# A note on ratliff-rush filtration, reduction number and postulation number
of $\mathfrak{m}$-primary ideals

Mousumi Mandal and Shruti Priya  Department of Mathematics, Indian Institute
of Technology Kharagpur, 721302, India<EMAIL_ADDRESS>Department
of Mathematics, Indian Institute of Technology Kharagpur, 721302, India
<EMAIL_ADDRESS>
###### ABSTRACT.

Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring of dimension $d\geq 2$
and $I$ an $\mathfrak{m}$-primary ideal. Let $\operatorname{rd}(I)$ be the
reduction number of $I$ and $\operatorname{n}(I)$ the postulation number. We
prove that for $d=2,$ if $\operatorname{n}(I)=\rho(I)-1,$ then
$\operatorname{rd}(I)\leq\operatorname{n}(I)+2$ and if
$\operatorname{n}(I)\neq\rho(I)-1,$ then
$\operatorname{rd}(I)\geq\operatorname{n}(I)+2.$ For $d\geq 3$, if $I$ is
integrally closed, $0pt\operatorname{gr}(I)=d-2$ and
$\operatorname{n}(I)=-(d-3).$ Then we prove that
$\operatorname{rd}(I)\geq\operatorname{n}(I)+d$. Our main result is to
generalize a result of T. Marley on the relation between the Hilbert-Samuel
function and the Hilbert-Samuel polynomial by relaxing the condition on the
depth of the associated graded ring with the good behaviour of the Ratliff-
Rush filtration with respect to $I$ mod a superficial element. From this
result, it follows that for a Cohen-Macaulay ring of dimension $d\geq 2$, if
$P_{I}(k)=H_{I}(k)$ for some $k\geq\rho(I)$, then $P_{I}(n)=H_{I}(n)$ for all
$n\geq k.$

AMS Classification 2010: 13H10, 13D40, 13A30.

Key words and phrases: Cohen-Macaulay local rings, reduction number,
postulation number, superficial elements, stability index, Ratliff-Rush
filtration, surjectivity index.

##  1\. INTRODUCTION

Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring of dimension $d$ with an
infinite residue field and $I$ an $\mathfrak{m}$-primary ideal. Pierre Samuel
proved that for sufficiently large $n$, the Hilbert-Samuel function
$H_{I}(n)=\lambda(R/I^{n})$ (where $\lambda()$ denotes the length) of an
$\mathfrak{m}$-primary ideal $I$ is a polynomial $P_{I}(n)$, known as the
“Hilbert-Samuel polynomial of I”. Since $P_{I}(n)$ takes only numerical values
for every integer $n$, by the theory on numerical polynomials we can write

$P_{I}(n)=e_{0}(I)\dbinom{n+d-1}{d}-e_{1}(I)\dbinom{n+d-2}{d-1}+\ldots+(-1)^{d}e_{d}(I)$

Here, $e_{0}(I),e_{1}(I),\ldots,e_{d}(I)$ are unique integers known as the
Hilbert coefficients. The leading coefficient $e_{0}(I)$ of the polynomial
$P_{I}(n)$ is called the multiplicity of the ideal $I$ and is denoted by
$e(I).$ Hilbert coefficients can be used to characterize the properties of the
ring $R$, the ideal $I$ and the blow-up algebras associated with $I$.

Marley revealed an interesting connection between the Hilbert coefficients and
the reduction number of an ideal $I$. Recall that a reduction of $I$ is an
ideal $J\subseteq I$ such that $JI^{n}=I^{n+1}$ for some $n\in\mathbb{N.}$ If
$J$ is a reduction of $I$, the reduction number of $I$ with respect to $J$
denoted by $\operatorname{rd}_{J}(I)$ is the smallest $n$ such that
$JI^{n}=I^{n+1}.$ A reduction is minimal if it is minimal with respect to
containment among all reductions. The reduction number of $I$ is defined as
$\operatorname{rd}(I):=\min\\{\operatorname{rd}_{J}(I):J$ is a minimal
reduction of $I\\}$. The concept of reduction and minimal reduction was first
introduced by Northcott and Rees. Reduction number is useful in studying the
Cohen-Macaulay properties of the associated graded ring
$\operatorname{gr}(I)=\displaystyle\bigoplus_{n\geq 0}I^{n}/I^{n+1}$ of $R$
with respect to $I.$ The reduction number $\operatorname{rd}(I)$ is said to be
independent if $\operatorname{rd}(I)=\operatorname{rd}_{J}(I)$ for every
minimal reduction $J$ of $I$. In the case of dimension one, the reduction
number $\operatorname{rd}(I)$ of an $\mathfrak{m}$-primary ideal $I$ is
independent. It was independently proved by Huckaba [6] and Trung [21] that if
$0pt\operatorname{gr}(I)\geq d-1$, then $\operatorname{rd}(I)$ is independent.
Sally raised the following question in [18]: If $(R,\mathfrak{m})$ is a Cohen-
Macaulay ring of dimension $d$, then is $\operatorname{rd}(\mathfrak{m})$
independent? As a natural extension, one can ask if $\operatorname{rd}(I)$ is
independent for any $\mathfrak{m}$-primary ideal $I.$ There are examples that
suggest that $\operatorname{rd}(I)$ is not independent in general. In this
paper, we have investigated the conditions under which $\operatorname{rd}(I)$
is independent (see corollaries 3.10, 3.11, 3.15, 3.17).

Define $\operatorname{n}(I):=\min\\{n\in\mathbb{Z}:P_{I}(t)=H_{I}(t)$ for
$t>n\\}$. Ooishi [12] called $\operatorname{n}(I)$ the postulation number of
$I$ and proved that if $\dim R=1$, then
$\operatorname{rd}(I)=\operatorname{n}(I)+1.$ Marley [10, Theorem 2.15] proved
that if $0pt\operatorname{gr}(I)\geq d-1$, then
$\operatorname{rd}(I)=\operatorname{n}(I)+d.$ Much of this paper was inspired
by the following question raised by Marley [10, Question 2.1]:

######  Question 1.1.

Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring of dimension $d$ with an
infinite residue field and $I$ an $\mathfrak{m}$-primary ideal. Is it always
the case that $\operatorname{rd}(I)\leq\operatorname{n}(I)+d$?

In this paper, we have answered the question proposed by Marley in a two
dimensional Cohen-Macaulay ring. We have observed that the stability index of
the Ratliff-Rush filtration $\rho(I)$ plays an important role in understanding
the relationship between the reduction number and postulation number of an
$\mathfrak{m}$-primary ideal $I$. Recall that the the stability index of the
Ratliff-Rush filtration with respect to $I$ is defined as:

$\rho(I):=\min\\{i\geq 1:\widetilde{I^{n}}=I^{n}$ for all $n\geq i\\}$.

Thus, we prove the following theorem:

######  Theorem 1.2.

Let $(R,\mathfrak{m})$ be a two dimensional Cohen-Macaulay local ring, $I$ an
$\mathfrak{m}$-primary ideal and $x\in I$ be a superficial element for $I$.
Let $S=R/(x)$ and $I^{\prime}=I/(x)$. If $\operatorname{n}(I)\neq\rho(I)-1$,
then $\operatorname{rd}(I)\geq\operatorname{n}(I)+2.$ Furthermore, if
$\operatorname{n}(I)=\rho(I)-1$, then
$\operatorname{rd}(I)\leq\operatorname{n}(I)+2$.

It is worth asking if we can generalize Theorem 1.2 to a higher dimensional
Cohen-Macaulay ring. The main difficulty in generalizing the result is that
there is no simple relation between $\rho(I)$ and $\rho\left(I/(x)\right)$.
Under some conditions on the depth of the associated graded ring, we prove the
following proposition:

######  Proposition 1.3.

Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring of dimension $d\geq 3$
and $I$ be an $\mathfrak{m}$-primary ideal such that $I$ is integrally closed
in $R$ and $0pt\operatorname{gr}(I)=d-2.$ If $\operatorname{n}(I)=-(d-3),$
then $\operatorname{rd}(I)\geq\operatorname{n}(I)+d.$

Another inspiration for this paper is to understand the relationship between
$\rho(I)$ and $\rho\left(I/(x)\right)$. If the Ratliff-Rush filtration with
respect to $I$ behaves well mod a superficial element $x$, then
$\rho(I)\geq\rho\left(I/(x)\right).$ This provides a necessary condition for
the Ratliff-Rush filtration with respect to $I$ to behave well mod a
superficial element. We observe that
$\frac{I^{n}+(x)}{(x)}=\frac{\widetilde{I^{n}}+(x)}{(x)}=\widetilde{\frac{I^{n}+(x)}{(x)}}$
for $n\gg 0.$ This fact is closely related to the surjectivity of the map
$\pi_{n}:\frac{\widetilde{I^{n}}}{I^{n}}\longrightarrow\frac{\widetilde{\frac{I^{n}+(x)}{(x)}}}{\frac{I^{n}+(x)}{(x)}}$
for $n\gg 0.$ Set $\omega(I):=\min\big{\\{}k\in\mathbb{N}:\pi_{n}$ is
surjective for all $n\geq k\big{\\}}.$ Note that if $\omega(I)=1,$ then the
Ratliff-Rush filtration with respect to $I$ behaves well mod a superficial
element. It is of some interest to find an upper bound for $\omega(I).$ We
prove the following Proposition:

######  Proposition 1.4.

Let $(R,\mathfrak{m})$ be a Cohen-Macaulay ring of dimension $d\geq 2,$ $I$ an
$\mathfrak{m}$-primary ideal and $x\in I$ be a superficial element for $I.$ If
$b_{1}^{I}\neq-1$, then
$\omega(I)=\max\\{1,b_{1}^{I}+2\\}\leq\max\\{1,a_{2}(\operatorname{gr}(I))+1\\}.$

As an immediate corollary to Proposition 1.4 we prove (see 4.2).

