Stability properties of powers of ideals over regular local rings of small dimension

TL;DR

This paper investigates the stability of associated primes and depth of powers of ideals in regular local rings, revealing that these properties coincide in low dimensions but can diverge arbitrarily in higher dimensions.

Contribution

It establishes the equality of stability indices for associated primes and depth in dimensions up to 2, and constructs examples showing divergence in higher dimensions.

Findings

In dimension ≤ 2, astab(I)=astab̄(I)=dstab(I).

In dimension 3, astab(I) and astab̄(I) can differ arbitrarily.

In dimension 4, there are ideals with arbitrarily large differences between astab and dstab.

Abstract

Let $(R,\mathfrak{m})$ be a regular local ring or a polynomial ring over a field, and let $I$ be an ideal of $R$ which we assume to be graded if $R$ is a polynomial ring. Let astab$(I)$ resp. $\overline{\rm astab}(I)$ be the smallest integer $n$ for which Ass$(I^n)$ resp. Ass$(\overline{I^n})$ stabilize, and dstab$(I)$ be the smallest integer $n$ for which depth$(I^n)$ stabilizes. Here $\overline{I^n}$ denotes the integral closure of $I^n$. We show that astab$(I)=\overline{\rm astab}(I)={\rm dstab}(I)$ if dim$\,R\leq 2$, while already in dimension $3$, astab$(I)$ and $\overline{\rm astab}(I)$ may differ by any amount. Moreover, we show that if dim$\,R=4$, then there exist ideals $I$ and $J$ such that for any positive integer $c$ one has ${\rm astab}(I)-{\rm dstab}(I)\geq c$ and ${\rm dstab}(J)-{\rm astab}(J)\geq c$.

Equations

$$\mathrm{astab}(I)=\overline{\mathrm{astab}}(I)=\mathrm{dstab}(I)=1.$$

$$I^n:I$$

$$I^n:x^{2c+2}=I^{n-1}+((y^{2c}z{)}^n(x,y^2{)}^n:x^{2c+2})=I^{n-1}+(y^{2c}z{)}^n(x,y^2{)}^{n-2(c+1)},$$

$$I^n:x{y}^{2c}z$$

$$I^n:y^{2c+2}z$$

$$I^{n-1}$$

$$I^n:x^{2c+2}$$

$$0⟶R/I^{n-1}\stackrel{x^{2c+2}}{⟶}R/I^n⟶R/(I^n,x^{2c+2})⟶0.$$

$$\mathrm{depth}R/I^n\ge \min \{\mathrm{depth}R/I^{n-1},\mathrm{depth}R/(I^n,x^{2c+2})\}=\min \{1,\mathrm{depth}R/(I^n,x^{2c+2})\}.$$

$$ux=x^{2c+2}(x{y}^{2c}z)(y^{2c+2}z{)}^c{y}^{2c},uy=x^{2c+2}(y^{2c+2}z{)}^{c+1}$$

$$uz=(x{y}^{2c}z{)}^{c+1}(x{y}^{2c}z)(y{x}^c).$$

$$y^{(c+1)(2c+2)-1}{z}^{c+1}\in (I^{c+2}:x^{2c+2})=I^{c+1}+(y^{2c}z{)}^{c+2},$$

$$\overline{I}=(I,(x^3{y}^{2c-1}z,x^4{y}^{2c-2}z,\dots ,x^{2c+1}yz)).$$

$$(x^i{y}^{2c-i+2}z{)}^{2c}$$

$$(\overline{I}{)}_P$$

$$\mathrm{Ass}(\overline{I})=\{(x,z),(x,y)\}.$$

$$\mathrm{Ass}(\overline{I^k})=\{(x,z),(x,y),(x,y,z)\}\text{for all}k\ge 2,$$

$$L$$

$$I^2=(x^{4c+4},x^2{y}^{4c}{z}^2,y^{4c+4}{z}^2,x^{2c+3}{y}^{2c}z,x^{2c+2}{y}^{2c+2}z,x{y}^{4c+2}{z}^2),$$

$$(x^i{y}^{4c-i+3}{z}^2{)}^{4c}{x}^{4ic}{y}^{4c(4c-i+3)}{z}^{8c}=x^{2(4c-i+3)}{y}^{4c(4c-i+3)}{z}^{2(4c-i+3)}{x}^{4c(i-2)+2i-6}{z}^{2i-6}$$

$$=(x^2{y}^{4c}{z}^2{)}^{4c-i+3}(x^{4c+4}{)}^{i-3}{x}^{4c-2i+6}{z}^{2i-6}$$

$$(x^{2c+i}{y}^{2c+4-i}z{)}^{4c}$$

$$\mathrm{Ass}(L)=\{(x,z),(x,y),(x,y,z)\}.$$

$$(I^{2n-1}:y^{2c+1}z)$$

Full text

We show that $\operatorname{astab}(I)=\overline{\operatorname{astab}}(I)=\operatorname{dstab}(I)$ if $\dim R\leq 2$, while already in dimension $3$, $\operatorname{astab}(I)$ and $\overline{\operatorname{astab}}(I)$ may differ by any amount. Moreover, we show that if $\dim R=4$, then there exist ideals $I$ and $J$ such that for any positive integer $c$ one has $\operatorname{astab}(I)-\operatorname{dstab}(I)\geq c$ and $\operatorname{dstab}(J)-\operatorname{astab}(J)\geq c$.

