# Maximal generating degrees of powers of homogeneous ideals

Le Tuan Hoa  Institute of Mathematics, VAST, 18 Hoang Quoc Viet, 10307 Hanoi,
Viet Nam<EMAIL_ADDRESS>Dedicated to Professor Nguyen Tu Cuong on the
occasion of his 70th birthday.

###### Abstract.

The degree excess function $\epsilon(I;n)$ is the difference between the
maximal generating degree $d(I^{n})$ of a homogeneous ideal $I$ of a
polynomial ring and $p(I)n$, where $p(I)$ is the leading coefficient of the
asymptotically linear function $d(I^{n})$. It is shown that any non-increasing
numerical function can be realized as a degree excess function, and there is a
monomial ideal $I$ whose $\epsilon(I;n)$ has exactly a given number of local
maxima. In the case of monomial ideals, an upper bound on $\epsilon(I;n)$ is
provided. As an application it is shown that in the worst case, the so-called
stability index of the Castelnuovo-Mumford regularity of a monomial ideal $I$
must be at least an exponential function of the number of variables.

###### Key words and phrases:

monomial ideal, integral closure, maximal generating degrees

###### 1991 Mathematics Subject Classification:

13F20, 13F55, 05E40, 90C10

## Introduction

In this paper we study the maximal generating degree $d(I^{n})$ of powers of a
homogeneous ideal $I$ of a polynomial $R=K[X_{1},...,X_{s}]$ of $s$ variables
over a field $K$. It is a trivial fact that this function is bounded by
$d(I)n$, but it is not obvious that $d(I^{n})=p(I)n+c(I)$ for some $p(I)>0$
and $c(I)\geq 0$, when $n\gg 0$. This fact was established in [5, Theorem
3.1]. It is natural to ask how is the initial behavior of this function and
when does this function become stable? Some recent results are related to this
kind of problems. A comparison on the numbers of generators $\mu(I)$ and
$\mu(I^{2})$ of $I$ and $I^{2}$, respectively, in [8] leads to the study of
the initial behavior of the function $\mu(I^{n})$. A somewhat surprising
result says that given a number $q$ there is a monomial ideal $I$ with the
property that the function which describes the number of generators of $I^{n}$
has at least $q$ local maxima, see [1]. It is even more surprising, that a
conjecture of Herzog and Hibi is recently proved in [7], which states that the
function $\operatorname{depth}(R/I^{n})$ can be any convergent numerical
function. The study of the maximal generating degree function
$d(\overline{I^{n}})$ of integral closures of a monomial ideal $I$ is carried
out in [11]. The readers may consult the survey paper [9] for other related
problems and an extended list of references.

Back to the maximal generating degrees, since $d(I^{n})\geq p(I)n$ by [14,
Proposition 4], it is equivalent to pose the above questions on the function
$\epsilon(I;n)=d(I^{n})-p(I)n$, which is called the degree excess function of
$I$ in this paper. Due to some obvious restrictions (see Lemma 1.3), not any
convergent numerical function can be realized as a degree excess function. For
an example, when $n$ is bigger than the so-called reduction number $r(I)$ (see
(1.2) for the definition), this function must be non-increasing. However, in
this paper, we show that in general degree excess functions could be quite
complicate. After showing that any non-increasing numerical function can be
realized as a degree excess function (Theorem 1.9), we show that there is an
ideal that has exactly a given number of local maxima (Theorem 1.10). Moreover
these ideals can be taken to be monomial.

In the case of monomial ideals we also give an effective way to compute the
number $p(I)$ mentioned above. It is the maximal length of vertices of the
Newton polyhedron associated to $I$ (see Theorem 2.4(i)). This is achieved by
using a technique from Linear Programming. Thank to it we are able to give
upper bounds for $r(I)$ and $c(I)$ in terms of generating degrees and the
number of generators, see Theorem 2.4. These bounds are close to be sharp (see
Proposition 2.6), where we also show that in the worst case, the stability
index $\operatorname{gstab}(I)$ of maximal generating degree (see (1.3) for
the exact definition) of a monomial ideal $I$ should be at least a function of
$d(I)$ with the number of variables in the exponent.

Our study of degree excess functions leads to some interesting consequences on
the Castelnuovo-Mumford regularities $\operatorname{reg}(I^{n})$ of $I^{n}$.
It was independently proved in [5] and [14], that
$\operatorname{reg}(I^{n})=p(I)n+e(I)$ for some $e(I)\geq 0$ and all $n\gg 0$.
It is an open problem to bound $e(I)$ and the so-called index of stability
$\operatorname{reg-stab}(I)$ (see (2.10) for the definition), even in the case
of monomial ideals, see, e.g. [2, 4, 6]. We show that even in the case of
monomial ideals, in the worst case, the invariants $e(I)$ and
$\operatorname{reg-stab}(I)$ should be at least exponential functions of the
number of variables (Theorem 2.8). This could be a reason why the study of
$e(I)$ and $\operatorname{reg-stab}(I)$ is so hard.

The paper has two sections. In Section 1, we construct some simple degree
excess functions. Then, using a quite simple observation to combine them
(Lemma 1.8), we are able to construct monomial ideals satisfying Theorem 1.9
and Theorem 1.10. In Section 2, we first use a geometric object, the Newton
polyhedron, to determine $p(I)$ and to give bounds on $r(I)$ and on
$\epsilon(I;n)$ (Theorem 2.4). Then we construct monomial ideals to show that
these bounds are close to be sharp. We conclude the paper with the proof of
Theorem 2.8.

##  1\. Degree excess function

The symbols $\mathbb{Z},\ \mathbb{N},\ \mathbb{Q}$, $\mathbb{R}$ and
$\mathbb{R}_{+}$ denote the sets of integers, non-negative integers,
rationals, real numbers and non-negative real numbers, respectively. Let
$R:=K[X_{1},...,X_{s}]$ be a polynomial ring over a field $K$ and
$\mathfrak{m}:=(X_{1},...,X_{s})$. Consider the standard grading in $R$, that
is $\deg(X_{i})=1$ for all $i=1,...,r.$ Given a vector
${\boldsymbol{\alpha}}\in\mathbb{N}^{r}$ we write
$\mathbf{X}^{\boldsymbol{\alpha}}:=X_{1}^{\alpha_{1}}\cdots
X_{r}^{\alpha_{r}}$. Then
$\deg(\mathbf{X}^{\boldsymbol{\alpha}})=|{\boldsymbol{\alpha}}|=\alpha_{1}+\cdots+\alpha_{r}$.
Let $I\subset R$ be a homogeneous non-zero ideal, which is minimally generated
by homogeneous polynomials $\\{f_{1},...,f_{t}\\}$. The number

$d(I)=\max\\{\deg(f_{1}),...,\deg(f_{t})\\}$

is called the maximal generating degree of $I$.

Kodiyalam’s construction: Let

$p(I):=\max\\{\deg(f_{i})|\ f_{i}^{n}\not\in\mathfrak{m}I^{n}\ \text{for all
}n\geq 1,\ i\leq t\\},$

and

$J=(f_{i}|\ \deg(f_{i})\leq p(I),\ \ i\leq t)=(I_{\leq p(I)}),$

where $I_{\leq q}$ denotes the set of all homogeneous elements of $I$ of
degrees at most $q$. Then $J$ is a reduction of $I$, i.e., $I^{n}=JI^{n-1}$
for $n\gg 0$.

The following result of Cutkosky-Herzog-Trung and Kodiyalam is a starting
point of our study.

######  Lemma 1.1.

([14, Corollary 3 and Proposition 4] and [5, Theorem 3.1]). Let $I$ be a non-
zero proper homogeneous ideal of $R$. Then

  * (i)

There is $q(I)\geq 0$ such that

(1.1) $p(I)n\leq d(I^{n})\leq p(I)n+q(I)\ \text{for all }n\geq 1.$

  * (ii)

There is $q_{0}(I)\geq 0$ such that $d(I^{n})=p(I)n+q_{0}(I)$ for all $n\gg
0$.

Because of this lemma, we can make the following definition:

######  Definition 1.2.

We call $\epsilon(I;n)=d(I^{n})-p(I)n\geq 0$ the degree excess function of
$I$.

If $p(I)=d(I)$, then $\epsilon(I;n)=0$ for all $n$. However, in the general
case, $\epsilon(I;n)$ could be quite complicate.

Recall that the reduction number of $I$ w.r.t. a reduction $J^{\prime}$ is
defined by

$r_{J^{\prime}}(I)=\min\\{n\geq 0|\ I^{n+1}=J^{\prime}I^{n}\\}.$

In this paper we set

(1.2) $r(I):=r_{J}(I),$

where $J$ is the minimal reduction in the Kodiyalam’s construction. Note that
$I$ has many reductions. In the literature, $r(I)$ usually means the least
number among the reduction numbers with respect to the so-called minimal
reductions. This is not the case here, and $r(I)$ in the above definition is
in general much smaller than the usual one.

Below are some elementary properties of a degree excess function.

######  Lemma 1.3.

Let $J$ be as above and $r(I)$ the reduction number of $I$ w.r.t. $J$. Then

  * (i)

$\epsilon(I;n)$ is non-increasing for $n\geq r(I)$ and is a constant for $n\gg
0$.

  * (ii)

$\epsilon(I;n)\leq\min\\{n,r(I)\\}(d(I)-p(I))$ for all $n\geq 1$.

  * (iii)

For all $n\leq r(I)$, $\epsilon(I;n)\geq n$.

  * (iv)

For all $m\leq n$, $n\epsilon(I;m)\geq m\epsilon(I;n)$.

###### Proof.

The fact that $\epsilon(I;n)$ is a constant for $n\gg 0$ is a consequence of
Lemma 1.1(ii). The rest of (i) and (ii) follows from the equality
$I^{n}=J^{n-r(I)}I^{r(I)}$ for all $n\geq r(I)+1$ and $d(J)=p(I)$.

