On the Hilbert coefficients, depth of associated graded rings and reduction numbers

TL;DR

This paper investigates the relationships between Hilbert coefficients, the depth of associated graded rings, and reduction numbers in Cohen-Macaulay local rings, providing new bounds and conditions for their behavior.

Contribution

It establishes new bounds on the depth of associated graded rings based on Hilbert coefficients and explores conditions under which reduction numbers are independent.

Findings

Depth of associated graded ring is bounded below by a function of Hilbert coefficients.

Specific formulas relate $e_2(I)$ and $e_1(I)$ to the structure of the ideal.

Reduction number $r(I)$ is shown to be independent under certain conditions.

Abstract

Let $(R,\mathfrak{m})$ be a $d$-dimensional Cohen-Macaulay local ring, $I$ an $\mathfrak{m}$-primary ideal of $R$ and $J=(x_1,...,x_d)$ a minimal reduction of $I$. We show that if $J_{d-1}=(x_1,...,x_{d-1})$ and $\sum\limits_{n=1}^\infty\lambda{({I^{n+1}\cap J_{d-1}})/({J{I^n} \cap J_{d-1}})=i}$ where i=0,1, then depth $G(I)\geq{d-i-1}$. Moreover, we prove that if $e_2(I) = \sum_{n=2}^\infty (n-1) \lambda (I^n/JI^{n-1})-2;$ or if $I$ is integrally closed and $e_2(I) = \sum_{n=2}^\infty (n-1)\lambda({{I^{n}}}/JI^{n-1})-i$ where $i=3,4$, then $e_1(I) = \sum_{n=1}^\infty \lambda(I^n / JI^{n-1})-1.$ In addition, we show that $r(I)$ is independent. Furthermore, we study the independence of $r(I)$ with some other conditions.

Equations

$$P_I(n)=\sum _{i=0}^d(-1{)}^i{e}_i(I)(\genfrac{}{}{0ex}{}{n+d-i-1}{d-i}),$$

$$0⟶R/I_{n-1}\stackrel{x_1}{⟶}R/I_n⟶0.$$

$$0⟶R/I_{n-l}⟶(R/I_{n-l+1}{)}^l⟶(R/I_{n-l+2}{)}^{\left(\genfrac{}{}{0ex}{}{l}{2}\right)}⟶...⟶R/I_n⟶0.$$

$$0⟶{C_{.}}^{\prime }(n)⟶C_{.}(n)⟶{C_{.}}^{\prime }(n-1)[-1]⟶0.$$

$$\cdots ⟶H_i({C_{.}}^{\prime }(n))⟶H_i(C_{.}(n))⟶H_{i-1}({C_{.}}^{\prime }(n-1))\stackrel{x_l}{⟶}{H}_{i-1}({C_{.}}^{\prime }(n))⟶\cdots .(\ast )$$

$$h_i(\underline{x},\mathcal{F}):=\sum _{n=1}^{\infty }\lambda (H_i(C_{.}(\underline{x},\mathcal{F},n)))$$

$$k_i(\underline{x},\mathcal{F}):=\sum _{n=2}^{\infty }(n-1)\lambda (H_i(C_{.}(\underline{x},\mathcal{F},n))).$$

$$h_i^{\prime }({\underline{x}}^{\prime },\mathcal{F}):=\sum _{n=1}^{\infty }\lambda (H_i(C_{.}({\underline{x}}^{\prime },\mathcal{F},n))).$$

$$0⟶H_l(C_{.}(n))⟶...⟶H_i({C_{.}}^{\prime }(n-1))\stackrel{x_l}{⟶}{H}_i({C_{.}}^{\prime }(n))⟶B_n⟶0.$$

$$\lambda (B_n)=\sum _{j=i+1}^l{(-1)}^{j-i-1}\lambda (H_j(C_{.}(n)))+\sum _{j=i}^l{(-1)}^{j-i}\Delta [\lambda (H_j({C_{.}}^{\prime }(n)))]$$

$$\sum _{n=2}^{\infty }(n-1)\Delta [\lambda (H_j({C^{\prime }}_{.}(n)))]=-\sum _{n=1}^{\infty }\lambda (H_j({C^{\prime }}_{.}(n)))=-h_j^{\prime }({\underline{x}}^{\prime },\mathcal{F}).$$

$$\sum _{n=2}^{\infty }(n-1)\lambda (B_n)=\sum _{j=i+1}^l{(-1)}^{j-i-1}{k}_j(\underline{x},\mathcal{F})-\sum _{j=i}^l{(-1)}^{j-i}{h^{\prime }}_j({\underline{x}}^{\prime },\mathcal{F}).(\ast \ast )$$

$$\sum _{j=i+1}^l{(-1)}^{j-i-1}{k}_j(\underline{x},\mathcal{F})=0.$$

$${\Delta }^d[P_{\mathcal{F}}(n)-H_{\mathcal{F}}(n)]=\lambda (I_n/J{I}_{n-1})-\sum _{i=2}^d{(-1)}^i\lambda (H_i(C_{.}(n)))$$

$$e_i(\mathcal{F})=\sum _{n=i}^{\infty }\left({}_{i-1}^{n-1}\right){\Delta }^d[P_{\mathcal{F}}(n)-H_{\mathcal{F}}(n)],$$

$$e_1(\mathcal{F})=\sum _{n=1}^{\infty }\lambda (I_n/J{I}_{n-1})-\sum _{i=2}^d{(-1)}^i{h}_i(\underline{x},\mathcal{F})$$

$$e_2(\mathcal{F})=\sum _{n=2}^{\infty }(n-1)\lambda (I_n/J{I}_{n-1})-\sum _{i=2}^d{(-1)}^i{k}_i(\underline{x},\mathcal{F}).$$

$$e_2(\mathcal{F})\le \sum _{n=2}^{\infty }(n-1)\lambda (I_n/J{I}_{n-1}).$$

$$e_2(\mathcal{F})=\sum _{n=2}^{\infty }(n-1)\lambda (I_n/J{I}_{n-1})-\sum _{i=2}^d{(-1)}^{i-2}{k}_i(J,\mathcal{F}),$$

$$\sum _{i=2}^d{(-1)}^{i-2}{k}_i(J,\mathcal{F})\ge 0.$$

$$e_2(\mathcal{F})\le \sum _{n=2}^{\infty }(n-1)\lambda (I_n/J{I}_{n-1}).$$

$$0⟶(I_{n-l+1}:(\underline{x}))/I_{n-l}⟶(I_{n-l+1}:({\underline{x}}^{\prime }))/I_{n-l}\stackrel{x_l}{⟶}(I_{n-l+2}:({\underline{x}}^{\prime }))/I_{n-l+1}⟶...$$