Introduction

Let $(R,{\mathfrak{m}})$ be a commutative Noetherian ring and $I$ be an ideal of $R$. Brodmann [3] proved that the set of associated prime ideals $\operatorname{Ass}(I^{k})$ stabilizes. In other words, there exists an integer $k_{0}$ such that $\operatorname{Ass}(I^{k})=\operatorname{Ass}(I^{k_{0}})$ for all $k\geq k_{0}$. The smallest such integer $k_{0}$ is called the index of Ass-stability of $I$, and denoted by $\operatorname{astab}(I)$. Moreover, $\operatorname{Ass}(I^{k_{0}})$ is called the stable set of associated prime ideals of $I$. It is denoted by $\operatorname{Ass}^{\infty}(I)$. For the integral closures $\overline{I^{k}}$ of the powers of $I$, McAdam and Eakin [13] showed that $\operatorname{Ass}({\overline{I^{k}}})$ stabilizes as well. We denote the index of stability for the integral closures of the powers of $I$ by $\overline{\operatorname{astab}}(I)$, and denote its stable set of associated prime ideals by $\overline{\operatorname{Ass}}^{\infty}(I)$.

Brodmann [4] also showed that $\operatorname{depth}R/I^{k}$ stabilizes. The smallest power of $I$ for which depth stabilizes is denoted by $\operatorname{dstab}(I)$. This stable depth is called the limit depth of $I$, and is denoted by $\lim_{k\rightarrow\infty}\operatorname{depth}R/I^{k}$. These indices of stability have been studied and compared to some extend in [8] and [9]. The purpose of this work is to compare once again these stability indices. The main result is that if $(R,{\mathfrak{m}})$ is a regular local ring with $\dim R\leq 2$, then all 3 stability indices are equal, but if $\dim R=3$, then we still have $\operatorname{astab}(I)=\operatorname{dstab}(I)$, while $\operatorname{astab}(I)$ and $\overline{\operatorname{astab}}(I)$ may differ by any amount. On the other hand, if $\dim R\geq 4$, we will show by examples that in general a comparison between these stability indices is no longer possible. In other words, any inequality between these invariants may occur.

Quite often, but not always, $\operatorname{depth}(R/I^{k})$ is a non-increasing function on $n$. In the last section we prove that if $(R,{\mathfrak{m}})$ is a $3$-dimensional regular local ring and $I$ satisfies $I^{k+1}:I=I^{k}$ for all $k$, then $\operatorname{depth}R/I^{k}$ is non-increasing. For any unexplained notion or terminology, we refer the reader to [5].

Several explicit example were performed with help of the computer algebra systems CoCoA [1] and Macaulay2 [6], as well as with the program in [2] which allows to compute $\operatorname{Ass}^{\infty}(I)$ of a monomial ideal $I$.

1. The case $\dim R\leq 3$

In this section we study the behavior of the stability indices for regular rings of dimension $\leq 3$. In the proofs we will use the following elementary and well-known fact: let $I\subset R$ be an ideal of height $1$ in the regular local ring $R$. Then there exists $f\in R$ such that $I=fJ$ where either $J=R$ or otherwise $\operatorname{height}J>1$. Indeed, let $I=(f_{1},\ldots,f_{m})$. Since $R$ is factorial, the greatest common divisor of $f_{1},\ldots,f_{m}$ exists. Let $f=\gcd(f_{1},\ldots,f_{m})$, and $g_{i}=f_{i}/f$ for $i=1,\ldots,m$. Then $I=fJ$, where $J=(g_{1},\ldots,g_{m})$. Suppose $\operatorname{height}J=1$, then there exist a prime ideal $P$ of height 1 with $J\subset P$. Since $R$ is regular, $P$ is a principal ideal, say $P=(g)$. It follows then that $g$ divides all $g_{i}$, but $\gcd(g_{1},\ldots,g_{m})=1$, a contradiction.

We first observe

Remark 1.1**.**

Let $(R,{\mathfrak{m}})$ be a regular local ring with $\dim R\leq 2$ and let $I$ be an ideal of $R$. Then

[TABLE]

Proof.