(iii) Let $f\in I^{n}\setminus JI^{n-1}$ be a minimal generator of $I^{n}$.
One can write $f=\sum_{k}f_{k1}\cdots f_{kn}g_{k},$ where
$f_{ki}\in\\{f_{1},...,f_{t}\\}$ for all $k,i$, $g_{k}$ is a homogeneous
polynomial and $\deg(f_{k1})\geq\cdots\geq\deg(f_{kn})$. Since $f\not\in
JI^{n-1}$ there is $k$ such that $\deg(f_{k1}),...,\deg(f_{kn})>p(I)$. Then

$d(I^{n})\geq\deg(f)\geq\deg(f_{k1})+\cdots+\deg(f_{kn})\geq n(p(I)+1),$

which implies $\epsilon(I;n)\geq n$.

(iv) Let $f\in I^{n}$ be a minimal generator such that $\deg(f)=d(I^{n})$. As
above we write $f=\sum_{k}f_{k1}\cdots f_{kn}g_{k},$ where
$f_{ki}\in\\{f_{1},...,f_{t}\\}$ for all $k,i$, $g_{k}$ is a homogeneous
polynomial and $\deg(f_{k1})\geq\cdots\geq\deg(f_{kn})$. Since $f$ is a
minimal generator, there is $k$ such that $g_{k}=1$ and for all $m<n$,
$f_{k1}\cdots f_{km}$ is a minimal generator of $I^{m}$. Hence

$\begin{array}[]{ll}n[p(I)m+\epsilon(I;m)]&=nd(I^{m})\geq
n[\deg(f_{k1})+\cdots+\deg(f_{km})]\\\ &\geq
m[\deg(f_{k1})+\cdots+\deg(f_{kn})]=m\deg(f)\\\
&=md(I^{n})=m[p(I)n+\epsilon(I;n)],\end{array}$

which yields $n\epsilon(I;m)\geq m\epsilon(I;n)$. ∎

In order to understand the degree excess functions, the following questions/
problems are quite natural to ask for.

Problems. 1. Characterize degree excess functions. In particular, how many
local maxima can the function $\epsilon(I;n)$ have?

2\. When does $\epsilon(I;n)$ become a constant, or equivalently give a bound
on the index of stability of the maximal generating degree defined by

(1.3) $\operatorname{gstab}(I)=\min\\{n_{0}|\ \epsilon(I;n)=\epsilon(I;n_{0})\
\ \text{for all}\ n\geq n_{0}\\}.$

3\. Give a bound on $r(I)$.

In the rest of paper we give some partial solutions to these problems. Note
that Vasconcelos already showed that for any homogeneous ideal of $R$ there is
a bound on any reduction number, but this bound is a non-elementary function
with four levels of exponentiation (see [16, Proposition 2.3]). We believe
that for the reduction number $r(I)$ in (1.2) the bound should be much
smaller. However, even in the case of monomial ideals, the bound in the worst
case should be at least an exponential function, see Proposition 2.6.

In the sequel we mainly consider monomial ideals. Recall that a monomial ideal
of $R$ is an ideal generated by monomials. If $I$ is a monomial ideal, $I$ has
the unique minimal generating system consisting of monomials which is denoted
by $G(I)$. An element of $G(I)$ is called a (monomial) minimal generator of
$I$.

Notation. We often consider the following special monomials in two variables:

$m_{i}=X_{1}^{p-i}X_{2}^{i},\ \ \text{where}\ p\geq 4,\ 0\leq i\leq p.$

Of course, these monomials depend on the exponent $p$. However, this number is
clearly given in the context, so that we omit it for the sake of simplicity.

######  Remark 1.4.

Assume that a monomial ideal $\mathfrak{a}$ is a reduction of $I$ and $f\in R$
is a monomial such that $I=(\mathfrak{a},f)$. Then

$r_{\mathfrak{a}}(I)=\min\\{t\geq 1|\ f^{t}\in\mathfrak{a}^{t}\\}-1.$

The following result shows that the bounds in Lemma 1.3 are sharp.

######  Lemma 1.5.

Given $p\geq 4$ and $2\leq a\leq p-2$ such that $p-1$ is not divisible by $a$.
Let

$J_{p,a}:=(m_{0},m_{p},m_{p-1},X_{1}^{p-1}X_{2}^{a})\subset K[X_{1},X_{2}].$

Then $r(J_{p,a})=r:=\lfloor\frac{p-1}{a}\rfloor$ and

$\epsilon(J_{p,a};n)=\begin{cases}n(a-1)&\ \text{if}\ 1\leq n\leq r-1,\\\
r(a-1)&\ \text{if}\ n\geq r.\end{cases}$

###### Proof.

(i) For short, we set $m=X_{1}^{p-1}X_{2}^{a}$, $I=J_{p,a}$ and
$J=(m_{0},m_{p},m_{p-1})$. Then $J$ is the reduction of $I$ in the Kodiyalam’s
construction, and $p(I)=p$.

Note that $r+1>p/a$, whence $(r+1)a>p$, or equivalently $(r+1)a-p>0$. Since
$a\geq 2$, $p\geq r+1$. Hence

$\begin{array}[]{ll}m^{r+1}&=X_{1}^{(r+1)(p-1)}X_{2}^{(r+1)a}=X_{1}^{rp}X_{2}^{p}X_{1}^{p-(r+1)}X_{2}^{(r+1)a-p}\\\
&=m_{0}m_{p}^{r}X_{1}^{p-(r+1)}X_{2}^{(r+1)a-p}\in J^{r+1}.\end{array}$

By Remark 1.4, $r(I)\leq r$, and

(1.4) $d(I^{n})\leq d(J^{n-r})+d(I^{r})\leq np+r(a-1)\ \text{for all}\ n\geq
r.$

(ii) Further, assume that $m^{t}\in J^{t}$ for some $0<t\leq r$. Then
$m^{t}=m^{\prime}m_{0}^{\alpha_{0}}m_{p}^{\alpha_{p}}m_{p-1}^{\alpha_{p-1}}$
for a monomial $m^{\prime}$ and some non-negative integers $\alpha_{i}$ such
that $\alpha_{0}+\alpha_{p}+\alpha_{p-1}=t$. Looking at the exponents of
$X_{1}$ on both sides, we get $(p-1)t\geq p\alpha_{0}$, whence $\alpha_{0}<t$.
Hence $\alpha_{p}+\alpha_{p-1}\geq 1$. Looking at the exponents of $X_{2}$, we
now obtain

(1.5) $at\geq
p\alpha_{p}+(p-1)\alpha_{p-1}\geq(p-1)(\alpha_{p}+\alpha_{p-1})\geq p-1.$

Since $p-1$ is not divisible by $a$, $r<(p-1)/a$, whence $ar<p-1$. Hence
$p-1>ar\geq at\geq p-1$, a contradiction. Thus,

(1.6) $m^{t}\not\in J^{t}\ \text{for all}\ 0<t\leq r.$

By Remark 1.4 this implies $r(I)\geq r$, which then yields $r(I)=r$.

(iii) Now let $t\leq r$ and $n\geq t$. Assume that
$m^{t}m_{0}^{n-t}\in\mathfrak{m}I^{n}$. Then one can write

$m^{t}m_{0}^{n-t}=m^{\prime}m_{0}^{\alpha_{0}}m_{p}^{\alpha_{p}}m_{p-1}^{\alpha_{p-1}}m^{\alpha},$

where $m^{\prime}\neq 1$ is a monomial and
$\alpha+\alpha_{0}+\alpha_{p}+\alpha_{p-1}=n$. Comparing the total degrees in
both sides we must have $\alpha<t$. Therefore

$m^{u}m_{0}^{n-t}=m^{\prime}m_{0}^{\alpha_{0}}m_{p}^{\alpha_{p}}m_{p-1}^{\alpha_{p-1}},$

where $1\leq u:=t-\alpha\leq r$. By virtue of (1.6), $\alpha_{0}<n-t$. Set
$v=n-t-\alpha_{0}>0$. Then

$m^{u}m_{0}^{v}=m^{\prime}m_{p}^{\alpha_{p}}m_{p-1}^{\alpha_{p-1}},$

where $\alpha_{p}+\alpha_{p-1}=u+v$. Comparing the exponents of $X_{2}$ in
both sides, we get
$au\geq\alpha_{p}p+\alpha_{p-1}(p-1)\geq(p-1)(\alpha_{p}+\alpha_{p-1})$. Since
$au\leq ar<p-1$, we must have $\alpha_{p}=\alpha_{p-1}=0$. Then $u+v=0$, a
contradiction.

Summing up, we have shown that $m^{t}m_{0}^{n-t}$ is a minimal generator of
$I^{n}$. Therefore $d(I^{n})\geq t(p+a-1)+(n-t)p=np+t(a-1)$. Since
$d(I^{t})\leq t(p+a-1)$ for all $t$, together with (1.4), this implies
$d(I^{t})=tp+t(a-1)$ for all $t\leq r$ and $d(I^{n})=np+r(a-1)$ for all $n\geq
r$, or equivalently $\epsilon(I;n)=n(a-1)$ for $n\leq r$ and
$\epsilon(I;n)=r(a-1)$ for $n\geq r$. ∎

The following result is a folklore. We give a proof for the sake of
completeness.

######  Lemma 1.6.

For given $p\geq 4$ and $2\leq i\leq\lfloor p/2\rfloor$, let

$\begin{array}[]{ll}I_{p}&=(m_{0},m_{p},m_{1},m_{p-1}),\ \text{and}\\\
I_{p,i,0}&=(m_{0},m_{p},m_{1},m_{p-1},m_{i}),\end{array}$

be two ideals of $K[X_{1},X_{2}]$. Then

  * (i)

For all $n\geq p-2$,

$(I_{p,i,0})^{n}=(I_{p})^{n}=(\mathbf{X}^{{\boldsymbol{\alpha}}}|\
|{\boldsymbol{\alpha}}|=pn).$

  * (ii)

For all $u\leq i-2$ and $v\leq p-i-2$ we have
$m_{i}m_{0}^{u}m_{p}^{v}\not\in(I_{p})^{1+u+v}$.

