grisalde, Reyes and Villarreal [23, Theorem 3.2]
shows how to compute the $\operatorname{v}$-number of a graded ideal. For a
finitely generated graded $S$-module $M=\bigoplus_{d\geq 0}M_{d}\neq 0$, we
call $\alpha(M)=\min\\{d:M_{d}\neq 0\\}$ the initial degree of $M$. In the
next theorem, 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
$\operatorname{v}(I)=\min\\{\alpha((I:\mathfrak{p})/I):\mathfrak{p}\in\operatorname{Ass}(I)\\}$.

##  2\. Asymptotic growth of the modules $(I^{k}:\mathfrak{p})/I^{k}$

Let $R$ be a commutative Noetherian domain and $I\subset R$ an ideal. It is
known by Brodmann [3] 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 $\mathfrak{p}\subset R$ 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
$\operatorname{Ass}^{\infty}(I)$. Likewise, $\operatorname{Max}^{\infty}(I)$
denotes the set of stable primes of $I$, maximal with respect to the
inclusion. The least integer $k_{0}$ such that
$\operatorname{Ass}(I^{k})=\operatorname{Ass}(I^{k_{0}})$ for all $k\geq
k_{0}$ is denoted by $\operatorname{astab}(I)$.

Now, let $S=K[x_{1},\dots,x_{n}]$ be the standard graded polynomial ring, with
$K$ a field, and unique graded maximal ideal
$\mathfrak{m}=(x_{1},\dots,x_{n})$. Let $I\subset S$ be a graded ideal and let
$\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$. In light of Theorem 1.1 and
Brodmann result, to understand the asymptotic behaviour of the function
$\operatorname{v}_{\mathfrak{p}}(I^{k})$, one has to understand the asymptotic
growth of the modules $(I^{k}:\mathfrak{p})/I^{k}$ for $k\gg 0$.

Let $M\neq 0$ be a finitely generated graded $S$-module. Let $\omega(M)$ be
the highest degree of a homogeneous element of the $K$-vector space
$M/\mathfrak{m}M$. Equivalently, the highest degree $j$ such that the graded
Betti number $\beta_{0,j}(M)$ is non-zero. Thus

$\omega(M)=\max\\{d:\beta_{0,d}(M)\neq
0\\}=\max\\{d:\operatorname{Tor}^{S}_{0}(S/\mathfrak{m},M)_{d}\neq 0\\}.$

Similarly, one has that
$\alpha(M)=\min\\{d:\operatorname{Tor}^{S}_{0}(S/\mathfrak{m},M)_{d}\neq
0\\}$.

The following theorem provides natural asymptotic upper and lower bounds for
the $\operatorname{v}$-function $\operatorname{v}(I^{k})$ which are linear
functions in $k$, for $k\gg 0$.

######  Theorem 2.1.

Let $I\subset S=K[x_{1},\dots,x_{n}]$ be a graded ideal, and let
$\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$. Then, the following holds.

  1. (a)

For all $k\geq 1$, we have

$\alpha((I^{k}:\mathfrak{p})/I^{k})\ \leq\
\operatorname{v}_{\mathfrak{p}}(I^{k})\ \leq\
\omega((I^{k}:\mathfrak{p})/I^{k}).$

  2. (b)

The functions $\alpha((I^{k}:\mathfrak{p})/I^{k})$,
$\omega((I^{k}:\mathfrak{p})/I^{k})$ are linear in $k$ for $k\gg 0$.

  3. (c)

There exist eventually linear functions $f(k)$ and $g(k)$ such that

$f(k)\leq\operatorname{v}(I^{k})\leq g(k),\ \ \textit{for all}\ \ k\gg 0.$

Statement (a) follows immediately from Theorem 1.1(a). Assume for a moment
that statement (b) holds, then (c) can be proved as follows. By Brodmann, we
have
$\operatorname{v}(I^{k})=\min\\{\operatorname{v}_{\mathfrak{p}}(I^{k}):\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)\\}$
for all $k\gg 0$. Thus, by (a), for all $k\gg 0$

$\min_{\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)}\alpha((I^{k}:\mathfrak{p})/I^{k})\leq\operatorname{v}(I^{k})\leq\min_{\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)}\omega((I^{k}:\mathfrak{p})/I^{k}).$

Setting
$f(k)=\min_{\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)}\alpha((I^{k}:\mathfrak{p})/I^{k})$
and
$g(k)=\min_{\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)}\omega((I^{k}:\mathfrak{p})/I^{k})$,
by statement (b) it follows that $f(k)$ and $g(k)$ are the required eventually
linear functions in $k$. Statement (c) follows.

To prove statement (b), we construct a suitable module that encodes the growth
of the modules $(I^{k}:\mathfrak{p})/I^{k}$. Indeed, we define it in the
following more general context. Let $I$ be an ideal of a commutative
Noetherian domain $R$ and let $\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$.
Then we set

$\operatorname{Soc}_{\mathfrak{p}}(I)\ =\ \bigoplus_{k\geq
0}(I^{k}:\mathfrak{p})/I^{k},$

and $\operatorname{Soc}_{\mathfrak{p}}(I)_{k}=(I^{k}:\mathfrak{p})/I^{k}$ for
all $k\geq 0$.

The symbol “$\operatorname{Soc}$” is used, because when $R=S$ or $R$ is local
and $\mathfrak{p}=\mathfrak{m}$ is the (graded) maximal ideal, then
$(I^{k}:\mathfrak{m})/I^{k}$ is the socle module of $S/I^{k}$, see [6].

The first step consists in showing that $\operatorname{Soc}_{\mathfrak{p}}(I)$
is a finitely generated graded module over a suitable ring. For this aim, we
introduce the following ring,

$\mathcal{F}_{\mathfrak{p}}(I)\ =\ \bigoplus_{k\geq
0}(I^{k}/\mathfrak{p}I^{k}),$

and we set $\mathcal{F}_{\mathfrak{p}}(I)_{k}=I^{k}/\mathfrak{p}I^{k}$. We
define addition in the obvious way and multiplication as follows. If $a\in
I^{k}/\mathfrak{p}I^{k}$ and $b\in I^{\ell}/\mathfrak{p}I^{\ell}$, then $ab\in
I^{k+\ell}/\mathfrak{p}I^{k+\ell}$. It is routine to check that this
multiplication is well–defined.

As before, we note that if $R=S$ or $R$ is local and
$\mathfrak{p}=\mathfrak{m}$ is the maximal ideal, then
$\mathcal{F}_{\mathfrak{m}}(I)=\bigoplus_{k\geq 0}(I^{k}/\mathfrak{m}I^{k})$
is the well–known fiber cone of $I$.

With the notation introduced, we have

######  Theorem 2.2.

Let $I$ be an ideal of a Noetherian commutative domain $R$ and let
$\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$. Then,
$\operatorname{Soc}_{\mathfrak{p}}(I)$ is a finitely generated graded
$\mathcal{F}_{\mathfrak{p}}(I)$-module.

###### Proof.

Firstly, we show that $\operatorname{Soc}_{\mathfrak{p}}(I)$ has the structure
of a graded $\mathcal{F}_{\mathfrak{p}}(I)$-module. For this purpose, let
$f\in I^{\ell}/\mathfrak{p}I^{\ell}$. It is clear that multiplication by $f$
induces a map
$(I^{k}:\mathfrak{p})/I^{k}\rightarrow(I^{k+\ell}:\mathfrak{p})/I^{k+\ell}$
for any $k\geq 0$. Hence
$\mathcal{F}_{\mathfrak{p}}(I)_{\ell}\operatorname{Soc}_{\mathfrak{p}}(I)_{k}\subseteq\operatorname{Soc}_{\mathfrak{p}}(I)_{k+\ell}$.

To prove that $\operatorname{Soc}_{\mathfrak{p}}(I)$ is a finitely generated
$\mathcal{F}_{\mathfrak{p}}(I)$-module, we consider

$J=(0:_{\operatorname{gr}_{I}(R)}\mathfrak{p})=\\{f\in\operatorname{gr}_{I}(R):f\mathfrak{p}=0\\},$

_i.e._ , the annihilator of $\mathfrak{p}$ in the associated graded ring of
$I$, $\operatorname{gr}_{I}(R)=\bigoplus_{k\geq 0}(I^{k}/I^{k+1})$. Recall
that $\operatorname{gr}_{I}(R)$ is a Noetherian ring [29, Proposition (10.D)].
Thus, as an ideal of $\operatorname{gr}_{I}(R)$, $J$ is a finitely generated
graded $\operatorname{gr}_{I}(R)$-module. Since $\mathfrak{p}$ annihilates
$J$, then $J$ has also the structure of a finitely generated graded
$\operatorname{gr}_{I}(R)/\mathfrak{p}\operatorname{gr}_{I}(R)$-module. But

$\displaystyle\operatorname{gr}_{I}(R)/\mathfrak{p}\operatorname{gr}_{I}(R)\ $
$\displaystyle=\ \frac{\bigoplus_{k\geq
0}(I^{k}/I^{k+1})}{\mathfrak{p}\bigoplus_{k\geq
0}(I^{k}/I^{k+1})}=\frac{\bigoplus_{k\geq 0}(I^{k}/I^{k+1})}{\bigoplus_{k\geq
0}(\mathfrak{p}I^{k}/I^{k+1})}$ $\displaystyle=\ \bigoplus_{k\geq
0}\frac{I^{k}/I^{k+1}}{\mathfrak{p}I^{k}/I^{k+1}}=\bigoplus_{k\geq
0}(I^{k}/\mathfrak{p}I^{k})$ $\displaystyle=\ \mathcal{F}_{\mathfrak{p}}(I).$

Consequently, $J$ is a finitely generated graded
$\mathcal{F}_{\mathfrak{p}}(I)$-module.

