$n$-absorbing monomial ideals in polynomial rings

TL;DR

This paper investigates the invariant (I) for monomial ideals in polynomial rings, calculating its value for specific cases, thereby enhancing understanding of n-absorbing ideals in algebraic structures.

Contribution

It provides explicit calculations of (I) for certain monomial ideals in polynomial rings, extending the theory of n-absorbing ideals.

Findings

(I) values are computed for specific monomial ideals

The work advances understanding of n-absorbing properties in polynomial rings

Results contribute to the classification of monomial ideals based on (I)

Abstract

In a commutative ring $R$ with unity, given an ideal $I$ of $R$, Anderson and Badawi in 2011 introduced the invariant $\omega(I)$, which is the minimal integer $n$ for which $I$ is an $n$-absorbing ideal of $R$. In the specific case that $R = k[x_{1}, \ldots, x_{n}]$ is a polynomial ring over a field $k$ in $n$ variables $x_{1},\ldots, x_{n}$, we calculate $\omega(I)$ for certain monomial ideals $I$ of $R$.

Equations

$$I\cap J=(\{lcm(u,v)\mid u\in G(I),v\in G(J)\}).$$

$$\omega (Q)={\omega }^{\bullet }(Q)=\underset{1\le i\le r}{\max }\{e(Q_i)\}.$$

$$\omega (I)={\omega }^{\bullet }(I)=\max \{1+2+2-3+1,4+1+2-3+1,1+3+1-3+1\}=5.$$

$$f:=x_{i_1}^{a_1-1}\cdot \cdot \cdot x_{i_l}^{a_l-1}(x_{i_1}+\cdot \cdot \cdot +x_{i_l}).$$

$$f_{i,j}=x_{i_{j}}+\sum\limits_{t\neq j}x_{i_{t}}^{2}\text{ and }f_{i}=\displaystyle\Big{(}\sum\limits_{l=1}^{s_{i}}x_{i_{l}}\Big{)}\Big{(}\prod_{j=1}^{s_{i}}f_{i,j}^{a_{j}-1}\Big{)}.$$

$$\omega (I)$$

$$\displaystyle\leq\omega\Big{(}Q_{1}\cap Q_{2}\Big{)}+\omega\Big{(}\bigcap_{i\neq 1,2}Q_{i}\Big{)}$$

$$\displaystyle=\max\{e(Q_{1}),e(Q_{2})\}+\omega\Big{(}\bigcap_{i\neq 1,2}Q_{i}\Big{)}$$

$$\le \max \{e(Q_1),e(Q_2)\}+\sum _{i\ne 1,2}e(Q_i)$$

$$<\sum _{i=1}^{r}e(Q_i).$$

$$\omega(I)=\omega^{\bullet}(I)=\left\{\begin{array}[]{ll}\max\{e(Q_{k}),\sum\limits_{i\neq k}e(Q_{i})\}&\text{ if }\sqrt{Q_{k}}=\mathfrak{m}\text{ for some }k\in\{1,\cdot\cdot\cdot,r\}.\\ \sum\limits_{i=1}^{r}e(Q_{i})&\text{ otherwise }.\\ \end{array}\right.$$

$$\omega (I)$$

$$=5+\max \{e((x^3,y^2,z^2,xy)),e((x^2,y))+e((y,z))\}$$

$$=5+\max \{4,2+1\}$$

$$=9.$$

$$\omega(J)=\omega^{\bullet}(J)=\left\{\begin{array}[]{ll}a_{1}+b_{1}&\text{ if }r=1.\\ \max_{1\leq i\leq r-1}\{a_{i}+b_{i+1}\}-1&\text{ if }r>1.\\ \end{array}\right.$$

$$\omega (J)={\omega }^{\bullet }(J)$$

$$=\max \{\underset{1\le i\le r-1}{\max }\{e((x^{a_i},y^{b_{i+1}}))\},a_r+b_1\}$$

$$=\max \{\underset{1\le i\le r-1}{\max }\{a_i+b_{i+1}-1\},a_r+b_1\}$$

$$=\underset{1\le i\le r-1}{\max }\{a_i+b_{i+1}\}-1.$$

$$\omega (I\cap J)$$

$$\le \max \{\underset{1\le i\le r}{\max }\{e(Q_i)\},\underset{1\le j\le s}{\max }\{e(T_j)\}\}$$

$$=\max \{\omega (I),\omega (J)\}.$$

$$I$$

$$J$$

$$I\cap J$$

$$I+J$$

$$\omega (I^n)$$

$$\iff n(deg(f))+\omega (I^n)$$

$$\iff deg(f^n)+\omega (I^n)$$

$$\iff \omega ((fI{)}^n)$$

$$\omega (IJ)$$

$$=\max \{\underset{1\le i\le r-1}{\max }\{a_i+b_{i+1}+c_1\}-1,\underset{1\le j\le s-1}{\max }\{c_j+d_{j+1}+b_r\}-1\}$$

$$=\max \{\omega (I)+c_1,\omega (J)+b_r\}$$

$$=\max \{a_1+c_1,b_r+d_s\}.$$

$$f^k+c_1{f}^{k-1}+\cdots +c_{k-1}f+c_k$$

$$I^m=\bigcap _{P\text{ is a minimal prime of I}}{P}^m$$

Full text

1. Introduction

Throughout this paper, we set $\mathbb{N}:=\{1,2,\ldots,\}$ and $\mathbb{N}_{0}:=\{0,1,2,\ldots\}$ and $R$ will denote a commutative ring with unity. Given a nonzero ideal $I$ of $R$, Ass$(R/I)$ will denote the set of associated primes of $I$ in $R$. The primary notion we are interested in this paper is the following:

Definition 1**.**

Let $n\in\mathbb{N}$, $R$ a commutative ring with unity, and $I$ an ideal of $R$. $I$ is said to be an n-absorbing ideal of a ring $R$ if for any $x_{1},\ldots,x_{n+1}\in R$ such that $x_{1}\cdots x_{n+1}\in I$, there is some $j$ with $1\leq j\leq n+1$ such that $x_{1}\cdots x_{j-1}x_{j+1}\cdots x_{n+1}\in I$. $I$ is said to be a strongly n-absorbing ideal of a ring $R$ if for any ideals $I_{1},\ldots,I_{n+1}$ of $R$ such that $I_{1}\cdots I_{n+1}\subseteq I$, there is some $j$ with $1\leq j\leq n+1$ such that $I_{1}\cdots I_{j-1}I_{j+1}\cdots I_{n+1}\subseteq I$.