$\rho(I)\geq\rho\left(I/(x)\right),$ if $\rho(I)\geq\omega(I).$

Northcott in his work has investigated the relation between the Hilbert-Samuel
polynomial and the Hilbert-Samuel function in a one dimensional Cohen-Macaulay
ring and has proved [11] that $P_{I}(n+1)-H_{I}(n+1)\geq P_{I}(n)-H_{I}(n)$
for all $n.$ However, there are examples that illustrate that when the
dimension of the ring is greater than one, the values of $P_{I}(n)-H_{I}(n)$
do not easily correlate with one another. Marley has proved a similar
statement [10, Theorem 2.4]: Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local
ring of dimension $d$ and $I$ an $\mathfrak{m}$-primary ideal such that
$0pt\operatorname{gr}(I)\geq d-1.$ Then for $0\leq i\leq d$ and for all
$n\in\mathbb{Z}$

$(-1)^{d-i}\Delta^{i}(P_{I}(n)-H_{I}(n))\geq 0.$

According to what we have observed, the condition on the depth of the
associated graded ring can be relaxed by the weaker condition on the good
behavior of the Ratliff-Rush filtration with respect to $I$ mod a superficial
sequence. We prove the following theorem:

######  Theorem 1.5.

Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring of dimension $d\geq 2,$
$I$ an $\mathfrak{m}$-primary ideal. Let the Ratliff-Rush filtration with
respect to $I$ behaves well mod a superficial sequence
$x_{1},x_{2},\ldots,x_{d-2}\in I.$ Then for all $0\leq i\leq d$ and for all
$n\geq\rho(I)$

$(-1)^{d-i}\Delta^{i}(P_{I}(n)-H_{I}(n))\geq 0.$

As a corollary to the above theorem, we have partially answered (see corollary
5.8) the following question posed by Marley [10, Question 2.3]:

######  Question 1.6.

Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring and $I$ an
$\mathfrak{m}$-primary ideal. Suppose $P_{I}(k)=H_{I}(k)$ for some $k\geq 1$.
Then is necessarily $\operatorname{n}(I)<k?$

This paper is organized into five sections. Definitions, introductory ideas,
notations, and a technical lemma are all covered in Section 2. In Section 3,
we have established the relationship between the postulation number of an
$\mathfrak{m}$-primary ideal $I$ and the postulation number of
$I^{\prime}=I/(x)$, where $x\in I$ is a superficial element for $I.$ We have
proved Theorem 1.2 and its consequences related to the independence of
reduction number $\operatorname{rd}(I).$ We also prove Proposition 1.3 in this
section. In Section 4, we prove Proposition 1.4. In Section 5, we prove
Theorem 1.5 and have partially answered Marley’s Question 1.6 in a
$d$-dimensional Cohen-Macaulay ring. The techniques used in sections 4 and 5
were inspired by Purthenpurakal [13, 14]. We provide examples in support of
our claims.

##  2\. PRELIMINARIES

In this section, we recall the basic definitions, and preliminaries and
introduce some notations which will be used throughout this paper. Throughout,
$(R,\mathfrak{m})$ will be a Cohen-Macaulay local ring with an infinite
residue field and $I$ an $\mathfrak{m}$-primary ideal of $R$. The Krull
dimension of the ring $R$ will be denoted by $d$ and will be always assumed to
be positive.

######  Definition 2.1.

Consider the increasing chain of ideals in $R$

$I\subseteq(I^{2}:I)\subseteq(I^{3}:I^{2})\subseteq\ldots\subseteq(I^{n+1}:I^{n})\subseteq\ldots$

Since, $R$ is Noetherian, this chain of ideal stabilizes to an ideal
$\widetilde{I}=\displaystyle\bigcup_{n\geq 1}(I^{n+1}:I^{n})$. This ideal
$\widetilde{I}$ is known as the Ratliff-Rush closure of $I$ or Ratliff-Rush
ideal associated with $I.$ An ideal $I$ is called Ratliff-Rush closed if
$\widetilde{I}=I.$ If $I$ is a regular ideal, then $\widetilde{I}$ is the
largest unique ideal such that $\widetilde{I^{n}}=I^{n}$ for $n\gg 0.$

######  Definition 2.2.

An element $x\in I/I^{2}$ is called a superficial element for $I$ if there
exist a positive integer $c$ such that $(I^{n+1}:x)\cap I^{c}=I^{n}$ for all
$n\geq c.$ A sequence $x_{1},x_{2},\ldots,x_{j}$ is said to be a superficial
sequence for $I$ if $x_{1}$ is a superficial element for $I$, and $x_{i}$ is a
superficial element for $I/(x_{1},x_{2},\ldots,x_{i-1}),$ for all
$i=2,3,\ldots,j.$

If $x\in I$ is a superficial element for $I$, we get that $(I^{n+1}:x)=I^{n}$
for $n\gg 0.$

######  Definition 2.3.

[13, 2.6] Let $x$ be any superficial element for $I,$ then the stability index
for $x$ is defined as

$\rho_{x}(I):=\min\\{\text{$k\in\mathbb{N}:(I^{n+1}:x)=I^{n}$ for all $n\geq
k$}\\}$

In [13, Corollary 2.7], Puthenpurakal has proved that $\rho(I)=\rho_{x}(I)$
and thus $\rho_{x}(I)$ is independent of the choice of the superficial
element. The following Lemma gives a relation between the Hilbert-Samuel
function and the Hilbert-Samuel polynomial of an ideal $I$ and $I/(x),$ where
$x$ is a superficial element in $I$. It was proved by Yinghwa Wu, we provide
an alternate proof.

######  Lemma 2.4.

[22, Lemma 2.3] Suppose that $x$ is a superficial element for $I$. Let
$S=R/(x)$ and $I^{\prime}=I/(x)$. Then for all $n\in\mathbb{Z},$

  1. (1)

$H_{I^{\prime}}(n)\geq H_{I}(n)-H_{I}(n-1)$ and

  2. (2)

$P_{I^{\prime}}(n)=P_{I}(n)-P_{I}(n-1)$

###### Proof.

Let us consider the following exact sequence:

$0\longrightarrow(I^{n+1}:x)/I^{n+1}\longrightarrow
R/I^{n+1}\xlongrightarrow{x}R/I^{n+1}\longrightarrow
R/(I^{n+1}+(x))\longrightarrow 0.$ (1)

Then from the exact sequence (1), we have
$\lambda((I^{n+1}:x)/I^{n+1})=\lambda(R/(I^{n+1}+(x))).$

Now

$\begin{split}H_{I^{\prime}}(n+1)&=\lambda(S/I^{\prime n+1})\\\
&=\lambda(R/(I^{n+1}+(x)))\\\ &=\lambda((I^{n+1}:x)/I^{n+1})\\\
&=\lambda((I^{n+1}:x)/I^{n})+\lambda(I^{n}/I^{n+1})\\\
&\geq\lambda(I^{n}/I^{n+1})\\\ &=\lambda(R/I^{n+1})-\lambda(R/I^{n})\\\
&=\text{$H_{I}(n+1)-H_{I}(n)$ for all $n$}\end{split}$ (2)

Therefore,

$H_{I^{\prime}}(n)\geq H_{I}(n)-H_{I}(n-1)$ for all $n$

Clearly, $P_{I^{\prime}}(n)=P_{I}(n)-P_{I}(n-1)$ for all $n$ ∎

The Rees Ring of $I$ is defined as
$\mathcal{R}(I)=\displaystyle\bigoplus_{n\geq 0}I^{n}t^{n}$ is the standard
graded $R_{0}$-algebra. Set $\mathcal{R}(I)^{+}=\displaystyle\bigoplus_{n\geq
1}I^{n}t^{n}$ and
$\mathfrak{M}_{\mathcal{R}(I)}=\mathfrak{m}\bigoplus\mathcal{R}(I)^{+}.$ For
the reference of local cohomology we use [1]. Let $M$ be a graded
$\mathcal{R}(I)$-module. Define
$\operatorname{end}(M)=\sup\\{n\in\mathbb{Z}:M_{n}\neq 0\\}$. Let
$H^{i}_{\mathfrak{M}_{\mathcal{R}(I)}}(M)$ denote the $i$-th local cohomology
module of $M$ with respect to $\mathfrak{M}_{\mathcal{R}(I)}.$ If $M$ is a
finitely generated $\mathcal{R}(I)$-module, then for each $i\geq 0$,
$H^{i}_{\mathfrak{M}_{\mathcal{R}(I)}}(M)_{n}=0$ for $n\gg 0.$

Define
$a_{i}(M)=\operatorname{end}\left(H^{i}_{{\mathcal{R}(I)^{+}}}(M)\right)$. The
Castelnuovo-Mumford regularity of $M$ is defined as
$\operatorname{reg}(M)=\max\\{a_{i}(M)+i:0\leq i\leq\dim M\\}.$

Set $L^{I}(R)=\displaystyle\bigoplus_{n\geq 0}R/I^{n+1}.$ Puthenpurakal in
[13] has proved that $L^{I}(R)$ is a $\mathcal{R}(I)$-module which is not
finitely generated. See [13, 14] for reference and further applications of
$L^{I}(R)$ module.