Let us show that $\operatorname{Soc}_{\mathfrak{p}}(I)_{k+1}=J_{k}$ for $k\gg
0$. For this purpose, we compute the $k$th graded component of $J$. We have

$\displaystyle J_{k}\ $ $\displaystyle=\
\\{f\in\operatorname{gr}_{I}(R)_{k}:f\mathfrak{p}=0\\}=\\{f\in I^{k}/I^{k+1}\
:f\mathfrak{p}=0\\}$ $\displaystyle=\ \\{f\in I^{k}:f\mathfrak{p}\in
I^{k+1}\\}/I^{k+1}=(\\{f\in R:f\mathfrak{p}\in I^{k+1}\\}\cap I^{k})/I^{k+1}$
$\displaystyle=\ ((I^{k+1}:\mathfrak{p})\cap I^{k})/I^{k+1}.$

By Ratliff [31, Corollary 4.2], there exists $r$ such that $(I^{k+1}:I)=I^{k}$
for all $k\geq r$. Whereas, by Brodmann [3], there exists $b$ such that
$\operatorname{Ass}(I^{k})=\operatorname{Ass}^{\infty}(I)$ for all $k\geq b$.
Let $k^{*}=\max\\{r,b\\}$. Next, we show that
$\operatorname{Soc}_{\mathfrak{p}}(I)_{k+1}=J_{k}$ for $k\geq k^{*}$.

Let $k\geq k^{*}$. We claim that $\mathfrak{p}$ contains $I$. Indeed,
$\mathfrak{p}\in\operatorname{Ass}(I^{k})$, hence
$I^{k}\subseteq\mathfrak{p}$. Let $a\in I$, then $a^{k}\in
I^{k}\subseteq\mathfrak{p}$. Since $\mathfrak{p}$ is prime, actually
$a\in\mathfrak{p}$ and so $I\subseteq\mathfrak{p}$. Therefore,
$(I^{k+1}:\mathfrak{p})\subseteq(I^{k+1}:I)=I^{k}$ by the Ratliff property.
Hence,

$J_{k}=((I^{k+1}:\mathfrak{p})\cap
I^{k})/I^{k+1}=(I^{k+1}:\mathfrak{p})/I^{k+1}=\operatorname{Soc}_{\mathfrak{p}}(I)_{k+1}.$

Consequently, we obtain that $\operatorname{Soc}_{\mathfrak{p}}(I)_{\geq
k^{*}+1}=J_{\geq k^{*}}$, where $M_{\geq\ell}$ denotes
$\bigoplus_{k\geq\ell}M_{\ell}$ if $M=\bigoplus_{k\geq 0}M_{k}$ is graded.
Since $J$ is finitely generated as a $\mathcal{F}_{\mathfrak{p}}(I)$-module,
it follows that $\operatorname{Soc}_{\mathfrak{p}}(I)$ is a finitely generated
$\mathcal{F}_{\mathfrak{p}}(I)$-module as well.

Now, we assume furthermore that $R=S=K[x_{1},\dots,x_{n}]$ is the standard
graded polynomial ring with $K$ a field, that $I$ is a graded ideal of $S$ and
$\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$ is a stable prime of $I$.
Then, $I^{k}/\mathfrak{p}I^{k}$ is a graded $S$-module, for all $k\geq 0$.
Therefore, $\mathcal{F}_{\mathfrak{p}}(I)$ is in a natural way a bigraded
ring:

$\mathcal{F}_{\mathfrak{p}}(I)\ =\ \bigoplus_{d,k\geq
0}(I^{k}/\mathfrak{p}I^{k})_{d}.$

In particular, we set
$\mathcal{F}_{\mathfrak{p}}(I)_{(d,k)}=(I^{k}/\mathfrak{p}I^{k})_{d}$ and
$\textup{bideg}(f)=(d,k)$ for $f\in\mathcal{F}_{\mathfrak{p}}(I)_{(d,k)}$.

Note that each module $(I^{k}:\mathfrak{p})/I^{k}$ is a graded $S$-module.
Thus, we can write

$\operatorname{Soc}_{\mathfrak{p}}(I)\ =\ \bigoplus_{d,k\geq
0}\operatorname{Soc}_{\mathfrak{p}}(I)_{(d,k)}$

where
$\operatorname{Soc}_{\mathfrak{p}}(I)_{(d,k)}=((I^{k}:\mathfrak{p})/I^{k})_{d}$.
Hence, $\operatorname{Soc}_{\mathfrak{p}}(I)$ is a bigraded
$\mathcal{F}_{\mathfrak{p}}(I)$-module, because
$\mathcal{F}_{\mathfrak{p}}(I)_{(d_{1},\ell)}\operatorname{Soc}_{\mathfrak{p}}(I)_{(d_{2},k)}\subseteq\operatorname{Soc}_{\mathfrak{p}}(I)_{(d_{1}+d_{2},k+\ell)}$.

Therefore, we have proved that

######  Corollary 2.3.

Let $I$ be a graded ideal of $S=K[x_{1},\dots,x_{n}]$ with $K$ a field and let
$\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$. Then,
$\operatorname{Soc}_{\mathfrak{p}}(I)$ is a finitely generated bigraded
$\mathcal{F}_{\mathfrak{p}}(I)$-module.

Let $u_{1},\dots,u_{m}$ be a minimal system of homogeneous generators of
$I\subset S$. It is well–known that the associated graded ring
$\operatorname{gr}_{I}(S)$ has a presentation

$\varphi:T=K[x_{1},\dots,x_{n},y_{1},\dots,y_{m}]\rightarrow\operatorname{gr}_{I}(S)$

defined by setting

$\displaystyle\varphi(x_{i})=x_{i}+I\in\operatorname{gr}_{I}(S)_{0}=S/I,\ \ $
$\displaystyle\text{for}\ \ 1\leq i\leq n,$
$\displaystyle\varphi(y_{i})=u_{i}+I^{2}\in\operatorname{gr}_{I}(S)_{1}=I/I^{2},\
\ $ $\displaystyle\text{for}\ \ 1\leq i\leq m.$

Since $I$ is graded, $\operatorname{gr}_{I}(S)$ is naturally bigraded, with
$\operatorname{gr}_{I}(S)_{(d,k)}=(I^{k}/I^{k+1})_{d}$. Moreover, $T$ can be
made into a bigraded ring by setting $\textup{bideg}(x_{i})=(1,0)$ for $1\leq
i\leq n$, and $\textup{bideg}(y_{i})=(\deg(u_{i}),1)$ for $1\leq i\leq m$,
where $\deg(u_{i})>0$ is the degree of $u_{i}$ in $S$. With these bigradings,
$\varphi$ is a bigraded surjective ring homomorphism.

In the proof of Theorem 2.2 we have seen that
$\mathcal{F}_{\mathfrak{p}}(I)=\operatorname{gr}_{I}(S)/\mathfrak{p}\operatorname{gr}_{I}(S)$.
Let $\pi:\operatorname{gr}_{I}(S)\rightarrow\mathcal{F}_{\mathfrak{p}}(I)$ be
the canonical epimorphism. Then, the composition map

$\psi=\pi\circ\varphi\ :\ T\rightarrow\mathcal{F}_{\mathfrak{p}}(I)$ (1)

is a surjective ring homomorphism. It is clear that $\psi$ preserves the
bigraded structure. Thus, $\operatorname{Soc}_{\mathfrak{p}}(I)$ has also the
structure of a bigraded $T$-module, if we set

$af=\psi(a)f\ \ \ \text{for all}\ \ a\in T\ \ \text{and all}\ \
f\in\operatorname{Soc}_{\mathfrak{p}}(I).$

Since $\psi$ is surjective and $\operatorname{Soc}_{\mathfrak{p}}(I)$ is a
finitely generated $\mathcal{F}_{\mathfrak{p}}(I)$-module, it follows that
$\operatorname{Soc}_{\mathfrak{p}}(I)$ is a finitely generated $T$-module, as
well.

The following lemma is required. For a bigraded $T$-module
$M=\bigoplus_{d,k}M_{d,k}$, we set $M_{(*,k)}=\bigoplus_{d}M_{(d,k)}$. Note
that $M_{(*,k)}$ becomes a graded $S$-module.

######  Lemma 2.4.

Let $T=K[x_{1},\dots,x_{n},y_{1},\dots,y_{m}]$ be a bigraded polynomial ring,
with $K$ a field, $\textup{bideg}(x_{i})=(1,0)$ for $1\leq i\leq n$ and
$\textup{bideg}(y_{i})=(d_{i},1)$ for $1\leq i\leq m$. Let
$\mathfrak{m}=(x_{1},\dots,x_{n})$ and $S=K[x_{1},\dots,x_{n}]\subset T$. Let
$M$ be a finitely generated bigraded $T$-module. Then,

$\operatorname{Tor}^{S}_{i}(S/\mathfrak{m},M_{(*,k)})\cong\operatorname{Tor}^{T}_{i}(T/\mathfrak{m},M)_{(*,k)}$

for all $i$ and $k$.

###### Proof.

Let ${\mathbb{F}}:0\rightarrow\cdots\rightarrow
F_{j}\rightarrow\cdots\rightarrow F_{1}\rightarrow F_{0}\rightarrow
M\rightarrow 0$ be a minimal bigraded $T$-resolution of $M$. Then,

${\mathbb{F}}_{k}:0\rightarrow\cdots\rightarrow(F_{j})_{(*,k)}\rightarrow\cdots\rightarrow(F_{1})_{(*,k)}\rightarrow(F_{0})_{(*,k)}\rightarrow
M_{(*,k)}\rightarrow 0$

is a graded (possibly non-minimal) free $S$-resolution of
$M_{(*,k)}=\bigoplus_{d}M_{(d,k)}$. Since
$\operatorname{Tor}^{T}_{i}(T/\mathfrak{m},M)=H_{i}({\mathbb{F}}/\mathfrak{m}{\mathbb{F}})$
we have that
$\operatorname{Tor}^{T}_{i}(T/\mathfrak{m},M)_{(*,k)}=H_{i}({\mathbb{F}}_{k}/\mathfrak{m}{\mathbb{F}}_{k})$
which in turn is isomorphic to
$\operatorname{Tor}^{S}_{i}(S/\mathfrak{m},M_{(*,k)})$. The desired conclusion
follows.