  * (iii)

$m_{i}^{t}\not\in I_{p}^{t}$ for all $1\leq t<\frac{p-1}{i}$.

###### Proof.

(i) Clearly

$(I_{p})^{n}\subseteq(I_{p,i,0})^{n}\subseteq(\mathbf{X}^{{\boldsymbol{\alpha}}}|\
|{\boldsymbol{\alpha}}|=pn).$

We first prove the reverse inclusions in the case $n=p-2$. Let
$\alpha_{1},\alpha_{2}\in\mathbb{N}$ such that $\alpha_{1}+\alpha_{2}=p(p-2)$.
Let $\alpha_{1}=ap+j$, where $0\leq j<p$ and $a\leq p-2$. If $j\leq p-a-2$,
then $p-2-a-j\geq 0$, and

$\begin{array}[]{ll}X_{1}^{\alpha_{1}}X_{2}^{\alpha_{2}}&=(X_{1}X_{2}^{p-1})^{j}(X_{1}^{p})^{a}(X_{2}^{p})^{p-2-a-j}\\\
&=m_{p-1}^{j}m_{0}^{a}m_{p}^{p-2-a-j}\in(I_{p})^{p-2}.\end{array}$

Assume that $j\geq p-a-1$. Then $a\leq p-3$, $a+j+1-p\geq 0$ and

$X_{1}^{\alpha_{1}}X_{2}^{\alpha_{2}}=m_{1}^{p-j}m_{0}^{a+j+1-p}m_{p}^{p-3-a}\in(I_{p})^{p-2}.$

In both cases $X_{1}^{\alpha_{1}}X_{2}^{\alpha_{2}}\in(I_{p})^{p-2}$, which
yields $(\mathbf{X}^{{\boldsymbol{\alpha}}}|\
|{\boldsymbol{\alpha}}|=p(p-2))\subseteq(I_{p})^{p-2}$, whence

$(I_{p})^{p-2}=(I_{p,i,0})^{p-2}=(\mathbf{X}^{{\boldsymbol{\alpha}}}|\
|{\boldsymbol{\alpha}}|=p(p-2)).$

For $n>p-2$, by induction we have

$(\mathbf{X}^{{\boldsymbol{\alpha}}}|\
|{\boldsymbol{\alpha}}|=pn)=(X_{1}^{p},X_{2}^{p})(\mathbf{X}^{{\boldsymbol{\alpha}}}|\
|{\boldsymbol{\alpha}}|=p(n-1))=(m_{0},m_{p})(I_{p})^{n-1}\subseteq(I_{p})^{n}.$

This implies (i).

(ii) Let $u\leq i-2$ and $v\leq p-i-2$. Assume that

(1.7)
$m_{i}m_{0}^{u}m_{p}^{v}=m_{0}^{\alpha_{0}}m_{p}^{\alpha_{p}}m_{1}^{\alpha_{1}}m_{p-1}^{\alpha_{p-1}},\
\alpha_{0}+\alpha_{1}+\alpha_{p}+\alpha_{p-1}=1+u+v.$

Note that $\alpha_{k}\leq p-3$ for all $k=0,1,p,p-1$. Comparing the exponents
of $X_{1}$ and $X_{2}$ in both sides of the above equality, we get

(1.8) $\displaystyle p-i+pu$ $\displaystyle=$
$\displaystyle\alpha_{0}p+\alpha_{1}(p-1)+\alpha_{p-1}=(\alpha_{0}+\alpha_{1})p+\alpha_{p-1}-\alpha_{1},$
(1.9) $\displaystyle i+pv$ $\displaystyle=$
$\displaystyle\alpha_{1}+\alpha_{p-1}(p-1)+\alpha_{p}p=(\alpha_{p}+\alpha_{p-1})p+\alpha_{1}-\alpha_{p-1}.$

Case 1. $\alpha_{1}\geq\alpha_{p-1}$. Since $\alpha_{1}<p$, from (1.9) it
follows that $\alpha_{1}-\alpha_{p-1}=i$. In particular $\alpha_{1}\geq i$.
Then (1.8) gives $i-1\geq u+1=\alpha_{0}+\alpha_{1}\geq i$, a contradiction.

Case 2. $\alpha_{1}<\alpha_{p-1}$. Then, by (1.8)
$\alpha_{p-1}-\alpha_{1}=p-i$, whence $\alpha_{p-1}\geq p-i$, and by (1.9),
$p-i-2\geq v=\alpha_{p}+\alpha_{p-1}-1\geq p-i-1$, a contradiction.

Summing up, (1.7) cannot happen.

(iii) The proof of this is similar to the proof of Lemma 1.5. ∎

By Lemma 1.3(i), $\epsilon(I;n)$ is non-increasing for $n\geq r(I)$. One may
expect that after $r(I)$, the function $\epsilon(I;n)$ becomes stable (i.e.,
equals a constant) very soon. However, using the above result we can now
construct ideals $I$ with small reduction number $r(I)$ and big index of
stability $\operatorname{gstab}(I)$. It shows that, in general, the reduction
number $r(I)$ and the number of variables are not enough to bound
$\operatorname{gstab}(I)$. A bound on $\operatorname{gstab}(I)$ must depend on
the generating degrees of $I$. The fact that
$r(I_{p,2,a})=\lceil\frac{p-1}{2}\rceil-1$ was pointed out by the referee.

######  Lemma 1.7.

Given $p\geq 4$, $2\leq i\leq\lfloor p/2\rfloor$ and $a\geq 1$. Let

$I_{p,i,a}=(m_{0},m_{p},m_{1},m_{p-1},m_{i}X_{3}^{a})\subset
K[X_{1},X_{2},X_{3}].$

Then $I_{p}=(m_{0},m_{p},m_{1},m_{p-1})$ is a reduction of $I_{p,i,a}$, and

  * (i)

$r(I_{p,i,a})\geq\lceil\frac{p-1}{i}\rceil-1$. The equality holds if $i=2$.

  * (ii)

$\epsilon(I_{p,i,a};n)=an$ for $1\leq n\leq\lceil\frac{p-1}{i}\rceil-1$ and
$\epsilon(I_{p,i,a};n)=0$ for $n\geq p-2$.

  * (iii)

(The case $i=\lfloor p/2\rfloor$): $r(I_{p,\lfloor p/2\rfloor,a})=1$, and

$\epsilon(I_{p,\lfloor p/2\rfloor,a};n)=\begin{cases}a&\ \text{if}\ 1\leq
n\leq p-3,\\\ 0&\ \text{if}\ n\geq p-2.\end{cases}$

###### Proof.

The fact that $I_{p}=(m_{0},m_{p},m_{1},m_{p-1})$ is a reduction of
$I_{p,i,a}$ in the Kodiyalam’s construction is clear. Hence $p(I_{p,i,a})=p$.

(i) Assume that $(m_{i}X_{3}^{a})^{t}\in(I_{p})^{t}$ for some $t\geq 1$. Since
no generator of $I_{p}$ contains $X_{3}$, $m_{i}^{t}\in(I_{p})^{t}$. By Lemma
1.6(iii), $t\geq\frac{p-1}{i}$, whence $t\geq\lceil\frac{p-1}{i}\rceil$. Since
$I_{p,i,a}=(I_{p},m_{i}X_{3}^{a})$, by Remark 1.4,
$r(I_{p,i,a})\geq\lceil\frac{p-1}{i}\rceil-1$.

Let $i=2$. If $p=2q$, then
$m_{2}^{q}=X_{1}^{q(2q-2)}X_{2}^{2q}=m_{0}^{q-1}m_{p}$. If $p=2q+1$, then
$m_{2}^{q}=X_{1}^{q(2q-1)}X_{2}^{2q}=m_{0}^{q-1}m_{p-1}$. In both cases
$m_{2}^{q}\in I_{p}^{q}$. By Remark 1.4, $r(I_{p,2,a})\leq
q=\lceil\frac{p-1}{2}\rceil-1$, whence
$r(I_{p,2,a})=\lceil\frac{p-1}{2}\rceil-1$.

(ii) Assume that $n\geq p-2$. By Lemma 1.6(i) we have

$(I_{p})^{n}\subseteq(I_{p,i,a})^{n}\subseteq(I_{p,i,0})^{n}=(I_{p})^{n}.$

Here we consider the ideals $I_{p}$ and $I_{p,i,0}$ defined in Lemma 1.6 as
ideals in $K[X_{1},X_{2},X_{3}]$. Hence $(I_{p,i,a})^{n}=(I_{p})^{n}$, which
yields $d((I_{p,i,a})^{n})=d((I_{p})^{n})=pn$, or equivalently
$\epsilon(I_{p,i,a};n)=0$ for all $n\geq p-2$.

Further, let $t\leq\lceil\frac{p-1}{i}\rceil-1$. Assume that
$(m_{i}X_{3}^{a})^{t}\in\mathfrak{m}(I_{p,i,a})^{t}$. Then

$(m_{i}X_{3}^{a})^{t}=m^{\prime}m_{0}^{\alpha_{0}}m_{p}^{\alpha_{p}}m_{1}^{\alpha_{1}}m_{p-1}^{\alpha_{p-1}}(m_{i}X_{3}^{a})^{\alpha},$

for some $\alpha,\alpha_{i}\in\mathbb{N}$ such that
$\alpha_{0}+\alpha_{p}+\alpha_{1}+\alpha_{p-1}+\alpha=t$ and $m^{\prime}\neq
1$ is a monomial. Comparing the total degrees in both sides we must have
$\alpha<t$. But then $m_{i}^{t-\alpha}\in(I_{p})^{t-\alpha}$, which is
impossible by Lemma 1.6(iii). Hence $(m_{i}X_{3}^{a})^{t}$ is a minimal
generator of $(I_{p,i,a})^{t}$, which yields $\epsilon(I_{p,i,a};t)\geq at$,
whence $\epsilon(I_{p,i,a};t)=at$.