If $\dim R\leq 1$, then either $R$ is a field or a principal ideal domain, and the statement is trivial. Now suppose that $\dim R=2$ and $I\neq 0$. If $\operatorname{height}I=2$, then ${\mathfrak{m}}$ belongs to $\operatorname{Ass}(I^{k})$ and $\operatorname{Ass}(\overline{I^{k}})$ for all $k$, and the assertion is trivial. Hence, we may assume that $\operatorname{height}(I)=1$. Then $I=fJ$ with $J=R$ or $\operatorname{height}J=2$. In the first case $I$ is a principal ideal, and the assertion is trivial. In the second case, $I^{k}=f^{k}J^{k}$ for all $k$, and $J^{k}$ is ${\mathfrak{m}}$-primary. Thus there exists $g\not\in J^{k}$ with $g{\mathfrak{m}}\in J^{k}$. Then $gf^{k}\not\in f^{k}J^{k}$ and $gf^{k}{\mathfrak{m}}\in f^{k}J^{k}$. This shows that in the second case ${\mathfrak{m}}\in\operatorname{Ass}(I^{k})$ for $k$, so that $\operatorname{astab}(I)=\operatorname{dstab}(I)=1$.

Finally observe that in the second case, $\overline{I^{k}}=f^{k}\overline{J^{k}}$ for all $k$. This shows that ${\mathfrak{m}}\in\operatorname{Ass}(\overline{I^{k}})$ for all $k$, so that also in this case $\overline{\operatorname{astab}}(I)=1$.

Theorem 1.2**.**

Let $(R,{\mathfrak{m}})$ be a regular local ring with $\dim R\leq 3$ and $I$ be an ideal of $R$. Then $\operatorname{astab}(I)=\operatorname{dstab}(I)$.

Proof.

By Remark 1.1, we may assume that $\dim R=3$. If $\operatorname{height}I\geq 2$, then $\operatorname{Ass}(I^{k})\subseteq\operatorname{Min}(I)\cup\{{\mathfrak{m}}\}$ for all $k$. This implies at once that $\operatorname{astab}(I)=\operatorname{dstab}(I)$. Now suppose that $\operatorname{height}I=1$. If $I$ is a principal ideal, then the assertion is again trivial. Otherwise, $I=fJ$ with $\operatorname{height}J\geq 2$. Since $I^{k}$ is isomorphic to $J^{k}$ as an $R$-module, it follows that $\operatorname{proj\,dim}I^{k}=\operatorname{proj\,dim}J^{k}$ for all $k$. This implies that $\operatorname{proj\,dim}R/I^{k}=\operatorname{proj\,dim}R/J^{k}$ for all $k$, and consequently $\operatorname{depth}R/I^{k}=\operatorname{depth}R/J^{k}$, by the Auslander–Buchsbaum formula. Thus, $\operatorname{dstab}(I)=\operatorname{dstab}(J)$.

We claim that $\operatorname{astab}(I)=\operatorname{astab}(J)$. Since we have already seen that $\operatorname{astab}(J)=\operatorname{dstab}(J)$ if $\operatorname{height}J\geq 2$, the claim then implies that $\operatorname{astab}(I)=\operatorname{dstab}(I)$, as desired.

The claim follows once we have shown that $\operatorname{Ass}(I^{k})=\operatorname{Ass}(f^{k}J^{k})=\operatorname{Min}(f)\cup\operatorname{Ass}(J^{k})$. For that we only need to prove the second equation. So let $P\in\operatorname{Spec}R$ with $f^{k}J^{k}\subset P$. Then $P\in\operatorname{Ass}(f^{k}J^{k})$ if and only if $\operatorname{depth}R_{P}/f^{k}J^{k}R_{P}=0$. If $J\not\subset P$, then $f^{k}J^{k}R_{P}=f^{k}R_{P}$, and hence $\operatorname{depth}R_{P}/f^{k}J^{k}R_{P}=0$ if and only if $\operatorname{depth}R_{P}/f^{k}R_{P}=0$, and this is the case if and only if $P\in\operatorname{Min}(f)$. If $J\subset P$, then the $R_{P}$-modules $f^{k}J^{k}R_{P}$ and $J^{k}R_{P}$ are isomorphic, so that with the arguments as above $\operatorname{depth}R_{P}/f^{k}J^{k}R_{P}=\operatorname{depth}R_{P}/J^{k}R_{P}$, which shows that in this case $P\in\operatorname{Ass}(f^{k}J^{k})$ if and only if $P\in\operatorname{Ass}(J^{k})$. This completes the proof.

The statements shown so far and its proofs made for ideals in a regular local ring are valid as well for any graded ideal in a polynomial ring.

We now turn to some explicit examples. In [10, Proposition 1.5] Hibi et al show that for any integer $t\geq 2$ the ideal $I=(x^{t},xy^{t-2}z,y^{t-1}z)\subset K[x,y,z]$ satisfies $\operatorname{dstab}(I)=t$. Since by Theorem 1.2, $\operatorname{astab}(I)=\operatorname{dstab}(I)$, this example shows that in a $3$-dimensional graded or local ring (we may pass to $K[|x,y,z|]$) both the index of depth stability as well as the index of Ass stability may be any given number.