Note that $T/\mathfrak{m}=K[y_{1},\dots,y_{m}]$ and that
$\operatorname{Tor}^{T}_{0}(T/\mathfrak{m},\operatorname{Soc}_{\mathfrak{p}}(I))$
is a finitely generated bigraded $T/\mathfrak{m}$-module. Therefore, by the
above lemma, we have

$\displaystyle\alpha((I^{k}:\mathfrak{p})/I^{k})$
$\displaystyle=\alpha(\operatorname{Tor}^{S}_{0}(S/\mathfrak{m},(I^{k}:\mathfrak{p})/I^{k}))=\alpha(\operatorname{Tor}^{S}_{0}(S/\mathfrak{m},\operatorname{Soc}_{\mathfrak{p}}(I)_{(*,k)}))$
$\displaystyle=\alpha(\operatorname{Tor}^{T}_{0}(T/\mathfrak{m},\operatorname{Soc}_{\mathfrak{p}}(I))_{(*,k)}).$

Similarly,
$\omega((I^{k}:\mathfrak{p})/I^{k})=\omega(\operatorname{Tor}^{T}_{0}(T/\mathfrak{m},\operatorname{Soc}_{\mathfrak{p}}(I))_{(*,k)})$.

From this discussion, Theorem 2.1(b) follows from the next more general
statement, which is a variation of [9, Theorem 3.4].

######  Proposition 2.5.

Let $V=K[y_{1},\dots,y_{s}]$ a polynomial ring, with
$\textup{bideg}(y_{i})=(d_{i},1)$, $d_{i}\geq 1$, for $1\leq i\leq s$ and $K$
a field, and let $M$ be a finitely generated bigraded $V$-module. Then,
$\alpha_{M}(k)=\min\\{d:M_{(d,k)}\neq 0\\}$ and
$\omega_{M}(k)=\max\\{d:M_{(d,k)}\neq 0\\}$ are linear functions in $k$ for
$k\gg 0$.

###### Proof.

The claim about the linearity of $\omega_{M}(k)$ follows from [9, Theorem
3.4]. The proof of the claim of the linearity of $\alpha_{M}(k)$ is similar,
but we include here all the details for the convenience of the reader.

For any exact sequence $0\rightarrow M\rightarrow N\rightarrow P\rightarrow 0$
of finitely generated bigraded $V$-modules we have
$\alpha_{N}(k)=\min\\{\alpha_{M}(k),\alpha_{P}(k)\\}$, for all $k$.

Since $M$ is a finitely generated $V$-module and $V$ is Noetherian, by the
bigraded version of [11, Proposition 3.7] there exists a sequence of bigraded
$V$-submodules

$0=M_{0}\subset M_{1}\subset\cdots\subset M_{i-1}\subset M_{i}=M$

of $M$ such that $M_{j}/M_{j-1}\cong V/\mathfrak{p}_{j}$, with
$\mathfrak{p}_{j}$ a bigraded prime ideal of $V$, for all $1\leq j\leq i$.
Hence, we may suppose that $M=V/J$ with $J$ a bigraded ideal of $V$. We show
that $J$ can be replaced by a monomial ideal. For this aim, let $>$ be a
monomial order on $V$, and let $\textup{in}(J)$ be the initial ideal of $J$
with respect to $>$. The natural $K$-basis of $V/J$ consists of all residue
classes (modulo $J$) of all monomials not belonging to $\textup{in}(J)$, see
[24, Proposition 2.2.5.(a)]. The same residue classes modulo $\textup{in}(J)$
form a $K$-basis for $V/\textup{in}(J)$. Thus
$\alpha_{M}(k)=\alpha_{V/J}(k)=\alpha_{V/\textup{in}(J)}(k)$, and we can
assume that $M=V/J$ with $J$ a monomial ideal of $V$.

Recall that $\textup{bideg}(y_{i})=(d_{i},1)$ for $1\leq i\leq s$. For later
use, we may suppose that $d_{1}\leq d_{2}\leq\cdots\leq d_{s}$. Furthermore,
we can assume that $J$ is minimally generated by the monomials ${\bf y}^{{\bf
c}_{i}}=y_{1}^{c_{i,1}}y_{2}^{c_{i,2}}\cdots y_{s}^{c_{i,s}}$, for $1\leq
i\leq r$.

Let ${\bf a}=(a_{1},a_{2},\dots,a_{s})\in{\mathbb{N}}^{s}$, we denote by
$\overline{{\bf y^{a}}}$ the residue class of ${\bf
y^{a}}=y_{1}^{a_{1}}y_{2}^{a_{2}}\cdots y_{s}^{a_{s}}$ in $V/J$. Let $k\geq
0$, by $B_{k}$ we denote the minimal basis of $(V/J)_{k}$. Then, we can write
$\alpha_{M}(k)=\min\\{v({\bf a}):\overline{{\bf y^{a}}}\in B_{k}\\}$, where
$v({\bf a})=\sum_{i=1}^{s}a_{i}\deg(y_{i})=\sum_{i=1}^{s}a_{i}d_{i}.$

Clearly, $\overline{{\bf y^{a}}}\in B_{k}$ if and only if
$\sum_{j=1}^{s}a_{j}=k$, and for all $i=1,\dots,s$, there exists $j$ such that
$a_{j}<c_{i,j}$. Denote by $L$ the set of all maps
$\\{1,\dots,r\\}\rightarrow\\{1,\dots,s\\}$. We can decompose the set $B_{k}$
as the union $\bigcup_{f\in L}B_{k,f}$, where

$B_{k,f}=\big{\\{}\overline{{\bf y^{a}}}\ :\ \sum_{j=1}^{s}a_{j}=k\
\text{and}\ a_{f(i)}<c_{i,f(i)},\ i=1,\dots,r\big{\\}}.$

With this in mind, we can write $\alpha_{M}(k)=\min_{f\in L}\alpha_{f}(k)$,
where $\alpha_{f}(k)$ is defined as $\alpha_{f}(k)=\min\\{v({\bf
a}):\overline{{\bf y^{a}}}\in B_{k,f}\\}$. Hence, it is enough to prove that
$\alpha_{f}(k)$ is a linear function with integer coefficients for all $f\in
L$ and all $k\gg 0$.

Fix $f\in L$. Let $\\{j_{1}<j_{2}<\dots<j_{t}\\}$ be the image of $f$. For
$h=1,\dots,t$, we set $c_{j_{h}}=\min\\{c_{i,j_{h}}:i=1,\dots,r\\}-1$. Then,
we have that

$B_{k,f}=\big{\\{}\overline{{\bf y^{a}}}\ :\ \sum_{j=1}^{s}a_{j}=k\
\text{and}\ a_{j_{h}}\leq c_{j_{h}},\ h=1,\dots,t\big{\\}}.$

Thus, $\alpha_{f}(k)$ is given by the maximum of the functional $v({\bf a})$
on the set

$C_{k,f}=\big{\\{}{\bf a}\ :\ \sum_{j=1}^{s}a_{j}=k\ \text{and}\ a_{j_{h}}\leq
c_{j_{h}},\ h=1,\dots,t\big{\\}}.$

Let $\ell$ be the smallest integer such that $j_{1}=1$, $\dots$,
$j_{\ell}=\ell$ and $j_{\ell+1}>\ell+1$. Thus, for ${\bf
a}=(a_{1},a_{2},\dots,a_{s})\in C_{k,f}$ we have $a_{1}<c_{j_{1}},\
a_{2}<c_{j_{2}},\ \dots,\ a_{\ell}<c_{j_{\ell}}$ and no bound on
$a_{j_{\ell+1}}$, except that $\sum_{j=1}^{s}a_{j}=k$. We distinguish the two
possible cases.
Case 1. Suppose ${\ell}=s$. Then $\sum_{j=1}^{s}a_{j}$ can be at most
$c_{j_{1}}+c_{j_{2}}+\dots+c_{j_{\ell}}$. Thus, for all $k\gg 0$,
$B_{k,f}=\emptyset$ and so $\alpha_{f}(k)=0$.
Case 2. Suppose $\ell<s$. We let $k$ such that $k\geq
c_{j_{1}}+c_{j_{2}}+\dots+c_{j_{\ell}}$. We claim that the functional $v$ has
its minimal value for ${\bf
a}_{*}=(c_{j_{1}},c_{j_{2}},\dots,c_{j_{\ell}},k-\sum_{p=1}^{\ell}c_{j_{p}},0,0,\dots,0)$.
Then, for all large $k\gg 0$, we have that

$\alpha_{f}(k)=v({\bf
a}_{*})=\sum_{p=1}^{\ell}c_{j_{p}}d_{j_{p}}+d_{j_{\ell+1}}(k-\sum_{p=1}^{\ell}c_{j_{p}}),$

which is a linear function in $k$ with integer coefficients, as desired.

Let ${\bf a}=(a_{1},a_{2},\dots,a_{s})\in C_{k,f}$. Assume that for some
$1\leq i<j\leq s$ we have $a_{i}<c_{i}$ and $a_{j}>0$. Then, ${\bf
a}^{\prime}=(a_{1},a_{2}\dots,a_{i}+1,\dots,a_{j}-1,\dots,a_{s})$ also belongs
to $C_{k,f}$ and $v({\bf a}^{\prime})\leq v({\bf a})$ because we have
$d_{i}\leq d_{j}$. Thus, we see that the minimal value of $v$ on $C_{k,f}$ is
achieved when we fill up the first “boxes” of ${\bf a}\in C_{k,f}$ as much as
possible. Finally, the functional $v$ reaches its minimal value when ${\bf
a}={\bf a}_{*}$.

Theorem 1.1(c) combined with Theorem 2.1(b) yields

######  Corollary 2.6.

Let $I\subset S=K[x_{1},\dots,x_{n}]$ be a graded ideal and let
$\mathfrak{p}\in\operatorname{Max}^{\infty}(I)$. Then,
$\operatorname{v}_{\mathfrak{p}}(I^{k})$ is a linear function in $k$, for
$k\gg 0$.

We conclude this section with the following result that gives an estimate on
the growth of the modules $(I^{k}:\mathfrak{p})/I^{k}$ for large $k$.

######  Proposition 2.7.

Let $I\subset S$ be a graded ideal, and let
$\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$. Then,

$(I^{k}:\mathfrak{p})/I^{k}\ =\ (I(I^{k-1}:\mathfrak{p}))/I^{k},\ \
\textit{for all}\ k\gg 0.$

In particular, $(I^{k}:\mathfrak{p})\ =\ I(I^{k-1}:\mathfrak{p})$, for all
$k\gg 0$.

In order to prove the above result, we need the next general property.