(iii) Let $p=2q+\varepsilon$, $\varepsilon=0,1$. This means $q=\lfloor
p/2\rfloor$. Since

$(m_{q}X_{3}^{a})^{2}=X_{1}^{2(q+\varepsilon)}X_{2}^{2q}X_{3}^{2a}=\begin{cases}m_{0}m_{p}X_{3}^{2a}&\
\text{if}\ \varepsilon=0,\\\ m_{0}m_{p-1}X_{3}^{2a}&\ \text{if}\
\varepsilon=1,\end{cases}$

$(m_{q}X_{3}^{a})^{2}\in(I_{p})^{2}$. Hence
$(I_{p,q,a})^{n}=(I_{p})^{n-1}I_{p,q,a}$ for all $n\geq 2$. This means $I_{p}$
is a reduction of $I_{p,q,a}$ and $r(I_{p,q,a})=1$. This also implies

(1.10) $d((I_{p,q,a})^{n})\leq p(n-1)+p+a=pn+a.$

Let $u\leq q-2$ and $v\leq p-q-2$. Assume that
$(m_{q}X_{3}^{a})m_{0}^{u}m_{p}^{v}\in\mathfrak{m}(I_{p,q,a})^{1+u+v}$. Then

(1.11)
$(m_{q}X_{3}^{a})m_{0}^{u}m_{p}^{v}=m^{\prime}m_{0}^{\alpha_{0}}m_{p}^{\alpha_{p}}m_{1}^{\alpha_{1}}m_{p-1}^{\alpha_{p-1}}(m_{q}X_{3}^{a})^{\alpha_{q}},$

where $1\neq m^{\prime}$ is a monomial of $K[X_{1},X_{2},X_{3}]$,
$\alpha_{i}\in\mathbb{N}$ and
$\alpha_{0}+\alpha_{p}+\alpha_{1}+\alpha_{p-1}+\alpha_{q}=1+u+v$. A comparison
of the total degrees in two sides gives $\alpha_{q}=0$, and (1.11) becomes

$m_{q}m_{0}^{u}m_{p}^{v}=m"m_{0}^{\alpha_{0}}m_{p}^{\alpha_{p}}m_{1}^{\alpha_{1}}m_{p-1}^{\alpha_{p-1}}\in(I_{p})^{1+u+v},$

which is impossible by virtue of Lemma 1.6(ii). This says that
$(m_{q}X_{3}^{a})m_{0}^{u}m_{p}^{v}$ is a minimal generator of
$(I_{p,q,a})^{1+u+v}$, whence $d((I_{p,q,a})^{n})\geq pn+a$ for all $n\leq
1+q-2+p-q-2=p-3$. Together with (1.10), we get $d((I_{p,qa})^{n})=pn+a$, and
so $\epsilon(I_{p,qa};n)=a$ for all $n\leq p-3$. ∎

Using the following simple observation and the ideals in Lemmas 1.5 and 1.7,
one can construct ideals with more complicate degree excess functions. The
following proof is suggested by the referee, which is much simpler than the
original one.

######  Lemma 1.8.

Let $I\subset K[\mathbf{X}]$ and $J\subset K[\mathbf{Y}]$ be non-zero proper
homogeneous ideals, where $\mathbf{X}$ and $\mathbf{Y}$ are disjoint sets of
variables. Assume that $\\{f_{1},...,f_{t}\\}\subset K[\mathbf{X}]$ and
$\\{g_{1},...,g_{u}\\}\subset K[\mathbf{Y}]$ are homogeneous polynomials which
form minimal bases of $I$ and $J$, respectively. Then the set $\\{f_{i}g_{j}|\
i=1,...,t;j=1,...,u\\}$ forms a minimal basis of the product
$(IJ):=IJK[\mathbf{X},\mathbf{Y}]$. It implies that the degree excess
functions has the following property

$\epsilon(IJ;n)=\epsilon(I;n)+\epsilon(J;n).$

###### Proof.

Applying the first claim to $I^{n}$ and $J^{n}$ for all $n$ we immediately get
$d((IJ)^{n})=d(I^{n})+d(J^{n})$, which then implies
$\epsilon(IJ;n)=\epsilon(I;n)+\epsilon(J;n).$

In order to prove the first claim, denote $\mathfrak{m}=(\mathbf{X})$,
$\mathfrak{n}=(\mathbf{Y})$ and $T=K[\mathbf{X},\mathbf{Y}]$. Note that

$\frac{I}{\mathfrak{m}I}\otimes_{K}\frac{J}{\mathfrak{n}J}\cong\frac{I\otimes_{K}J}{I\otimes_{K}(\mathfrak{n}J)+(\mathfrak{m}I)\otimes_{K}J}\cong\frac{IJ}{I\mathfrak{n}J+\mathfrak{m}IJ}=\frac{IJ}{(\mathfrak{m}T+\mathfrak{n}T)IJ}.$

Therefore $\mu(IJ)=\mu(I)\mu(J)$, where $\mu(.)$ denotes the minimal number of
generators. Since the set $\\{f_{i}g_{j}|\ i=1,...,t;j=1,...,u\\}$ generates
$IJ$, it must be a minimal basis of $IJ$. ∎

The following two results say that a degree excess function could be quite
complicate.

######  Theorem 1.9.

Let $f:\ \mathbb{N}\rightarrow\mathbb{N}$ be any non-increasing function. Then
there is a monomial ideal $I$ such that $\epsilon(I;n)=f(n)$ for all $n\geq
1$.

###### Proof.

Assume that

$f(n)=\begin{cases}d_{1}&\ \text{if}\ 1=n_{1}\leq n<n_{2},\\\ d_{2}&\
\text{if}\ n_{2}\leq n<n_{3},\\\ &...\\\ d_{k-1}&\ \text{if}\ n_{k-1}\leq
n<n_{k},\\\ d_{k}&\ \text{if}\ n\geq n_{k},\end{cases}$

where $d_{1}>d_{2}>\cdots>d_{k}$ and $1=n_{1}<n_{2}<\cdots<n_{k}$. If $k=1$ we
may assume that $d_{1}>0$. Then the ideal $I=J_{d_{1}+3,d_{1}+1}$ defined in
Lemma 1.5 has $\epsilon(I,n)=d_{1}$ for all $n\geq
r(I)=\lfloor(d_{1}+2)/(d_{1}+1)\rfloor=1=n_{1}$.

Let $k\geq 2$. Set $p_{i}=n_{i+1}+2$ and $a_{i}=d_{i}-d_{i+1}\geq 1$, where
$i=1,...,k-1$. Consider the ideals $I_{p_{i},\lfloor p_{i}/2\rfloor,a_{i}}$ of
the ring $K[\mathbf{X}_{(i)}]:=K[X_{i1},X_{i2},X_{i3}]$ defined in Lemma 1.7,
where $\mathbf{X}_{(1)},...,\mathbf{X}_{(k-1)}$ are disjoint sets of
variables. Let

$I=\prod_{i=1}^{k-1}I_{p_{i},\lfloor p_{i}/2\rfloor,a_{i}}\subset
K[\mathbf{X}_{(1)},...,\mathbf{X}_{(k-1)}].$

Assume that $n_{j}\leq n<n_{j+1}$ for some $j=1,...,k$, where
$n_{k+1}:=\infty$. If $i\leq j-1$, then $p_{i}-2=n_{i+1}\leq n_{j}\leq n$. By
Lemma 1.7, $\epsilon(I_{p_{i},\lfloor p_{i}/2\rfloor,a_{i}};n)=0$. If $i\geq
j$, then $p_{i}-2=n_{i+1}\geq n_{j+1}>n$. By Lemma 1.7,
$\epsilon(I_{p_{i},\lfloor p_{i}/2\rfloor,a_{i}};n)=a_{i}$. Therefore, by
Lemma 1.8,

$\begin{array}[]{ll}\epsilon(I;n)&=\sum_{i=1}^{k-1}\epsilon(I_{p_{i},\lfloor
p_{i}/2\rfloor,a_{i}};n)=a_{j}+a_{j+1}+\cdots+a_{k-1}\\\
&=(d_{j}-d_{j+1})+(d_{j+1}-d_{j+2})+\cdots+(d_{k-1}-d_{k})=d_{j}-d_{k}.\end{array}$

Thus, $\epsilon(I;n)=f(n)-d_{k}$ for all $n\geq 1$. If $d_{k}=0$ then we are
done. If $d_{k}>0$, we set $I^{\prime}=IJ\subset
K[\mathbf{X}_{(1)},...,\mathbf{X}_{(k)}]$, where $J=J_{d_{k}+3,d_{k}+1}$ is
the ideal of $K[\mathbf{X}_{(k)}:=K[X_{k1},X_{k2}]$ defined in Lemma 1.5. By
this lemma, $r(J)=1$ and $\epsilon(J;n)=d_{k}$ for all $n\geq 1$. Using again
Lemma 1.8, we then get

$\epsilon(I^{\prime};n)=\epsilon(I;n)+\epsilon(J;n)=f(n)-d_{k}+d_{k}=f(n),$

for all $n\geq 1$. ∎

Remark. If $r(I)=0$, then $\epsilon(I;n)=0$ for all $n\geq 1$. If $r(I)=1$, by
Lemma 1.3, $\epsilon(I;n)$ is non-increasing for all $n$. Both ideals $I$ and
$I^{\prime}$ in the proof of the above theorem have reduction number one. So
Theorem 1.9 implies that a numerical function
$f:\mathbb{N}\rightarrow\mathbb{N}$ is a degree excess functions of a
homogeneous ideal $I$ with $r(I)=1$ if and only if it is non-increasing.