The following example shows that already for an ideal $I$ in a $3$-dimensional polynomial ring the invariants $\operatorname{astab}(I)$ and $\overline{\operatorname{astab}}(I)$ may differ.

Example 1.3**.**

Let $R=K[x,y,z]$ be a polynomial ring over a field $K$ and let $I=((xy)^{2},(xz)^{2},(yz)^{2})\subset R$. Then $\operatorname{astab}(I)=2$ and $\overline{\operatorname{astab}}(I)=1$.

Proof.

We first claim that $I^{n}:(xy)^{2}=I^{n-1}+z^{2n}(x^{2},y^{2})^{n-2}$. Indeed, let $J=((xz)^{2},(yz)^{2})$. Then $I^{n}=J^{n}+(xy)^{2}I^{n-1}$, and hence $I^{n}:(xy)^{2}=J^{n}:(xy)^{2}+I^{n-1}$. Since $J^{n}:(xy)^{2}=z^{2n}(x^{2},y^{2})^{n}:(xy)^{2}=z^{2n}(x^{2},y^{2})^{n-2}$, the assertion follows.

By symmetry, we also have $I^{n}:(xz)^{2}=I^{n-1}+y^{2n}(x^{2},z^{2})^{n-2}$ and $I^{n}:(yz)^{2}=I^{n-1}+x^{2n}(y^{2},z^{2})^{n-2}$. Thus, for all $n\geq 1$ we obtain

[TABLE]

In other words, $I$ satisfies strong persistence in the sense of [8]. In particular, $\operatorname{Ass}(I^{n})\subset\operatorname{Ass}(I^{n+1})$ for all $n\geq 1$. Now since $\operatorname{Ass}(I)=\{(x,y),(x,z),(y,z)\}$ and $\operatorname{Ass}(I^{2})=\{(x,y),(x,z),(y,z),(x,y,z)\}$, we deduce from this that $\operatorname{astab}(I)=\operatorname{dstab}(I)=2$.

With Macaulay2 it can be checked that $\overline{I}=\{(xy)^{2},(xz)^{2},(yz)^{2},xyz^{2},xy^{2}z,x^{2}yz\}$ and that $\operatorname{Ass}(\overline{I})=\{(x,y),(x,z),(y,z),(x,y,z)\}$. By [12, Corollary 11.28], one has $\operatorname{Ass}(\overline{I})\subset\operatorname{Ass}(\overline{I^{2}})\subset\cdots\subset\overline{\operatorname{Ass}}^{\infty}(I)$. Since $\operatorname{Ass}(\overline{I^{n}})$ is a subset of the monomial prime ideals containing $I$, and since this set is $\{(x,y),(x,z),(y,z),(x,y,z)\}$, we see that $\operatorname{Ass}(\overline{I})=\operatorname{Ass}(\overline{I^{n}})$ for all $n$. Hence, $\overline{\operatorname{astab}}(I)=1$.

The next result says that the difference $\operatorname{astab}(I)-\overline{\operatorname{astab}}(I)$ may indeed be as big as we want.

Theorem 1.4**.**

Let $R=k[x,y,z]$ be the polynomial ring over a field $K$, $c$ a positive integer and $I=(x^{2c+2},xy^{2c}z,y^{2c+2}z)$. Then $\operatorname{astab}(I)=c+2$ and $\overline{\operatorname{astab}}(I)=2$.

Proof.

Note that $I=(x^{2c+2},z)\cap(x,y^{2c+2})\cap(y^{2c},x^{2c+2})$, from which it follows that $\dim R/I=\operatorname{depth}R/I=1.$

In the next step we prove that $I^{n}:I=I^{n-1}$ for all $n$. Then [8, Theorem 1.3] implies that $\operatorname{Ass}(I^{n})\subseteq\operatorname{Ass}(I^{n+1})$ for all $n$. In particular, if $\operatorname{depth}(R/I^{k})=0$ for some $k$, then $\operatorname{depth}(R/I^{r})=0$ for all $r\geq 0$. Since $\operatorname{depth}R/I^{k}\leq 1$ for all $k$, it then follows that $\operatorname{depth}(R/I^{k})\geq\operatorname{depth}(R/I^{k+1})$ for all $k$.

In order to show that $I^{n}:I=I^{n-1}$, observe that

[TABLE]

and that

[TABLE]

Similarly we have

[TABLE]

Now since

[TABLE]

it follows $I^{n}:I=I^{n-1}$ for all $n$, as desired.

Next we claim that $I^{n}:x^{2c+2}=I^{n-1}$ for all $n\leq{c+1}$.