######  Proposition 2.8.

Let $T=K[x_{1},\dots,x_{n},y_{1},\dots,y_{m}]$ be a bigraded polynomial ring,
with $K$ a field, $\textup{bideg}(x_{i})=(1,0)$ for $1\leq i\leq n$ and
$\textup{bideg}(y_{i})=(d_{i},1)$, $d_{i}\geq 1$, for $1\leq i\leq m$. Let $M$
be a finitely generated bigraded $T$-module. Then

$M_{(*,k)}\ =\ \\{y_{1},\dots,y_{m}\\}M_{(*,k-1)},\ \ \textit{for all}\ k\gg
0.$

Here $\\{y_{1},\dots,y_{m}\\}M_{(*,k-1)}$ denotes the set
$\\{\sum_{i=1}^{m}y_{i}f_{i}:f_{i}\in M_{(*,k-1)}\\}$.

###### Proof.

Since $M$ is finitely generated, there exists bihomogeneous elements
$g_{1},\dots,g_{r}$ that generate $M$ as a bigraded $T$-module. Let
$\textup{bideg}(g_{i})=(p_{i},q_{i})$ for $1\leq i\leq r$.

For vectors ${\bf a}=(a_{1},\dots,a_{n})\in{\mathbb{Z}}_{\geq 0}^{n}$ and
${\bf b}=(b_{1},\dots,b_{m})\in{\mathbb{Z}}_{\geq 0}^{m}$, we set

${\bf x^{a}y^{b}}\ =\ x_{1}^{a_{1}}\cdots x_{n}^{a_{n}}y_{1}^{b_{1}}\cdots
y_{m}^{b_{m}}.$

Then, these elements form a $K$-basis of $T$, and $\textup{bideg}({\bf
x^{a}y^{b}})=(|{\bf a}|+\sum_{i=1}^{m}b_{i}d_{i},|{\bf b}|)$, where $|{\bf
c}|=c_{1}+\dots+c_{t}$ is the modulus of ${\bf c}$, if ${\bf
c}=(c_{1},\dots,c_{t})\in{\mathbb{Z}}_{\geq 0}^{t}$ . Hence, any bihomogeneous
element $f\in M$ with $\textup{bideg}(f)=(d,k)$ can be written as

$\displaystyle f\ =\ \sum_{i=1}^{r}(\sum_{{\bf a},{\bf b}}k_{{\bf a},{\bf
b},i}{\bf x}^{{\bf a}}{\bf y}^{{\bf b}})g_{i},$ (2)

where the sum is taken over all ${\bf
a}=(a_{1},\dots,a_{n})\in{\mathbb{Z}}_{\geq 0}^{n}$, ${\bf
b}=(b_{1},\dots,b_{m})\in{\mathbb{Z}}_{\geq 0}^{m}$ such that
$\textup{bideg}({\bf x^{a}y^{b}})=(|{\bf a}|+\sum_{j=1}^{m}b_{j}d_{j},|{\bf
b}|)=(d-p_{i},k-q_{i})\in{\mathbb{Z}}_{\geq 0}^{2}$, and $k_{{\bf a},{\bf
b},i}\in K$.

Now, let $k>\max\\{q_{1},\dots,q_{r}\\}$ be an integer. It is clear that
$M_{(*,k)}$ contains $\\{y_{1},\dots,y_{m}\\}M_{(*,k-1)}$. For the reverse
inclusion, take $f\in M_{(*,k)}$ bihomogeneous of degree $(d,k)$. Writing $f$
as in (2), since $k>q_{i}$ for all $i$, we see that for all ${\bf a},{\bf b}$
such that the corresponding summand $k_{{\bf a},{\bf b},i}{\bf x^{a}y^{b}}$ in
(2) is non-zero, we have $|{\bf b}|=k-q_{i}>0$ for all $i$. Hence, ${\bf
y}^{{\bf b}}=y_{j_{{\bf a,b},i}}{\bf y}^{{\bf b}-{\bf e}_{j_{{\bf a,b},i}}}$,
for some $1\leq j_{{\bf a,b},i}\leq m$. Here ${\bf e}_{\ell}$ is the vector
with all entries zero except for the $\ell$th one which is equal to one.

Therefore, we can write

$f\ =\ \sum_{i=1}^{r}(\sum_{{\bf a},{\bf b}}k_{{\bf a},{\bf b},i}{\bf
x^{a}y^{b}})g_{i}\ =\ \sum_{i=1}^{r}\sum_{{\bf a},{\bf b}}y_{j_{{\bf
a,b},i}}(k_{{\bf a},{\bf b},i}{\bf x}^{{\bf a}}{\bf y}^{{\bf b}-{\bf
e}_{j_{{\bf a,b},i}}}g_{i}).$

Since each element $k_{{\bf a},{\bf b},i}{\bf x}^{{\bf a}}{\bf y}^{{\bf
b}-{\bf e}_{j_{{\bf a,b},i}}}g_{i}$ has bidegree $(d-d_{j_{{\bf
a,b},i}},k-1)$, from the above equation we see that
$f\in\\{y_{1},\dots,y_{m}\\}M_{(*,k-1)}$. Hence,
$M_{(*,k)}\subseteq\\{y_{1},\dots,y_{m}\\}M_{(*,k-1)}$ and the proof is
complete.

We are now ready for the proof of the proposition.

###### Proof of Proposition 2.7.

Taking into account equation (1) and that
$\operatorname{Soc}_{\mathfrak{p}}(I)$ is a finitely generated bigraded
$T$-module, by applying Proposition 2.8 we obtain that

$\operatorname{Soc}_{\mathfrak{p}}(I)_{(*,k)}\ =\
\\{y_{1},\dots,y_{m}\\}\operatorname{Soc}_{\mathfrak{p}}(I)_{(*,k-1)},\ \
\textup{for all}\ k\gg 0.$

Note that $\operatorname{Soc}_{\mathfrak{p}}(I)_{(*,k)}=\bigoplus_{d\geq
0}((I^{k}:\mathfrak{p})/I^{k})_{d}=(I^{k}:\mathfrak{p})/I^{k}$, for all $k$.
Hence,

$(I^{k}:\mathfrak{p})/I^{k}\ =\
\\{y_{1},\dots,y_{m}\\}(I^{k-1}:\mathfrak{p})/I^{k-1},\ \ \textup{for all}\
k\gg 0.$

Now, since
$\mathcal{F}_{\mathfrak{p}}(I)_{(*,1)}\operatorname{Soc}_{\mathfrak{p}}(I)_{(*,k-1)}\subseteq\operatorname{Soc}_{\mathfrak{p}}(I)_{(*,k)}$
and $\psi(y_{i})=u_{i}+\mathfrak{p}I\in\mathcal{F}_{\mathfrak{p}}(I)_{(*,1)}$
for all $1\leq i\leq m$, we obtain that

$(I^{k}:\mathfrak{p})/I^{k}\ =\ (I(I^{k-1}:\mathfrak{p}))/I^{k},\ \
\textup{for all}\ k\gg 0.$

Finally, lifting this equation to $S$ yields
$(I^{k}:\mathfrak{p})=I(I^{k-1}:\mathfrak{p})$, for all $k\gg 0$.

With a similar argument, we can show the next analogue of Proposition 2.7.
This result complements [31, Corollary 4.2] due to Ratliff. To the best of our
knowledge, it was previously unknown.

######  Corollary 2.9.

Let $R$ be a commutative Noetherian domain, $I\subset R$ be an ideal, and
$\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$ be a stable prime of $I$.
Then,

$(I^{k}:\mathfrak{p})\ =\ I(I^{k-1}:\mathfrak{p}),\ \ \textit{for all}\ k\gg
0.$

##  3\. Asymptotic behaviour of the $\operatorname{v}$-number

In this section, we prove the main result in the article:
$\operatorname{v}(I^{k})$ is indeed linear function in $k$ for all $k\gg 0$,
and all graded ideals $I\subset S$.

######  Theorem 3.1.

Let $I\subset S=K[x_{1},\dots,x_{n}]$ be a graded ideal. Then,
$\operatorname{v}_{\mathfrak{p}}(I^{k})$, for all for
$\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$, and $\operatorname{v}(I^{k})$
are linear functions in $k$, for $k\gg 0$.

In order to prove the theorem, we construct a new module starting from
$\operatorname{Soc}_{\mathfrak{p}}(I)$.

Let $I\subset S$ be a graded ideal, and let
$\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$. For all $k$, let
$\overline{g}_{k,1},\dots,\overline{g}_{k,r_{k}}$ be a minimal homogeneous
generating set of
$\operatorname{Soc}_{\mathfrak{p}}(I)_{(*,k)}=(I^{k}:\mathfrak{p})/I^{k}$. We
set

$\displaystyle A_{k}\ $ $\displaystyle=\ \\{\overline{g}_{k,j}\ :\
(I^{k}:g_{k,j})=\mathfrak{p}\\},$ $\displaystyle B_{k}\ $ $\displaystyle=\
\\{\overline{g}_{k,j}\ :\ (I^{k}:g_{k,j})\neq\mathfrak{p}\\}.$

Let $\mathcal{N}$ be the submodule of $\operatorname{Soc}_{\mathfrak{p}}(I)$
generated by $\bigcup_{k}B_{k}$. Since each element of the set
$\bigcup_{k}B_{k}$ is bihomogeneous, $\mathcal{N}$ is a bigraded submodule of
$\operatorname{Soc}_{\mathfrak{p}}(I)$. We define the following
${\mathcal{F}}_{\mathfrak{p}}(I)$-module:

$\operatorname{Soc}_{\mathfrak{p}}^{*}(I)\ =\
\operatorname{Soc}_{\mathfrak{p}}(I)/\mathcal{N}.$

By Theorem 2.2 and the fact that $\mathcal{N}$ is a bigraded submodule of
$\operatorname{Soc}_{\mathfrak{p}}(I)$, it follows that
$\operatorname{Soc}^{*}_{\mathfrak{p}}(I)$ is a finitely generated bigraded
${\mathcal{F}}_{\mathfrak{p}}(I)$-module.

Moreover,

$\operatorname{Soc}_{\mathfrak{p}}^{*}(I)_{(*,k)}\ =\
\dfrac{(I^{k}:\mathfrak{p})/I^{k}}{\mathcal{N}_{(*,k)}},$

for all $k\geq 0$.