In Lemma 1.5 we already construct ideals with a non-decreasing degree excess
function. Combining this lemma with Lemma 1.8, one can construct ideals with
more complicate non-decreasing degree excess functions. However, unlike the
case of non-increasing functions, due to the restriction given in Lemma
1.3(iv), not any convergent non-decreasing function could be realized as a
degree excess function (of a homogeneous ideal). For an example, if $f(n)=1$
and $f(n+1)>1$ for some $n\geq 2$, then $f(n)$ cannot be a degree excess
function. However, the following result says that the degree excess function
could get local maxima at any given collection of positions. Here, a numerical
function $f:\ \mathbb{N}\rightarrow\mathbb{Z}$ is said to have a local maxima
at $n\geq 2$ if $f(n)>f(n-1)$ and $f(n)>f(n+1)$. It is of interest to mention
that, inspired of the comparison of numbers of generators of $I$ and $I^{2}$
in [8], it was shown in [1] that given a number $q$ there is a monomial ideal
$I$ with the property that the function which describes the number of
generators of $I^{n}$ has at least $q$ local maxima.

######  Theorem 1.10.

Given any sequence of positive numbers $2\leq n_{1}<n_{2}<\cdots<n_{k}$
($k\geq 1$) such that $n_{i+1}-n_{i}\geq 2$. Then there is a monomial ideal
such that its degree excess function gets local maxima exactly at points of
the set $\\{n_{1},...,n_{k}\\}$.

###### Proof.

Let $J=J_{2(n_{k}+2),2}$ be the ideal of $K[Y_{1},Y_{2}]$ defined in Lemma
1.5. Then $r(J)=\lfloor(2n_{k}+3)/2\rfloor=n_{k}+1$, and

$\epsilon(J;n)=\begin{cases}n&\ \text{if}\ n\leq n_{k},\\\ n_{k}+1&\
\text{if}\ n\geq n_{k}+1.\end{cases}$

Consider the following function

$f(n)=2(k+1-i)\ \text{if}\ n_{i-1}<n\leq n_{i},\ (i=1,...,k+1),$

where we set $n_{0}=0$ and $n_{k+1}:=\infty$. By Theorem 1.9, there is a
monomial ideal $I\subset K[X_{1},...,X_{s}]$ such that $\epsilon(I;n)=f(n)$
for all $n\geq 1$. Consider the product $IJ\subset
K[X_{1},...,X_{s},Y_{1},Y_{2}]$. By Lemma 1.8, we have

$\begin{array}[]{ll}\epsilon(IJ;n_{i})&=f(n_{i})+\epsilon(J,n_{i})=2(k+1-i)+n_{i}=2(k-i)+n_{i}+2,\\\
\epsilon(IJ;n_{i}-1)&=f(n_{i}-1)+\epsilon(J,n_{i}-1)=2(k+1-i)+n_{i}-1=2(k-i)+n_{i}+1,\\\
\epsilon(IJ;n_{i}+1)&=f(n_{i}+1)+\epsilon(J,n_{i}+1)=2(k+1-i-1)+n_{i}+1=2(k-i)+n_{i}+1,\end{array}$

for any $i=1,...,k$. Thus, $n_{i}$ is a local maximum point of
$\epsilon(IJ;n)$. Since $f(n)$ is a constant and $\epsilon(J;n)$ is non-
decreasing in any interval $n_{i-1}<n<n_{i}$, where $i=1,...,k+1$, there is no
other local maximum points of $\epsilon(IJ;n)$. ∎

Remark. In the proof of Theorem 1.10, if instead of $J_{2(n_{k}+2),2}$ we take
$J=I_{2n_{k}+4,2,1}$ defined in Lemma 1.7, then by Lemma 1.7 and Lemma 1.3(i),
we get

$\epsilon(J;n)=\begin{cases}n&\ \text{if}\ n\leq n_{k}+1,\\\ &\ \ \text{non-
increasing for}\ n_{k}+1\leq n\leq 2n_{k}+1,\\\ 0&\ \text{if}\ n\geq
2n_{k}+2.\end{cases}$

With the same choice of $I$, then $\epsilon(IJ;n)$ gets local maxima at points
$\\{n_{1},...,n_{k}\\}$.

This degree excess function has a given number of local maxima and drops to
$0$ when $n\gg 0$, while the one in the proof of Theorem 1.10 has a big limit
value.

##  2\. Maximal generating degrees of powers of a monomial ideal

In this section we consider powers of monomial ideals. Using a geometric
interpretation, we give an explicit way to compute the coefficient $p(I)$ in
the Kodiyalam’s construction. Moreover we can give a bound on the reduction
number $r(I)$.

For a subset $A\subseteq R$, the exponent set of $A$ is
$E(A):=\\{{\boldsymbol{\alpha}}\mid\mathbf{X}^{{\boldsymbol{\alpha}}}\in
A\\}\subseteq\mathbb{N}^{s}$. Recall that the Newton polyhedron of a monomial
ideal $I$ is $NP(I):=\operatorname{conv}(E(I))$, the convex hull of the
exponent set of $I$ in the space $\mathbb{R}^{s}$. If
$G(I)=\\{\mathbf{X}^{{\boldsymbol{\alpha}}_{1}},...,\mathbf{X}^{{\boldsymbol{\alpha}}_{t}}\\}$
is the set of minimal generating monomials of $I$, then
$E(G(I))=\\{{\boldsymbol{\alpha}}_{1},...,{\boldsymbol{\alpha}}_{t}\\}$ and

$NP(I):=\operatorname{conv}({\boldsymbol{\alpha}}_{1},...,{\boldsymbol{\alpha}}_{t})+\mathbb{R}_{+}^{s}.$

A useful property of $NP(I)$ is that it is a pointed polyhedron, that is
$NP(I)$ contains no non-trivial linear subspace of $\mathbb{R}^{s}$. Hence
each minimal face of $NP(I)$ consists of just one point which is called a
vertex of $NP(I)$. The set of vertices of $NP(I)$, denoted by $V(I)$, is
uniquely defined by $NP(I)$ and is a subset of $E(G(I))$. The decomposition
theorem of polyhedra says that

(2.1) $NP(I)=\operatorname{conv}(V(I))+\mathbb{R}_{+}^{s},$

(see, e.g., (29) on p. 106 in [15]). Algebraically, $V(I)$ is the minimal
subset of $E(G(I))$ satisfying the equation (2.1). In the sequel, the unit
vectors of $\mathbb{R}^{s}$ are denoted by
$\mathbf{e}_{1},...,\mathbf{e}_{s}$.

######  Example 2.1.

Let $I=(X_{1}^{2},X_{2}^{3},X_{1}X_{2}^{2}X_{3})$. Since

$(1,2,3)=\frac{1}{2}(2,0,0)+\frac{1}{2}(0,3,0)+\frac{1}{2}\mathbf{e}_{2}+\mathbf{e}_{3},$

$V(I)=\\{(2,0,0),(0,3,0)\\}$, while $E(G(I))=\\{(2,0,0),(0,3,0),(1,2,1)\\}$.
The presentation of $(1,2,3)$ in
$\operatorname{conv}(V(I))+\mathbb{R}_{+}^{s}$ is not unique. For an example,
here is another presentation:

$(1,2,3)=\frac{1}{3}(2,0,0)+\frac{2}{3}(0,3,0)+\frac{1}{3}\mathbf{e}_{1}+\mathbf{e}_{3}.$

The set $V(I)$ was used in [11] to study the maximal generating degrees of the
integral closure $\overline{I^{n}}$ of $I^{n}$. We set

$\delta(I)=\max\\{|\mathbf{v}||\ \mathbf{v}\in V(I)\\}.$

In [11] it is shown that $d(\overline{I^{n}})=\delta(I)n+c$ for all $n\gg 0$,
where $c\in\mathbb{N}$ is a constant. We now show that like the case of
integral closures, this number $\delta(I)$ is exactly $p(I)$.

Since $NP(I^{n})=nNP(I)$, the set of vertices of $NP(I^{n})$ is the set of
$n$-multiples of vertices of $V(I)$, i.e. $V(I^{n})=nV(I)$. On the other hand,
by the definition, $V(I^{n})\subseteq E(G(I^{n}))$. Hence

######  Lemma 2.2.

For all $n>0$, $d(I^{n})\geq\delta(I)n$.

We need the following technical result from Linear Programming.

######  Lemma 2.3.

Let $U$ be a finite subset of non-negative integral vectors in
$\mathbb{N}^{s}$ of lengths at most $\delta$ and
$\mathbf{v}\in\mathbb{N}^{s}$. Assume that
$\mathbf{v}\in\operatorname{conv}(U)+\mathbb{R}^{s}_{+}$. Then one can find a
positive integer $N\leq s^{s/2}\delta^{s}$, non-negative integers
$\alpha_{1},...,\alpha_{m},\beta_{1},...,\beta_{s}$ and vectors
$\mathbf{u}_{1},...,\mathbf{u}_{m}\in U$ such that $\sum\alpha_{i}=N$ and

$N\mathbf{v}=\sum_{i=1}^{m}\alpha_{i}\mathbf{u}_{i}+\sum_{j=1}^{s}\beta_{j}\mathbf{e}_{j}.$

###### Proof.

By Carathéodory’s Theorem, one can find $m\leq s+1$ points
$\mathbf{u}_{1},...,\mathbf{u}_{m}\in U$ and non-negative numbers
$\alpha^{\prime}_{1},...,\alpha^{\prime}_{m},\beta^{\prime}_{1},...,\beta^{\prime}_{s}$
such that $\sum\alpha^{\prime}_{i}=1$, and

(2.2)
$\mathbf{v}=\sum_{i=1}^{m}\alpha^{\prime}_{i}\mathbf{u}_{i}+\sum_{j=1}^{s}\beta^{\prime}_{j}\mathbf{e}_{j}.$

Consider the following system of linear constrains

(2.3)
$\begin{cases}(u_{11}-u_{m1})x_{1}+\cdots+(u_{(m-1)1}-u_{m1})x_{m-1}+y_{1}=v_{1},\\\
\cdots\\\
(u_{1s}-u_{ms})x_{1}+\cdots+(u_{(m-1)s}-u_{ms})x_{m-1}+y_{s}=v_{s},\\\
x_{1}+\cdots+x_{m-1}\leq 1,\\\ x_{1},...,x_{m-1},y_{1},...,y_{s}\geq
0.\end{cases}$