(Strongly) $2$-absorbing ideals were initially defined and investigated by Badawi in [3] as a generalization of prime ideals, which are precisely the proper $1$-absorbing ideals. In 2011, Anderson and Badawi together generalized this further to the notion of a (strongly) $n$-absorbing ideal for any $n\in\mathbb{N}$ defined above in [2]. For an ideal $I$ in a ring $R$, we let $\omega(I)$ denote the minimal integer $n\in\mathbb{N}$ such that $I$ is $n$-absorbing. In a general ring, $I$ may not be $n$-absorbing for any $n\in\mathbb{N}$, in which case we set $\omega(I)=\infty$. Similarly, we can define the invariant $\omega^{\bullet}(I)$ to be the smallest integer $n\in\mathbb{N}$ for which an ideal $I$ is strongly $n$-absorbing, and set $\omega^{\bullet}(I)=\infty$ if no such integer exists. We set $\omega(R)=\omega^{\bullet}(R)=0$. It is easy to see that $\omega(I)\leq\omega^{\bullet}(I)$ holds for each ideal $I$ of $R$. In fact, Anderson and Badawi in Conjecture $1$ of [2, page 1669] postulate that $\omega(I)=\omega^{\bullet}(I)$ holds for any ideal $I$ in arbitrary ring $R$; that is, they conjecture that the notion of an $n$-absorbing ideal and strongly $n$-absorbing ideal coincide. As of this writing, this problem remains open. However, it is known that the conjecture holds true for any $n\in\mathbb{N}$ if $R$ is a Prüfer domain ([2, Corollary 6.9]) or a commutative algebra over an infinite field([7]), and for any ring $R$ if $n=2$ ([3, Theorem 2.13]). The interested reader may refer to the survey article [4, Section 5] for further information on strongly $n$-absorbing ideals. Recall that for an ideal $I$ in a ring $R$, the Noether exponent of $I$, denoted by $e(I)$, is the minimal integer $\mu\in\mathbb{N}$ such that $(\sqrt{I})^{\mu}\subseteq I$. If such an integer does not exist, we set $e(I)=\infty$. We also set $e(R)=0$. In a Noetherian ring, since $\sqrt{I}$ is finitely generated for any ideal $I$, $e(I)<\infty$. Anderson and Badawi in [2] establish a connection between $\omega^{\bullet}(I)$ and Noether exponents:

Theorem 1** (Remark 2.2, Theorem 5.3, Section 6, Paragraph 2 on page 1669, [2]).**

Let $I_{1},\cdot\cdot\cdot,I_{r}$ be ideals of a ring $R$. Then $\omega(I_{1}\cap\cdot\cdot\cdot\cap I_{r})\leq\omega(I_{1})+\cdot\cdot\cdot+\omega(I_{r})$ and $\omega^{\bullet}(I_{1}\cap\cdot\cdot\cdot\cap I_{r})\leq\omega^{\bullet}(I_{1})+\cdot\cdot\cdot+\omega^{\bullet}(I_{r})$. In particular, let $I$ be an ideal in a Noetherian ring $R$. If $I=Q_{1}\cap\cdots\cap Q_{n}$, where the $Q_{i}$ are primary ideals, then $\omega(I)\leq\omega^{\bullet}(I)\leq\sum\limits\limits_{i=1}^{n}e(Q_{i})$. Thus every ideal in a Noetherian ring is $n$-absorbing for some $n\in\mathbb{N}$.

On the other hand, $e(I)$ is actually a lower bound of $\omega(I)$.

Lemma 1**.**

Given an ideal $I$ of a ring $R$, $e(I)\leq\omega(I)$. If $Q$ is a primary ideal of $R$, then $\omega(Q)=\omega^{\bullet}(Q)=e(Q)$.

Proof.

The first statement follows either [6, Corollary 3] or [8]. The second statements follows from [2, Theorem 6.3(c), Theorem 6.6]. ∎

This raises the question then if for an arbitrary ideal $I$ whether $\omega(I)$ can be described purely in terms of Noether exponents or possibly other well-known ring-theoretic invariants. This has been investigated to some extent by others in at least one case. Namely, Moghimi and Naghani [11, Theorem 2.21(1)] show that in a discrete valuation ring $R$, $\omega(I)$ is precisely the length of the $R$-module $R/I$.

In this spirit, we attempt to give in this paper a description of $\omega(I)$ in terms of other ring-theoretic invariants in the special case that $I$ is a monomial ideal of a polynomial ring over a field. In some cases, our arguments are general enough to also give the same results for $\omega^{\bullet}(I)$, and thus as a side-effect we can show that in some cases the notion of $n$-absorbing ideal and strongly $n$-absorbing ideal coincide as Anderson and Badawi conjecture.

The present paper is divided into two parts. In section 2, we review the definitions and facts concerning $n$-absorbing ideals and monomial ideals. Using these we calculate $\omega(I)$ for primary monomial ideals by computing Noether exponents and the standard primary decomposition of monomial ideals. These results lead to the study of how $\omega(I)$ can be explicitly computed from the generating set of $I$ when $I$ is a monomial ideal of $R=k[x_{1},\cdot\cdot\cdot,x_{n}]$ with $n\leq 3$ in the following section.

The second part is section 4, where we define and investigate $\omega$-linear monomial ideals, i.e., monomial ideals $I$ such that $\omega(I^{m})=m\omega(I)$ for each $m\in\mathbb{N}$. We give a characterization theorem for primary $\omega$-linear monomial ideals, and in particular show that every integrally closed monomial ideal in $R=k[x,y]$ and the edge ideal of a cycle is $\omega$-linear.

2. Some Background

As a prerequisite of the main section of this paper, we briefly review some of the basic material excerpted from [10] regarding monomial ideals, and show that $\omega(I)$ can be directly calculated from the generators of $I$ when $I$ is a primary monomial ideal.