Finally, we recall the following basic rules. Let
$I,I_{1},I_{2},\\{J_{i}\\}_{i}$ be ideals of a commutative Noetherian ring $R$
and let $\mathfrak{p}$ be a prime ideal of $R$. Then,

  1. (i)

$(I:\sum_{i}J_{i})=\bigcap_{i}(I:J_{i})$,

  2. (ii)

$((I:I_{1}):I_{2})=(I:I_{1}I_{2})$,

  3. (iii)

if $\mathfrak{p}=\bigcap_{i}J_{i}$, then $\mathfrak{p}=J_{i}$ for some $i$.

###### Proof of Theorem 3.1.

Let $\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$. By Theorem 1.1, for all
$k\gg 0$ we have

$\operatorname{v}_{\mathfrak{p}}(I^{k})=\min_{\overline{g}_{k,j}\in
A_{k}}\deg(g_{k,j}).$

Next, note that $A_{k}\cup B_{k}$ generates
$\operatorname{Soc}_{\mathfrak{p}}(I)_{(*,k)}$ and
$B_{k}\subseteq\mathcal{N}$. It follows that a minimal homogeneous generating
set of $\operatorname{Soc}_{\mathfrak{p}}^{*}(I)_{(*,k)}$ is given by the non-
zero residue classes, modulo $\mathcal{N}_{(*,k)}$, of the elements of
$A_{k}$. We claim that each of these classes is non-zero modulo
$\mathcal{N}_{(*,k)}$. Suppose for a contradiction that some
$\overline{g}_{k,j}\in A_{k}$ belongs to $\mathcal{N}_{(*,k)}$. Since
$\mathcal{N}$ is generated by $\bigcup_{\ell}B_{\ell}$, we can write

$\overline{g}_{k,j}\ =\ \sum_{\begin{subarray}{c}1\leq\ell\leq k\\\
\overline{g}_{\ell,p}\in
B_{\ell}\end{subarray}}\overline{a}_{k-\ell,p}\overline{g}_{\ell,p},$ (3)

with each
$\overline{a}_{k-\ell,p}\in\mathcal{F}_{\mathfrak{p}}(I)_{(*,k-\ell)}=I^{k-\ell}/\mathfrak{p}I^{k-\ell}$.
Since $\overline{g}_{k,j}\in A_{k}$ we have $(I^{k}:g_{k,j})=\mathfrak{p}$. By
equation (3) we can write $g_{k,j}=\sum a_{k-\ell,p}g_{\ell,p}+z$ where $z$ is
a suitable element of $I^{k}$. Note that
$\bigcap(I^{k}:a_{k-\ell,p}g_{\ell,p})$ is contained in
$(I^{k}:g_{k,j})=\mathfrak{p}$. Indeed, take
$h\in\bigcap(I^{k}:a_{k-\ell,p}g_{\ell,p})$. Then $ha_{k-\ell,p}g_{\ell,p}\in
I^{k}$ for all terms in the intersection. Hence $hg_{k,j}=\sum
ha_{k-\ell,p}g_{\ell,p}+hz\in I^{k}$. Since
$(I^{\ell}:g_{\ell,p})\subseteq(I^{k}:a_{k-\ell,p}g_{\ell,p})$ for all $\ell$
and $p$,

$\displaystyle\mathfrak{p}\ \subseteq\ \bigcap(I^{\ell}:g_{\ell,p})\
\subseteq\ \bigcap(I^{k}:a_{k-\ell,p}g_{\ell,p})\ \subseteq\ \mathfrak{p}.$

Hence $\bigcap(I^{\ell}:g_{\ell,p})=\mathfrak{p}$. By rule (iii), we have
$(I^{\ell}:g_{\ell,p})=\mathfrak{p}$ for some $\ell$ and $p$. But this is a
contradiction because $\overline{g}_{\ell,p}\in B_{\ell}$. This proves our
claim. As a consequence, we obtain that for all $k\gg 0$

$\alpha(\operatorname{Soc}_{\mathfrak{p}}^{*}(I)_{(*,k)})\ =\
\min_{\overline{g}_{k,j}\in A_{k}}\deg(g_{k,j})\ =\
\operatorname{v}_{\mathfrak{p}}(I^{k}).$

Using the map (1), we see that $\operatorname{Soc}_{\mathfrak{p}}^{*}(I)$ is a
finitely generated bigraded $T$-module. Moreover,
$\alpha(\operatorname{Soc}_{\mathfrak{p}}^{*}(I)_{(*,k)})=\alpha(\operatorname{Tor}^{T}_{0}(T/\mathfrak{m},\operatorname{Soc}_{\mathfrak{p}}^{*}(I))_{(*,k)})$.
Applying Proposition 2.5, we conclude that
$\operatorname{v}_{\mathfrak{p}}(I^{k})=\alpha(\operatorname{Soc}_{\mathfrak{p}}^{*}(I)_{(*,k)})$
is a linear function in $k$ for $k\gg 0$.

Finally,
$\operatorname{v}(I)=\min_{\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)}\operatorname{v}_{\mathfrak{p}}(I^{k})$
is a linear function in $k$ for $k\gg 0$, for it is given by the minimum of
finitely many eventually linear functions in $k$.

We conclude the section with some results which estimate the growth of the
functions $\operatorname{v}_{\mathfrak{p}}(I^{k})$, when $k\gg 0$.

######  Proposition 3.2.

Let $I\subset S$ be a graded ideal and let
$\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$. Then,

$\operatorname{v}_{\mathfrak{p}}(I^{k-1})+\alpha(I)\ \leq\
\operatorname{v}_{\mathfrak{p}}(I^{k})\ \leq\
\operatorname{v}_{\mathfrak{p}}(I^{k-1})+\omega(I),\ \ \textit{for all}\ k\gg
0.$

In particular,

$\operatorname{v}(I^{k-1})+\alpha(I)\leq\operatorname{v}(I^{k})\leq\operatorname{v}(I^{k-1})+\omega(I),\
\ \textit{for all}\ k\gg 0.$

###### Proof.

We first prove that
$\operatorname{v}_{\mathfrak{p}}(I^{k-1})+\alpha(I)\leq\operatorname{v}_{\mathfrak{p}}(I^{k})$,
for $k\gg 0$. By Proposition 2.7, there exists $k_{0}>0$ such that
$(I^{k}:\mathfrak{p})/I^{k}\ =\ (I(I^{k-1}:\mathfrak{p}))/I^{k}$, for all
$k\geq k_{0}$. Let $k\geq k_{0}$. Take
$\overline{f}\in(I^{k}:\mathfrak{p})/I^{k}$ such that $(I^{k}:f)=\mathfrak{p}$
and $\deg(f)=\operatorname{v}_{\mathfrak{p}}(I^{k})$. Since
$(I^{k}:\mathfrak{p})=I(I^{k-1}:\mathfrak{p})$, we can write $f=ug$ where
$u\in I$ and $g\in(I^{k-1}:\mathfrak{p})$. Then

$\mathfrak{p}\subseteq(I^{k-1}:g)\subseteq(uI^{k-1}:ug)=(uI^{k-1}:f)\subseteq(I^{k}:f)=\mathfrak{p}.$

Here we used that $g\in(I^{k-1}:\mathfrak{p})$ and $uI^{k-1}\subseteq I^{k}$.
Hence, $(I^{k-1}:g)=\mathfrak{p}$ and so

$\operatorname{v}_{\mathfrak{p}}(I^{k})=\deg(f)=\deg(ug)=\deg(g)+\deg(u)\geq\operatorname{v}_{\mathfrak{p}}(I^{k-1})+\alpha(I).$

Now, we prove the inequality
$\operatorname{v}_{\mathfrak{p}}(I^{k})\leq\operatorname{v}_{\mathfrak{p}}(I^{k-1})+\omega(I)$,
for $k\gg 0$. By Brodmann and Ratliff, there exists $k^{*}>0$ such that
$\operatorname{Ass}^{\infty}(I)=\operatorname{Ass}(I^{k})$ and
$(I^{k}:I)=I^{k-1}$ for all $k\geq k^{*}$. Fix $k\geq k^{*}$ and let
$\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$.

Let $f\in S$ be a homogeneous element with $(I^{k-1}:f)=\mathfrak{p}$ and
$\deg(f)=\operatorname{v}_{\mathfrak{p}}(I^{k-1})$. Let $u_{1},\dots,u_{m}$ be
a minimal homogeneous generating set of $I$. By rules (ii) and (i),

$\displaystyle\mathfrak{p}=(I^{k-1}:f)\ $ $\displaystyle=\
(I^{k}:I):f=(I^{k}:fI)$ $\displaystyle=\
(I^{k}:\sum_{i=1}^{m}(fu_{i}))=\bigcap_{i=1}^{m}(I^{k}:fu_{i}).$

Hence, by rule (iii), we have $\mathfrak{p}=(I^{k}:fu_{i})$ for some $i$. By
the definition of $\operatorname{v}_{\mathfrak{p}}(I^{k})$, this means that
$\operatorname{v}_{\mathfrak{p}}(I^{k})\leq\deg(fu_{i})=\deg(f)+\deg(u_{i})\leq\operatorname{v}_{\mathfrak{p}}(I^{k-1})+\omega(I)$.

We have the next nice consequences.

######  Corollary 3.3.

Let $I\subset S$ be a graded ideal, and let
$\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$. Then,

  1. (a)

The function $\operatorname{v}_{\mathfrak{p}}(I^{k})$ is strictly increasing,
for all $k\gg 0$.

  2. (b)

The function $\operatorname{v}(I^{k})$ is strictly increasing, for all $k\gg
0$.

  3. (c)

$\omega((I^{k}:\mathfrak{p})/I^{k})\geq\alpha((I^{k-1}:\mathfrak{p})/I^{k-1})+\alpha(I)$,
for all $k\gg 0$.

  4. (d)

$\alpha((I^{k}:\mathfrak{p})/I^{k})\leq\omega((I^{k-1}:\mathfrak{p})/I^{k-1})+\omega(I)$,
for all $k\gg 0$.