$$⟶(I_{n-1}\cap ({\underline{x}}^{\prime }))/({\underline{x}}^{\prime })I_{n-2}\stackrel{x_l}{⟶}(I_n\cap ({\underline{x}}^{\prime }))/({\underline{x}}^{\prime })I_{n-1}⟶(I_n\cap (\underline{x}))/(\underline{x})I_{n-1}⟶$$

$$R/(I_{n-1},({\underline{x}}^{\prime }))\stackrel{x_l}{⟶}R/(I_n,({\underline{x}}^{\prime }))⟶R/(I_n,(\underline{x}))⟶0.$$

$$\begin{array}[]{ll}{I^{n+1}}\cap J_{d-1}&=J{I^{n}}\cap J_{d-1}\\ &=(J_{d-1}{I^{n}}+{x_{d}}{I^{n}})\cap J_{d-1}\\ &=J_{d-1}{I^{n}}+({x_{d}}{I^{n}}\cap J_{d-1})\\ &=J_{d-1}{I^{n}}+{x_{d}}({I^{n}}\cap J_{d-1})\\ &=J_{d-1}{I^{n}}+{x_{d}}(J_{d-1}{I^{n-1}})=J_{d-1}{I^{n}}.\end{array}$$

$$0⟶H_d(C_{.}(n+1))⟶...⟶H_1({C_{.}}^{\prime }(n))\stackrel{x_d}{⟶}{H}_1({C_{.}}^{\prime }(n+1))⟶B_{n+1}⟶0.$$

$$...⟶I^n\cap J_{d-1}/J_{d-1}{I}^{n-1}\stackrel{x_d}{⟶}{I}^{n+1}\cap J_{d-1}/J_{d-1}{I}^n⟶B_{n+1}⟶0$$

$$\sum _{n=1}^{\infty }\lambda (B_{n+1})=\sum _{n=1}^{\infty }\lambda ((I^{n+1}\cap J_{d-1})/(J{I}^n\cap J_{d-1}))=\sum _{i=1}^d{(-1)}^{i-2}{h}_i=1$$

$$0\to H_d(C_{.}(n))\to ...\to H_1({C_{.}}^{\prime }(n-1))\stackrel{x_d}{\to }{H}_1({C_{.}}^{\prime }(n))\to B_n\to 0.$$

$$\lambda (B_n)=\sum _{i=2}^d{(-1)}^{i-2}\lambda (H_i(C_{.}(n)))+\sum _{i=1}^d{(-1)}^{i-1}\Delta [\lambda (H_i({C_{.}}^{\prime }(n)))].$$

Full text

${e_{2}}(I)=\sum\limits_{n=2}^{\infty}{(n-1)\lambda({{I^{n}}}/{J{I^{n-1}}})}-3$, then ${e_{1}}(I)=\sum\limits_{n=1}^{\infty}{\lambda({{I^{n}}}/{J{I^{n-1}}})}-1,$ where the integers $e_{i}$ are the Hilbert coefficients of $I$. In addition, if $J$ is a minimal reduction of $I$ then we prove that the reduction number $r_{J}(I)$ is independent of $J$.

1. Introduction

Throughout the paper we will assume that $(R,\mathfrak{m})$ is a $d$-dimensional Cohen-Macaulay local ring having an infinite residue field and $I$ is an $\mathfrak{m}$-primary ideal of $R$. An ideal $J\subseteq I$ is called a reduction of $I$ if $I^{r+1}=JI^{r}$ for some nonnegative integer $r$ (see [21]). The least such $r$ is called the reduction number of $I$ with respect to $J$ and is denoted by $r_{J}(I)$. A reduction $J$ is called a minimal reduction if it does not properly contain a reduction of $I$. Under our assumption every minimal reduction is generated by a regular sequence. The reduction number of $I$ is defined as $r(I)=\min\{r_{J}(I):J$ is a minimal reduction of $I\}$. The reduction number $r(I)$ is said to be independent if $r(I)=r_{J}(I)$ for all minimal reductions $J$ of $I$. Sally in [27] raised the following question: If $(R,\mathfrak{m})$ is a $d$-dimensional Cohen-Macaulay local ring having an infinite residue field, then is $r(\mathfrak{m})$ independent? A natural extension of this question is to replace $r(\mathfrak{m})$ with $r(I)$. Let $G(I)=\bigoplus_{n\geq 0}I^{n}/I^{n+1}$ be the associated graded ring of $I$. Huckaba in [11] and Trung in [30] independently proved that if $0ptG(I)\geq{d-1}$, then $r(I)$ is independent(see also [19], [9], [10] and [29]). Moreover, Wu in [36] with some conditions proved that if $0ptG(I)\geq{d-2}$, then $r(I)$ is independent. However, if $d\geq 2$ and $0ptG(I)\leq{d-2}$, then $r(I)$ is not independent in general. Counter-examples have been obtained in [11], [19] and [18].

The Hilbert function of $I$ is the numerical function $H_{I}(n)=\lambda(R/I^{n})$ (where $\lambda(.)$ denotes length) that measures the growth of the length of $R/I^{n}$ for all $n\geq 1$. It is well known that for $n\gg 0$, $H_{I}(n)$ is a polynomial in $n$ of degree $d$

[TABLE]

called the Hilbert polynomial of $I$, whose coefficients $e_{0}(I),e_{1}(I),...,e_{d}(I)$ are uniquely determined by $I$ and called the Hilbert coefficients of $I$.

Valabrega and Valla in [31] obtained that $G(I)$ is Cohen-Macaulay if and only if there exists a minimal reduction $J$ of $I$ such that $I^{n}\cap J=I^{n-1}J$ for all $n$. Later on, Guerrieri in [5] asked that if $J$ is a minimal reduction of $I$ such that $\sum_{n\geq 1}\lambda({I^{n}\cap J}/{I^{n-1}J})=t$, then is $0ptG(I)={d-t}$? The case $t=0$ is simply a restatement of the Valabrega-Vall theorem, whereas the case $t=1$ was proved in [5]. Some partial answers in the cases $t=2$ or $t=3$ were also proved in [6] and [8]. Huckaba and Marley in [14] and Vaz Pinto in [32] independently showed that $e_{1}(I)\leq\sum_{n\geq 1}\lambda({I^{n}}/{I^{n-1}J})$ and equality holds for some minimal reduction $J$ of $I$ if and only if $0ptG(I)\geq{d-1}$. Another closely related conjecture was raised by Wang in [34] while attempting to prove Guerrieri’s conjecture. Namely, he asked whether the difference $\sum_{n\geq 1}\lambda({I^{n}}/{I^{n-1}J})-e_{1}(I)=s\geq 0$, implies $0ptG(I)\geq{d-s-1}$. Wang first showed that an affirmative answer to his conjecture implies the validity of Guerrieri’s conjecture. Then in [34] he settled the case $s=1$ (see also [23]). Unfortunately, both conjectures fail in general as shown in [35].