Let $k$ be a field and $R=k[x_{1},\cdot\cdot\cdot,x_{n}]$ be the polynomial ring with $n$ variables over $k$. An element of $R$ of the form $x_{1}^{a_{1}}\cdot\cdot\cdot x_{n}^{a_{n}}$ with $a_{i}\in\mathbb{N}_{0}$ is called a monomial, and an ideal of $R$ generated by monomials is called a monomial ideal. The degree of $f=x_{1}^{a_{1}}\cdot\cdot\cdot x_{n}^{a_{n}}$, denoted by $\deg(f)$, is defined to be $a_{1}+\cdot\cdot\cdot+a_{n}$. $G(I)$ will denote the set of monomials in $I$ which are minimal with respect to divisibility. Any element of $R$ can be written uniquely as a $k$-linear combination of monomials; that is, given $f\in R$, we may write $f=\sum a_{u}u$ where the sum is taken over the monomial ideals of $R$ and $a_{u}\in k$ for each monomial $u$. Then the support of $f$, denoted by $\text{supp}(f)$, is the set of monomials $u$ such that $a_{u}\neq 0$. An ideal $I$ of a ring $R$ is irreducible if there are no ideals $I_{1},I_{2}$ of $R$ such that $I=I_{1}\cap I_{2}$ and $I\subsetneq I_{1}$, $I\subsetneq I_{2}$. We denote by $\mathfrak{m}$ the unique maximal homogeneous ideal of $R$.

Lemma 2**.**

([10, Chapter 1]) Let $R=k[x_{1},\ldots,x_{n}]$ and $I$ a monomial ideal of $R$ generated by monomials $u_{1},\cdot\cdot\cdot,u_{r}$ of $R$. Then the following hold:

1. (i)

Given a monomial $f\in I$, there exists $i\in\{1,\cdot\cdot\cdot,r\}$ so $u_{i}|f$. 2. (ii)

$G(I)$* is the unique minimal set of monomial generators of $I$.* 3. (iii)

$I$* can be written as a finite intersection of ideals of the form $(x_{i_{1}}^{d_{1}},\ldots,x_{i_{m}}^{d_{m}})$. An irredundant presentation of this form is unique ($I=Q_{1}\cap\cdots\cap Q_{r}$ is irredundant if none of the ideals $Q_{i}$ can be omitted).* 4. (iv)

$I$* is irreducible if and only if $I$ is of the form $(x_{i_{1}}^{d_{1}},\ldots,x_{i_{m}}^{d_{m}})$. Moreover, every irreducible monomial ideal of the form $(x_{i_{1}}^{d_{1}},\ldots,x_{i_{m}}^{d_{m}})$ is $(x_{i_{1}},\ldots,x_{i_{m}})$-primary.* 5. (v)

If $J$ is another monomial ideal of $R$, then

[TABLE]

In particular, if $a$ and $b$ are coprime monomials of $R$ and $I$ is a monomial ideal of $R$, then $(ab,I)=(a,I)\cap(b,I)$. 6. (vi)

An ideal $I^{\prime}$ of $R$ is monomial if and only if for each $f\in I^{\prime}$, $\mbox{{supp}}(f)\subseteq I^{\prime}$.

By Lemma 2(iv), the irredundant unique decomposition of Lemma 2(iii) is also a primary decomposition of $I$, which is known as the standard decomposition of $I$ (see [10, P. 12]). We will also need the following characterization of primary monomial ideals:

Lemma 3**.**

[9, Exercise 3.6]** Let $R=k[x_{1},\cdot\cdot\cdot,x_{n}]$ and $P=(x_{i_{1}},\cdot\cdot\cdot,x_{i_{r}})$ a monomial prime ideal of $R$. Then given a $P$-primary monomial ideal $Q$, there exists $a_{1},\cdot\cdot\cdot,a_{r}\in\mathbb{N}$ and monomials $f_{1},\cdot\cdot\cdot,f_{s}\in k[x_{i_{1}},\cdot\cdot\cdot,x_{i_{r}}]$ so $G(Q)=\{x_{1}^{a_{1}},\cdot\cdot\cdot,x_{i_{r}}^{a_{r}},f_{1},\cdot\cdot\cdot,f_{s}\}$. Conversely, every monomial ideal of this form is a $P$-primary ideal.

Corollary 1**.**

Let $P$ be a prime monomial ideal and $I,J$ be $P$-primary monomial ideals of $R$. Then both $I\cap J$ and $IJ$ are $P$-primary monomial ideals. Moreover, $I:J$ is a $P$-primary monomial ideal provided $J\not\subset I$.

Proof.

This is an immediate consequence of Lemma 2.(v) and Lemma 3. ∎

We can now calculate $\omega(I)$, where $I$ is an irreducible monomial ideal.

Lemma 4**.**

Let $R=k[x_{1},\ldots,x_{n}]$ denote a polynomial ring over a field $k$. Let $I=(x^{d_{1}}_{i_{1}},\ldots,x_{i_{m}}^{d_{m}})$, where $d_{1},\ldots,d_{n}\in\mathbb{N}$ and $1\leq i_{1}<i_{2}<\cdots<i_{m}\leq n$. Then $\omega(I)=\omega^{\bullet}(I)=e(I)=d_{1}+\cdots+d_{m}-m+1$.

Proof.

Since $I$ is a $(x_{i_{1}},x_{i_{2}},\cdot\cdot\cdot,x_{i_{m}})$-primary ideal by Lemma 3, the first two equalities follow from Lemma 1. Thus it suffices to show that $e(I)=r$, where $r=d_{1}+\cdots+d_{m}-m+1$. We have $\sqrt{I}=(x_{i_{1}},\ldots,x_{i_{m}})$. For $N\in\mathbb{N}$, $(\sqrt{I})^{N}\subseteq I$ if and only if for every $c_{1},\ldots,c_{m}\in\mathbb{N}_{0}$ with $c_{1}+\cdots+c_{m}=N$, we have $x^{c_{1}}_{i_{1}}\cdots x^{c_{m}}_{i_{m}}\in I$. By Lemma 2(i), the latter happens precisely when $c_{i}\geq d_{i}$ for some $1\leq i\leq m$. Thus $e(I)=r$. ∎

Next, we produce a way to calculate $\omega(I)$ when $I$ is a monomial primary ideal not necessarily generated by pure powers.