If $n=1$, there is nothing to prove. Let $1<n\leq{c+1}$. By a calculation as before we see that

[TABLE]

We proceed by induction $n$ to show that $\operatorname{depth}S/I^{n}=1$ for $n\leq c+1$. We observed already that $\operatorname{depth}S/I=1$, Now let $1<n\leq c+1$. Then, since $I^{n}:x^{2c+2}=I^{n-1}$, we obtain the exact sequence

[TABLE]

Since by induction hypothesis $\operatorname{depth}R/I^{n-1}=1$, it follows that

[TABLE]

Note that $(I^{n},x^{2c+2})=((y^{2c}z(x,y^{2})^{n},x^{2c+2})$, which implies that $\operatorname{depth}R/(I^{n},x^{2c+2})=1$. Thus we have $\operatorname{depth}R/I^{n}\geq 1$. On the other hand, we have seen before that $\operatorname{depth}R/I^{n}\leq\operatorname{depth}R/I=1$, and so $\operatorname{depth}R/I^{n}=1$ for all $n\leq{c+1}$.

In the next step we show that $\operatorname{depth}R/I^{c+2}=0$, which then implies that $\operatorname{depth}R/I^{n}=0$ for all $n\geq c+2$. In particular, it will follows that $\operatorname{astab}(I)=c+2$.

In order to prove that $\operatorname{depth}R/I^{c+2}=0$, we show that $x^{2c+2}y^{(c+1)(2c+2)-1}z^{c+1}\in(I^{c+2}:\mathfrak{m})\setminus I^{c+2}$.

Indeed, let $u=x^{2c+2}y^{(c+1)(2c+2)-1}z^{c+1}$. Then

[TABLE]

and

[TABLE]

This shows that $u\in(I^{c+2}:\mathfrak{m}).$

Assume that $x^{2c+2}y^{(c+1)(2c+2)-1}z^{c+1}\in I^{c+2}$. Then

[TABLE]

and so $y^{(c+1)(2c+2)-1}z^{c+1}\in I^{c+1}$. Since $I^{c+1}=(x^{2c+2},y^{2c}z(x,y^{2}))^{c+1}$, expansion of this power implies that $y^{(c+1)(2c+2)-1}\in\sum_{i=0}^{c+1}(x^{2c+2})^{i}(y^{2c}(x,y^{2}))^{c+1-i}$. It follow that $y^{(c+1)(2c+2)-1}\in(y^{2c}(x,y^{2}))^{c+1}$, which is a contradiction.

Now we compute $\overline{\operatorname{astab}}(I)$, and first prove that

[TABLE]

Let $J=(I,(x^{3}y^{2c-1}z,x^{4}y^{2c-2}z,\ldots,x^{2c+1}yz))$. For all integers $i$ with $3\leq i\leq{2c+1}$, we have

[TABLE]

Thus $J\subseteq\overline{I}$. We have $\operatorname{Ass}(\overline{I}/J)\subseteq\operatorname{Ass}(J)$. The primary decomposition of $J$ shows that $\operatorname{Ass}(J)=\{(x,z),(x,y)\}$. Let $P=(x,z)$. Then $(\overline{I})_{P}=\overline{(I_{P})}=\overline{(x^{2c+2},z)_{P}}=(x^{2c+2},z)_{P}$. The last equality follows by [11, Proposition 1.3.5], and so $(\overline{I}/J)_{P}=0$. Hence $P\notin\operatorname{Ass}(\overline{I}/J)$. Now let $P=(x,y)$. Then

[TABLE]

The second equality follows by [11, Exercise 1.19]. Thus we have $(\overline{I}/J)_{P}=0$. This shows that $\operatorname{Ass}(\overline{I}/J)=\emptyset$, and hence $\overline{I}=J$, as desired. In particular we see that

[TABLE]

Since $\operatorname{Ass}(\overline{I})\subseteq\operatorname{Ass}(\overline{I^{k}})$ for all $k$, it follows that $\{(x,z),(x,y)\}\subset\operatorname{Ass}(I^{k})$ for all $k$. Suppose that $(y,z)\in\operatorname{Ass}(\overline{I^{k}})$ for some $k$. Then $(y,z)$ is a minimal prime ideal of $I$. However, this is not the case, as can be seen from the primary decomposition of $I$.

Next we show that ${\mathfrak{m}}=(x,y,z)$ belongs to $\operatorname{Ass}(\overline{I^{2}})$. Then it follows that

[TABLE]

thereby showing that $\overline{\operatorname{astab}}(I)=2$.

In order to prove that ${\mathfrak{m}}\in\operatorname{Ass}(\overline{I^{2}})$, we first show that the ideal

[TABLE]

is contained in $\overline{I^{2}}$.

Since

[TABLE]

it follows that for all integers $i$ with $4\leq i\leq{2c+2}$ the element

[TABLE]

belongs to $(I^{2})^{4c}$. Also, for all integers $i$ with $5\leq i\leq{2c-2}$, the element

[TABLE]

belongs to $(I^{2})^{4c}$, This shows $L\subseteq\overline{I^{2}}$.

By using primary decomposition for the ideal $L$, we see that

[TABLE]

On the other hand, by easy calculation, one verifies that $L:(x^{2c+2}y^{2c+1}z)=\mathfrak{m}$. Finally we show that $x^{2c+2}y^{2c+1}z\notin\overline{I^{2}}$, which then implies that ${\mathfrak{m}}\in\operatorname{Ass}(\overline{I^{2}})$, as desired.