Let $x\in I$ be superficial for $I.$ Set $S=R/(x)$ and $I^{\prime}=I/(x).$ We
recall the second fundamental exact sequence from the literature [13, 6.2,
6.3], for each $n\geq 1$ we have the following exact sequence of $R$-modules:

$0\longrightarrow\frac{(I^{n+1}:x)}{I^{n}}\longrightarrow\frac{R}{I^{n}}\xlongrightarrow{\psi_{n}}\frac{R}{I^{n+1}}\longrightarrow\frac{S}{I^{\prime
n+1}}\longrightarrow 0,$

where $\psi_{n}(a+I^{n})=xa+I^{n+1}$ for all $a\in R.$ This sequence induces
an exact sequence of $\mathcal{R}(I)$-modules, called the second fundamental
exact sequence:

$0\longrightarrow\mathcal{B}(x,R)\longrightarrow
L^{I}(R)(-1)\xlongrightarrow{\Psi_{X}}L^{I}(R)\longrightarrow
L^{I^{\prime}}(S)\longrightarrow 0,$ (3)

where $\Psi_{X}$ is multiplication by $X$ and

$\mathcal{B}(x,R)=\bigoplus_{n\geq 0}\frac{(I^{n+1}:x)}{I^{n}}.$

Note that for $n\gg 0,$ $(I^{n+1}:x)=I^{n}.$ Therefore, $\mathcal{B}(x,R)$ has
finite length. So we observe that,
$H^{0}_{\mathfrak{M}_{\mathcal{R}(I)}}\left(\mathcal{B}(x,R)\right)$
$=\mathcal{B}(x,R)$ The sequence (3) induces a long exact sequence of local
cohomology modules. For convenience from now onwards, we will write
$H^{i}_{\mathfrak{M}_{\mathcal{R}(I)}}\left(L^{I}(R)\right)=H^{i}\left(L^{I}(R)\right)$
for all $i.$

$\displaystyle 0\longrightarrow\mathcal{B}(x,R)$ $\displaystyle\longrightarrow
H^{0}\left(L^{I}(R)\right)(-1)\longrightarrow
H^{0}\left(L^{I}(R)\right)\longrightarrow H^{0}\left(L^{I^{\prime}}(S)\right)$
(4) $\displaystyle\longrightarrow
H^{1}\left(L^{I}(R)\right)(-1)\longrightarrow
H^{1}\left(L^{I}(R)\right)\longrightarrow H^{1}\left(L^{I^{\prime}}(S)\right)$
$\displaystyle\ldots\ldots$ $\displaystyle\longrightarrow
H^{n}\left(L^{I}(R)\right)(-1)\longrightarrow
H^{n}\left(L^{I}(R)\right)\longrightarrow
H^{n}\left(L^{I^{\prime}}(S)\right)\ldots$

A graded $\mathcal{R}(I)$ module $M$ is called $*$-$Artinian$ if every
descending chain of graded submodules of $M$ terminates. From [13, Proposition
4.4, 5.4] we get that, $H^{i}(L^{I}(R))$ are $*$-Artinian for every
$i=0,1,2,\ldots d-1.$ Define
$b^{I}_{i}(R)=\operatorname{end}\left(H^{i}(L^{I}(R))\right)$ for all
$i=0,1,2,\ldots d-1.$ From [13, 5.4] we get, $b^{I}_{i}(R)\leq
a_{i+1}(\operatorname{gr}(I))-1.$ Set
$b^{I}=\max\\{b^{I}_{i}:i=0,1,\ldots,d-1\\}.$ By [13, Proposition 4.7],
$H^{0}\left(L^{I}(R)\right)=\displaystyle\bigoplus_{n\geq
0}\frac{\widetilde{I^{n+1}}}{I^{n+1}}.$

######  Observation 2.5.

The zeroth local cohomology of $L^{I}(R)$ i.e., $H^{0}\left(L^{I}(R)\right)$
does not have negatively graded components. From the long exact sequence in
local cohomology (4), we have

$\longrightarrow\left(H^{0}\left(L^{I}(R)\right)\right)_{n}\longrightarrow\left(H^{0}\left(L^{I^{\prime}}(S)\right)\right)_{n}\longrightarrow\left(H^{1}\left(L^{I}(R)\right)\right)_{n-1}\longrightarrow\left(H^{1}\left(L^{I}(R)\right)\right)_{n}$

If possible, let $b_{1}<-1$, then at $n=b_{1}+1$ we have

$0\longrightarrow\left(H^{1}\left(L^{I}(R)\right)\right)_{b_{1}}\longrightarrow\left(H^{1}\left(L^{I}(R)\right)\right)_{b_{1}+1}$

which implies that $\left(H^{1}\left(L^{I}(R)\right)\right)_{b_{1}}=0$, which
is a contradiction. Therefore, $b_{1}\geq-1.$ Note that if $I$ is an
integrally closed ideal in $R$, then $b_{1}\geq 0.$

##  3\. Relation between the reduction number and the postulation number of an
$\mathfrak{m}$-primary ideal

Marley in [10] has proved that if $(R,\mathfrak{m})$ is a Cohen-Macaulay local
ring of dimension $d$ with an infinite residue field and $I$ an
$\mathfrak{m}$-primary ideal of $R$ such that $0pt\operatorname{gr}(I)\geq
d-1$ then $\operatorname{n}(I)+1=\operatorname{n}(I^{\prime}),$ where
$I^{\prime}=I/(x),$ and $x\in I$ is a superficial element for $I$. He has also
proved with the above hypothesis that
$\operatorname{rd}(I)=\operatorname{n}(I)+d.$ Inspired by Marley’s work, one
can ask whether these results can be extended without any condition on depth
on the associated graded ring. In this section, we attempt to extend Marley’s
results.

In the next proposition, we observe that the stability index $\rho(I)$ of the
Ratliff-Rush filtration with respect to $I$ plays an important role to
describe the behaviour of the postulation number of $I$ and the postulation
number of $I$ going mod a superficial element.

######  Proposition 3.1.

Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring of dimension $d$ with an
infinite residue field and $I$ an $\mathfrak{m}$-primary ideal. Let $x\in I$
be a superficial element for $I$. Set $S=R/(x)$ and $I^{\prime}=I/(x)$.

  1. (1)

If $\operatorname{n}(I)>\rho(I)-1$, then
$\operatorname{n}(I^{\prime})=\operatorname{n}(I)+1.$

  2. (2)

If $\operatorname{n}(I)=\rho(I)-1$, then
$\operatorname{n}(I^{\prime})\leq\operatorname{n}(I)+1.$

  3. (3)

If $\operatorname{n}(I)<\rho(I)-1$, then
$\operatorname{n}(I^{\prime})\geq\operatorname{n}(I)+1.$

###### Proof.

Since $x$ is a superficial element for $I$, therefore, $(I^{n+1}:x)=I^{n}$ for
all $n\geq\rho(I)$. By Lemma 2.4, we have
$H_{I^{\prime}}(n)=H_{I}(n)-H_{I}(n-1)$ for all $n\geq\rho(I)+1.$ By equation
(1) at $\rho(I)$, we have
$H_{I^{\prime}}(\rho(I))>H_{I}(\rho(I))-H_{I}(\rho(I)-1).$

  1. (1)

Suppose $\operatorname{n}(I)>\rho(I)-1.$ For any
$m>\operatorname{n}(I)+1\geq\rho(I)+1$, we have
$H_{I^{\prime}}(m)=H_{I}(m)-H_{I}(m-1)=P_{I}(m)-P_{I}(m-1)=P_{I^{\prime}}(m)$
so, $\operatorname{n}({I^{\prime}})\leq\operatorname{n}(I)+1$. If
$\operatorname{n}(I^{\prime})<\operatorname{n}(I)+1,$ then
$H_{I^{\prime}}(\operatorname{n}(I)+1)=P_{I^{\prime}}(\operatorname{n}(I)+1),$
which implies
$H_{I}(\operatorname{n}(I)+1)-H_{I}(\operatorname{n}(I))=P_{I}(\operatorname{n}(I)+1)-P_{I}(\operatorname{n}(I))$.
Thus, $H_{I}(\operatorname{n}(I))=P_{I}(\operatorname{n}(I))$, a
contradiction. Hence, $\operatorname{n}(I^{\prime})=\operatorname{n}(I)+1.$

  2. (2)

Suppose $\operatorname{n}(I)=\rho(I)-1.$ Then for any $m\geq\rho(I)+1$, we
have
$H_{I^{\prime}}(m)=H_{I}(m)-H_{I}(m-1)=P_{I}(m)-P_{I}(m-1)=P_{I^{\prime}}(m)$.
Therefore, we get $\operatorname{n}(I^{\prime})<\rho(I)+1$ which implies
$\operatorname{n}(I^{\prime})\leq\operatorname{n}(I)+1.$

  3. (3)

Suppose $\operatorname{n}(I)<\rho(I)-1$. We have
$H_{I^{\prime}}(\rho(I))>P_{I^{\prime}}(\rho(I))$ as
$H_{I^{\prime}}(\rho(I))>H_{I}(\rho(I))-H_{I}(\rho(I)-1)=P_{I}(\rho(I))-P_{I}(\rho(I)-1)=P_{I^{\prime}}(\rho(I)).$
Therefore, we get
$\operatorname{n}(I^{\prime})\geq\rho(I)=\operatorname{n}(I)+1$ which implies
$\operatorname{n}(I^{\prime})\geq\operatorname{n}(I)+1.$

∎

######  Remark 3.2.

From Lemma 3.1, we observe that for $\operatorname{n}(I)=\rho(I)-1$,
$H_{I}(\rho(I)-1)\neq P_{I}(\rho(I)-1)$ which implies
$H_{I}(\rho(I)-1)=P_{I}(\rho(I)-1)+l$, for some non-zero integer $l.$ If
$l<0$, then $H_{I}(\rho(I)-1)<P_{I}(\rho(I)-1).$ At $\rho(I)$,
$H_{I^{\prime}}(\rho(I))>P_{I^{\prime}}(\rho(I)).$ Thus,
$\operatorname{n}(I^{\prime})\geq\rho(I)=\operatorname{n}(I)+1.$ Also, for
$m\geq\rho(I)+1$, we have $H_{I^{\prime}}(m)=P_{I^{\prime}}(m).$ Thus,
$\operatorname{n}(I^{\prime})<\rho(I)+1$. Therefore,
$\operatorname{n}(I^{\prime})\leq\operatorname{n}(I)+1.$ Thus, if
$H_{I}(n)<P_{I}(n)$ for $n=\rho(I)$, we have
$\operatorname{n}(I^{\prime})=\operatorname{n}(I)+1$.