Lemma 5**.**

Let $I$ be an ideal of a ring $R$. Suppose there is $P\in\mbox{{Spec}}(R)$ such that $I=J_{1}\cap\cdot\cdot\cdot\cap J_{r}$, where $J_{i}$ are ideals of $R$ with $\sqrt{J_{i}}=P$ for each $i\in\{1,\ldots,r\}$. Then $e(I)=\max_{1\leq i\leq r}\{e(J_{i})\}$.

Proof.

Note that $\sqrt{I}=\sqrt{J_{1}}\cap\cdot\cdot\cdot\cap\sqrt{J_{r}}=P$. Thus given $\mu\in\mathbb{N}$, $(\sqrt{I})^{\mu}\subseteq I$ if and only if $(\sqrt{J_{i}})^{\mu}\subseteq J_{i}$ for each $i\in\{1,\ldots,r\}$, from which the conclusion of the lemma follows. ∎

Corollary 2**.**

Let $R=k[x_{1},\ldots,x_{n}]$ denote a polynomial ring over a field $k$. If $Q$ is a monomial primary ideal of $R$ and $Q=\cap_{i=1}^{r}Q_{i}$ is its standard decomposition, then

[TABLE]

Example 1**.**

Let $R=k[x,y,z]$ with a field $k$ and $I=(x^{4},y^{3},z^{2},xy,y^{2}z)$. Then repeatedly applying Lemma 2(v), we obtain the standard decomposition $I=(x,y^{2},z^{2})\cap(x^{4},y,z^{2})\cap(x,y^{3},z)$. Thus by Lemma 4 and Corollary 2,

[TABLE]

3. When $I$ is a monomial ideal of $R=k[x_{1},\cdot\cdot\cdot,x_{n}]$ with $n\leq 3$

In this section we show that when $I$ is a monomial ideal of $R=k[x_{1},\cdot\cdot\cdot,x_{n}]$ with $n\leq 3$, then $\omega(I)$ can be explicitly calculated from $G(I)$. We first prove a theorem analogous to [1, Theorem 2.5].

Lemma 6**.**

Let $R$ be a UFD and $p$ an irreducible element of $R$. Then given $n\in\mathbb{N}$, $I$ is an $n$-absorbing ideal of $R$ if and only if $pI$ is an $(n+1)$-absorbing ideal of $R$. In particular, $\omega(pI)=\omega(I)+1$.

Proof.

Suppose that $I$ is $n$-absorbing. Let $f_{1},\ldots,f_{n+2}\in R$ and $f_{1}\cdots f_{n+2}\in pI$. Then since $p$ is irreducible, $p\mid f_{i}$ for some $i$. Without loss of generality, suppose that $p\mid f_{1}$. Then $f_{1}/p\in R$, and so $(f_{1}/p)f_{2}\cdots f_{n+2}\in I$. Since $I$ is $n$-absorbing, and hence $(n+1)$-absorbing as well, we have that either $(f_{1}/p)f_{2}\cdots\widehat{f_{i}}\cdots f_{n+2}\in I$ for some $i\in\{2,\ldots,n+2\}$, in which case $f_{1}f_{2}\cdots\widehat{f_{i}}\cdots f_{n+2}\in pI$ and we’re done, or $f_{2}\cdots f_{n+2}\in I$. This is a product of length $n+1$, so that since $I$ is $n$-absorbing, for some $j$ with $2\leq j\neq n+1$, we have $f_{2}\cdots\widehat{f_{j}}\cdots f_{n+2}\in I$. Thus $pf_{2}\cdots\widehat{f_{j}}\cdots f_{n+2}\in pI$, and so $f_{1}f_{2}\cdots\widehat{f_{j}}\cdots f_{n+2}\in pI$. This shows that $pI$ is then $(n+1)$-absorbing, and $\omega(pI)\leq\omega(I)+1$.

To show the converse, suppose that $pI$ is an $(n+1)$-absorbing ideal. If $I$ is not an $n$-absorbing ideal, then there exists $f_{1},\cdot\cdot\cdot,f_{n+1}\in R$ such that $f=f_{1}\cdot\cdot\cdot f_{n+1}\in I$ but $f_{1}\cdot\cdot\cdot\widehat{f_{i}}\cdot\cdot\cdot f_{n+1}\not\in I$ for each $i$. Since $pI$ is $(n+1)$-absorbing and $pf\in pI$, it follows that either $pf_{1}\cdot\cdot\cdot\widehat{f_{i}}\cdot\cdot\cdot f_{n+1}\in pI$ for some $i$ or $f\in pI$. But the former is impossible by our choice of $f_{i}$’s, and without loss of generality we may assume that $p|f_{1}$. Now $(f_{1}/p)f_{2}\cdot\cdot\cdot f_{n}\in I$, and neither $\widehat{(f_{1}/p)}f_{2}\cdot\cdot\cdot f_{n+1}$ nor $(f_{1}/p)f_{2}\cdot\cdot\cdot\widehat{f_{i}}\cdot\cdot\cdot f_{n+1}$ is an element of $I$ for each $i\geq 2$. Therefore, since $R$ is a UFD, we may assume that none of $f_{i}$ are divisible by $p$. Now $pf_{1}\cdot\cdot\cdot f_{n+1}\in pI$, but $pf_{1}\cdot\cdot\cdot\widehat{f_{i}}\cdot\cdot\cdot f_{n+1}\not\in pI$ and $f_{1}\cdot\cdot\cdot f_{n+1}\not\in pI$, which contradicts the assumption that $pI$ is an $(n+1)$-absorbing ideal. Hence $I$ is an $n$-absorbing ideal and $\omega(pI)\geq\omega(I)+1$. ∎

The following corollary is now immediate.

Corollary 3**.**

Given a monomial $f$ and an ideal $I$ of $R=k[x_{1},\cdot\cdot\cdot,x_{n}]$, $\omega(fI)=$deg$(f)+\omega(I)$. In particular, $\omega(fR)=$deg$(f)$.