Then
$\alpha^{\prime}_{1},...,\alpha^{\prime}_{m-1},\beta^{\prime}_{1},...,\beta^{\prime}_{s}$
is a feasible solution of the system (2.3). Note that
$\alpha^{\prime}_{1},...,\alpha^{\prime}_{m-1}$ is a feasible solution of the
following system of linear constrains of $m-1$ variables:

(2.4) $\begin{cases}(u_{11}-u_{m1})x_{1}+\cdots+(u_{(m-1)1}-u_{m1})x_{m-1}\leq
v_{1},\\\ \cdots\\\ (u_{1s}-u_{ms})x_{1}+\cdots+(u_{(m-1)s}-u_{ms})x_{m-1}\leq
v_{s},\\\ x_{1}+\cdots+x_{m-1}\leq 1,\\\ x_{1},...,x_{m-1}\geq 0.\end{cases}$

Conversely, if $\alpha^{\prime}_{1},...,\alpha^{\prime}_{m-1}$ is a feasible
solution of (2.4), then
$\alpha^{\prime}_{1},...,\alpha^{\prime}_{m-1},\beta^{\prime}_{1},...,\beta^{\prime}_{s}$
is a solution of (2.3), where
$\beta^{\prime}_{j}=v_{j}-[(u_{1j}-u_{mj})\alpha^{\prime}_{1}+\cdots+(u_{(m-1)j}-u_{mj})\alpha^{\prime}_{m-1}]$,
and the numbers
$\alpha^{\prime}_{1},...,\alpha^{\prime}_{m-1},\alpha^{\prime}_{m},\beta^{\prime}_{1},...,\beta^{\prime}_{s}$
satisfy the relation (2.2), where
$\alpha^{\prime}_{m}=1-(\alpha^{\prime}_{1}+\cdots+\alpha^{\prime}_{m-1})$.

Let $\mathcal{P}\subset\mathbb{R}_{+}^{m-1}$ be the set of solutions of (2.4).
By virtue of (2.2), $\mathcal{P}$ is a non-empty pointed polyhedron. Let
${\boldsymbol{\alpha}}^{*}=(\alpha^{*}_{1},...,\alpha^{*}_{m-1})$ be a vertex
of $\mathcal{P}$. By a result of Hoffman and Kruskal [13] (see also, e.g.,
[15, Theorem 8.4]), one may assume that ${\boldsymbol{\alpha}}^{*}$ is the
unique solution of a system of $m-1$ equations $A\mathbf{x}=\mathbf{b}$ for
some subsystem $A\mathbf{x}\leq\mathbf{b}$ of the system of constrains (2.4).
Let $N$ be the determinant of $A$. Then, by Cramer’s rule,
$\alpha^{*}_{i}=\alpha_{i}/N$, where $\alpha_{i}\in\mathbb{N}$. Since each
entry of $A$ is either $u_{ij}-u_{mj}$, or $1$ or $0$, its absolute value is
at most $\delta$. Hence, by Hadamard’s inequality we get

$N\leq(m-1)^{(m-1)/2}\delta^{m-1}\leq s^{s/2}\delta^{s}.$

Then

$\begin{array}[]{ll}\alpha^{*}_{m}&=1-(\alpha^{*}_{1}+\cdots+\alpha^{*}_{m-1})=\alpha_{m}/N,\\\
\beta^{*}_{j}&=v_{j}-[(u_{1j}-u_{mj})\alpha^{*}_{1}+\cdots+(u_{(m-1)j}-u_{mj})\alpha^{*}_{m-1}]=\beta_{j}/N,\
(j=1,...,s),\end{array}$

for some non-negative integers $\alpha_{m},\beta_{1},...,\beta_{s}$. By the
above argument, we get

$\mathbf{v}=\sum_{i=1}^{m}\alpha^{*}_{i}\mathbf{u}_{i}+\sum_{j=1}^{s}\beta^{*}_{j}\mathbf{e}_{j},$

or equivalently,

$N\mathbf{v}=\sum_{i=1}^{m}\alpha_{i}\mathbf{u}_{i}+\sum_{j=1}^{s}\beta_{j}\mathbf{e}_{j}.$

∎

We can now state and prove the main result of this section.

######  Theorem 2.4.

Let $I$ be a monomial ideal. Let $q$ be the number of minimal generators from
$G(I)$ of degree bigger than $\delta(I)$. Then:

  * (i)

$p(I)=\delta(I)$. That means there is a non-negative integer $q_{0}(I)$ such
that $d(I^{n})=\delta(I)n+q_{0}(I)$ for all $n\gg 0.$

  * (ii)

$r(I)\leq qs^{s/2}\delta(I)^{s}-q$.

  * (iii)

$\epsilon(I;n)\leq q[s^{s/2}\delta(I)^{s}-1][d(I)-\delta(I)]$ for all $n\geq
1$.

###### Proof.

Set $\delta=\delta(I)$. Assume that $I=(\mathbf{X}^{\mathbf{v}_{i}}|\
i=1,...,p,..,t)$, such that $V(I)=\\{\mathbf{v}_{1},...,\mathbf{v}_{p}\\}$,
$|\mathbf{v}_{i}|\leq\delta$ for all $i\leq t-q$ and $|\mathbf{v}_{i}|>\delta$
for all $i\geq t-q+1$, where $q\leq t-p$. Let

$\mathcal{J}:=(\mathbf{X}^{\mathbf{v}_{1}},...,\mathbf{X}^{\mathbf{v}_{p}})\subseteq
J:=(\mathbf{X}^{\mathbf{v}_{1}},...,\mathbf{X}^{\mathbf{v}_{t-q}}).$

For each $j=t-q+1,...,t$,
$\mathbf{v}_{j}\in\operatorname{conv}(V(I))+\mathbb{R}_{+}^{s}$. By Lemma 2.3,
there is a positive integer $N_{j}\leq s^{s/2}\delta^{s}$ and non-negative
integers $\alpha_{j,i},\gamma_{j,k}$ such that
$\sum_{i=1}^{p}\alpha_{j,i}=N_{j}$ and

$N_{j}\mathbf{v}_{j}=\sum_{i=1}^{p}\alpha_{j,i}\mathbf{v}_{i}+\sum\gamma_{j,k}\mathbf{e}_{k}.$

Then $\mathbf{X}^{N_{j}\mathbf{v}_{j}}\in\mathcal{J}^{N_{j}}$. If $q=0$, let
$N=0$. Otherwise, let

(2.5) $N=(N_{t-q+1}-1)+\cdots+(N_{t}-1)\leq q(s^{s/2}\delta^{s}-1).$

Then

(2.6) $I^{n+1}=JI^{n}\ \text{ for all}\ n\geq N.$

Indeed, assume by contrary that there is $m\in I^{n+1}\setminus JI^{n}$. Then
$m=\prod_{j=t-q+1}^{t}\mathbf{X}^{n_{j}\mathbf{v}_{j}}$ for some non-negative
$n_{j}$ with $\sum n_{j}=n+1$. Assume $n_{k}\geq N_{k}$ for some $k$. Then

$\mathbf{X}^{n_{k}\mathbf{v}_{k}}=\mathbf{X}^{N_{k}\mathbf{v}_{k}}\mathbf{X}^{(n_{k}-N_{k})\mathbf{v}_{k}}\in\mathcal{J}^{N_{k}}I^{n_{k}-N_{k}}\subseteq
J^{N_{k}}I^{n_{k}-N_{k}}.$

Hence $m\in J^{N_{k}}I^{n+1-N_{k}}\subseteq JI^{n}$, a contradiction. So all
$n_{j}\leq N_{j}-1$ and $n+1\leq N$, which contradicts to the condition $n\geq
N$. Hence (2.6) holds true. Now for all $n\geq N$ we have

$d(I^{n})\leq d(J^{n-N})+d(I^{N})\leq\delta(n-N)+Nd(I)=\delta
n+N(d(I)-\delta).$

On the other hand, by Lemma 2.2, $d(I^{n})\geq\delta n$. From (1.1), it
follows that $p(I)=\delta$, and $J$ is exactly the reduction of $I$ in the
Kodiyalam’s construction. By virtue of Lemma 1.1, $d(I^{n})=\delta n+q_{0}(I)$
for some $q_{0}(I)$ and all $n\gg 0$. This shows (i). The statement (ii)
follows from (2.6) and (2.5), and (iii) follows from (ii) and Lemma 1.3(ii). ∎

In [11], we can give a number $n_{0}$ in terms of $s$ and $\delta(I)$ such
that $d(\overline{I^{n}})=\delta(I)n+c(I)$ for some $c(I)\geq 0$ and all
$n\geq n_{0}$. Unfortunately we cannot solve the similar problem for
$\operatorname{gstab}(I)$. Lemma 1.7 already shows that in the worst case,
$\operatorname{gstab}(I)$ must depend on $\delta(I)$. The following example
not only shows that the bound on the reduction number in the above theorem is
close to be optimal, but also shows that in the worst case, the constant value
of $\epsilon(I;n)$ as well as the index of stability $\operatorname{gstab}(I)$
are at least exponential functions of $s$, namely
$\epsilon(I;n),\operatorname{gstab}(I)>(\delta(I)-1)^{s-3}$.

######  Example 2.5.

Given $s\geq 3$ and $\delta\geq s+1$. Set

$\begin{array}[]{ll}\mathbf{v}_{1}&=(\delta-1,1,0,0,\cdots,0,0),\\\
\mathbf{v}_{2}&=(0,\delta-1,1,0,\cdots,0,0),\\\ &\cdots\\\
\mathbf{v}_{s-1}&=(0,0,0,0,\cdots,\delta-1,1),\\\
\mathbf{v}_{s}&=(1,0,0,0,\cdots,0,\delta-1),\end{array}$

and

$\mathbf{v}=(\delta-s+1,1,...,1).$

Let $a$ be a positive integer. Consider the following ideal in $s+1$
variables:

$I=(M_{1}:=\mathbf{X}^{\mathbf{v}_{1}},...,M_{s}:=\mathbf{X}^{\mathbf{v}_{s}},M:=\mathbf{X}^{\mathbf{v}}Y^{a})\subset
K[\mathbf{X},Y].$

Then $V(I)=\\{(\mathbf{v}_{1},0),...,(\mathbf{v}_{s},0)\\}$ and the reduction
$J$ of $I$ in the Kodiyalam’s construction is
$J=(\mathbf{X}^{\mathbf{v}_{1}},...,\mathbf{X}^{\mathbf{v}_{s}})$. Hence
$\delta(I)=\delta$.