The following examples illustrate the above theorem.

######  Example 3.3.

[19, Example 5.1] Let $R=\mathbb{Q}[[x,y]]$ and
$I=(x^{6},x^{4}y,xy^{5},y^{6})$. The Hilbert polynomial for $I$ is

$P_{I}(n)=30\dbinom{n+1}{2}-10n+3.$

Then $P_{I}(0)\neq H_{I}(0)$ and $P_{I}(n)=H_{I}(n)$ for all $n\geq 1$. Thus,
$\operatorname{n}(I)=0.$ Note that $p=x^{4}y+y^{6}$ is a superficial element
for $I$ as $e_{0}(I)=30=e_{0}(I/(p)),e_{1}(I)=10=e_{1}(I/(p))$ and
$\rho(I)\geq 2$. Here, $n(I)\neq\rho(I)-1.$ Let $S=R/(p)$ and
$I^{\prime}=I/(p).$ The Hilbert polynomial for $I^{\prime}$ is

$P_{I^{\prime}}(n)=30n-10.$

Then $P_{I^{\prime}}(0)\neq H_{I^{\prime}}(0)$, $P_{I^{\prime}}(1)\neq
H_{I^{\prime}}(1)$, $P_{I^{\prime}}(2)\neq H_{I^{\prime}}(2)$ and
$P_{I^{\prime}}(n)=H_{I^{\prime}}(n)$ for all $n\geq 3.$ Thus,
$\operatorname{n}(I^{\prime})=2\geq\operatorname{n}(I)+1.$

######  Example 3.4.

[10, Example 2.7] Let $R=\mathbb{Q}[[x,y]]$ and
$I=(x^{7},x^{6}y,xy^{6},y^{7})$. The Hilbert polynomial for $I$ is

$P_{I}(n)=49\dbinom{n+1}{2}-21n.$

Then $P_{I}(0)\neq H_{I}(0)$, $P_{I}(1)\neq H_{I}(1)$, $P_{I}(2)\neq
H_{I}(2)$, $P_{I}(3)\neq H_{I}(3)$, $P_{I}(4)\neq H_{I}(4)$ and
$P_{I}(n)=H_{I}(n)$ for all $n\geq 5.$ Thus, $\operatorname{n}(I)=4.$ Note
that $p=x^{7}+y^{7}$ is a superficial element for $I$ as
$e_{0}(I)=49=e_{0}(I/(p))$ and $e_{1}(I)=21=e_{1}(I/(p))$ with $\rho(I)=5.$
Here, $\operatorname{n}(I)=\rho(I)-1.$ Let $S=R/(p)$ and $I^{\prime}=I/(p).$
The Hilbert polynomial for $I^{\prime}$ is

$P_{I^{\prime}}(n)=49n-21.$

Then $P_{I^{\prime}}(0)\neq H_{I^{\prime}}(0)$, $P_{I^{\prime}}(1)\neq
H_{I^{\prime}}(1)$, $P_{I^{\prime}}(2)\neq H_{I^{\prime}}(2)$,
$P_{I^{\prime}}(3)\neq H_{I^{\prime}}(3)$ and
$P_{I^{\prime}}(n)=H_{I^{\prime}}(n)$ for all $n\geq 4.$ Thus,
$\operatorname{n}(I^{\prime})=3\leq\operatorname{n}(I)+1.$

We now answer Marley’s question 1.1 in a two dimensional Cohen-Macaulay ring.
We observe that both inequalities are possible under certain conditions.

######  Theorem 3.5.

Let $(R,\mathfrak{m})$ be a two dimensional Cohen-Macaulay local ring, $I$ an
$\mathfrak{m}$-primary ideal in $R$ and $x\in I$ be a superficial element for
$I$. If $\operatorname{n}(I)\neq\rho(I)-1,$ then
$\operatorname{rd}(I)\geq\operatorname{n}(I)+2.$

###### Proof.

Suppose $\operatorname{n}(I)\neq\rho(I)-1$. Set $S=R/(x)$ and
$I^{\prime}=I/(x).$ By proposition 3.1,
$\operatorname{n}(I^{\prime})\geq\operatorname{n}(I)+1$. Here, $S$ a one-
dimensional Cohen-Macaulay local ring, hence
$\operatorname{n}(I^{\prime})+1=\operatorname{rd}(I^{\prime}).$ Therefore,
$\operatorname{n}(I)+1\leq\operatorname{n}(I^{\prime})=\operatorname{rd}(I^{\prime})-1\leq\operatorname{rd}(I)-1$.
Thus, $\operatorname{rd}(I)\geq\operatorname{n}(I)+2.$ ∎

We recall example 3.3 to illustrate the above proposition.

######  Example 3.6.

[19, Example 5.1] Let $R=\mathbb{Q}[[x,y]]$ and
$I=(x^{6},x^{4}y,xy^{5},y^{6})$. Since $x^{3}y^{4}\in(I^{2}:I)$ but
$x^{3}y^{4}\notin I$, therefore $0pt\operatorname{gr}(I)=0.$ Note that
$J=(x^{4}y,x^{6}+xy^{5}+y^{6})$ is a minimal reduction for $I$ with
$\operatorname{rd}_{J}(I)=3\geq\operatorname{n}(I)+2.$ Yinghwa Wu [22, Theorem
3.3’] proved that if there exist a minimal reduction $J^{\prime}$ of $I$ such
that $\operatorname{rd}_{J^{\prime}}(I)\geq\operatorname{n}(I)+2,$ then
$\operatorname{rd}(I)$ is independent. Therefore,
$\operatorname{rd}(I)\geq\operatorname{n}(I)+2.$

######  Remark 3.7.

Yinghwa Wu [22, Remark 1.(b)] has proved that if $(R,\mathfrak{m})$ is a
Cohen-Macaulay ring of dimension two and $I$ is an $\mathfrak{m}$-primary
ideal such that $J=(x,y)$ is a minimal reduction of $I$ and $x$ a superficial
element for $I$ such that $\operatorname{rd}_{J}(I)\leq\operatorname{n}(I)+1$
then $\operatorname{n}(I)=\rho(I)-1.$ The next proposition gives us the
converse of the above-stated result.

######  Theorem 3.8.

Let $(R,\mathfrak{m})$ be a two dimensional Cohen-Macaulay local ring, and $I$
an $\mathfrak{m}$-primary ideal. Let $J=(x,y)$ be a minimal reduction of $I$
such that $x$ be a superficial element for $I$. If
$\operatorname{n}(I)=\rho(I)-1$, then
$\operatorname{rd}(I)\leq\operatorname{n}(I)+2.$

###### Proof.

As $J=(x,y)$ is the minimal reduction of $I$, we have
$\operatorname{rd}_{J^{\prime}}(I^{\prime})\leq\operatorname{rd}_{J}(I)$ where
$J^{\prime}=J/(x)$ and $I^{\prime}=I/(x).$ Since
$\operatorname{n}(I)=\rho(I)-1$, by Proposition 3.1,
$\operatorname{n}(I^{\prime})\leq\operatorname{n}(I)+1.$ Now $S=R/(x)$ is a
one dimensional Cohen-Macaulay ring, we have
$\operatorname{rd}(I^{\prime})=\operatorname{n}(I^{\prime})+1$ which implies
$\operatorname{rd}(I^{\prime})\leq\rho(I)+1.$ If possible let
$\operatorname{rd}_{J}(I)>\rho(I)+1.$ By [8, lemma 3.1], we have
$\widetilde{I^{\rho(I)+1}}\neq I^{\rho(I)+1}$ as
$\operatorname{rd}(I^{\prime})\leq\rho(I)+1<\operatorname{rd}_{J}(I)$, which
is a contradiction. Therefore, $\operatorname{rd}_{J}(I)\leq\rho(I)+1$ which
implies $\operatorname{rd}_{J}(I)\leq\operatorname{n}(I)+2.$ Also
$\operatorname{rd}(I)\leq\operatorname{rd}_{J}(I)$ for any minimal reduction
$J$ of $I$. Thus, $\operatorname{rd}(I)\leq\operatorname{n}(I)+2.$ ∎

The following example illustrates the above proposition.