Given a monomial ideal $I$ with the standard decomposition $I=\cap_{\ell=1}^{t}T_{\ell}$, we can define an equivalence relation on $\{1,\cdot\cdot\cdot,t\}$ by defining $i\sim j$ iff $\sqrt{T_{i}}=\sqrt{T_{j}}$, and set $\{S_{i}\}_{i=1}^{r}$ to be the corresponding equivalence classes. Then $Q_{i}=\cap_{\ell\in S_{i}}T_{\ell}$ is a monomial primary ideal for each $i\in\{1,\ldots,r\}$, and $I=\cap_{i=1}^{r}Q_{i}$ is an irredundant primary decomposition of $I$. We will call this decomposition the canonical primary decomposition of $I$.

Lemma 7**.**

Let $R=k[x_{1},\cdot\cdot\cdot,x_{n}]$. Let $I$ be a monomial ideal with canonical primary decomposition $I=\cap_{i=1}^{r}{Q_{i}}$. If there exists $k\in\{1,\cdot\cdot\cdot,r\}$ such that $\sqrt{Q_{i}}\subseteq\sqrt{Q_{k}}$ for all $i\in\{1,\cdot\cdot\cdot,r\}$, then $\omega(I)=\max\{e(Q_{k}),\omega(\cap_{1\leq i\leq r,i\neq k}Q_{i})\}$ and $\omega^{\bullet}(I)=\max\{e(Q_{k}),\omega^{\bullet}(\cap_{1\leq i\leq r,i\neq k}Q_{i})\}$.

Proof.

Let $t=\max\{e(Q_{k}),\omega(\cap_{1\leq i\leq r,i\neq k}Q_{i})\}$. We will first show that $I$ is $t$-absorbing. If not, then there are $f_{1},\ldots,f_{t+1}\in R$ such that $f=\prod_{j=1}^{t+1}f_{j}\in I$ but $g_{j}:=f/f_{j}\not\in I$ for each $j\in\{1,\ldots,t+1\}$. Hence given any $i\in\{1,\ldots,t+1\}$, there exists $\ell\in\{1,\ldots,r\}$ such that $g_{i}\not\in Q_{\ell}$, and since $f_{i}g_{i}=f\in I\subseteq Q_{\ell}$, we must have $f_{i}\in\sqrt{Q_{\ell}}\subseteq\sqrt{Q_{k}}$. Therefore, $g_{j}\in(\sqrt{Q_{k}})^{t}\subseteq(\sqrt{Q_{k}})^{e(Q_{k})}\subseteq Q_{k}$ for all $j\in\{1,\ldots,t+1\}$. On the other hand, $\cap_{1\leq i\leq r,i\neq k}Q_{i}$ is $t$-absorbing and $f\in\cap_{1\leq i\leq r,i\neq k}Q_{i}$, so that we conclude $g_{j}\in\cap_{1\leq i\leq r,i\neq k}Q_{i}$ for some $j\in\{1,\ldots,t+1\}$ and thereby $g_{j}\in I$, a contradiction. Thus $\omega(I)\leq t$. Next, we show that $\omega(I)\geq t$; that is, $I$ is not $(t-1)$-absorbing. We now consider two cases.

Case $1$: $t=\omega(\cap_{1\leq i\leq r,i\neq k}Q_{i})$. Since $\cap_{1\leq i\leq r,i\neq k}Q_{i}$ is not $(t-1)$-absorbing, there are $h_{1},\ldots,h_{t}\in R$ such that $h=\prod_{i=1}^{t}h_{i}\in\cap_{1\leq i\leq r,i\neq k}Q_{i}$ and $\ell_{j}:=h/h_{j}\not\in\cap_{1\leq i\leq r,i\neq k}Q_{i}$ for each $j\in\{1,\ldots,t\}$. By an argument similar to the first paragraph of this proof, $h_{i}\in\sqrt{Q_{k}}$ for each $i\in\{1,\ldots,t\}$, and so $h\in Q_{k}$. Hence $h\in I$ and $\ell_{j}\not\in I$ for each $j\in\{1,\ldots,t\}$, so that $I$ is not $(t-1)$-absorbing.

Case $2$: $t=e(Q_{k})$. Consider the standard decomposition of $I$, and choose an irreducible component $T$ of $I$ such that $e(T)=e(Q_{k})$ and $\sqrt{T}=\sqrt{Q_{k}}$. Since we obtained the canonical primary decomposition $I=\cap_{i=1}^{r}Q_{i}$ from the standard decomposition, we can choose a monomial $g\in(\cap_{1\leq i\leq r,i\neq k}Q_{i})\setminus T$ by Lemma 2(vi). Now $T=(x_{i_{1}}^{a_{1}},\cdot\cdot\cdot,x_{i_{l}}^{a_{l}})$ for some $a_{j}\in\mathbb{N}$ and $1\leq i_{1}<\cdot\cdot\cdot<i_{l}\leq n$. Note that we may assume that $g=\prod_{j=1}^{l}x_{i_{j}}^{c_{j}}$ for some $c_{j}\in\mathbb{N}_{0}$ such that $c_{j}<a_{j}$ for each $j\in\{1,\cdot\cdot\cdot,l\}$. Set

[TABLE]

Then $f$ is a product of $e(T)$ elements of $\sqrt{T}$ by Lemma 4, and so $f\in(\sqrt{T})^{e(T)}=(\sqrt{Q_{k}})^{e(Q_{k})}\subseteq Q_{k}$. Since $g\mid f$ it also follows that $f\in\cap_{1\leq i\leq r,i\neq k}Q_{i}$. Hence $f\in I$. However, given $j\in\{1,\cdot\cdot\cdot,l\}$, $\cfrac{f}{x_{i_{j}}}\not\in T$. Indeed, $x_{i_{1}}^{a_{1}-1}\cdot\cdot\cdot x_{i_{l}}^{a_{l}-1}\in\mbox{{supp}}(\cfrac{f}{x_{i_{j}}})\setminus T$ by Lemma 2.(i), and $\cfrac{f}{x_{i_{j}}}\not\in T$ by Lemma 2.(vi). Similarly $x_{i_{1}}^{a_{1}-1}\cdot\cdot\cdot x_{i_{l}}^{a_{l}-1}=\cfrac{f}{x_{i_{1}}+\cdot\cdot\cdot+x_{i_{l}}}\not\in T$. Therefore $I$ is not $(e(Q_{k})-1)$-absorbing, and $\omega(I)\geq e(Q_{k})=t$. Hence we have shown that $\omega(I)=\max\{e(Q_{k}),\omega(\cap_{1\leq i\leq r,i\neq k}Q_{i})\}$. The proof of $\omega^{\bullet}(I)=\max\{e(Q_{k}),\omega^{\bullet}(\cap_{1\leq i\leq r,i\neq k}Q_{i})\}$ can be obtained in a similar manner, and is omitted. ∎

The following corollary is immediate.