Corso, Polini and Rossi in [2] established a general upper bound on $e_{2}(I)$, which is reminiscent of the bound on $e_{1}(I)$ due to Huckaba and Marley in [14] and Vaz Pinto in [32]. Namely, it holds that $e_{2}(I)\leq\sum_{n\geq 2}{(n-1)}\lambda({I^{n}}/{I^{n-1}J})$ for any minimal reduction reduction $J$ of $I$. Furthermore, the upper bound is attained if and only if $0ptG(I)\geq{d-1}$. In addition, if $e_{2}(I)\geq\sum_{n\geq 2}{(n-1)}\lambda({I^{n}}/{I^{n-1}J})-2$ or if $I$ is integrally closed and $e_{2}(I)\geq\sum_{n\geq 2}{(n-1)}\lambda({I^{n}}/{I^{n-1}J})-4$, then $0ptG(I)\geq{d-2}$.

In this paper we prove the following results.

Theorem 1.1**.**

*Let $J=(x_{1},...,x_{d})$ be a minimal reduction of $I$.

(1) Suppose that one of the following conditions holds:*

• (i)

${e_{2}}(I)=\sum\limits_{n=2}^{\infty}{(n-1)\lambda({{I^{n}}}/{J{I^{n-1}}})}-2;$**

• (ii)

$I$* is integrally closed and ${e_{2}}(I)=\sum\limits_{n=2}^{\infty}{(n-1)\lambda({{I^{n}}}/{J{I^{n-1}}})}-3$.*

*Then ${e_{1}}(I)=\sum\limits_{n=1}^{\infty}{\lambda({{I^{n}}}/{J{I^{n-1}}})}-1.$ Moreover, we give a counterexample showing that ${e_{1}}(I)=\sum\limits_{n=1}^{\infty}{\lambda({{I^{n}}}/{J{I^{n-1}}})}-1$ but the conditions (i) and (ii) do not hold.

(2) If $J_{d-1}=(x_{1},...,x_{d-1})$ and $\sum\limits_{n=1}^{\infty}\lambda{({{I^{n+1}}\cap J_{d-1}})/({J{I^{n}}\cap J_{d-1}})=i}$ where $i=0$ or $i=1$, then $0ptG(I)\geq{d-i-1}$.*

Theorem 1.2**.**

*Let $J=(x_{1},...,x_{d})$ be a minimal reduction of $I$.

(1) Suppose that one of the following conditions holds:*

• (i)

${e_{2}}(I)=\sum\limits_{n=2}^{\infty}{(n-1)\lambda({{I^{n}}}/{J{I^{n-1}}})}-2;$**

• (ii)

$I$* is integrally closed and ${e_{2}}(I)=\sum\limits_{n=2}^{\infty}{(n-1)\lambda({{I^{n}}}/{J{I^{n-1}}})}-3$.*

Then $r(I)$ is independent. Also, we study the independence of $r(I)$ with some other conditions.

For any unexplained notation or terminology, we refer the reader to [1] and [26].

2. Preliminaries

In this section we recall some known results which is studied in [14]. An element $x\in I\setminus I^{2}$ is said to be superficial for $I$ if there is an integer $c$ such that $(I^{n+1}:x)\cap I^{c}=I^{n}$ for all $n\geq c$. If $\operatorname{grade}I\geq 1$ and $x$ is a superficial element, then $x$ is a regular element of $R$ and by Artin-Rees Theorem $I^{n+1}:x=I^{n}$ for all $n$ sufficiently large. If $R/{\mathfrak{m}}$ is infinite, then a superficial element for $I$ always exists. A sequence $x_{1},...,x_{s}$ is called a superficial sequence for $I$ if $x_{1}$ is superficial for $I$ and $x_{i}$ is superficial for $I/(x_{1},...,x_{i-1})$ for $2\leq i\leq s$. If $J$ is a minimal reduction of $I$, then there is a superficial sequence $x_{1},...,x_{d}$ in $I$ such that $J=(x_{1},...,x_{d})$. For any element $x\in I$ we let $x^{*}$ denote the image of $x$ in $I/I^{2}$. We note that if $x^{*}$ is a regular element of $G(I)$, then $x$ is a regular element of $R$ and $G(I/{(x)})\cong G(I)/{(x^{*})}$ (see [14]).

A set of ideals $\mathcal{F}={\left\{{{I_{n}}}\right\}_{n\in\mathbb{Z}}}$ of $R$ where $I_{n}=R$ for all $n\leq 0$ and $I_{1}=I$ is called a Hilbert filtration if we have $(i)$ $I_{n+1}\subseteq I_{n}$ for all $n\geq 0$, $(ii)$ $I_{n}I_{m}\subseteq I_{n+m}$ for all $n,m\geq 0$, and $(iii)$ there is a $k\geq 0$ such that $I^{n}\subseteq I_{n}\subseteq I^{n-k}$ for all $n\geq 0$. An element $x\in I_{1}$ is called superficial for $\mathcal{F}$ if there exists an integer $c$ such that $(I_{n+1}:x)\cap I_{c}=I_{n}$ for all $n\geq c$. A sequence $x_{1},\ldots,x_{l}$ is called a superficial sequence for $\mathcal{F}$ if $x_{1}$ is superficial for $\mathcal{F}$ and $x_{i}$ is superficial for $\mathcal{F}/(x_{1},\ldots,x_{i-1})$ for $2\leq i\leq l$ (see [14]). Let $\mathcal{F}$ be a Hilbert filtration and $\underline{x}={x_{1}},...,{x_{l}}\in{I_{1}}$. For any integer $n$, Huckaba and Marley in [14] constructed the complex $C_{.}(\underline{x},\mathcal{F},n)$ as follows: for $l=1$ they define $C_{.}(x_{1},\mathcal{F},n)$ to be the complex

[TABLE]

For $l>1$, assume that $C_{.}(x_{1},\ldots,x_{l-1},\mathcal{F},n)$ has been defined and consider the chain map $f:C_{.}(x_{1},\ldots,x_{l-1},\mathcal{F},n-1)\longrightarrow C_{.}(x_{1},\ldots,x_{l-1},\mathcal{F},n)$ given by multiplication by $x_{l}$. Define $C_{.}(x_{1},\ldots,x_{l-1},\mathcal{F},n)$ to be the mapping cylinder of $f$. Thus one can show that $C_{.}(x_{1},\ldots,x_{l-1},\mathcal{F},n)$ has the following form

[TABLE]