###### Proof.

(a) By the previous Proposition 3.2,
$\operatorname{v}_{\mathfrak{p}}(I^{k})\geq\operatorname{v}_{\mathfrak{p}}(I^{k-1})+\alpha(I)$,
for all $k\gg 0$. Since $\alpha(I)\geq 1$, we get
$\operatorname{v}_{\mathfrak{p}}(I^{k})>\operatorname{v}_{\mathfrak{p}}(I^{k-1})$
for $k\gg 0$, as wanted. Statement (b) follows from (a) and Theorem 1.1(b).
Finally, statements (c) and (d) follow from Theorems 1.1(a) and 2.1(a)
combined with Proposition 3.2.

The next elementary lemma will be needed several times in the sequel.

######  Lemma 3.4.

Let $\\{a_{k}\\}_{k}$ be a numerical sequence. The following statements hold.

  1. (a)

Suppose that $a_{k}\geq a_{k-1}+a$, for all $k\gg 0$ and some integer $a$.
Then, there exists a constant $c$ such that $a_{k}\geq ak+c$, for all $k\gg
0$.

  2. (b)

Suppose that $a_{k}\leq a_{k-1}+a$, for all $k\gg 0$ and some integer $a$.
Then, there exists a constant $c$ such that $a_{k}\leq ak+c$, for all $k\gg
0$.

  3. (c)

Suppose that $a_{k}=a_{k-1}+a$, for all $k\gg 0$ and some integer $a$. Then,
there exists a constant $c$ such that $a_{k}=ak+c$, for all $k\gg 0$.

  4. (d)

Suppose that $ak+b\leq a_{k}\leq ak+c$, for all $k\gg 0$ and some integers
$a,b,c$. Then, $\lim\limits_{k\rightarrow\infty}\frac{a_{k}}{k}=a$.

###### Proof.

Suppose that $a_{k}\geq a_{k-1}+a$, for all $k\geq k_{0}$ and a certain
$k_{0}>0$. Then

$\displaystyle a_{k}\ $ $\displaystyle\geq\ a_{k-1}+a\ \geq\ a_{k-2}+2a\ \geq\
\cdots\ \geq\ a_{k_{0}-1}+(k-k_{0}+1)a$ $\displaystyle=\
ak+(a_{k_{0}-1}-(k_{0}-1)a),$

for all $k\geq k_{0}$. Setting $c=a_{k_{0}-1}-(k_{0}-1)a$, claim (a) follows.
Replacing “$\geq$” with “$\leq$” or “$=$”, we see that (b) and (c) hold as
well. For the proof of (d), note that

$\frac{ak+b}{k}\ \leq\ \frac{a_{k}}{k}\ \leq\ \frac{ak+c}{k}$

for all $k\gg 0$. Thus, taking the limit for $k\rightarrow\infty$, the Squeeze
Theorem for numerical sequences implies that
$\lim\limits_{k\rightarrow\infty}\frac{a_{k}}{k}=a$, as wanted.

##  4\. The limit behaviour of the sequence
$\frac{\operatorname{v}(I^{k})}{k}$

Let $I\subset S$ be a graded ideal. Due to Theorem 3.1, there exist integers
$a$ and $b$ such that $\operatorname{v}(I^{k})=ak+b$ for $k\gg 0$. As a
consequence, the limit $\lim_{k\rightarrow\infty}\operatorname{v}(I^{k})/k$
exists and it is equal to $a$. In this section, we prove the following
surprising fact.

######  Theorem 4.1.

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

$\lim_{k\rightarrow\infty}\frac{\operatorname{v}(I^{k})}{k}\ =\ \alpha(I).$

In view of statements (a), (b) and (d) of Lemma 3.4, to prove the theorem, it
is enough to find lower and upper bounds for $\operatorname{v}(I^{k})$, which
are linear functions in $k$ of the type $\alpha(I)k+c$, for some constant $c$,
and for all $k\gg 0$.

We begin by providing the linear upper bound.

######  Lemma 4.2.

Let $I\subset S$ be a graded ideal. Then, there exists a constant $c$ such
that

$\operatorname{v}(I^{k})\ \leq\ \alpha(I)k+c,\ \ \textit{for all}\ k\gg 0.$

The proof of this lemma is based upon the next bound which generalizes a
result of Saha and Sengupta [33, Proposition 3.11].

######  Proposition 4.3.

Let $I\subset S$ be a graded ideal and let $f\in S$ be a homogeneous element
not belonging to $I$. Then

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

###### Proof.

If $(I:f)\in\operatorname{Ass}(I)$, then $\operatorname{v}(I)\leq\deg(f)$.
Since, also $\operatorname{v}(I:f)\geq 0$, the asserted inequality follows.
Suppose that $(I:f)\notin\operatorname{Ass}(I)$. Note that $(I:f)$ is a proper
ideal, because $f\notin I$. From the short exact sequence

$0\rightarrow S/(I:f)\rightarrow S/I\rightarrow S/(I,f)\rightarrow 0$

it follows that $\operatorname{Ass}(I:f)\subseteq\operatorname{Ass}(I)$. Let
$g\in S$ be a homogeneous element such that
$((I:f):g)\in\operatorname{Ass}(I:f)$ and $\deg(g)=\operatorname{v}(I:f)$.
Then

$((I:f):g)=(I:(fg))\in\operatorname{Ass}(I).$

Thus
$\operatorname{v}(I)\leq\deg(fg)=\deg(f)+\deg(g)=\deg(f)+\operatorname{v}(I:f)$.

Now, we are ready for the proof of Lemma 4.2 which follows closely the
argument given in [2, Theorem 4.7].

###### Proof of Lemma 4.2.

Let $f\in I$ with $\deg(f)=\alpha(I)$. Then for any $k\geq 2$, $f^{k-1}\notin
I^{k}$ by degree reasons. By the previous proposition, we have
$\operatorname{v}(I^{k})\leq\operatorname{v}(I^{k}:f)+\alpha(I)$. Applying
again the proposition we obtain
$\operatorname{v}(I^{k}:f)\leq\operatorname{v}(I^{k}:f^{2})+\alpha(I)$ and
hence $\operatorname{v}(I^{k})\leq\operatorname{v}(I^{k}:f^{2})+2\alpha(I)$.
Iterating this reasoning, we get

$\displaystyle\operatorname{v}(I^{k})\ \leq\
\alpha(I)k+(\operatorname{v}(I^{k}:f^{k-1})-\alpha(I)).$ (4)

On the other hand, consider the ascending chain

$I\subseteq(I^{2}:f)\subseteq(I^{3}:f^{2})\subseteq\cdots\subseteq(I^{k}:f^{k-1})\subseteq\cdots.$

Since $S$ is Noetherian, there exists $k_{0}>0$ such that
$(I^{k}:f^{k-1})=(I^{k_{0}}:f^{k_{0}-1})$, for all $k\geq k_{0}$. Thus for all
$k\geq k_{0}$, the number $\operatorname{v}(I^{k}:f^{k-1})-\alpha(I)$ is a
constant $c$. Hence, by equation (4) we see that
$\operatorname{v}(I^{k})\leq\alpha(I)k+c$, for all $k\gg 0$.

Next, combining the second lower bound in Proposition 3.2 with Lemma 3.4(a) we
obtain the next lower linear bound for $\operatorname{v}(I^{k})$.

######  Lemma 4.4.

Let $I\subset S$ be a graded ideal. Then, there exists a constant $c$ such
that $\operatorname{v}(I^{k})\geq\alpha(I)k+c$, for all $k\gg 0$.

Now, we are ready for the proof of Theorem 4.1.

###### Proof of Theorem 4.1.

It follows by combining Lemmas 4.2, 4.4 and 3.4(d).

Let $I\subset S$ be a graded ideal, and let
$\mathfrak{p}\in\operatorname{Ass}^{\infty}(I)$. By Theorem 3.1,
$\operatorname{v}_{\mathfrak{p}}(I^{k})$ is also an eventually linear function
of the form $a_{\mathfrak{p}}k+b_{\mathfrak{p}}$ for $k\gg 0$. By Proposition
3.2, we obtain that $\alpha(I)\leq a_{\mathfrak{p}}\leq\omega(I)$. However, in
contrast to Theorem 4.1, $a_{\mathfrak{p}}$ needs not to be equal to
$\alpha(I)$.

For instance, consider the monomial ideal $I=(x^{5},x^{4}y^{3},x^{2}y^{4})$ of
$S=K[x,y]$. By Proposition 5.1(a) we have that
$\mathfrak{p}_{x}=(x)\in\operatorname{Ass}(I^{k})=\operatorname{Ass}^{\infty}(I)$
for all $k$. However, by Corollary 5.4(a) we have that
$\operatorname{v}_{\mathfrak{p}_{x}}(I^{k})=\deg(x^{2}y^{4})k-1=6k-1$, for all
$k\geq 1$, and the slope of this linear function is $6>5=\alpha(I)$.

##  5\. The $\operatorname{v}$-number of monomial ideals in two variables

In this section, we consider monomial ideals of the polynomial ring in two
variables $S=K[x,y]$. Let $I\subset S$ be a monomial ideal. As customary, we
denote by $G(I)$ the unique minimal monomial generating set of $I$. Then

$G(I)\ =\
\\{x^{a_{1}}y^{b_{1}},x^{a_{2}}y^{b_{2}},\dots,x^{a_{m}}y^{b_{m}}\\},$

where ${\bf a}:a_{1}>a_{2}>\cdots>a_{m}\geq 0$ and ${\bf b}:0\leq
b_{1}<b_{2}<\cdots<b_{m}$. Conversely, given any two such sequences ${\bf a}$
and ${\bf b}$, the set
$\\{x^{a_{1}}y^{b_{1}},x^{a_{2}}y^{b_{2}},\dots,x^{a_{m}}y^{b_{m}}\\}$ is the
minimal monomial generating set of a monomial ideal of $S$.

Therefore, the monomial ideals of $S=K[x,y]$ are in bijection with all pairs
$({\bf a},{\bf b})$ of sequences ${\bf a}:a_{1}>a_{2}>\cdots>a_{m}\geq 0$ and
${\bf b}:0\leq b_{1}<b_{2}<\cdots<b_{m}$ as above. Hereafter, we write
$I=I_{\bf a,b}$ for
$I=(x^{a_{1}}y^{b_{1}},x^{a_{2}}y^{b_{2}},\dots,x^{a_{m}}y^{b_{m}})$.