######  Example 3.9.

[17, Example 4.6] Let $R=\mathbb{Q}[[x,y]]$ and
$I=(x^{7},x^{6}y,x^{2}y^{5},y^{7})$. Since $x^{5}y^{4}\in(I^{2}:I)$ but
$x^{5}y^{4}\notin I$, therefore, $\widetilde{I}\neq I.$ Thus,
$0pt\operatorname{gr}(I)=0.$ The Hilbert polynomial for $I$ is

$P_{I}(n)=49\dbinom{n+1}{2}-21n+3.$

Then $P_{I}(0)\neq H_{I}(0)$, $P_{I}(1)\neq H_{I}(1)$, $P_{I}(2)\neq
H_{I}(2)$, $P_{I}(3)\neq H_{I}(3)$ and $P_{I}(n)=H_{I}(n)$ for all $n\geq 4$.
Thus, $\operatorname{n}(I)=3.$ Here $p=x^{7}+y^{7}$ is a superficial element
for $I$ as $e_{0}(I)=49=e_{0}(I/(p)),e_{1}(I)=21=e_{1}(I/(p))$ and $\rho(I)=4$
which is equal to $n(I)+1.$ The reduction number of $I$ with respect to the
minimal reduction $J=(x^{7},y^{7})$ is
$\operatorname{rd}_{J}(I)=4\leq\operatorname{n}(I)+2.$ Therefore,
$\operatorname{rd}(I)\leq n(I)+2.$

As an immediate application of the above theorem, we give an alternative proof
for the proposition by Le Tuan Hoa regarding the independence of reduction
number.

######  Corollary 3.10.

[5, Proposition 3.7] [With the hypothesis as in 3.5] If
$\operatorname{n}(I)\neq\rho(I)-1$, then the reduction number is independent.

###### Proof.

Suppose $\operatorname{n}(I)\neq\rho(I)-1$, by Theorem 3.5, we have
$\operatorname{rd}(I)\geq\operatorname{n}(I)+2.$ Thus, the conclusion follows
from [22, Theorem 3.3]. ∎

######  Corollary 3.11.

[With the hypothesis as in 3.8] If $\operatorname{n}(I)=\rho(I)-1$, then the
reduction number is independent if
$\operatorname{rd}(I)=\operatorname{n}(I)+2$, otherwise if
$\operatorname{rd}(I)\leq\operatorname{n}(I)+2,$ then
$\operatorname{rd}_{J}(I)\leq\operatorname{n}(I)+2,$ for every minimal
reduction $J$ if $I.$ ∎

Next, we generalize Theorem 3.5 for a Cohen-Macaulay local ring of dimension
$d\geq 3$ with $0pt\operatorname{gr}(I)=d-2$ under certain conditions on $I$
and $\operatorname{n}(I).$ It is very well known from the literature that the
integral closure of an ideal behaves well reducing modulo a superficial
element. This was proved by Shiroh Itoh that:

######  Lemma 3.12.

[7, Page 648] If $I$ is an $\mathfrak{m}$-primary ideal which is integrally
closed in a Cohen–Macaulay local ring $(R,\mathfrak{m})$ of dimension $d\geq
2$, then (at least after passing to a faithfully flat extension) there exists
a superficial element $x\in I$ such that $I/(x)$ is integrally closed in
$R/(x)$.

In the next two propositions, we establish the relation between the
postulation number and reduction number in a Cohen-Macaulay local ring of
dimension $d\geq 3$ with $0pt\operatorname{gr}(I)=d-2.$

######  Proposition 3.13.

Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring of dimension $d\geq 3$
and $I$ be an $\mathfrak{m}$-primary ideal such that $I$ is integrally closed
in $R$ and $0pt\operatorname{gr}(I)=d-2.$ If $\operatorname{n}(I)=-(d-3),$
then $\operatorname{rd}(I)\geq\operatorname{n}(I)+d.$

###### Proof.

By Lemma 3.12, we choose a superficial sequence $x_{1},x_{2},\ldots,x_{d-2}$
for $I$ such that $I^{\prime}$ is integrally closed in $R^{\prime}$, where
$I^{\prime}=I/(x_{1},x_{2},\ldots,x_{d-2})$ and
$R^{\prime}=R/(x_{1},x_{2},\ldots,x_{d-2}).$ Since the
$0pt\operatorname{gr}(I)=d-2$, therefore,
$\operatorname{n}(I^{\prime})=\operatorname{n}(I)+d-2.$ Now
$\operatorname{n}(I)=-(d-3)$, implies $\operatorname{n}(I^{\prime})=1.$ Note
that $R^{\prime}$ is a two-dimensional Cohen-Macaulay ring with
$0pt\operatorname{gr}(I^{\prime})=0$. We claim that $\rho(I^{\prime})\geq 3$.
As $0pt\operatorname{gr}(I^{\prime})=0$, by [4, 1.3] $\rho(I^{\prime})>1.$
Suppose $\rho(I^{\prime})=2$, then $I^{\prime}\neq\widetilde{I^{\prime}}$, but
$I^{\prime}$ is integrally closed in $R^{\prime}$ which implies
$I^{\prime}=\widetilde{I^{\prime}}=\bar{I^{\prime}},$ this implies that
$I^{\prime n}=\widetilde{I^{\prime n}}$ for all $n\geq 1$. By [4, 1.3]
$0pt\operatorname{gr}(I^{\prime})>0,$ which is a contradiction. Therefore,
$\rho(I^{\prime})\geq 3$ and
$\operatorname{n}(I^{\prime})\neq\rho(I^{\prime})-1$. Hence, by Theorem 3.5,
$\operatorname{rd}(I^{\prime})\geq\operatorname{n}(I^{\prime})+2.$ Therefore,
$\operatorname{rd}(I)=\operatorname{rd}(I^{\prime})\geq\operatorname{n}(I^{\prime})+2=\operatorname{n}(I)+d.$
∎

######  Remark 3.14.

In the hypothesis of Proposition 3.13, $I$ being integrally closed in $R$ is a
necessary condition. Let $R=\mathbb{Q}[[x,y,z]]$ and
$I=(x^{3},y^{3},z^{3},xy^{2},x^{2}z,xz^{2},y^{2}z,yz^{2},xyz)$. $I$ is not
integrally closed as $x^{2}y\in\bar{I}$ but $x^{2}y\notin I.$ Note that
$0pt\operatorname{gr}(I)=1$ by [16, Proposition 1.12]). The Hilbert polynomial
for $I$ is

$P_{I}(n)=27\dbinom{n+2}{3}-18\dbinom{n+1}{2}+n+1.$

Here $P_{I}(0)\neq H_{I}(0)$, so $\operatorname{n}(I)=0.$ Using Macaulay 2,
$K=(\frac{5}{3}x^{3}+6xy^{2}+\frac{7}{6}y^{3}+\frac{1}{10}x^{2}z+xyz+\frac{3}{10}y^{2}z+\frac{7}{4}xz^{2}+\frac{1}{4}yz^{2}+z^{3},\frac{6}{5}x^{3}+\frac{1}{4}xy^{2}+\frac{3}{5}y^{3}+\frac{1}{4}x^{2}z+\frac{3}{8}xyz+\frac{6}{5}y^{2}z+\frac{3}{10}xz^{2}+\frac{7}{9}yz^{2}+z^{3},\frac{1}{6}x^{3}+\frac{2}{3}xy^{2}+5y^{3}+\frac{1}{5}x^{2}z+\frac{7}{8}xyz+\frac{3}{4}y^{2}z+xz^{2}+\frac{5}{3}yz^{2},3z^{3})$
is a minimal reduction of $I$ and
$2=\operatorname{rd}_{K}(I)<\operatorname{n}(I)+3$ also
$\operatorname{rd}_{K}(I)\geq\operatorname{rd}(I),$ thus
$\operatorname{rd}(I)<\operatorname{n}(I)+3.$

######  Corollary 3.15.

[With the hypothesis as in 3.13] If $\operatorname{n}(I)=-(d-3),$ then
reduction number is independent.

###### Proof.

As $\operatorname{n}(I)=-(d-3),$ by Proposition 3.13 we have
$\operatorname{rd}(I)\geq\operatorname{n}(I)+d.$ Then the conclusion follows
from [22, Theorem 4.3]. ∎

######  Proposition 3.16.

Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring of dimension $d\geq 3$
and $I$ be an $\mathfrak{m}$-primary ideal such that
$0pt\operatorname{gr}(I)=d-2.$ If $\operatorname{n}(I)<-(d-3),$ then
$\operatorname{rd}(I)\geq\operatorname{n}(I)+d.$

###### Proof.

Let $x_{1},x_{2},\ldots,x_{d-2}$ be a superficial sequence for $I$ such that
$I^{\prime}=I/(x_{1},\ldots,x_{d-2})$ and
$R^{\prime}=R/(x_{1},\ldots,x_{d-2}).$ Since the
$0pt\operatorname{gr}(I)=d-2$, therefore,
$\operatorname{n}(I^{\prime})=\operatorname{n}(I)+d-2.$ It is given that
$\operatorname{n}(I)<-(d-3)$ which implies $\operatorname{n}(I^{\prime})<1.$
Since $0pt\operatorname{gr}(I^{\prime})=0$, by [4, 1.3] we have
$\rho(I^{\prime})>1.$ Which implies
$\operatorname{n}(I^{\prime})\neq\rho(I^{\prime})-1.$ Hence, by Theorem 3.5,
we have $\operatorname{rd}(I^{\prime})\geq\operatorname{n}(I^{\prime})+2.$
Therefore,
$\operatorname{rd}(I)=\operatorname{rd}(I^{\prime})\geq\operatorname{n}(I^{\prime})+2=\operatorname{n}(I)+d.$
∎

######  Corollary 3.17.

[With the hypothesis as in 3.16] If $\operatorname{n}(I)<-(d-3),$ then
reduction number is independent. ∎

##  4\. Ratliff-Rush filtration mod a superficial element

Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring of dimension $d$. Let
$x\in I$ be a superficial element for $I,$ set $S=R/(x).$ Consider the natural
map $\pi:R\longrightarrow S.$ It is easy to see that
$\pi(\widetilde{I^{n}})\subseteq\widetilde{\frac{I^{n}+(x)}{(x)}}$ for all
$n\geq 1.$ We briefly recall the good behaviour of the Ratliff-Rush filtration
with respect to $I$ mod a superficial element from the literature [14, Section
2]. The Ratliff-Rush filtration of $I$ behaves well mod $x$ if
$\pi(\widetilde{I^{n}})=\widetilde{\frac{I^{n}+(x)}{(x)}}$ for all $n\geq 1,$
i.e., $\frac{\widetilde{I^{n}}+(x)}{(x)}=\widetilde{\frac{I^{n}+(x)}{(x)}}$
for all $n\geq 1.$ This natural map $\pi$ induces the map
$\pi_{n}:\frac{\widetilde{I^{n}}}{I^{n}}\longrightarrow\frac{\widetilde{\frac{I^{n}+(x)}{(x)}}}{\frac{I^{n}+(x)}{(x)}}$
for all $n\geq 1.$ And for a fixed $n\geq 1,$
$\frac{\widetilde{I^{n}}+(x)}{(x)}=\widetilde{\frac{I^{n}+(x)}{(x)}}$ if and
only in $\pi_{n}$ is surjective.