Let ${C_{.}}(n)={C_{.}}({x_{1}},{x_{2}},...,{x_{l}},\mathcal{F},n)$ and ${C_{.}}^{\prime}(n)={C_{.}}({x_{1}},{x_{2}},...,{x_{l-1}},\mathcal{F},n)$. For any $n$ there is an exact sequence of complexes

[TABLE]

Thus, we have the corresponding long exact sequence on homology:

[TABLE]

Since $\mathcal{F}$ is a Hilbert filtration, ${H_{i}}({C_{.}}(\underline{x},\mathcal{F},n))$ has finite length for all $i$ and $n$. For $i\geq 1$, consider

[TABLE]

and

[TABLE]

These integers are well-defined by [14, Lemma 3.6]. Also note that, ${h_{i}}(\underline{x},\mathcal{F})={k_{i}}(\underline{x},\mathcal{F})=0$ for all $i>l$. For ${\underline{x}^{\prime}}={x_{1}},...,{x_{l-1}}$ we define

[TABLE]

3. The associated graded ring and the first and second Hilbert coefficients

Lemma 3.1**.**

([14, Theorem 3.7]) Let $\mathcal{F}$ be a Hilbert filtration and $\underline{x}={x_{1}},...,{x_{l}}$ a regular sequence on R and a superficial sequence for $\mathcal{F}$. Then for each $i\geq 1$ $\sum\limits_{j\geq i}{{{(-1)}^{j-i}}{h_{j}}(\underline{x}},\mathcal{F})\geq 0$. Moreover, equality occurs if and only if $\operatorname{grade}(\underline{x}^{*})\geq{l-i+1}$.

Lemma 3.2**.**

Let $\mathcal{F}$ be a Hilbert filtration and $\underline{x}={x_{1}},...,{x_{l}}$ be a regular sequence on $R$ and a superficial sequence for $\mathcal{F}$. Then, for each $i\geq 1$, $\sum\limits_{j\geq i+1}{{{(-1)}^{j-i-1}}{k_{j}}(\underline{x}},\mathcal{F})\geq 0$. Moreover, equality occurs if and only if $\operatorname{grade}(\underline{x}^{*})\geq{l-i}$.

Proof.

Fix $i\geq 1$ and for each $n$ let $B_{n}$ be the kernel of the map ${H_{i}}({C_{.}}(n))\longrightarrow{H_{i-1}}({C_{.}}^{\prime}(n-1))$ given in $(*)$. Then, for each $n$, we have the exact sequence

[TABLE]

Therefore, for each $n$, we have

[TABLE]

and also by using [12, Lemma 2.7] we have

[TABLE]

Thus we see that

[TABLE]

By Lemma 3.1, we have $\sum\limits_{j=i}^{l}{{{(-1)}^{j-i}}{{h^{\prime}}_{j}(\underline{x}^{\prime},\mathcal{F})}}\geq 0$ and so $\sum\limits_{j=i+1}^{l}{{{(-1)}^{j-i-1}}{k_{j}(\underline{x},\mathcal{F})}}\geq 0$ for each $i\geq 1$.

By [14, Proposition 3.3], if $\operatorname{grade}(\underline{x}^{*})\geq{l-i}$ then $H_{j}({C_{.}}(n))=0$ for all $n$ and $j\geq i+1$. Thus, $k_{j}(\underline{x},\mathcal{F})=0$ for $j\geq i+1$.

Conversely, suppose for $i\geq 1$

[TABLE]

So by $(**)$, $\sum\limits_{n=2}^{\infty}{(n-1)\lambda({B_{n}})}=0$. Then $\sum\limits_{n=1}^{\infty}{\lambda({B_{n}})}=\sum\limits_{j=i+1}^{l}{{{(-1)}^{j-i-1}}{h_{j}(\underline{x},\mathcal{F})}}=0$ and by Lemma 3.1 we obtain $\operatorname{grade}(\underline{x}^{*})\geq{l-(i+1)+1=l-i}.$ ∎

Following Marley [18], given a function $f:\mathbb{Z}\longrightarrow\mathbb{Z}$, define the first difference function of $f$, $\Delta^{1}(f):\mathbb{Z}\longrightarrow\mathbb{Z}$ by $\Delta^{1}(f)(n)=f(n+1)-f(n)$. We usually write $\Delta^{1}(f(n))$ for $\Delta^{1}(f)(n)$. Inductively define the ith difference function of $f$, $\Delta^{i}(f):\mathbb{Z}\longrightarrow\mathbb{Z}$ , by $\Delta^{i}(f(n))=\Delta^{1}(\Delta^{i-1}(f(n)))$. Following Huckaba and Marley [14], a reduction of a filtration $\mathcal{F}$ is an ideal $J\subseteq I$ such that $JI_{n}=I_{n+1}$ for all large $n$. A minimal reduction of $\mathcal{F}$ is a reduction of $\mathcal{F}$ minimal with respect to containment.

Remark 3.3*.*

By [14, §4.] we have

[TABLE]

and

[TABLE]

where $J$ is a minimal reduction of $\mathcal{F}$ and ${e_{i}}(\mathcal{F})$ is the Hilbert coefficients of $\mathcal{F}$. Therefore we obtain the following

[TABLE]

and

[TABLE]

Proposition 3.4**.**

(compare with [2, Theorem 3.1]) Let $\mathcal{F}$ be a Hilbert filtration and $J=(x_{1},...,x_{d})$ be a minimal reduction of $\mathcal{F}$. Then

[TABLE]

The equality holds if and only if $0ptG(\mathcal{F})\geq{d-1}.$

Proof.

By Remark 3.3 and Lemma 3.2 we have the following

[TABLE]

and

[TABLE]

Thus

[TABLE]

Also the equality follows by Lemma 3.2. ∎

Remark 3.5*.*

By [14, Lemma 3.2] and $(*)$ we obtain the following exact sequence

[TABLE]

[TABLE]

[TABLE]

If $A_{n}$ is the kernel of the map $R/{({I_{n-1}},(\underline{x}^{\prime}))}\overset{x_{l}}{\longrightarrow}R/{({I_{n}},(\underline{x}^{\prime}))}$, then $A_{n}=(({I_{n}},(\underline{x}^{\prime})):{x_{l}})/{({I_{n-1}},(\underline{x}^{\prime}))}$.