In order to prove this we show by induction on $n$ that $(x^{2c+2}y^{2c+1}z)^{n}\notin(I^{2})^{n}$ for all $n$. Indeed, let $n=1$, and assume that $x^{2c+2}y^{2c+1}z\in I^{2}$. Then $y^{2c+1}z\in I^{2}:x^{2c+2}=I+(y^{2c}z)^{2}=I$, and this is contradiction.

Now let $n>1$. By induction hypothesis we may assume that $(x^{2c+2}y^{2c+1}z)^{n-1}\notin(I^{2})^{n-1}$. If $(x^{2c+2}y^{2c+1}z)^{n}\in(I^{2})^{n}$, then $x^{(2c+2)(n-1)}(y^{2c+1}z)^{n}\in(I^{2n}:x^{2c+2})=I^{2n-1}+(y^{2c}z)^{2n}(x,y^{2})^{2n-2(c+1)}$, and so $x^{(2c+2)(n-1)}(y^{2c+1}z)^{n}\in I^{2n-1}$.

It follows that $x^{(2c+2)(n-1)}(y^{2c+1}z)^{n-1}\in(I^{2n-1}:y^{2c+1}z)$. Since

[TABLE]

we see that $x^{(2c+2)(n-1)}(y^{2c+1}z)^{n-1}\in y(I^{2})^{n-1}$, a contradiction.

Thus $(x^{2c+2}y^{2c+1}z)^{n}\notin(I^{2})^{n}$ for all $n$, as desired.

The theorem says that for any positive integer $c$ there exists a monomial ideal in $K[x,y,z]$ with $\operatorname{astab}(I)-\overline{\operatorname{astab}}(I)=c$. However we do not know whether for all ideals in $I\subset K[x,y,z]$ one has $\overline{\operatorname{astab}}(I)\leq\operatorname{astab}(I)$.

2. The case $\dim R>3$

The purpose of this section is to show that for a polynomial ring $S$ in more than $3$ variables, for a graded ideal $I\subset S$ the invariants $\operatorname{astab}(I)$ and $\operatorname{dstab}(I)$ may differ by any amount.

We begin with two examples.

Example 2.1**.**

Let $R=k[x,y,z,u]$ be the polynomial ring over a field $k$ and consider the ideal $I=(xy,yz,zu)$ of $R$. Then $\operatorname{astab}(I)=1$ and $\operatorname{dstab}(I)=2$.

Proof.

We have $\operatorname{Ass}(I)=\operatorname{Min}(I)$, and since I may be viewed as the edge ideal of a bipartite graph it follows from [7, Definition 1.4.5, Corollary 10.3.17] that $\operatorname{Ass}(I)=\operatorname{Ass}(I^{n})$ for all $n\in\mathbb{N}$. Therefore $\operatorname{astab}(I)=1$. By [7, Corollary 10.3.18], $\lim_{k\rightarrow\infty}\operatorname{depth}R/I^{k}=1$. Moreover, it can be seen that $\operatorname{depth}R/I=2$ and $\operatorname{depth}R/I^{2}=1$. Since $I$ has a linear resolution, [7, Theorem 10.2.6] implies that for all $k\geq 1$, $I^{k}$ has a linear resolution as well. Therefore, by [9, Proposition 2.2] we have $\operatorname{depth}R/I^{k+1}\leq\operatorname{depth}R/I^{k}$ for all $k\in\mathbb{N}$. Hence $\operatorname{depth}R/I^{k}=1$ for all $k\geq 2$, and so $\operatorname{dstab}(I)=2$.

Example 2.2**.**

Let $R=K[x,y,z,u]$ be the polynomial ring in 4 variables over a field $K$, and let $I=(x^{2}z,uyz,u^{3})$. Then $\operatorname{astab}(I)=2$ and $\operatorname{dstab}(I)=1$.

Proof.

Set $J=(uyz,u^{3})$ and so for all $n\in\mathbb{N}$, it therefore follows that $I^{n}:x^{2}z=(J^{n}+x^{2}zI^{n-1}):x^{2}z=I^{n-1}+(J^{n}:x^{2}z)=I^{n-1}$. Hence, for all $n\in\mathbb{N}$, $\operatorname{Ass}(I^{n})\subseteq\operatorname{Ass}(I^{n+1})$. By using Macaulay2 [6] and the program [2], we see that $\operatorname{Ass}^{\infty}(I)=\operatorname{Ass}(I^{2})=\{(x,u),(z,u),(x,y,u),(x,z,u)\}$. Therefore $\operatorname{astab}(I)=2$. As $\operatorname{Ass}(I^{n})\subseteq\operatorname{Ass}(I^{n+1})$ for all $n\in\mathbb{N}$, it follows that $\mathfrak{m}=(x,y,z,u)\notin\operatorname{Ass}(I^{n})$ and so we have $\operatorname{depth}R/I^{n}\geq 1$. $y-z\in\mathfrak{m}$ is a nonzerodivisor on $R/I^{n}$ for all $n\in\mathbb{N}$. Set $\overline{R}=R/(y-z)$. Thus by [5, Lemma 4.2.16] we have $\overline{(R/I^{n})}=\overline{R}/{\overline{I^{n}}}\cong K[x,z,u]/(x^{2}z,uz^{2},u^{3})^{n}$. Since $xzu^{3n-1}\in(\overline{I^{n}}):\overline{\mathfrak{m}}\setminus\overline{I^{n}}$, it follows $\operatorname{depth}\overline{R}/{\overline{I^{n}}}=0$ and so $\operatorname{depth}R/I^{n}=1$ for all $n\in\mathbb{N}$. Therefore $\operatorname{dstab}(I)=1$.