######  Proposition 2.6.

Let $I$ be the monomial ideal given in Example 2.5. Then:

  * (i)

$(\delta-1)^{s-2}<r(I)\leq(\delta-1)^{s}$;

  * (ii)

$\epsilon(I;n)=an$ for all $n\leq r(I)-1$ and
$\epsilon(I;n)=r(I)a>(\delta-1)^{s-2}$ for all $n\geq r(I)$. In particular
$\operatorname{gstab}(I)=r(I)>(\delta-1)^{s-2}$.

###### Proof.

Step 1: To show $V(I)=\\{(\mathbf{v}_{1},0),...,(\mathbf{v}_{s},0)\\}$.

First, we find $\alpha_{1},...,\alpha_{s}$ such that

(2.7) $\alpha_{1}\mathbf{v}_{1}+\cdots+\alpha_{s}\mathbf{v}_{s}=\mathbf{v}.$

Of course $(\alpha_{1},...,\alpha_{s})$ is a solution of the following system
of equations

$\begin{cases}\begin{array}[]{ccccccccccc}(\delta-1)x_{1}&&&&&&&+&x_{s}&=&\delta-s+1,\\\
x_{1}&+&(\delta-1)x_{2}&&&&&&&=&1,\\\ &&x_{2}&+&(\delta-1)x_{3}&&&&&=&1,\\\
&&&&&\cdots&&&&&\\\
&&&&&&x_{s-1}&+&(\delta-1)x_{s}&=&1.\end{array}\end{cases}$

Using the Laplace expansion along the first row we can calculate the
determinant of this system:

$D=\begin{vmatrix}\delta-1&0&0&\cdots&0&1\\\ 1&\delta-1&0&\cdots&0&0\\\
0&1&\delta-1&\cdots&0&0\\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
0&0&0&\cdots&1&\delta-1\end{vmatrix}=(\delta-1)^{s}+(-1)^{s-1}>0.$

So, the system (2.7) has the unique solution.

We show that all $\alpha_{i}>0$. In order to compute the solution of (2.7),
let us first compute the following determinant of size $t\geq 2$:

$B_{t,c}=\begin{vmatrix}c&\delta-1&0&\cdots&0\\\ 1&1&\delta-1&\cdots&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\ 1&0&0&\cdots&\delta-1\\\
1&0&0&\cdots&1\end{vmatrix},$

where $c$ is any number. Using the Laplace expansion along the last row, we
get the recursion relation $B_{t,c}=(-1)^{t-1}(\delta-1)^{t-1}+B_{(t-1),c}$.
Hence

$\begin{array}[]{ll}B_{t,c}&=(-1)^{t-1}(\delta-1)^{t-1}+\cdots+(-1)^{2}(\delta-1)^{2}+\begin{vmatrix}c&\delta-1\\\
1&1\end{vmatrix}\\\
&=(-1)^{t-1}\frac{\delta-1}{\delta}[(\delta-1)^{t-1}+(-1)^{t}]+c.\end{array}$

Denote by $D_{i}$ the determinant obtained from $D$ by replacing the $i$-th
column by $(\delta-s+1,1,\cdots,1)^{T}$. Then, using the Laplace expansion
along the first row, we now get

$\begin{array}[]{ll}D_{1}&=\begin{vmatrix}\delta-s+1&0&0&\cdots&0&1\\\
1&\delta-1&0&\cdots&0&0\\\ 1&1&\delta-1&\cdots&0&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
1&0&0&\cdots&1&\delta-1\end{vmatrix}\\\
&=(\delta-s+1)(\delta-1)^{s-1}+(-1)^{s-1}B_{(s-1),1}\\\
&=(\delta-s+1)(\delta-1)^{s-1}+(-1)^{s-1}\\{(-1)^{s-2}\frac{\delta-1}{\delta}[(\delta-1)^{s-2}+(-1)^{s-1}]+1\\}.\\\
&=(\delta-s+1)(\delta-1)^{s-1}-\frac{\delta-1}{\delta}[(\delta-1)^{s-2}+(-1)^{s-1}]+(-1)^{s-1}.\end{array}$

Clearly $D_{1}>0$, so $\alpha_{1}=D_{1}/D>0$.

For $1<i<s$, using the Laplace expansion along either last or first columns
several times, we obtain:

$\begin{array}[]{ll}D_{i}&=\begin{vmatrix}&&&&\text{i-th}&&&\\\
\delta-1&0&0&\cdots&\delta-s+1&\cdots&0&1\\\
1&\delta-1&0&\cdots&1&\cdots&0&0\\\ 0&1&\delta-1&\cdots&1&\cdots&0&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\ddots&\vdots&\vdots\\\
0&0&0&\cdots&1&\cdots&1&\delta-1\end{vmatrix}\\\
&=(-1)^{s-1}\begin{vmatrix}&&&&\text{i-th}&&\\\
1&\delta-1&0&\cdots&1&\cdots&0\\\ 0&1&\delta-1&\cdots&1&\cdots&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\ddots&\vdots\\\
0&0&0&\cdots&1&\cdots&1\end{vmatrix}+\end{array}$ $\begin{array}[]{ll}&\hskip
28.45274pt+(\delta-1)\begin{vmatrix}&&&&\text{i-th}&&\\\
\delta-1&0&0&\cdots&\delta-s+1&\cdots&0\\\ 1&\delta-1&0&\cdots&1&\cdots&0\\\
0&1&\delta-1&\cdots&1&\cdots&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\ddots&\vdots\\\
0&0&0&\cdots&1&\cdots&\delta-1\end{vmatrix}\\\
&=(-1)^{s-1}B_{(s-i),1}+(\delta-1)^{s-i}\begin{vmatrix}&&&&&\text{i-th}\\\
\delta-1&0&0&\cdots&0&\delta-s+1\\\ 1&\delta-1&0&\cdots&0&1\\\
0&1&\delta-1&\cdots&0&1\\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
0&0&0&\cdots&1&1\end{vmatrix}\\\
&=(-1)^{s-1}B_{(s-i),1}+(\delta-1)^{s-i}(-1)^{i-1}B_{i,\delta-s+1}.\end{array}$

To get the last equality we have moved the last column of the determinant in
the preceding line to the first position. So,

$\begin{array}[]{ll}D_{i}&=(-1)^{i}\frac{\delta-1}{\delta}[(\delta-1)^{s-i-1}+(-1)^{s-i}]+(-1)^{s-1}+\\\
&\hskip
28.45274pt\frac{\delta-1}{\delta}[(\delta-1)^{s-1}+(-1)^{i}(\delta-1)^{s-i}]+(-1)^{i-1}(\delta-1)^{s-i}(\delta-s+1).\end{array}$

It is clear that $D_{i}>0$ for $i\geq 3$. When $i=2$, since

$\frac{\delta-1}{\delta}[(\delta-1)^{s-1}+(\delta-1)^{s-2}]-(\delta-1)^{s-2}(\delta-s+1)=(\delta-1)^{s-2}(s-2)>0,$

also $D_{2}>0$. Hence $\alpha_{i}>0$ for all $1<i<s$. Finally, notice that
$D_{s}=(-1)^{s-1}B_{s,\delta-s+1}$ (moving the last column of $D_{s}$ to the
first position). Hence

(2.8) $\displaystyle D_{s}$ $\displaystyle=$
$\displaystyle\frac{\delta-1}{\delta}[(\delta-1)^{s-1}+(-1)^{s}]+(-1)^{s-1}(\delta-s+1)$
$\displaystyle=$
$\displaystyle\frac{(\delta-1)^{s}+(-1)^{s-1}(\delta^{2}-\delta
s+1)}{\delta}>0.$

Summing up, we have showed that all $\alpha_{i}>0$. Since
$|\mathbf{v}_{1}|=\cdots=|\mathbf{v}_{s}|=|\mathbf{v}|=\delta$, it follows
that $\alpha_{1}+\cdots+\alpha_{s}=1$ and $D_{1}+\cdots+D_{s}=D$. Hence
$\mathbf{v}\in\operatorname{conv}(\mathbf{v}_{1},...,\mathbf{v}_{s})$ and
$(\mathbf{v},a)\in\operatorname{conv}((\mathbf{v}_{1},0),...,(\mathbf{v}_{s},0))+\mathbb{R}_{+}^{s+1}$,
which yields $V(I)=\\{(\mathbf{v}_{1},0),...,(\mathbf{v}_{s},0)\\}$.

Step 2: To show (i).

Since $I=(J,M)$, by Remark 1.4,

(2.9) $r(I)=\min\\{t>0|\ M^{t}\in J^{t}\\}-1.$

As it is shown above

$\frac{D_{1}}{D}\mathbf{v}_{1}+\cdots+\frac{D_{s}}{D}\mathbf{v}_{s}=\mathbf{v},$

and $D_{1}+\cdots D_{s}=D$. Hence

$M^{D}=Y^{Da}(M_{1}^{D_{1}}\cdots M_{s}^{D_{s}})\in J^{D}.$

Therefore $r(I)<D=(\delta-1)^{s}+(-1)^{s+1}$, whence $r(I)\leq(\delta-1)^{s}$.