If $C_{n}$ is the kernel of the map ${({I_{n}}\cap(\underline{x}))}/{(\underline{x}){I_{n-1}}}\longrightarrow R/{({I_{n-1}},(\underline{x}^{\prime}))}$ or the cokernel of the map ${({I_{n-1}\cap(\underline{x}^{\prime})})}/{(\underline{x}^{\prime}){I_{n-2}}}\overset{x_{l}}{\longrightarrow}{({I_{n}}\cap(\underline{x}^{\prime}))}/{{(\underline{x}^{\prime}){I_{n-1}}}},$ then

$C_{n}={({I_{n}}\cap(\underline{x}^{\prime}))}/{((\underline{x}^{\prime}){I_{n-1}}+x_{l}(I_{n-1}\cap(\underline{x}^{\prime})))}$. Thus if $\operatorname{grade}(\underline{x}^{*})\geq l-1$, then $C_{n}=0$ for all $n$ and ${(({I_{n}},(\underline{x}^{\prime})):{x_{l}})}/{({I_{n-1}},(\underline{x}^{\prime}))}\cong{({I_{n}}\cap(\underline{x}))}/{(\underline{x}){I_{n-1}}}$.

From now on, we will assume that the filtration $\mathcal{F}=\{I^{n}\}_{n=0}^{\infty}$ is $I$-adic filtration. Let $J=({x_{1}},...,{x_{d}})$, where ${x_{1}},...,{x_{d}}$ is a superficial sequence in $I$, i.e., $J$ is a minimal reduction of $I$. For $i\leq d-1$, set $J_{i}=({x_{1}},...,{x_{i}})$ (with the convention $J_{i}=(0)$ if $i\leq 0$ ), and we denote $h_{i}(\underline{x},\mathcal{F})$, $k_{i}(\underline{x},\mathcal{F})$ and $h^{\prime}_{i}(\underline{x}^{\prime},\mathcal{F})$ for $I$-adic filtration by $h_{i}$ ,$k_{i}$ and $h^{\prime}_{i}$, respectively.

Proposition 3.6**.**

Let $d\geq 2$ and $J$ be a minimal reduction of $I$ such that $\sum\limits_{i=2}^{d}{{{(-1)}^{i-2}}{h_{i}}=1}$. Then $0ptG(I)\geq d-2.$

Proof.

If $\sum\limits_{i=2}^{d}{{{(-1)}^{i-2}}{h_{i}}=1}$, then by Remark 3.3, $\sum\limits_{n=0}^{\infty}\lambda{({{I^{n+1}}}/{J{I^{n}}})}-{e_{1}}(I)=1$ and so by [34, Theorem 3.1] we have $0ptG(I)\geq d-2.$ ∎

Proposition 3.7**.**

*Let $J$ be a minimal reduction of $I$ such that

$\sum\limits_{n=1}^{\infty}\lambda({({{I^{n+1}}\cap J_{d-1}})/({J{I^{n}}\cap J_{d-1}}))=0}$. Then $0ptG(I)\geq{d-1}.$*

Proof.

By using induction on $n$, we prove that ${I^{n+1}}\cap J_{d-1}=J_{d-1}{I^{n}}$ for every $n\geq 0$. For $n=0$, there is nothing to prove. Assume that $n\geq 1$ and ${I^{n}}\cap J_{d-1}=J_{d-1}{I^{n-1}}$. From the following equalities:

[TABLE]

The fourth equality follows from the fact that $x_{1},x_{2},..,x_{d}$ form a regular sequence. Then by using Valabrega and Valla’s theorem (see also [33, Theorem 5.16]) we have $x_{1}^{*},x_{2}^{*},...,x_{d-1}^{*}$ are regular sequence and $0pt(G(I))\geq d-1$.

∎

Proposition 3.8**.**

*Let $J$ be a minimal reduction of $I$ such that

$\sum\limits_{n=1}^{\infty}\lambda(({{I^{n+1}}\cap J_{d-1}})/({J{I^{n}}\cap J_{d-1}}))=1.$ Then $0ptG(I)\geq{d-2}.$*

Proof.

Let $B_{n+1}$ be the kernel of the map ${H_{1}}({C_{.}}(n+1))\to{H_{0}}({C_{.}}^{\prime}(n))$ given in $(*)$. Consider the following exact sequence

[TABLE]

Therefore by [14, Lemma 3.2] we have the following exact sequence

[TABLE]

and $B_{n+1}={{I^{n+1}}\cap J_{d-1}}/{J{I^{n}}\cap J_{d-1}}.$ Thus

[TABLE]

and so by Proposition 3.6 we have $0ptG(I)\geq d-2$. ∎

Remark 3.9*.*

Let $B_{n}$ be the kernel of the map ${H_{1}}({C_{.}}(n))\to{H_{0}}({C_{.}}^{\prime}(n-1))$ given in $(*)$. Then for each $n$ we have the exact sequence

[TABLE]

Therefore, for each $n$ we have

[TABLE]

By using [12, Lemma 2.7]

[TABLE]

and

[TABLE]

Since $\sum\limits_{n=2}^{\infty}{(n-1)\lambda({B_{n}})=\sum\limits_{i=2}^{d}{{{(-1)}^{i-2}}{k_{i}}}}-\sum\limits_{i=1}^{d}{{{(-1)}^{i-1}}{{h^{\prime}}_{i}}}$ and $\sum\limits_{n=1}^{\infty}{\lambda({B_{n}})}$ is equal to ${\sum\limits_{i=2}^{d}{{{(-1)}^{i-2}}{h_{i}}}}$, then by Lemma 3.1, $\sum\limits_{i=1}^{d}{{{(-1)}^{i-1}}{{h^{\prime}}_{i}}}=0$ if and only if $0ptG(I)\geq d-1$ if and only if $\sum\limits_{n=2}^{\infty}{(n-1)\lambda({B_{n}})}=0.$ Thus $\sum\limits_{i=2}^{d}{{{(-1)}^{i-2}}{k_{i}}}=1$ cannot be happen. Now assume that $I$ is integrally closed ideal. Since $I^{2}\cap J=JI$ (see [17, Lemma 2.1]), then by [14, Lemma 3.2] we have $H_{1}(C.(2))=0$ and $B_{2}=0$. So if $I$ is integrally closed ideal, then $\sum\limits_{i=2}^{d}{{{(-1)}^{i-2}}{k_{i}}}=2$ cannot happen, because in this case $\lambda(B_{2})=1$ and this is a contradiction. So we have the following three cases:

If $\sum\limits_{i=2}^{d}{{{(-1)}^{i-2}}{k_{i}}}=2$, then we have $\sum\limits_{i=1}^{d}{{{(-1)}^{i-1}}{{h^{\prime}}_{i}}}=1$ and $\sum\limits_{n=2}^{\infty}{(n-1)\lambda({B_{n}})}=1$. In this case we obtain $\lambda(B_{2})=1$ and $\lambda(B_{n})=0$ for any $n\neq 2$. Therefore $\sum\limits_{n=1}^{\infty}{\lambda({B_{n}})=\sum\limits_{i=2}^{d}{{{(-1)}^{i-2}}{h_{i}}}}=1$ and by Proposition 3.6 $0ptG(I)\geq d-2$.