The natural $K$-basis of $S/I_{\bf a,b}$ consists of the residue classes
(modulo $I_{\bf a,b}$) of the monomials not belonging to $I_{\bf a,b}$. These
basis elements can be represented by the lattice points
$(c,d)\in{\mathbb{Z}}_{\geq 0}\times{\mathbb{Z}}_{\geq 0}$ such that
$x^{c}y^{d}\notin I_{\bf a,b}$, as in the next picture.

$\succ$

$\succ$$\succ$$(a_{n},b_{n})$$(a_{2},b_{2})$$(a_{1},b_{1})$$I$

In this section we investigate the $\operatorname{v}$-function of $I_{\bf
a,b}$ in terms of ${\bf a}$ and ${\bf b}$.

The associated prime ideals of a monomial ideal $I$ are monomial prime ideals,
that is, ideals generated by a subset of the variables [24, Corollary 1.3.9].
Thus, in our case $\operatorname{Ass}(I)\subseteq\\{(x),(y),(x,y)\\}$. We set
$\mathfrak{p}_{x}=(x)$, $\mathfrak{p}_{y}=(y)$ and $\mathfrak{m}=(x,y)$.

We can compute $\operatorname{Ass}(I_{\bf a,b})$ in terms of the sequences
${\bf a}$ and ${\bf b}$.

######  Proposition 5.1.

Let $I=I_{\bf a,b}\subset S$ be a monomial ideal. Then,

  1. (a)

$\mathfrak{p}_{x}\in\operatorname{Ass}(I)$ if and only if $a_{m}>0$.

  2. (b)

$\mathfrak{p}_{y}\in\operatorname{Ass}(I)$ if and only if $b_{1}>0$.

  3. (c)

$\mathfrak{m}\in\operatorname{Ass}(I)$ if and only if $m>1$, i.e., $I$ is not
a principal ideal.

###### Proof.

If $\mathfrak{p}_{x}\in\operatorname{Ass}(I)$ then
$I\subseteq\mathfrak{p}_{x}=(x)$. Hence $x$ divides all minimal monomial
generators of $I$. In particular, $x$ divides $x^{a_{m}}y^{b_{m}}$ and
$a_{m}>0$.

Conversely, let $a_{m}>0$. Then $x$ divides all minimal monomial generators of
$I$. Hence $I\subseteq\mathfrak{p}_{x}=(x)$. Since $\mathfrak{p}_{x}$ is of
height one, it follows that $\mathfrak{p}_{x}\in\operatorname{Ass}(I)$.

This proves (a), statement (b) can be proved similarly.

Finally, for the proof of (c), suppose $\mathfrak{m}\in\operatorname{Ass}(I)$.
If $I$ is a principal ideal, then for some $c$ and $d$,
$I=(x^{c}y^{d})=(x^{c})\cap(y^{d})=\mathfrak{p}_{x}^{c}\cap\mathfrak{p}_{y}^{d}$
is the primary decomposition of $I$, which contradicts our assumption. Hence
$I$ is not principal.

Conversely, suppose $I$ is not principal, but
$\mathfrak{m}\notin\operatorname{Ass}(I)$. Then
$\operatorname{Ass}(I)\subseteq\\{\mathfrak{p}_{x},\mathfrak{p}_{y}\\}$. Since
$\mathfrak{p}_{x}$ and $\mathfrak{p}_{y}$ are height one prime ideals,
$I=\mathfrak{p}_{x}^{c}\cap\mathfrak{p}_{y}^{d}=(x^{c})\cap(y^{d})=(x^{c}y^{d})$
for some $c$ and $d$, against our assumption. The assertion follows.

The following lemma is required.

######  Lemma 5.2.

Let $I=I_{\bf a,b}\subset S$ be a monomial ideal. Then,

$x^{ka_{1}}y^{kb_{1}},x^{ka_{m}}y^{kb_{m}}\in G(I^{k}),\ \ \textit{for all}\ \
k\geq 1.$

###### Proof.

Let $k\geq 1$. We know that

$I^{k}\ =\ (\prod_{i=1}^{m}(x^{a_{i}}y^{b_{i}})^{k_{i}}\ :\
\sum_{i=1}^{m}k_{i}=k).$

Let $x^{r}y^{s}$ be an arbitrary generator of $I^{k}$ different from
$x^{ka_{1}}y^{kb_{1}}$. Then, we have $r=k_{1}a_{1}+\dots k_{m}a_{m}$,
$s=k_{1}b_{1}+\dots k_{m}b_{m}$, $\sum_{i=1}^{m}k_{i}=k$ and $k_{i}>0$ for
some $i\neq 1$. Thus,

$ka_{1}=k_{1}a_{1}+k_{2}a_{1}+\dots k_{m}a_{1}>k_{1}a_{1}+k_{2}a_{2}+\dots
k_{m}a_{m}=r$

and

$kb_{1}=k_{1}b_{1}+k_{2}b_{1}+\dots k_{m}b_{1}<k_{1}b_{1}+k_{2}b_{2}+\dots
k_{m}b_{m}=s.$

Therefore, $x^{r}y^{s}$ does not divide $x^{ka_{1}}y^{kb_{1}}$. This shows
that $x^{ka_{1}}y^{kb_{1}}$ is a minimal generator of $I^{k}$. By a similar
argument we obtain that $x^{ka_{m}}y^{kb_{m}}\in G(I^{k})$.

######  Corollary 5.3.

Let $I=I_{\bf a,b}\subset S$ be a monomial ideal. Then
$\operatorname{Ass}(I^{k})=\operatorname{Ass}^{\infty}(I)$, for all $k\geq 1$.
In particular, $\textup{astab}(I)=1$.

###### Proof.

Let us prove that $\operatorname{Ass}(I)=\operatorname{Ass}(I^{k})$ for all
$k\geq 2$. By the previous proposition,
$\mathfrak{p}_{x}\in\operatorname{Ass}(I)$ if and only if $x$ divides all
minimal monomial generators of $I$. Hence, if
$\mathfrak{p}_{x}\in\operatorname{Ass}(I)$, then
$\mathfrak{p}_{x}\in\operatorname{Ass}(I^{k})$ for all $k\geq 2$, as well.

Now, suppose that $\mathfrak{p}_{x}\in\operatorname{Ass}(I^{k})$ for some
$k\geq 2$, but $\mathfrak{p}_{x}\notin\operatorname{Ass}(I)$. Then, by
Proposition 5.1(a) we have $a_{m}=0$. Hence $y^{b_{m}}\in G(I)$. By the
previous corollary, $y^{kb_{m}}\in G(I^{k})$, as well. But this is impossible,
because $y^{kb_{m}}\notin\mathfrak{p}_{x}$, but by assumption
$\mathfrak{p}_{x}$ contains $I^{k}$. Hence $a_{m}>0$ and
$\mathfrak{p}_{x}\in\operatorname{Ass}(I)$, as wanted.

The same reasoning can be applied to show that
$\mathfrak{p}_{y}\in\operatorname{Ass}(I)$ if and only if
$\mathfrak{p}_{y}\in\operatorname{Ass}(I^{k})$, for any $k\geq 2$.

Finally, by the previous proposition, $\mathfrak{m}\in\operatorname{Ass}(I)$
if and only $I$ is not principal. Lemma 5.2 implies that $I$ is not principal
if and only if $I^{k}$ is not principal for any $k\geq 2$. Thus
$\mathfrak{m}\in\operatorname{Ass}(I)$ if and only if
$\mathfrak{m}\in\operatorname{Ass}(I^{k})$ for any $k\geq 2$.

Next, we compute the functions $\operatorname{v}_{\mathfrak{p}_{x}}(I_{\bf
a,b}^{k})$, $\operatorname{v}_{\mathfrak{p}_{y}}(I_{\bf a,b}^{k})$.

######  Corollary 5.4.

Let $I=I_{\bf a,b}\subset S$ be a monomial ideal. The following holds.

  1. (a)

If $a_{m}>0$, then
$\operatorname{v}_{\mathfrak{p}_{x}}(I^{k})=k(a_{m}+b_{m})-1$, for all $k\geq
1$.

  2. (b)

If $b_{1}>0$, then
$\operatorname{v}_{\mathfrak{p}_{y}}(I^{k})=k(a_{1}+b_{1})-1$, for all $k\geq
1$.

###### Proof.

The only generator $\overline{u}\in(I:\mathfrak{p}_{x})/I$ such that
$(I:u)=\mathfrak{p}_{x}$ is $x^{a_{m}-1}y^{b_{m}}$, for it has the largest
$y$-degree. For each $k\geq 1$, from Lemma 5.2, $x^{ka_{m}}y^{kb_{m}}\in
G(I^{k})$ and such generator has the highest $y$-degree. Thus,
$\overline{u}=\overline{x^{ka_{m}-1}y^{kb_{m}}}$ is the only generator of
$(I^{k}:\mathfrak{p}_{x})/I^{k}$ such that $(I^{k}:u)=\mathfrak{p}_{x}$.
Similarly, one can prove (b).

In the next proposition, we show how to compute
$\operatorname{v}_{\mathfrak{m}}(I)$ for a non principal monomial ideal
$I=I_{\bf a,b}\subset S$. For our convenience, if $c\geq 1$, in the proof of
the next result we regard $x^{-c}$ and $y^{-c}$ as $1$.

######  Proposition 5.5.

Let $I=I_{\bf a,b}\subset S$ be a non principal monomial ideal. Then,

$(I:\mathfrak{m})/I=(x^{a_{j}-1}y^{b_{j+1}-1}\ :\ 1\leq j\leq m-1)/I.$ (5)

In particular,

$\operatorname{v}_{\mathfrak{m}}(I)=\min\\{a_{j}+b_{j+1}-2\ :\ 1\leq j\leq
m-1\\}.$

###### Proof.

Firstly, we compute $I:\mathfrak{m}$. We have

$\displaystyle I:\mathfrak{m}$ $\displaystyle=\
(I:\mathfrak{p}_{x})\cap(I:\mathfrak{p}_{y})$ (6) $\displaystyle=\
(x^{a_{1}-1}y^{b_{1}},\dots,x^{a_{m}-1}y^{b_{m}})\cap(x^{a_{1}}y^{b_{1}-1},\dots,x^{a_{m}}y^{b_{m}-1})$
$\displaystyle=\ (\text{lcm}(x^{a_{i}-1}y^{b_{i}},x^{a_{j}}y^{b_{j}-1})\ :\
1\leq i\leq m,\ 1\leq j\leq m)$ $\displaystyle=\
(\text{lcm}(x^{a_{i}-1}y^{b_{i}},x^{a_{j+1}}y^{b_{j+1}-1})\ :\ 1\leq i\leq m,\
0\leq j\leq m-1)$ $\displaystyle=\
(x^{\max\\{a_{i}-1,a_{j+1}\\}}y^{\max\\{b_{i},b_{j+1}-1\\}}\ :\ 1\leq i\leq
m,\ 0\leq j\leq m-1).$

Fix $j\in\\{0,\dots,m-1\\}$ and let $i\in\\{1,\dots,m\\}$.