Now we come to the main result of this section

Theorem 2.3**.**

Let $R=k[x,y,z,u]$ be the polynomial ring over a field $k$. Then for any non-negative integer $c$, there exist two ideals $I$ and $J$ of $R$ such that the following statements hold:

• (i)

$\operatorname{astab}(I)-\operatorname{dstab}(I)\geq c$.

• (ii)

$\operatorname{dstab}(J)-\operatorname{astab}(J)\geq c$.

Proof.

We may assume that $c$ is a positive integer. Let $I=(x^{c+1}z^{c},u^{2c-1}yz,u^{2c+1})$ and $J=(x^{c}y^{c-1},y^{c-1}x^{c-1}z,z^{c}u^{c})$. We claim that ${\operatorname{astab}(I)=\operatorname{dstab}(J)=c+1}$ and ${\operatorname{astab}(J)=\operatorname{dstab}(I)=1}$.

(i) In this case, by using Example 2.2, we can assume that $c\geq 2$. For all $n\in\mathbb{N}$, we have $(I^{n}:x^{c+1}z^{c})=(((u^{2c-1}yz,u^{2c+1})^{n}+x^{c+1}z^{c}I^{n-1}):x^{c+1}z^{c})=I^{n-1}+((u^{2c-1}yz,u^{2c+1})^{n}:x^{c+1}z^{c})$. Since $((u^{2c-1}yz,u^{2c+1})^{n}:x^{c+1}z^{c})=((u^{2c-1}yz,u^{2c+1})^{n}:z^{c})\subseteq I^{n-1}$, it follows $(I^{n}:x^{c+1}z^{c})=I^{n-1}$ and so $\operatorname{Ass}(I^{n})\subseteq\operatorname{Ass}(I^{n+1})$. By using Macaulay2 [6] and [2], we have $\operatorname{Ass}(I)=\{(x,u),(z,u),(y,z,u),(x,y,u)\}$ and $\operatorname{Ass}^{\infty}(I)=\{(x,u),(z,u),(y,z,u),(x,z,u),(x,y,u)\}$. Set ${\mathfrak{p}=(x,z,u)}$. It is easily seen that $I^{i}:\mathfrak{p}=I^{i}$ for all ${i\leq c}$ and $x^{c}y^{c+1}z^{c}u^{(2c+1)c}\in(I_{\mathfrak{p}}^{c+1}:\mathfrak{p})\setminus I_{\mathfrak{p}}^{c+1}$. Hence $\operatorname{Ass}(I)=\operatorname{Ass}(I^{2})=...=\operatorname{Ass}(I^{c})$, $\operatorname{Ass}(I^{c+1})=\operatorname{Ass}^{\infty}(I)$ and so ${\operatorname{astab}(I)=c+1}$. By the same argument as used in the proof of Example 2.2, that $\mathfrak{m}=(x,y,z,u)\notin\operatorname{Ass}(I^{n})$ for all $n\in\mathbb{N}$ and so we have $\operatorname{depth}R/I^{n}\geq 1$ and $x-y-z\in\mathfrak{m}$ is a nonzerodivisor on $R/I^{n}$ for all $n\in\mathbb{N}$. Therefore $\overline{(R/I^{n})}=\overline{R}/{\overline{I^{n}}}\cong K[y,z,u]/((y+z)^{c+1}z^{c},u^{2c-1}yz,u^{2c+1})^{n}$, where $\overline{R}=R/(x-y-z)$. Since $z^{2c}u^{(2c+1)n-1}\in(\overline{I^{n}}):\overline{\mathfrak{m}}\setminus\overline{I^{n}}$, it follows $\operatorname{depth}\overline{R}/{\overline{I^{n}}}=0$ and so $\operatorname{depth}R/I^{n}=1$ for all $n\in\mathbb{N}$. Therefore $\operatorname{dstab}(I)=1$.