If we assume $I$ to be integrally closed ideal and $\sum\limits_{i=2}^{d}{{{(-1)}^{i-2}}{k_{i}}}=3,$ then we have $\sum\limits_{n=2}^{\infty}(n-1)\lambda({B_{n}})=1~{}or~{}2$. The case $\sum\limits_{n=2}^{\infty}(n-1)\lambda({B_{n}})=1$ cannot happen because $\lambda(B_{2})=0$. If $\sum\limits_{n=2}^{\infty}(n-1)\lambda({B_{n}})=2$, then $\lambda(B_{3})=1$ and $\lambda(B_{n})=0$ for all $n\neq 3$, so $\sum\limits_{i=2}^{d}{{{(-1)}^{i-2}}{h_{i}}}=1$ and $0ptG(I)\geq d-2$.

If we assume $I$ to be integrally closed ideal and $\sum\limits_{i=2}^{d}{{{(-1)}^{i-2}}{k_{i}}}=4$, then we have $\sum\limits_{n=2}^{\infty}(n-1)\lambda({B_{n}})=2~{}or~{}3$. If $\sum\limits_{n=2}^{\infty}(n-1)\lambda({B_{n}})=2$, then $0ptG(I)\geq d-2$. If $\sum\limits_{n=2}^{\infty}(n-1)\lambda({B_{n}})=3$, then $\lambda(B_{4})=1$ and $\lambda(B_{n})=0$ for any $n\neq 4$. Thus $\sum\limits_{i=2}^{d}{{{(-1)}^{i-2}}{h_{i}}}=1$ and $0ptG(I)\geq d-2$ .

In the following result we compare [2, Theorem 3.1 and 3.3] with [34, Theorem 3.1].

Theorem 3.10**.**

Let $J$ be a minimal reduction of $I$. Suppose that one of the following conditions holds:

1. (1)

${e_{2}}(I)=\sum\limits_{n=2}^{\infty}{(n-1)\lambda({{I^{n}}}/{J{I^{n-1}}})}-2;$** 2. (2)

$I$* is integrally closed and ${e_{2}}(I)=\sum\limits_{n=2}^{\infty}{(n-1)\lambda({{I^{n}}}/{J{I^{n-1}}})}-i$, where $i=3,4$.*

Then ${e_{1}}(I)=\sum\limits_{n=1}^{\infty}{\lambda({{I^{n}}}/{J{I^{n-1}}})}-1.$

Proof.

$(1)$ If ${e_{2}}(I)=\sum\limits_{n=2}^{\infty}{(n-1)\lambda({{I^{n}}}/{J{I^{n-1}}})}-2$, then $\sum\limits_{i=2}^{d}{{{(-1)}^{i-2}}{k_{i}}}=2$ and by Remark 3.9 $\sum\limits_{n=1}^{\infty}{\lambda({B_{n}})=\sum\limits_{i=2}^{d}{{{(-1)}^{i-2}}{h_{i}}}}=1$. Therefore by Remark 3.3 we have ${e_{1}}(I)=\sum\limits_{n=1}^{\infty}\lambda{({{I^{n}}}/{J{I^{n-1}}})}-1$.

$(2)$ If I is integrally closed and ${e_{2}}(I)=\sum\limits_{n=2}^{\infty}{(n-1)\lambda({{I^{n}}}/{{J{I^{n-1}}}})}-i$ where $i=3,4$ then $\sum\limits_{i=2}^{d}{{{(-1)}^{i-2}}{k_{i}}}=3~{}~{}or~{}4$. Therefore by Remark 3.9 $\sum\limits_{n=1}^{\infty}{\lambda({B_{n}})=\sum\limits_{i=2}^{d}{{{(-1)}^{i-2}}{h_{i}}}}=1$ and so ${e_{1}}(I)=\sum\limits_{n=1}^{\infty}\lambda{({{I^{n}}}/{{J{I^{n-1}}}})}-1.$

∎

The following example shows that the converse of Theorem 3.10 in general is not true.

Example 3.11*.*

Let $R=k[x,y]_{(x,y)}$, where $k$ is a field and $I=(x^{6},y^{6},x^{5}y+x^{2}y^{4})$. Then by using Macaulay 2 [4] we obtain the following Hilbert polynomial

[TABLE]

and ${e_{1}}(I)=\sum\limits_{n=1}^{\infty}\lambda{({{I^{n}}}}/{{J{I^{n-1}}})}-1$ but ${e_{2}}(I)=\sum\limits_{n=2}^{\infty}{(n-1)\lambda({{I^{n}}}/{J{I^{n-1}}})}-3$ and $I$ is not integrally closed.

4. The depth of associated graded ring and the reduction number

Let $A=\bigoplus_{m\geq 0}A_{m}$ be a Notherian graded ring where $A_{0}$ is an Artinian local ring, $A$ is generated by $A_{1}$ over $A_{0}$ and $A_{+}=\bigoplus_{m>0}A_{m}$. Let $H_{A_{+}}^{i}(A)$ denote the i-th local cohomology module of $A$ with respect to the graded ideal $A_{+}$ and set $a_{i}(A)=\max\{m|\ \ [H_{A_{+}}^{i}(A)]_{m}\neq 0\}$ with the convention $a_{i}(A)=-\infty$, if $H_{A_{+}}^{i}(A)=0$. The Castelnuovo-Mumford regularity is defined by $\operatorname{reg}(A):=\max\{a_{i}(A)+i|\ \ i\geq 0\}.$ In the following theorem, for simplicity, we use $a_{i}$ instead of $a_{i}(G(I))$.

Theorem 4.1**.**

Let $J$ denote a minimal reduction of $I$. Suppose that one of the following conditions holds:

1. (1)

${e_{2}}(I)=\sum\limits_{n=2}^{\infty}{(n-1)\lambda({{I^{n}}}/{J{I^{n-1}}})}-2;$** 2. (2)

I is integrally closed and ${e_{2}}(I)=\sum\limits_{n=2}^{\infty}{(n-1)\lambda({{I^{n}}}/{J{I^{n-1}}})}-3$.

Then $r(I)$ is independent.

Proof.