If $i\leq j$, we have $a_{i}\geq a_{j}>a_{j+1}$ and $b_{i}\leq b_{j}<b_{j+1}$.
Therefore $x^{a_{j+1}}|x^{a_{i}-1}$ and $y^{b_{i}}|y^{b_{j+1}-1}$. Hence,

$x^{a_{j}-1}y^{b_{j+1}-1}\in(I:\mathfrak{m})\ \ \text{divides}\ \
x^{\max\\{a_{i}-1,a_{j+1}\\}}y^{\max\\{b_{i},b_{j+1}-1\\}}\ \ \text{for}\ \
i\leq j.$ (7)

If $i>j$, we have $a_{i}-1\leq a_{j+1}$ and $b_{i}\geq b_{j+1}-1$, so
$x^{a_{i}-1}|x^{a_{j+1}}$ and $y^{b_{j+1}-1}|y^{b_{i}}$. Hence,

$x^{a_{j+1}}y^{b_{i}}\in(I:\mathfrak{m})\ \ \text{divides}\ \
x^{\max\\{a_{i}-1,a_{j+1}\\}}y^{\max\\{b_{i},b_{j+1}-1\\}}\ \ \text{for}\ \
i>j.$ (8)

Thus, by equations (6), (7) and (8) we have

$I:\mathfrak{m}\ =\ (x^{a_{j}-1}y^{b_{j+1}-1},x^{a_{j+1}}y^{b_{i}}\ :\ 1\leq
j\leq m-1,\ j+1\leq i\leq m).$

Note that for each $i\geq j+1$ we have $x^{a_{j+1}}y^{b_{i}}\in I$. It is
clear that $x^{a_{j}-1}y^{b_{j+1}-1}\notin I$, for all $j=1,\dots,m-1$. Hence,
equation (5) follows.

The claim about $\operatorname{v}_{\mathfrak{m}}(I)$ follows from (5) and
Theorem 1.1(c).

As a consequence of our discussion, we obtain the next formula that shows us
how to compute the $\operatorname{v}$-number of $I_{\bf a,b}$ solely in terms
of the sequences ${\bf a}$ and ${\bf b}$.

######  Theorem 5.6.

Let $I=I_{\bf a,b}\subset S$ be a monomial ideal. Then

$\operatorname{v}(I)=\begin{cases}\min\\{a_{i}+b_{i+1}-2\ :\ 1\leq i\leq
m-1\\},\ \text{if}\ b_{1}=0\ \text{and}\ a_{m}=0,\\\
\min\\{a_{1}+b_{1}-1,a_{i}+b_{i+1}-2\ :\ 1\leq i\leq m-1\\},\ \text{if}\
b_{1}\neq 0\ \text{and}\ a_{m}=0,\\\ \min\\{a_{m}+b_{m}-1,a_{i}+b_{i+1}-2\ :\
1\leq i\leq m-1\\},\ \text{if}\ b_{1}=0\ \text{and}\ a_{m}\neq 0,\\\
\min\\{a_{1}+b_{1}-1,a_{m}+b_{m}-1,a_{i}+b_{i+1}-2\ :\ 1\leq i\leq m-1\\},\
\text{otherwise.}\end{cases}$

###### Proof.

Suppose $I$ is non principal. Then, the statement follows by combining
Proposition 5.1, Corollary 5.4 and Proposition 5.5. Now, if $I$ is principal,
the above formulas also hold. Indeed, in this case $m=1$ and in the above last
three minimums one does not have to consider the terms $a_{i}+b_{i+1}-2$
because $m-1=0$.

Let $I\subset S=K[x_{1},\dots,x_{n}]$ be a monomial ideal. Then

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

for all $k\gg 0$. By [13, Theorem 2.3(c)] we have that $b\geq-1$. Note also
that $a=\alpha(I)\geq 1$. On the other hand, let $f(k)=ak+b$ be any linear
function in $k$, with $a\geq 1$ and $b\geq-1$. In the next theorem, we prove
that there exists a monomial ideal $I\subset K[x,y]$ such that
$\operatorname{v}(I^{k})$ agrees with $f(k)$ for all $k\geq 1$.

######  Theorem 5.7.

Let $a\geq 1$ and $b\geq-1$ be integers. Then, there exists a monomial ideal
$I\subset S=K[x,y]$ such that

$\operatorname{v}(I^{k})=ak+b,\ \ \textit{for all}\ \ k\geq 1.$

###### Proof.

We claim that $I=(x^{a},x^{a-1}y^{b+2})=x^{a-1}(x,y^{b+2})$ satisfies our
assertion. For this aim, let us show that $I^{k}=(x^{ka-i}y^{i(b+2)}:0\leq
i\leq k)$ for all $k\geq 1$. Indeed,

$\displaystyle I^{k}$ $\displaystyle=$ $\displaystyle
x^{k(a-1)}(x,y^{b+2})^{k}=x^{k(a-1)}\sum_{i=0}^{k}(x^{k-i}y^{i(b+2)})$
$\displaystyle=$ $\displaystyle(x^{ka-i}y^{i(b+2)}:0\leq i\leq k).$

Since $ka>ka-1>\dots>ka-k$ and $b+2<2(b+2)<\dots<k(b+2)$, it follows that
$G(I^{k})=\\{x^{ka-i}y^{i(b+2)}:0\leq i\leq k\\}$.

Note that $\operatorname{Ass}^{\infty}(I)=\\{\mathfrak{p}_{x},\mathfrak{m}\\}$
if $a>1$, and $\operatorname{Ass}^{\infty}(I)=\\{\mathfrak{m}\\}$ if $a=1$.

If $a>1$, by Corollary 5.4(a) we have
$\operatorname{v}_{\mathfrak{p}_{x}}(I^{k})=k(a+b+1)-1$.

Whereas, by Proposition 5.5,

$\displaystyle\operatorname{v}_{\mathfrak{m}}(I^{k})\ $ $\displaystyle=\
\min\\{(ka-i)+(i+1)(b+2)-2:0\leq i\leq k-1\\}$ $\displaystyle=\
\min\\{ka+(i+1)b+i:0\leq i\leq k-1\\}$ $\displaystyle=\ ak+b.$

If $a>1$, then
$\operatorname{v}(I^{k})=\min\\{\operatorname{v}_{\mathfrak{p}_{x}}(I^{k}),\operatorname{v}_{\mathfrak{m}}(I^{k})\\}=\min\\{k(a+b+1)-1,ak+b\\}=ak+b$.
Otherwise, if $a=1$, then
$\operatorname{v}(I^{k})=\operatorname{v}_{\mathfrak{m}}(I^{k})=ak+b$ once
again.

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|  Elena Sorina Lupu is a PhD candidate in the Aerospace Department at
California Institute of Technology and an affiliate of the Keck Institute of
Space Sciences. She has master degrees from Caltech in Space Engineering and
from the Swiss Federal Institute of Technology, Lausanne (EPFL) in Robotics
and Autonomous Systems. Her current research focuses on autonomy, control, and
machine learning applied to robotics and spacecraft. She has won awards and
fellowships including the AI4Science Fellowship from Amazon and the Anita Borg
Women Techmakers from Google.
---|---
|  Fengze Xie is a PhD student in the Department of Computing and Mathematical
Sciences at the California Institute of Technology. He received a M.S. in
Electrical Engineering from the California Institute of Technology in 2021,
following the completion of a B.S. in Computer Engineering and a B.S. in
Engineering Physics from the University of Illinois Urbana-Champaign in 2020.
His research interests mainly focus on the intersection between machine
learning and robotics, including control and planning. He was the recipient of
the Simoudis Discovery Prize.
---|---
|  James Alan Preiss received the Ph.D. degree in computer science from the
University of Southern California in 2022 and the B.A./B.S. degree with
concentrations in mathematics and photography from The Evergreen State College
in 2010. He is currently a Postdoctoral Scholar Research Associate in the
Department of Computing + Mathematical Sciences at the California Institute of
Technology, Pasadena, CA, USA. His research interests focus on developing
rigorous foundations for the application of machine learning to planning and
control in robotics.
---|---
|  Jedidiah Alindogan is a Research Engineer at Caltech. He received his B.S.
in mechanical engineering from Caltech in 2023. At Caltech, he along with a
team of undergraduates submitted to the NASA 2022 BIG Idea challenge and was
awarded the Best Visionary Concept Award. Finally, he was granted membership
to the Tau Beta Phi engineering honor society for academic achievement as well
as a commitment to personal and professional integrity.
---|---
|  Matthew Anderson received a B.Eng (Aeronautical, Hons I) from The
University of Sydney in 2010 and a jointly awarded Ph.D from The University of
Sydney and Université libre de Bruxelles in 2018. He is currently a Staff
Scientist in the Computing + Mathematical Sciences (CMS) department at the
California Institute of Technology and supports a wide variety of robotics
research with a fleet of UAVs and UGVs. His primary areas of research focus on
UAV operations, aircraft modelling, and extreme environment robotics.
---|---
|  Soon-Jo Chung (M’06–SM’12) received the S.M. degree in aeronautics and
astronautics and the Sc.D. degree in estimation and control from Massachusetts
Institute of Technology, Cambridge, MA, USA, in 2002 and 2007, respectively.
He is currently Bren Professor of Control and Dynamical Systems, and a Jet
Propulsion Laboratory Senior Research Scientist at the California Institute of
Technology, Pasadena, CA, USA. He was an Associate Editor of IEEE Transactions
on Automatic Control and IEEE Transactions on Robotics, and the Guest Editor
of a Special Section on Aerial Swarm Robotics published in IEEE Transactions
on Robotics.
---|---