On the other hand, by virtue of (2.8), the greatest common divisor of $D$ and
$D_{s}$ is at most

$\begin{array}[]{l}\gcd((\delta-1)^{s}+(-1)^{s-1}(\delta^{2}-\delta
s+1),(\delta-1)^{s}+(-1)^{s+1})\\\ \hskip
14.22636pt=\gcd((-1)^{s-1}(\delta^{2}-\delta
s+1)-(-1)^{s+1},(\delta-1)^{s}+(-1)^{s+1})\\\ \hskip
14.22636pt\leq\delta^{2}-3\delta+2=(\delta-1)(\delta-2).\end{array}$

Hence $\alpha_{s}=\frac{n_{1}}{n_{2}}$ with $\gcd(n_{1},n_{2})=1$, and

$\begin{array}[]{ll}n_{2}&=\frac{D}{\gcd(D_{s},D)}\geq\frac{(\delta-1)^{s}+(-1)^{s+1}}{(\delta-1)(\delta-2)}\geq\frac{(\delta-1)^{s}-1}{(\delta-1)(\delta-2)}\\\
&=\frac{(\delta-1)^{s-1}}{\delta-2}-\frac{1}{(\delta-1)(\delta-2)}=(\delta-1)^{s-2}+\frac{(\delta-1)^{s-2}}{\delta-2}-\frac{1}{(\delta-1)(\delta-2)}\\\
&\geq(\delta-1)^{s-2}+\frac{\delta-1}{\delta-2}-\frac{1}{(\delta-1)(\delta-2)}=(\delta-1)^{s-2}+1+\frac{1}{\delta-1}.\end{array}$

This implies $n_{2}\geq(\delta-1)^{s-2}+2$.

Now let $t>0$ be a number such that $M^{t}\in J^{t}$. Then there are
$t_{1},...,t_{s}\in\mathbb{N}$ with $t_{1}+\cdots+t_{s}=t$ and
$\mathbf{u}\in\mathbb{N}^{s}$ such that
$t\mathbf{v}=t_{1}\mathbf{v}_{1}+\cdots+t_{s}\mathbf{v}_{s}+\mathbf{u}$. Since
$|\mathbf{v}_{1}|=\cdots=|\mathbf{v}_{s}|=|\mathbf{v}|$, it follows that
$\mathbf{u}=0$. Then

$\mathbf{v}=\frac{t_{1}}{t}\mathbf{v}_{1}+\cdots+\frac{t_{s}}{t}\mathbf{v}_{s}.$

Since the system (2.7) has exactly one solution, we must have
$\frac{t_{s}}{t}=\alpha_{s}=\frac{n_{1}}{n_{2}}$. By the choice of
$n_{1},n_{2}$, this implies $t\geq n_{2}$. Hence, by (2.9), $r(I)\geq
n_{2}-1>(\delta-1)^{s-2}$, which yields (i).

Step 3: To show (ii).

Let $1\leq t\leq r(I)$ and $n\geq 0$. Assume that
$M^{t}M_{1}^{n}\in\mathfrak{m}I^{t+n}$. Then one can find
$t^{\prime},n_{1},...,n_{s}\in\mathbb{N}$ and $m^{\prime}\neq 1$ a monomial in
$K[\mathbf{X},Y]$ such that $t^{\prime}+n_{1}+\cdots+n_{s}=t+n$ and

$M^{t}m_{1}^{n}=m^{\prime}M^{t^{\prime}}M_{1}^{n_{1}}\cdots M_{s}^{n_{s}}.$

Looking at the exponents of $Y$ in both sides, we get $t^{\prime}\leq t$. If
$t^{\prime}=t$, then from $\deg(M_{1})=\cdots=\deg(M_{s})$ we must have
$m^{\prime}=1$, a contradiction. Hence $t^{\prime}<t$. Dividing both sides by
$M^{t^{\prime}}$ we can assume that $t^{\prime}=0$, i.e.
$M^{t}m_{1}^{n}=m^{\prime}M_{1}^{n_{1}}\cdots M_{s}^{n_{s}}$, where
$n_{1}+\cdots+n_{s}=n+t$. Then
$t\mathbf{v}+n\mathbf{v}_{1}=\mathbf{u}+n_{1}\mathbf{v}_{1}+\cdots+n_{s}\mathbf{v}_{s}$
for some $\mathbf{u}\in\mathbb{N}^{s}$. Since
$|\mathbf{v}|=|\mathbf{v}_{1}|=\cdots=|\mathbf{v}_{s}|$, we must have
$\mathbf{u}=0$. Hence

$\mathbf{v}=\frac{n_{1}-n}{t}\mathbf{v}_{1}+\frac{n_{2}}{t}\mathbf{v}_{2}+\cdots+\frac{n_{s}}{t}\mathbf{v}_{s}.$

This means $(\frac{n_{1}-n}{t},\frac{n_{2}}{t},...,\frac{n_{s}}{t})$ is a
solution of (2.7). Hence $\frac{n_{1}-n}{t}>0$, or equivalently $n_{1}>n$, and
$t\mathbf{v}=(n_{1}-n)\mathbf{v}_{1}+\cdots+n_{s}\mathbf{v}_{s}$. This implies
$M^{t}\in J^{t}$. By virtue of (2.9) this contradicts to the assumption $t\leq
r(I)$. Hence $M^{t}M_{1}^{n}\not\in\mathfrak{m}I^{t+n}$, that means
$M^{t}M_{1}^{n}$ is a minimal generator of $I^{t+n}$, which yields
$d(I^{t+n})\geq\deg(M^{t}M_{1}^{n})=\delta n+ta$, or equivalently
$\epsilon(I;t+n)\geq ta$ for all $1\leq t\leq r$ and $n\geq 0$. Combining with
the bound in Lemma 1.3(ii) this gives the statement (ii) of the proposition. ∎

Our study of degree excess functions gives interesting consequences on the
Castelnuovo-Mumford regularity. For a finitely generated graded $R$-module
$E$, the Castelnuovo-Mumford regularity of $E$ is defined by

$\operatorname{reg}(E)=\max\\{t|\ \text{there is}\ i\ \text{such that}\
H_{\mathfrak{m}}^{i}(E)_{t-i}\neq 0\\},$

where $H_{\mathfrak{m}}^{i}(E)$ is the local cohomology module with the
support $\mathfrak{m}$.

In general, the Castelnuovo-Mumford regularity $\operatorname{reg}(I)$ of a
homogeneous ideal of $R$ could be much larger than $d(I)$. However, it was
independently proved in [5] and [14] that $reg(I^{n})=p(I)n+e(I)$ for some
non-negative integer $e(I)$ and all $n\gg 0$. As an immediate consequence of
Theorem 2.4, we obtain:

######  Corollary 2.7.

Let $I$ be a monomial ideal. Then there is a non-negative integer $e(I)$ such
that $\operatorname{reg}(I^{n})=\delta(I)n+e(I)$ for all $n\gg 0$.

Even in the case of monomial ideals, it is of great interest to give a bound
on the index of stability of $\operatorname{reg}(I^{n})$ defined as

(2.10) $\operatorname{reg-
stab}(I):=\min\\{t|\operatorname{reg}(I^{n})=\delta(I)n+e(I)\ \text{for all}\
n\geq t\\}.$

However, until now very small progress is achieved. The only existing bound is
established for monomial ideals of dimension zero, see [2, Theorem 3.1]. On
the other hand, a similar problem for the Castelnuovo-Mumford regularity
$\operatorname{reg}(\overline{I^{n}})$ of integral closures $\overline{I^{n}}$
of $I^{n}$ is solved in a recent paper [10]. In [11] we show that
$\delta(I)n\leq\operatorname{reg}(\overline{I^{n}})\leq\delta(I)n+\dim R/I$.
In particular, $\operatorname{reg}(\overline{I^{n}})=\delta(I)n+e^{\prime}(I)$
for some $0\leq e^{\prime}(I)\leq\dim R/I$ for all $n\gg 0$. Unfortunately we
cannot use the technique presented in this paper to give a bound on $e(I)$ of
Corollary 2.7 and on $\operatorname{reg-stab}(I)$. However, our study on
degree excess functions leads to the following somewhat unexpected result,
which says that in the worst case, the invariants $e(I)$ and
$\operatorname{reg-stab}(I)$ should be at least exponential functions of the
number of variables.

######  Theorem 2.8.

Let $I$ be a homogeneous ideal of a polynomial ring $K[X_{1},...,X_{m}]$ of
$m\geq 4$ variables. Assume that $d(I)\geq 3$. Then, in the worst case we must
have

  * (i)

$e(I)>(d(I)-2)^{m-3}$, and

  * (ii)

$\operatorname{reg-stab}(I)>\frac{1}{m-1}(d(I)-2)^{m-4}$.

###### Proof.

Consider the ideal $I$ in Example 2.5 with $a=1$, $s=m-1$, $\delta\geq m$ and
$Y=X_{m}$. For short, let $t:=\operatorname{reg-stab}(I)$ and $e:=e(I)$. Then
$\operatorname{reg}(I^{t})=\delta t+e$. The greatest common divisor of minimal
generators of $I^{t}$ is $F=X_{1}^{(\delta-1)t}\cdots
X_{s}^{(\delta-1)t}X_{m}^{t}$. By [3, Theorem 3.1(a)] (see also [12, Theorem
3.1]), $\operatorname{reg}(I^{t})<\deg(F)=s(\delta-1)t+t$. Hence

$e<(s-1)(\delta-1)t+t<s(\delta-1)t.$

On the other hand, since $\delta
n+\epsilon(I;n)=d(I^{n})\leq\operatorname{reg}(I^{n})$ for all $n$, taking
$n\gg 0$ we get $\epsilon(I;n)\leq e$. By Proposition 2.6, it implies

$e\geq r(I)>(\delta-1)^{s-2}=(d(I)-2)^{m-3},$

whence $s(\delta-1)t>e>(\delta-1)^{s-2}$. Hence

$t>\frac{1}{s}(\delta-1)^{s-3}=\frac{1}{m-1}(d(I)-2)^{m-4}.$

∎

Acknowledgement. I would like to thank the referee for careful reading and
useful suggestions. In particular, the proof of Lemma 1.8 is due to him/her.

This work is partially supported by NAFOSTED (Vietnam) under the grant number
101.04-2018.307 and the Program for Research Activities of Senior Researchers
of VAST under the grant number NVCC01.11/21-21.

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