(ii) For all $n\in\mathbb{N}$, we have $(J^{n}:z^{c}u^{c})=(((x^{c}y^{c-1},y^{c-1}x^{c-1}z)^{n}+z^{c}u^{c}J^{n-1}):z^{c}u^{c})=J^{n-1}+((x^{c}y^{c-1},y^{c-1}x^{c-1}z)^{n}:z^{c}u^{c})=J^{n-1}+((x^{c}y^{c-1},y^{c-1}x^{c-1}z)^{n}:z^{c})$. Since $((x^{c}y^{c-1},y^{c-1}x^{c-1}z)^{n}:z^{c})\subseteq J^{n-1}$, for all $n\in\mathbb{N}$ we have $(J^{n}:z^{c}u^{c})=J^{n-1}$. Therefore, for all $n\in\mathbb{N}$, $\operatorname{Ass}(J^{n})\subseteq\operatorname{Ass}(J^{n+1})$. By using [6] and [2] we have $\operatorname{Ass}^{\infty}(J)=\{(x,z),(x,u),(y,z),(y,u)\}=\operatorname{Min}(J)$ and so $\operatorname{astab}(J)=1$. Since $\mathfrak{m}\notin\operatorname{Ass}(J^{n})$ for all $n\in\mathbb{N}$, we have $2=\dim R/J\geq\operatorname{depth}R/J^{n}\geq 1$ and $x-y\in\mathfrak{m}$ is a nonzerodivisor on $R/J^{n}$ for all $n\in\mathbb{N}$. Again by the above argument, $\overline{(R/J^{n})}=\overline{R}/{\overline{J^{n}}}\cong K[x,z,u]/(x^{2c-1},x^{2c-2}z,z^{c}u^{c})^{n}$, where $\overline{R}=R/(x-y)$. Since $\overline{J^{i}}:\overline{\mathfrak{m}}=\overline{J^{i}}$ for all $i\leq c$ and $x^{(2c-1)n}z^{n-1}u^{c-1}\in\overline{J^{n}}:\overline{\mathfrak{m}}\setminus\overline{J^{n}}$ for all ${n\geq c+1}$. It therefore follows $\operatorname{depth}R/J=\operatorname{depth}R/J^{2}=...=\operatorname{depth}R/J^{c}=2$ and $\operatorname{depth}R/J^{n}=1$ for all ${n\geq c+1}$. Hence ${\operatorname{dstab}(J)=c+1}$.

3. Non-increasing depth functions

Theorem 3.1**.**

Let $(R,{\mathfrak{m}})$ be a regular local ring with $\dim R=3$ and $I$ be an ideal of $R$. If $I^{n+1}:I=I^{n}$ for all $n\in\mathbb{N}$, then $\operatorname{depth}R/I^{n}$ is non-increasing.

Proof.

Suppose $\operatorname{height}(I)\geq 2$. Since $I^{n+1}:I=I^{n}$ for all $n\in\mathbb{N}$, it follows that $\operatorname{depth}R/I^{n+1}\leq\operatorname{depth}R/I^{n}$. Now, let $\operatorname{height}(I)=1$. Then there exists an ideal $J$ of $R$ and an element $f\in R$ such that $I=f{J}$ and $\operatorname{height}(J)\geq 2$. As in the proof of Theorem 1.2, $\operatorname{depth}R/I^{n}=\operatorname{depth}R/J^{n}$ for all $n\in\mathbb{N}$. Since $I^{n+1}:I=I^{n}$ for all $n\in\mathbb{N}$, we have $J^{n+1}:J=J^{n}$. Thus $\operatorname{depth}R/J^{n+1}\leq\operatorname{depth}R/J^{n}$ and so $\operatorname{depth}R/I^{n+1}\leq\operatorname{depth}R/I^{n}$. This completes the proof.

Corollary 3.2**.**

• (i)

Let $(R,{\mathfrak{m}})$ be regular local ring with $\dim R=3$. Then $\operatorname{depth}R/\overline{I^{n}}$ is non-increasing.

• (ii)

Let $R=k[x,y,z]$ be a polynomial ring in $3$ indeterminates over a field $k$. If $I$ is an edge ideal of $R$, then $\operatorname{depth}R/I^{n}$ is non-increasing.

Example 3.3**.**

Let $R=k[x,y,z,u]$ be a polynomial ring and $I=(xy^{2}z,yz^{2}u,zu^{2}(x+y+z+u),xu(x+y+z+u)^{2},x^{2}y(x+y+z+u))$ be an ideal of $R$. Then $\operatorname{depth}R/I=\operatorname{depth}R/I^{4}=0$, $\operatorname{depth}R/I^{2}=\operatorname{depth}R/I^{3}=1$. Thus the depth function is neither non-increasing nor non-decreasing.**

In view of Theorem 3.1 one may ask whether in a regular local ring (of any dimension), $\operatorname{depth}R/I^{n}$ is a non-increasing function of $n$, if $I^{n+1}:I=I^{n}$ for all $n$.

Acknowledgements

The second author would like to thank Universität Duisburg-Essen, especially to the Department of Mathematics for its hospitality during the preparation of this work.