(1) By Remark 3.9, $\sum\limits_{i=2}^{d}{{{(-1)}^{i-2}}{k_{i}}}=2$ and $0ptG(I)\geq d-2$. If $0ptG(I)\geq d-1$, then by [18, Theorem 2] $r_{J}(I)=\operatorname{reg}(G(I))$ and so the result in this case follows. Now we assume that $0ptG(I)=d-2$ and $\sum\limits_{i=2}^{d}{{{(-1)}^{i-2}}{k_{i}}}=2$. By Remark 3.9, $\lambda(B_{2})=1$ and $\lambda(B_{n})=0$ for any $n\neq 2$. Therefore we have $\lambda({{I^{2}}\cap{J_{d-1}}}/{{J_{d-1}}I})=1$, $\lambda({{I^{n}}\cap{J_{d-1}}}/{{{J_{d-1}}{I^{n-1}}}})=0$ for any $n\neq 2$,

$\lambda({({I^{2}}+{J_{d-2}}):{x_{d-1}}}/{I})=1$ and $\lambda({({I^{n}}+{J_{d-2}}):{x_{d-1}}}/{{{I^{n-1}}+{J_{d-2}}}})=0$ for any $n\neq 2$. If $0ptG(I)=d-2$, then by applying [19, Theorem 2.1] there are two cases:

(i) If $a_{d-1}\leq a_{d}$, then $r_{J}(I)=a_{d}+d=reg(G(I))$.

(ii)If $a_{d}<a_{d-1}$, then $r_{J}(I)\leq a_{d-1}+d-1=reg(G(I))$ and ${a_{d-1}}=\max\{n|{(I^{n+d-1}}+{J_{d-2})}:{x_{d-1}}\neq{I^{n+d-2}}+{J_{d-2}}\}$. Therefore by the above process ${a_{d-1}}=3-d$ and so for any reduction $J$ of $I$, $r_{J}(I)\leq 2$. Hence $r(I)$ is independent.

(2) Let $i=3$. Then by Remark 3.9, $\lambda(B_{3})=1$ and $\lambda(B_{n})=0$ for any $n\neq 3$. Thus

[TABLE]

and

[TABLE]

for any $n\neq 3$. If $0ptG(I)=d-2$, then by [19, Theorem 2.1] there are two cases:

(i) If $a_{d-1}\leq a_{d}$, then $r_{J}(I)=a_{d}+d=reg(G(I))$.

(ii) If $a_{d}<a_{d-1}$, then $r_{J}(I)\leq a_{d-1}+d-1=reg(G(I))$ and ${a_{d-1}}=\max\{n|{(I^{n+d-1}}+{J_{d-2})}:{x_{d-1}}\neq{I^{n+d-2}}+{J_{d-2}}\}$. Therefore by the above process ${a_{d-1}}=4-d$ and so for any reduction $J$ of $I$, $r_{J}(I)\leq 3$. Since $I$ is integrally closed, then $r(I)$ is independent. ∎

Let $\mathfrak{a}$ be an ideal of grade at least $1$ in a Noetherian ring $R$. The Ratliff-Rush closure of $\mathfrak{a}$ is defined as the ideal $\widetilde{\mathfrak{a}}=\cup_{n\geq 1}({\mathfrak{a}}^{n+1}:{\mathfrak{a}}^{n}).$ It is a refinement of the integral closure of $\mathfrak{a}$ and $\widetilde{\mathfrak{a}}=\mathfrak{a}$ if $\mathfrak{a}$ is integrally closed (see [24]).

Proposition 4.2**.**

(compare with [2, Theorem 3.3(b)]) Let $d=3$ and $J$ be a minimal reduction of $I$. If $\widetilde{I}=I$ and ${e_{2}}(I)=\sum\limits_{n=2}^{\infty}{(n-1)\lambda({{I^{n}}}/{J{I^{n-1}}})}-3$ then $0ptG(I)\geq 1.$

Proof.

If ${e_{2}}(I)=\sum\limits_{n=2}^{\infty}{(n-1)\lambda({{I^{n}}}/{J{I^{n-1}}})}-3,$ then $\sum\limits_{i=2}^{d}{{{(-1)}^{i-2}}{k_{i}}}=3$. Hence by Remark 3.9, $\sum\limits_{i=1}^{d}{{{(-1)}^{i-1}}{{h^{\prime}}_{i}}}=1$ and $\sum\limits_{n=2}^{\infty}(n-1)\lambda({B_{n}})=2.$ Therefore we have the following cases:

$\lambda(B_{3})=1$ and $\lambda(B_{n})=0$ for any $n\neq 3$ .

or 2. 2)

$\lambda(B_{2})=2$ and $\lambda(B_{n})=0$ for any $n\neq 2$.

For the first case, by Proposition 3.6. $0ptG(I)\geq 1$ and the result follows. Let us consider the second case. Since $\widetilde{I}=I$, we obtain that $I^{2}:x=I$ for every superficial element $x$. Since $\lambda(B_{2})=2$, it follows $\lambda(I^{2}\cap J_{2}/J_{2}I)=2$. It therefore follows $\lambda(I^{2}\cap J_{1}/J_{1}I)=0$. Hence $\sum\limits_{i=1}^{d}{{{(-1)}^{i-1}}{{h^{\prime}}_{i}}}\geq 2$ and this is a contradiction.

∎

Northcott in [20] proved that ${e_{1}(I)}\geq{e_{0}(I)}-\lambda({R}/{I})$ and after that Huneke in [15] showed that ${e_{1}(I)}={e_{0}(I)}-\lambda({R}/{I})$ if and only if ${I^{2}}=JI$. When this is the case, $G(I)$ is Cohen-Macaulay. Sally in [28] proved that if $d\geq 2$, ${e_{1}(I)}-{e_{0}(I)}+\lambda({R}/{I})=1$ and $e_{2}(I)\neq 0$, then $0ptG(I)\geq d-1$ (see also [14, Corollary 4.15] and [7, Proposition 3.1]). Also Itoh in [16] with this conditions proved that if $I$ is integrally closed, then $G(I)$ is Cohen-Macaulay. In the following example we show that the integrally closedness of $I$ is essential for the Cohen-Macaulayness of $G(I)$.

The following example appears in [2].

Example 4.3*.*

Let $R=k[x,y,z]_{(x,y,z)}$, where $k$ is a field, and $I=(x^{2}-y^{2},y^{2}-z^{2},xy,yz,xz)$. Then, by Macaulay 2, we have ${e_{0}}(I)=8$, ${e_{1}}(I)=4$, ${e_{2}}(I)=0$ and $0ptG(I)=0.$

Lemma 4.4**.**

Let $J$ be a minimal reduction of $I$. If ${e_{1}(I)}-{e_{0}(I)}+\lambda({R}/{I})=2$ and $I=\bar{I}$, then $0ptG(I)\geq d-1.$

Proof.