Corollary 4**.**

*Let $R=k[x_{1},\cdot\cdot\cdot,x_{n}]$ and $I$ be a monomial ideal of $R$ with standard decomposition $I=\bigcap_{i=1}^{r}T_{i}$. Then $\omega(I)=\omega^{\bullet}(I)=\max_{1\leq i\leq r}\{e(T_{i})\}$ if Ass$(R/I)$ is totally ordered under set inclusion. *

In the next lemma, we give a characterization of when the upper bound of $\omega(I)$ from Theorem 1 is sharp.

Lemma 8**.**

Let $I$ be a monomial ideal of $R=k[x_{1},\cdot\cdot\cdot,x_{n}]$ with an irredundant primary decomposition $I=Q_{1}\cap\cdots\cap Q_{r}$. Then $\omega(I)=\omega^{\bullet}(I)=\sum\limits_{i=1}^{r}e(Q_{i})$ if and only if $I$ has no embedded associated primes.

Proof.

Set $P_{i}=\sqrt{Q_{i}}$ for each $i=1,\cdot\cdot\cdot,r$.

$\Leftarrow$: Assume that $P_{1},\cdot\cdot\cdot,P_{r}$ are incomparable prime ideals. The case when $r=1$ follows from Lemma 1, so we may assume that $r\geq 2$. Since $\omega(I)\leq\omega^{\bullet}(I)\leq\sum\limits_{i=1}^{r}e(Q_{i})$ by Theorem 1, it suffices to show that $I$ is not $(\sum\limits_{i=1}^{r}e(Q_{i})-1)$-absorbing. Now given $i\in\{1,\cdot\cdot\cdot,r\}$, choose $T_{i}$ to be an irreducible component of $I$ with $\sqrt{T_{i}}=P_{i}$ and $e(T_{i})=e(Q_{i})$. Write $T_{i}=(x_{i_{1}}^{a_{1}},\cdot\cdot\cdot,x_{i_{s_{i}}}^{a_{s_{i}}})$ with $1\leq i_{1}<\cdot\cdot\cdot<i_{s_{i}}\leq n$ and $a_{1},\cdot\cdot\cdot,a_{s_{i}}\in\mathbb{N}$. For $i\in\{1,\ldots,r\}$ and $j\in\{1,\ldots,s_{i}\}$, set

[TABLE]

It follows that $f_{i}\in P_{i}^{e(T_{i})}=(\sqrt{Q_{i}})^{e(Q_{i})}\subseteq Q_{i}$. Thus $f:=\prod_{i=1}^{r}f_{i}\in I$, and $f$ is a product of $\sum\limits_{i=1}^{r}e(Q_{i})$ elements of $R$. We wish to show that $\cfrac{f}{\sum\limits_{l=1}^{s_{i}}x_{i_{l}}}\not\in I$ and $\cfrac{f}{f_{i,j}}\not\in I$ for each $i\in\{1,\cdot\cdot\cdot,r\}\text{ and }j\in\{1,\cdot\cdot\cdot,s_{i}\}$. Without loss of generality, we let $i=1$. Note that $\cfrac{f_{1}}{f_{1,j}}\not\in T_{1}$, since \prod_{t=1}^{s_{1}}x_{1_{t}}^{a_{t}-1}\in\mbox{{supp}}\Big{(}\cfrac{f_{1}}{f_{1,j}}\Big{)}\setminus T_{1}. On the other hand, $\sum\limits_{l=1}^{s_{i}}x_{i_{l}}\not\in P_{1}$ and $f_{i,l}\not\in P_{1}$ for each $i\neq 1$ and $l\in\{1,\cdot\cdot\cdot,s_{i}\}$. Therefore $f_{i}\not\in P_{1}$ for each $i\neq 1$, and $\cfrac{f}{f_{1}}=\prod_{i=2}^{r}f_{i}\not\in P_{1}$. Hence $\cfrac{f}{f_{1,j}}=(\cfrac{f}{f_{1}})(\cfrac{f_{1}}{f_{1,j}})\not\in Q_{1}$. The proof that $\cfrac{f}{\sum\limits_{l=1}^{s_{1}}x_{1_{l}}}\not\in Q_{1}$ follows similarly. Hence we have $\omega(I)=\sum\limits_{i=1}^{r}e(Q_{i})$.

$\Rightarrow$: We prove the contrapositive; assume that $P_{1},\cdot\cdot\cdot,P_{r}$ are not incomparable prime ideals. Then without loss of generality we may assume that $P_{1}\subsetneq P_{2}$, and we have $\omega(Q_{1}\cap Q_{2})=\max\{e(Q_{1}),e(Q_{2})\}$ by Corollary 4. Therefore by Theorem 1 we have

[TABLE]

∎

Lemma 7 and Lemma 8 yield the following corollary.

Corollary 5**.**

Let $I$ is a monomial ideal of $R=k[x_{1},\cdot\cdot\cdot,x_{n}]$ with dim$(R/I)=1$. Let $I=\cap_{i=1}^{r}Q_{i}$ be the canonical primary decomposition of $I$. Then

[TABLE]

Corollary 6**.**

Let $f$ be a monomial of $R$. Then $\omega^{\bullet}(fR)=deg(f)$. In particular, $\omega(fR)=\omega^{\bullet}(fR)$.

Proof.