Define
$\rho(I/(x)):=\min\\{k\in\mathbb{N}:\frac{I^{n}+(x)}{(x)}=\widetilde{\frac{I^{n}+(x)}{(x)}}$
for all $n\geq k\\}$. Then for all $n\geq\rho(I/(x))$,
$\frac{\widetilde{I^{n}}+(x)}{(x)}=\widetilde{\frac{I^{n}+(x)}{(x)}}.$
Therefore, $\pi_{n}$ is surjective for all $n\geq\rho(I/(x))$. Define
$\omega(I):=\min\\{k\in\mathbb{N}:\pi_{n}$ is surjective for all $n\geq k\\}$.
We call this $\omega(I)$ as the surjectivity index of the Ratliff-Rush
filtration with respect to $I.$ Note that if the Ratliff-Rush filtration with
respect to $I$ mod a superficial element behaves well then $\omega(I)=1.$

######  Proposition 4.1.

Let $(R,\mathfrak{m})$ be a Cohen-Macaulay ring of dimension $d\geq 2,$ $I$ be
an $\mathfrak{m}$-primary ideal and $x\in I$ be a superficial element for $I.$
If $b^{I}_{1}\neq-1$, then
$\omega(I)=\max\\{1,b_{1}^{I}(R)+2\\}\leq\max\\{1,a_{2}(\operatorname{gr}(I))+1\\}.$

###### Proof.

Set $S=R/(x)$, $\mathfrak{M}=\mathfrak{M}_{\mathcal{R}(I)}$. Using the short
exact sequence of $\mathcal{R}(I)$-modules [13, 5]

$0\longrightarrow\operatorname{gr}(I)\longrightarrow L^{I}(R)\longrightarrow
L^{I}(R)(-1)\longrightarrow 0$

and the corresponding long exact sequence in local cohomolgy, we have that
$b_{i}^{I}(R)\leq a_{i+1}(\operatorname{gr}(I))-1$ for $i=0,1,\ldots,d-1$.
From [14, 2.1] note that, $\pi$ induces a natural $\mathcal{R}(I)$-linear map
$\pi^{*}:L^{I}(R)\longrightarrow L^{I/(x)}(S).$ This induces a map on the
local cohomology modules

$H^{0}\left(\pi^{*}\right):H^{0}\left(L^{I}(R)\right)\longrightarrow
H^{0}\left(L^{I/(x)}(S)\right)$

By [13, proposition 4.7] $H^{0}\left(\pi^{*}\right)_{n}=\pi_{n}$ for all
$n\in\mathbb{Z}.$ Note that $H^{1}\left(L^{I}(R)\right)_{n}=0$ for all $n\geq
b_{1}^{I}(R)+1$. Using the the second fundamental exact sequence [13, 6.2] and
[13, 6.3] we get,

$H^{0}\left(L^{I}(R)\right)_{n}\xrightarrow{\pi_{n}}H^{0}\left(L^{I/(x)}(S)\right)_{n}\longrightarrow
0,$ for all $n\geq b_{1}^{I}(R)+2$

Thus, $\pi_{n}$ is surjective for all $n\geq b_{1}^{I}(R)+2$, which implies
$\frac{\widetilde{I^{n}}+(x)}{(x)}=\widetilde{\frac{I^{n}+(x)}{(x)}}$ for all
$n\geq b_{1}^{I}(R)+2.$ Therefore, $\omega(I)\leq\max\\{1,b_{1}^{I}(R)+2\\}.$

Let if possible $b_{1}^{I}+1\geq\omega(I).$ Then, $\pi_{n}$ is surjective for
all $n\geq b_{1}^{I}+1.$ By the long exact sequence in local cohomology in
(4), we get that the $\mathcal{R}(I)$-linear map

$0\longrightarrow\left(H^{1}\left(L^{I}(R)\right)\right)_{n-1}\longrightarrow\left(H^{1}\left(L^{I}(R)\right)\right)_{n}$
is injective for all $n\geq b_{1}^{I}+1.$

Therefore, at $n=b_{1}^{I}+1$, we have
$\left(H^{1}\left(L^{I}(R)\right)\right)_{n}=0,$ which implies that
$\left(H^{1}\left(L^{I}(R)\right)\right)_{b_{1}^{I}}=0,$ which is a
contradiction. Thus, $b_{1}^{I}+1<\omega(I),$ which implies
$b_{1}^{I}+2\leq\omega(I).$ Therefore,
$\omega(I)=\max\\{1,b_{1}^{I}(R)+2\\}\leq\max\\{1,a_{2}(\operatorname{gr}(I))+1\\}.$
∎

As an immediate corollary we get

######  Corollary 4.2.

[With the hypothesis as in 4.1] If $\rho(I)\geq\omega(I),$ then
$\rho(I)\geq\rho(I/(x)).$

###### Proof.

Since $\rho(I)\geq\omega(I)$, we have
$\frac{\widetilde{I^{n}}+(x)}{(x)}=\widetilde{\frac{I^{n}+(x)}{(x)}}$ for all
$n\geq\rho(I).$ Also,
$\frac{I^{n}+(x)}{(x)}=\frac{\widetilde{I^{n}}+(x)}{(x)}$ for all
$n\geq\rho(I).$ Therefore,
$\frac{I^{n}+(x)}{(x)}=\widetilde{\frac{I^{n}+(x)}{(x)}}$ for all
$n\geq\rho(I).$ Thus, $\rho(I)\geq\rho(I/(x)).$ ∎

######  Corollary 4.3.

Let $(R,\mathfrak{m})$ be a Cohen-Macaulay ring of dimension two, $I$ an
$\mathfrak{m}$-primary ideal and $x\in I$ be a superficial element for $I.$ If
$\rho(I)\geq\operatorname{rd}(I)-1,$ then $\rho(I)\geq\rho(I/(x)).$

###### Proof.

Suppose $\rho(I)\geq\operatorname{rd}(I)-1.$ Then by [9, Lemma 1.2], we have
$a_{2}(\operatorname{gr}(I))+2\leq\operatorname{rd}(I).$ This implies
$\rho(I)\geq\max\\{1,a_{2}(\operatorname{gr}(I))+1\\}\geq\omega(I).$
Therefore, by Corollary 4.2 it follows that $\rho(I)\geq\rho(I/(x)).$ ∎

######  Remark 4.4.

As a particular case, if the Ratliff-Rush filtration with respect to $I$ mod a
superficial element $x$ behaves well then $\omega(I)=1$. Also, $\rho(I)\geq
1$, thus from Corollary 4.2, we have $\rho(I/(x))\leq\rho(I).$

The following example illustrates Corollary 4.3.

######  Example 4.5.

[15, Exapmle 3.3] Let $R=\mathbb{Q}[[x,y]]$ and
$I=(x^{4},x^{3}y,xy^{3},y^{4}).$ Since $x^{2}y^{2}\in(I^{2}:I)$ and
$x^{2}y^{2}\notin I$, therefore, $\tilde{I}\not=I$ and hence
$0pt\operatorname{gr}(I)=0.$ Note that $p=x^{4}+y^{4}$ is the superficial
element for $I$, as $e_{0}(I)=16=e_{0}(I/(p))$ and $e_{1}(I)=6=e_{1}(I/(p))$
and $J=(x^{4},y^{4})$ is a minimal reduction of $I.$ Using Macaulay 2, we get
that, $\rho(I)=2$ and $\operatorname{rd}(I)=2.$ This implies
$\rho(I)>\operatorname{rd}(I)-1.$ Also $\operatorname{rd}_{J/(p)}(I/(p))=2,$
by [16, Proposition 4.2]
$\rho(I/(p))\leq\operatorname{rd}_{J/(p)}(I/(p))=\operatorname{rd}(I)<\rho(I)+1.$
Thus, $\rho(I/(p))\leq\rho(I).$

We end this section with a corollary, which gives us the class of
$\mathfrak{m}$-primary ideals in a two dimensional Cohen-Macaulay local ring
for which the Ratliff-Rush filtration of $I$ behaves well mod a superficial
element.

######  Corollary 4.6.

Let $(R,\mathfrak{m})$ be a Cohen-Macaulay ring of dimension two, $I$ an
$\mathfrak{m}$-primary ideal such that $I$ is integrally closed and $x\in I$
be a superficial element for $I.$ If $\operatorname{rd}(I)=2,$ then the
Ratliff-Rush filtration of $I$ behaves well mod x.