By using the Sally machine and the good behaviour of $e_{1}(I)$ modulo superficial elements (see [3, Proposition 1.2] and [14, Lemma 2.2]), we may reduce the statement to dimension two. If ${e_{1}(I)}-{e_{0}(I)}+\lambda({R}/{I})=2$, by [25, Corollary 1.5] we have ${r_{J}}(I)\leq 3$ for any minimal reduction $J$ of $I$. If there exist a minimal reduction $J$ of $I$ such that ${r_{J}}(I)=2$, then by [17, Lemma 2.1] $G(I)$ is Cohen-Macaulay. If for any minimal reduction $J$ of $I$, ${r_{J}}(I)=3$ then by [25, Remark 1.8] ${r_{J}}(I)\leq{e_{1}(I)}-{e_{0}(I)}+\lambda({R}/{I})+2-\lambda({{I^{2}}}/{{JI}})$. Hence $r(I)=3\leq 4-\lambda({{I^{2}}}/{JI})$ and $\lambda({{I^{2}}}/{JI})\leq 1$ and so by [25, Corollary 1.7] $0ptG(I)\geq 1$. ∎

Proposition 4.5**.**

Let $J$ be a minimal reduction of $I$. If ${e_{1}(I)}-{e_{0}(I)}+\lambda({R}/{I})\leq 3$ and $I=\bar{I}$, then $r(I)$ is independent.

Proof.

By Lemma 3.15 and the above explanation, we can assume ${e_{1}(I)}-{e_{0}(I)}+\lambda({R}/{I})=3$ and also by [22, Corollary 4.7] and [11, Lemma 1.1] we may assume that $d=2$. If $0ptG(I)\geq 1$, then by [19, Corollary 2.2] we have $r_{J}(I)=\operatorname{reg}(G(I))$ and so $r(I)$ is independent. Now we may assume that $0ptG(I)=0$. Since ${e_{1}(I)}-{e_{0}(I)}+\lambda({R}/{I})=3$, by [25, Corollary 1.5] ${r_{J}}(I)\leq 4$ for any minimal reduction $J$ of $I$. If for some $J$, ${r_{J}}(I)=2$ then by [17, Lemma 2.1] $G(I)$ is Cohen-Macaulay and this is a contradiction with $0ptG(I)=0$. If for some $J$, ${r_{J}}(I)=4$ then by [25, Remark 1.8] ${r_{J}}(I)\leq{e_{1}(I)}-{e_{0}(I)}+\lambda({R}/{I})+2-\lambda({{I^{2}}}/{{JI}})$. Therefore $\lambda({{I^{2}}}/{{JI}})\leq 1$ and so $0ptG(I)\geq 1$ and this is also a contradiction. Hence we can assume that for any minimal reduction $J$ of $I$, ${r_{J}}(I)=3$ and so $r(I)$ is independent. ∎

Corollary 4.6**.**

Let $J$ be a minimal reduction of $I$. If ${e_{1}(I)}-{e_{0}(I)}+\lambda({R}/{I})\leq{r_{J}}(I)-1$ , $I=\bar{I}$ and $0ptG(I)\geq d-2$, then $r(I)$ is independent.

Proof.

By using the Sally machine and the good behaviour of $e_{1}(I)$ modulo superficial elements, we may reduce the statement to dimension two. If ${e_{1}(I)}-{e_{0}(I)}+\lambda({R}/{I})\leq{r_{J}}(I)-1$, then by [25, Remark 1.8] ${r_{J}}(I)\leq{e_{1}(I)}-{e_{0}(I)}+\lambda({R}/{I})+2-\lambda({{I^{2}}}/{{JI}})$. Hence $\lambda({{I^{2}}}/{{JI}})\leq 1$ and so $0ptG(I)\geq d-1$. Therefore the result follows by [18, Theorem 2]. ∎

Example 4.7*.*

Let $R=k[x,y]_{(x,y)}$ where $k$ is a field, and $I=(x^{6},y^{6},x^{5}y,x^{3}y^{3},x^{2}y^{4},xy^{5})$. Then by Macaulay 2 we have ${e_{0}}(I)=36$ , ${e_{1}}(I)=15$ and $\lambda(R/I)=22$. Hence ${e_{1}}(I)-{e_{0}}(I)+\lambda(R/I)=1$ but $0ptG(I)=0$ and $r_{J}(I)=2=reg(G(I))$ for all minimal reduction $J$ of $I$.

Example 4.8*.*

Let $R=k[x,y]_{(x,y)}$ where $k$ is a field, and $I=(x^{6},y^{6},x^{5}y,x^{3}y^{3},x^{2}y^{4})$. Then by Macaulay 2 we have ${e_{0}}(I)=36$ , ${e_{1}}(I)=15$ and $\lambda(R/I)=23$. Hence ${e_{1}}(I)-{e_{0}}(I)+\lambda(R/I)=2$ but $0ptG(I)=0$ and $r_{J}(I)=2=reg(G(I))$ for all minimal reduction $J$ of $I$.

Example 4.9*.*

Let $R=k[x,y]_{(x,y)}$ where $k$ is a field, and $I=(x^{6},y^{6},x^{5}y,x^{3}y^{3},xy^{5})$. Then by Macaulay 2 we have ${e_{0}}(I)=36$, ${e_{1}}(I)=15$ and $\lambda(R/I)=23$. Hence ${e_{1}}(I)-{e_{0}}(I)+\lambda(R/I)=3$ but $0ptG(I)=0$ and $r_{J}(I)=2=reg(G(I))$ for all minimal reduction $J$ of $I$.

Example 4.10*.*

Let $R=k[x,y]_{(x,y)}$ where $k$ is a field, and $I=(x^{6},y^{6},x^{5}y,x^{2}y^{4},xy^{5})$. Then by Macaulay 2 we have ${e_{0}}(I)=36$, ${e_{1}}(I)=15$ and $\lambda(R/I)=24$. Hence ${e_{1}}(I)-{e_{0}}(I)+\lambda(R/I)=3$ but for two minimal reduction $J_{1}=(x^{6},x^{5}y+y^{6})$ and $J_{2}=(x^{6},y^{6})$ we have $r_{J_{1}}(I)=2$ and $r_{J_{2}}(I)=3$ and $0ptG(I)=0$ because $I$ is not integrally closed.

The following example is due to Huckaba and Huneke [13].

Example 4.11*.*

Let $R=k[x,y,z]_{(x,y,z)}$ where $k$ is a field of characteristic $\neq 3$. Let $\mathfrak{a}=(x^{4},x(y^{3}+z^{3}),y(y^{3}+z^{3}),z(y^{3}+z^{3}))$ and set $I=\mathfrak{a}+{\mathfrak{m}}^{5}$ where $\mathfrak{m}$ is the maximal ideal of $R$. The ideal $I$ is a integral closer $m$-primary ideal whose associated graded ring $gr_{I}(R)$ has depth 2. We checked that ${e_{0}}(I)=76$ , ${e_{1}}(I)=48$ and $\lambda(R/I)=31$ so ${e_{1}}(I)-{e_{0}}(I)+\lambda(R/I)=3$ and $r(I)$ is independent.

*Acknowledgement *.

We would like to thank deeply grateful to the referee for the careful reading of the manuscript and the helpful suggestions