Let $f=\prod_{k=1}^{r}x_{i_{k}}^{a_{k}}$ for some $a_{1},\ldots,a_{r}\in\mathbb{N}$ and $1\leq i_{1}<i_{2}<\cdots<i_{r}\leq n$. Then $fR=x_{i_{1}}^{a_{1}}R\cap\cdot\cdot\cdot\cap x_{i_{r}}^{a_{r}}R$, and by Lemma 4 and Lemma 8 we have $\omega(fR)=\omega^{\bullet}(fR)=\sum_{i=1}^{r}e(x_{i_{k}}^{a_{k}}R)=\sum_{i=1}^{r}a_{k}=$deg$(f)$. ∎

Given a monomial ideal $I$ of $R=k[x,y,z]$, we can produce an algorithm that can compute $\omega(I)$. If $I$ is principal, then Corollary 6 says that $\omega(I)$ is equal to the degree of a generator for $I$. Otherwise, $I=hJ$ for some monomial $h$ and a monomial ideal $J$ with dim$(R/J)\leq 1$. Now, $\omega(J)$ can be calculated explicitly using Corollary 2 or Corollary 5 after obtaining a canonical primary decomposition of $I$, and we have $\omega(I)=deg(h)+\omega(J)$ by Corollary 3.

Example 2**.**

Let $R=k[x,y,z]$ and $I=(x^{3}y^{4},x^{2}y^{5},x^{4}y^{3}z^{2},x^{5}y^{3}z,x^{2}y^{4}z^{2})$. Then $I=x^{2}y^{3}J$ with canonical primary decomposition $J=(x^{2},y)\cap(y,z)\cap(x^{3},y^{2},z^{2},xy)$. By Lemma 4 and Corollary 2, the standard decomposition $(x^{3},y^{2},z^{2},xy)=(x,y^{2},z^{2})\cap(x^{3},y,z^{2})$ yields that $e((x^{3},y^{2},z^{2},xy))=4$. Thus by Corollary 5,

[TABLE]

Another interesting corollary of Lemma 6 and Lemma 7 is a formula of $\omega(I)$ and $\omega^{\bullet}(I)$ for monomial ideals of $R=k[x,y]$ where $k$ is a field and $x,y$ are indeterminates over $k$.

Theorem 2**.**

Let $R=k[x,y]$ and $J$ a monomial ideal of $R$. Write $J=(x^{a_{1}}y^{b_{1}},\cdot\cdot\cdot,x^{a_{r}}y^{b_{r}})$, where $\{a_{i}\}$ is strictly decreasing and $\{b_{i}\}$ is strictly increasing. Then

[TABLE]

Proof.

The case when $r=1$ follows from Corollary 6. For $r>1$, first observe the standard decomposition of $J$ is $J=x^{a_{r}}R\cap y^{b_{1}}R\cap(x^{a_{1}},y^{b_{2}})\cap(x^{a_{2}},y^{b_{3}})\cap\cdot\cdot\cdot\cap(x^{a_{r-1}},y^{b_{r}})$ ([12, Proposition 3.2]). The case $b_{1}=a_{r}=0$ follows from Corollary 2. Suppose that at least one of $a_{r}$ and $b_{1}$ is nonzero. Thus by Lemma 4 and Corollary 5,

[TABLE]

∎

Example 3**.**

*If $R=k[x,y]$ and $J=(x^{11}y^{4},x^{8}y^{5},x^{7}y^{9},x^{4}y^{10},x^{2}y^{16})$, then by Theorem 2,

$\omega(J)=\omega^{\bullet}(J)=\max\{11+5,8+9,7+10,4+16\}-1=19$. *

4. $\omega$-linear ideals

Given an ideal $I$ of a ring $R$, we will say that $I$ is an $\omega$-linear ideal if $\omega(I^{m})=m\omega(I)$ for each $m\in\mathbb{N}$. Perhaps the most common example of $\omega$-linear ideals can be found amongst those $P\in\mbox{{Spec}}(R)$ which $P^{n}$ is $P$-primary for each $n\in\mathbb{N}$ ([2, Theorem 3.1, Theorem 5.7]). For instance,

1. $R$ is a Prüfer domain and $P^{2}\neq P$.

2. $R$ is a Noetherian ring and $P$ is a maximal ideal that contains a nonzerodivisor.

3. $R=k[x_{1},\cdot\cdot\cdot,x_{n}]$ and $P$ is a monomial ideal.

In this section, we investigate the properties of $\omega$-linear ideals. Again, we will restrict our concern to monomial ideals of a polynomial ring $R=k[x_{1},\cdot\cdot\cdot,x_{n}]$ where $k$ is a field.

We first consider a few useful inequalities regarding monomial ideals.

Lemma 9**.**

Let $I$ be a monomial ideal of $R=k[x_{1},\ldots,x_{n}]$. Then $\omega(I)\geq\max\{\deg(f)\mid f\in G(I)\}$.

Proof.

Let $f\in G(I)$. Then $f=\prod_{k=1}^{r}x_{i_{k}}^{a_{k}}$ for some $a_{1},\ldots,a_{r}\in\mathbb{N}$ and $1\leq i_{1}<i_{2}<\cdots<i_{r}\leq n$. Since $f\in I$ but $\cfrac{f}{x_{i_{k}}}\not\in I$ for each $k\in\{1,\cdot\cdot\cdot,r\}$ by minimality of $G(I)$, we have that $I$ is not $($deg$(f)-1)$-absorbing. Hence $\omega(I)\geq$ deg$(f)$, and since $f$ was chosen arbitrarily, we have the desired conclusion. ∎

Lemma 10**.**

Let $I\subseteq J$ be ideals of a ring $R$. If $\sqrt{I}=\sqrt{J}$, then $e(J)\leq\omega(I)$. In particular, if $I$ and $J$ are both $P$-primary ideals of a prime ideal $P$ of $R$, then $\omega(J)\leq\omega(I)$.

Proof.