###### Proof.

From [9, Lemma 1.2] we observe that
$a_{2}(\operatorname{gr}(I))+2\leq\operatorname{rd}(I)$, also by Proposition
4.1 we have, $\omega(I)\leq\max\\{1,a_{2}(\operatorname{gr}(I))+1\\}$, which
implies $\omega(I)\leq\max\\{1,\operatorname{rd}(I)-1\\}$. Since
$\operatorname{rd}(I)=2$, thus $\omega(I)\leq 1$, which implies $\omega(I)=1$.
Therefore, the Ratliff-Rush filtration of $I$ behaves well mod $x.$ ∎

##  5\. Ratliff-Rush filtration mod a superficial sequence

In this section, we extend the theorem by Marley [10, Theorem 2.3] in a Cohen-
Macaulay local ring of dimension $d$. We observe that the stronger condition
on the depth of the associated graded ring can be replaced by the weaker
condition on the good behaviour of the Ratliff-Rush filtration with respect to
$I$. First, we recall the behaviour of the Ratliff-Rush filtration of an ideal
$I$ mod a superficial sequence [14, Section 4].

Assume $s\leq d-1$. Let $x_{1},x_{2},\ldots,x_{s}\in I/I^{2}$ be a superficial
sequence for $I$. Set $R_{i}=R/(x_{1},x_{2},\ldots,x_{i})$ for all
$i=1,2,\ldots,s.$ We have the natural map of quotients
$\pi_{i,j}:R_{i}\longrightarrow R_{j}$, for every $0\leq i<j\leq s.$ Note that
$\pi_{i,j}\left(\widetilde{I^{n}R_{i}}\right)\subseteq\widetilde{I^{n}R_{j}}$
for all $n\geq 1.$

This map induces the following map:
$\pi_{i,j}^{n}:\frac{\left(\widetilde{I^{n}R_{i}}\right)}{\left(I^{n}R_{i}\right)}\longrightarrow\frac{\left(\widetilde{I^{n}R_{j}}\right)}{\left(I^{n}R_{j}\right)}$
for every $n\geq 1.$ For a fixed $n\geq 1,$ we observe that
$\pi_{i,j}\left(\widetilde{I^{n}R_{i}}\right)=\widetilde{I^{n}R_{j}}$ if and
only in $\pi_{i,j}^{n}$ is surjective. For $0\leq i<j<k\leq s$, we have the
following commutative diagram:

${{\widetilde{I^{n}R_{i}}/I^{n}R_{i}}}$${{\widetilde{I^{n}R_{j}}/I^{n}R_{j}}}$${{\widetilde{I^{n}R_{k}}/I^{n}R_{k}}}$$\scriptstyle{\pi_{i,j}^{n}}$$\scriptstyle{\pi_{i,k}^{n}}$$\scriptstyle{\pi_{j,k}^{n}}$

######  Remark 5.1.

If $\pi_{i,j}^{n}$ and $\pi_{j,k}^{n}$ are surjective then $\pi_{i,k}^{n}$ is
surjective.

######  Definition 5.2.

[14, Definition 4.4] Ratliff-Rush filtration of $I$ behaves well mod
superficial sequence $x_{1},x_{2},\ldots,x_{s}$ if
$\pi_{0,s}(\widetilde{I^{n}})=\widetilde{I^{n}R_{s}}$ for all $n\geq 1.$

######  Definition 5.3.

Let $x_{1},x_{2},\ldots,x_{s}\in I/I^{2}$ be a superficial sequence for $I$,
then for each $i=1,2,\ldots,s$

$\rho\left(I/(x_{1},x_{2},\ldots,x_{i})\right):=\min\\{k\in\mathbb{N}:I^{n}R_{i}=\widetilde{I^{n}R_{i}}$
for all $n\geq k\\}.$

######  Proposition 5.4.

Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring of dimension $d\geq 2,$
$I$ an $\mathfrak{m}$-primary ideal and $x_{1},x_{2},\ldots,x_{s}\in I/I^{2}$
be a superficial sequence for $I.$ Then $\pi_{0,s}^{n}$ is surjective for all
$n\geq\max\\{1,b^{I}+2\\}.$

###### Proof.

We will prove by induction on $s.$ If $s=1,$ we have proved in Proposition
4.1. Let $s\geq 2,$ we assume that $\pi_{0,s-1}^{n}$ is surjective for all
$n\geq b^{I}+2,$ and will prove that $\pi_{0,s}^{n}$ is surjective for all
$n\geq b^{I}+2.$

By $s=1$ case $\pi_{0,1}^{n}$ is surjective for all $n\geq b^{I}+2.$ Using the
second fundamental exact sequence (3) and the corresponding long exact
sequence in local cohomolgy in (4), we get
$H^{i}\left(L^{I/(x)}(R/(x))\right)_{n}=0$ for all $n\geq b^{I}+2$ and for all
$i=1,2,\ldots,s-1.$ Thus, by induction hypothesis $\pi_{1,s}^{n}$ is
surjective for all $n\geq b^{I}+2.$ Now, using Remark 5.1, and the commutative
diagram, we have $\pi_{0,s}^{n}$ is surjective for all $n\geq b^{I}+2.$ ∎

As an easy consequence of the preceding proposition, we have the following
corollary.

######  Corollary 5.5.

[With the hypothesis as in 5.4] If $\rho(I)\geq b^{I}+2$, then
$\rho(I)\geq\rho\left(\frac{I}{(x_{1},x_{2,}\ldots,x_{s})}\right).$

###### Proof.

For all $n\geq\rho(I),$ we have $I^{n}=\widetilde{I^{n}}.$ Since $\rho(I)\geq
b^{I}+2,$ therefore, $\pi_{0,s}^{n}$ is surjective for all $n\geq\rho(I),$
which implies $\pi_{0,s}(\widetilde{I^{n}})=\widetilde{I^{n}R_{s}}$ for all
$n\geq\rho(I).$ Thus, $\pi_{0,s}(I^{n})=\widetilde{I^{n}R_{s}}$ for all
$n\geq\rho(I),$ which implies
$\rho(I)\geq\rho\left(\frac{I}{(x_{1},x_{2,}\ldots,x_{s})}\right).$ ∎

As a generalization of Marley’s result [10, Theorem 2.3], we now establish
Theorem 5.7. Before we prove the following lemma.

######  Lemma 5.6.

Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring of dimension $d\geq 2,$
$I$ an $\mathfrak{m}$-primary ideal. If the Ratliff-Rush filtration with
respect to $I$ behaves well mod the superficial sequence
$x_{1},x_{2},\ldots,x_{d-1}\in I,$ then

$\rho(I)\geq\rho\left(I/(x_{1})\right)\geq\ldots\geq\rho\left(I/(x_{1},x_{2},\ldots,x_{d-1})\right).$

###### Proof.

We prove by induction on $d$. Suppose $d=2.$ Let $x_{1}\in I$ be superficial
for $I.$ The Ratliff-Rush filtration with respect to $I$ behaves well mod
$x_{1},$ thus,
$\frac{\widetilde{I^{n}}+(x_{1})}{(x_{1})}=\frac{\widetilde{I^{n}+(x_{1})}}{(x_{1})}$
for all $n\geq 1.$ Also $\widetilde{I^{n}}=I^{n}$ for all $n\geq\rho(I)$.
Therefore,
$\frac{I^{n}+(x_{1})}{(x_{1})}=\frac{\widetilde{I^{n}+(x_{1})}}{(x_{1})}$ for
all $n\geq\rho(I),$ which implies $\rho\left(I/(x)\right)\leq\rho(I).$

We assume the lemma for $d-2$ and prove for $d-1.$ Since the Ratliff-Rush
filtration with respect to $I$ behaves well mod the superficial sequence
$x_{1},x_{2},\ldots,x_{d-1}\in I,$ then $\pi_{d-2,d-1}$ is surjective.
Therefore,
$\pi_{d-2,d-1}\left(\widetilde{I^{n}R_{d-2}}\right)=\widetilde{I^{n}R_{d-1}}$
for all $n\geq 1.$ Also $I^{n}R_{d-2}=\widetilde{I^{n}R_{d-2}}$ for all
$n\geq\rho\left(I/(x_{1},x_{2},\ldots,x_{d-2})\right).$ Thus,
$I^{n}R_{d-1}=\widetilde{I^{n}R_{d-1}}$ for all
$n\geq\rho\left(I/(x_{1},x_{2},\ldots,x_{d-2})\right).$ Therefore,
$\rho\left(I/(x_{1},x_{2},\ldots,x_{d-2})\right)\geq\rho\left(I/(x_{1},x_{2},\ldots,x_{d-1})\right).$
∎

Now we generalize the proof of [10, Theorem 2.3] by relaxing the condition on
the depth of the associated graded ring.

######  Theorem 5.7.

Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring of dimension $d\geq 2,$
$I$ an $\mathfrak{m}$-primary ideal. Let the Ratliff-Rush filtration with
respect to $I$ behave well mod a superficial sequence
$x_{1},x_{2},\ldots,x_{d-2}\in I.$ Then for all $0\leq i\leq d$ and for all
$n\geq\rho(I)$

$(-1)^{d-i}\Delta^{i}(P_{I}(n)-H_{I}(n))\geq 0.$

###### Proof.

Note that it suffices to prove the case $i=d.$ Let $g(n)=P_{I}(n)-H_{I}(n).$

Suppose $(-1)^{d-i}\Delta^{i}(g(n))\geq 0$ for all $n\geq\rho(I)$ and for some
$i>0.$ Then
$(-1)^{d-i}\Delta^{i}(g(n))=\Delta^{1}[(-1)^{d-i}\Delta^{i-1}(g(n))]\geq 0$
for all $n\geq\rho(I).$ However, $g(n)=0$ for sufficiently large $n$ so
$(-1)^{d-i}\Delta^{i-1}(g(n))=0$ for sufficiently large $n.$ By the property
of difference function we have $(-1)^{d-i}\Delta^{i-1}(g(n))\leq 0$ for all
$n\geq\rho(I).$ Therefore, $(-1)^{d-(i-1)}\Delta^{i-1}(g(n))\geq 0$ for all
$n\geq\rho(I),$ which gives the theorem for $i-1.$ Hence it is enough to prove
$\Delta^{d}(P_{I}(n)-H_{I}(n))\geq 0$ for all $n\geq\rho(I).$

We prove this by induction on d.

Case-1: Suppose $d=2.$ Let $x\in I$ be a superficial element for $I.$ Set
$S=R/(x)$ and $I^{\prime}=I/(x),$ then $S$ is a one dimensional Cohen-Macaulay
ring. By Lemma 2.4 and Proposition 3.1, we have,

$H_{I^{\prime}}(n+1)=H_{I}(n+1)-H_{I}(n)$ for all $n\geq\rho(I)$

$P_{I^{\prime}}(n+1)=P_{I}(n+1)-P_{I}(n)$ for all $n$

Consider

$\begin{split}\Delta^{2}(P_{I}(n)-H_{I}(n))&=\Delta^{1}[\Delta^{1}(P_{I}(n)-H_{I}(n))]\\\
&=\Delta^{1}[(P_{I}(n+1)-P_{I}(n))-(H_{I}(n+1)-H_{I}(n))]\\\
&=\text{$\Delta^{1}[P_{I^{\prime}}(n+1)-H_{I^{\prime}}(n+1)]$ for all
$n\geq\rho(I)$}\\\ &\geq\text{0 for all $n\geq\rho(I)$}.\end{split}$

Case-2: Suppose $d>2.$ Set $S=R/(x_{1})$ and $I_{1}=I/(x_{1})$. Since the
Ratliff-Rush filtration of $I$ behaves well mod a superficial sequence
$x_{1},x_{2},\ldots,x_{d-2}\in,$ by Lemma 5.6, we have
$\rho(I)\geq\rho\left(I/(x_{1})\right)\geq\ldots\geq\rho\left(I/(x_{1},x_{2},\ldots,x_{d-2})\right).$
We assume $\Delta^{d-1}(P_{I_{1}}(n)-H_{I_{1}}(n))\geq 0$ for all
$n\geq\rho(I_{1}).$

Consider

$\begin{split}\Delta^{d}(P_{I}(n)-H_{I}(n))&=\Delta^{d-1}[\Delta^{1}(P_{I}(n)-H_{I}(n))]\\\
&=\Delta^{d-1}[(P_{I}(n+1)-P_{I}(n))-(H_{I}(n+1)-H_{I}(n))]\\\
&=\text{$\Delta^{d-1}[P_{I_{1}}(n+1)-H_{I_{1}}(n+1)]$ for all
$n\geq\rho(I)$}\\\ &\geq\text{0 for all $n\geq\rho(I)-1$}.\end{split}$

Therefore, we have $\Delta^{d}(P_{I}(n)-H_{I}(n))\geq 0$ for all
$n\geq\rho(I).$ ∎

The following corollary partially answers Marley’s Question 1.6.

######  Corollary 5.8.

[With the hypothesis as in 5.7] If $P_{I}(k)=H_{I}(k)$ for some
$k\geq\rho(I)$, then $P_{I}(n)=H_{I}(n)$ for all $n\geq k.$

###### Proof.

By Proposition, 5.7 we have that

$(-1)^{d-1}\Delta^{1}(P_{I}(n)-H_{I}(n))\geq 0$ for all $n\geq\rho(I).$

Hence

$(-1)^{d-1}(P_{I}(n+1)-H_{I}(n+1))\geq(-1)^{d-1}(P_{I}(n)-H_{I}(n))$ for all
$n\geq\rho(I).$

Also $P_{I}(n)-H_{I}(n)=0$ for sufficiently large $n.$ Thus we get

$0\geq(-1)^{d-1}(P_{I}(n)-H_{I}(n))\geq(-1)^{d-1}(P_{I}(k)-H_{I}(k))$ for all
$n\geq k.$

It is given that $P_{I}(k)=H_{I}(k)$ for some $k\geq\rho(I),$ then
$P_{I}(n)=H_{I}(n)$ for all $n\geq k.$ ∎

Consequently, as a corollary we obtain Marley’s result.

######  Corollary 5.9.

Let $(R,\mathfrak{m})$ be a Cohen-Macaulay local ring of dimension $d\geq 2,$
$I$ an $\mathfrak{m}$-primary ideal such that $0pt\operatorname{gr}(I)\geq
d-1.$ Then for all $0\leq i\leq d$ and for all $n\geq 1$

$(-1)^{d-i}\Delta^{i}(P_{I}(n)-H_{I}(n))\geq 0.$

###### Proof.

Since the $0pt\operatorname{gr}(I)\geq d-1,$ from the proof of 5.7 and [4,
1.3] we have
$\rho(I)=\rho\left(I/(x_{1})\right)=\ldots=\rho\left(I/(x_{1},x_{2},\ldots,x_{d-2})\right)=1.$
Thus, $\Delta^{d}(P_{I}(n)-H_{I}(n))\geq 0$ for all $n\geq 1.$

∎

## Acknowledgement

The authors would like to thank Prof. T. J. Puthenpurakal and Prof.
Ramakrishna Nanduri for their helpful discussions and valuable suggestions.

## References

  * [1] M.P. Brodmann, R.Y. Sharp, Local Cohomology: An Algebraic Introduction with Geometric Applications, Cambridge Studies in Advanced Mathematics, vol. 60, Cambridge University Press, Cambridge, 1998.
  * [2] CoCoATeam, CoCoA: a system for doing computations in commutative algebra (2008). Available at http:// cocoa.dima.unige.it
  * [3] D.R. Grayson, M.E. Stillman, Macaulay2, a software system for research in algebraic geometry, available at http:// www .math .uiuc .edu /Macaulay2/.
  * [4] W. Heinzer, B. Johnston, D. Lantz and K. Shah, Coefficient ideals in and blowups of a commutative Noetherian domain, J. Algebra, 162 (1993), 355-391.
  * [5] Le Tuan Hoa, A note on the Hilbert-Samuel function in a two-dimensional local ring, Acta Math. Vietnam. 21 (1996), no. 2, 335–347.
  * [6] S. Huckaba, Reduction numbers for ideals of higher analytic spread, Proc. Camb. Phil. Soc. 102 (1987), 49-57.
  * [7] S. Itoh, Hilbert coefficients of integrally closed ideals, J. Algebra 176 (1995), 638–652.
  * [8] M. Mandal and K. Saloni, Bounds for the reduction number of primary ideal in dimension three, arXiv:2209.13319 [math.AC].
  * [9] T. Marley, The reduction number of an ideal and the local cohomology of the associated graded ring, Proc. Amer. Math. Soc. 117 (1993), 335–341.
  * [10] T. J. Marley, Hilbert functions of Ideals in Cohen-Macaulay Rings, Ph.D Thesis.
  * [11] D. G. Northcott, A note on the coefficients of the abstract Hilbert functions, J. London Math. Soc. 35 (1960), 209-214.
  * [12] A. Ooishi, Genera and arithmetical genera of local rings, Hiroshima Math. J. 17 (1987), 47-66.
  * [13] Tony J. Puthenpurakal, Ratliff–Rush filtration, regularity and depth of higher associated graded modules. I, J. Pure Appl. Algebra 208(1)(2007), 159–176.
  * [14] Tony J. Puthenpurakal, Ratliff–Rush filtration, regularity and depth of higher associated graded modules. II, J. Pure Appl. Algebra 221(3)(2017), 611–631.
  * [15] M.E. Rossi and G. Valla, Hilbert functions of filtered modules, Lecture Notes of the Unione Matematica Italiana, 9. Springer-Verlag, Berlin; UMI, Bologna, 2010. xviii+100 pp.
  * [16] M. E. Rossi and I. Swanson, Notes on the behavior of the Ratliff-Rush filtration, Contemp. Math. 331 (2001), 313-328.
  * [17] M. E. Rossi,D. T. Trung, N. V. Trung,  Castelnuovo-Mumford regularity and Ratliff-Rush closure, J.Algebra 504(2018), 568–586.
  * [18] J. D. Sally, Reductions, local cohomology and Hilbert functions of local ring, Commutative Algebra: Durham 1981, London Math. Soc. Lecture Notes Series no. 72 (Cambridge University Press, 1982), 231-241.
  * [19] J. D. Sally, Ideals Whose Hilbert Function and Hilbert Polynomial Agree at n = 1, J. Algebra 157 (1993), 534–547.
  * [20] K. Saloni and Anoot Kr. Yadav, Ratliff-Rush filtration, Hilbert coefficients and the reduction number of integrally closed ideals, arXiv:2304.04524 [math.AC].
  * [21] N. V. Trung. Reduction exponent and degree bound for the defining equations of graded rings, Proc. Amer. Math. Soc. 101 (1987), 229-236.
  * [22] Y. Wu, Reduction numbers and Hilbert polynomials of ideals in higher dimensional Cohen-Macaulay local rings, Mathematical Proceedings of the Cambridge Philosophical Society (1992), 111(1), 47-56.