Since $\sqrt{I}=\sqrt{J}$, $(\sqrt{J})^{\omega(I)}\subseteq(\sqrt{I})^{e(I)}\subseteq I\subseteq J$ by Lemma 1 and $e(J)\leq\omega(I)$. The second statement follows from that $e=\omega$ for primary ideals. ∎

Lemma 11**.**

Let $P$ be a prime monomial ideal of $R=k[x_{1},\ldots,x_{n}]$. If $I,J$ are $P$-primary monomial ideals of $R$, then $\omega(I+J)\leq\min\{\omega(I),\omega(J)\}\leq\max\{\omega(I),\omega(J)\}=\omega(I\cap J)\leq\omega(IJ)\leq\omega(I)+\omega(J)$. Moreover, $\omega(I:J)\geq\omega(I)-\omega(J)$.

Proof.

Note that by Corollary 1, $IJ\subseteq I\cap J\subseteq I+J$ are all $P$-primary monomial ideals. Therefore $\omega(I+J)\leq\min\{\omega(I),\omega(J)\}\leq\max\{\omega(I),\omega(J)\}\leq\omega(I\cap J)\leq\omega(IJ)$ by Lemma 10. On the other hand, let $I=\bigcap_{i=1}^{r}Q_{i}$ and $J=\bigcap_{j=1}^{s}T_{j}$ be standard decomposition of $I$ and $J$, respectively. Then $I\cap J=(\bigcap_{i=1}^{r}Q_{i})\cap(\bigcap_{j=1}^{s}T_{j})$ is an irreducible decomposition of $I\cap J$, and by throwing away any redundant component, there exists $A\subseteq\{1,\cdot\cdot\cdot,r\}$ and $B\subseteq\{1,\cdot\cdot\cdot,s\}$ so $I\cap J=(\bigcap_{i\in A}Q_{i})\cap(\bigcap_{j\in B}T_{j})$ is the standard decomposition of $I\cap J$. Thus by Corollary 2,

[TABLE]

Moreover, $(\sqrt{IJ})^{e(I)+e(J)}=P^{e(I)+e(J)}=P^{e(I)}P^{e(J)}=(\sqrt{I})^{e(I)}(\sqrt{J})^{e(J)}\subseteq IJ$, and so $e(IJ)\leq e(I)+e(J)$. Combined with Lemma 1, we have $\omega(IJ)\leq\omega(I)+\omega(J)$. It remains to show that $\omega(I:J)\geq\omega(I)-\omega(J)$. When $J\subseteq I$, then we have $I:J=R$ and $\omega(I:J)=0\geq\omega(I)-\omega(J)$ by Lemma 10. If $J\not\subseteq I$, then $I:J$ is $P$-primary by Corollary 1, and since $J(I:J)\subseteq I$, we have $\omega(I:J)+\omega(J)\geq\omega(I)$ by the first part of this lemma, hence the claim. ∎

As Anderson and Badawi pointed out ([2, Example 2.7]), the conclusion of Lemma 11 does not hold in every ring $R$. We add, that even in a polynomial ring over a field, the conclusion of the above lemma may fail if we drop any part of the hypothesis.

Example 4**.**

Let $R=k[x,y,z]$ and $I=(x^{2},xy,y^{2},xz^{2})$ and $J=(x^{2},xy,y^{2},yz^{3})$, so that neither $I$ nor $J$ are primary ideals. The standard decompositions of $I,J$ and $I\cap J$ are

[TABLE]

Thus we have $\omega(I)=3$, $\omega(J)=4$, $\omega(I\cap J)=2$ and $\omega(I+J)=4$, so that $\omega(I\cap J)<\omega(I+J)=\max\{\omega(I),\omega(J)\}$.

Example 5**.**

Let $R=k[x,y,z]$ and $I=(x,y)$ and $J=(y,z^{2})$, so that $I$ and $J$ are both primary, but $\sqrt{I}\neq\sqrt{J}$. Then we have $\omega(I)=1$, $\omega(J)=2$ and $\omega(I\cap J)=3$, so that $\omega(I\cap J)>\max\{\omega(I),\omega(J)\}$.

Corollary 7**.**

Let $I$ be a primary monomial ideal of $R=k[x_{1},\ldots,x_{n}]$. Then for each $m\in\mathbb{N}$ we have $\omega(I^{m})\leq m\omega(I)$.

Proof.

Follows immediately by induction on $m$ and Lemma 11. ∎

Next, we derive a characterization of primary monomial $\omega$-linear ideals.

Lemma 12**.**

Let $R=k[x_{1},\cdot\cdot\cdot,x_{n}]$ and $Q$ a primary monomial ideal of $R$, so that $G(Q)=\{x_{i_{1}}^{a_{1}},\cdot\cdot\cdot,x_{i_{r}}^{a_{r}},f_{1},\cdot\cdot\cdot,f_{t}\}$ where $a_{1},\ldots,a_{r}\in\mathbb{N}$, $1\leq i_{1}<i_{2}<\cdots<i_{r}\leq n$ and $f_{1},\cdot\cdot\cdot,f_{t}$ are monomials in $k[x_{i_{1}},\cdot\cdot\cdot,x_{i_{r}}]$. Choose $s\in\{1,\cdot\cdot\cdot,r\}$ so $a_{s}=\max_{1\leq j\leq r}\{a_{j}\}$.

1. (1)

If $f_{1}=f_{2}=\cdot\cdot\cdot=f_{t}=0$, then $\omega(Q^{m})=(m-1)a_{s}+\omega(Q)$ for each $m\in\mathbb{N}$. 2. (2)

$Q$* is $\omega$-linear if and only if $\omega(Q)=a_{s}$.*

Proof.

1. (1)

Let $Q=(x_{i_{1}}^{a_{1}},\cdot\cdot\cdot,x_{i_{r}}^{a_{r}})$. Then given $m\in\mathbb{N}$, set $S_{m}=\{(k_{1},\cdot\cdot\cdot,k_{r})\in\mathbb{N}^{r}\mid\sum\limits_{j=1}^{r}k_{j}=m+r-1\}$ and $Q_{k}=(x_{i_{1}}^{k_{1}a_{1}},\cdot\cdot\cdot,x_{i_{r}}^{k_{r}a_{r}})$ for each $k=(k_{1},\cdot\cdot\cdot,k_{r})\in S_{m}$. Then $Q^{m}=\cap_{k\in S_{m}}Q_{k}$ ([13, Theorem 6.2.4]). Now by Corollary 2 and Lemma 4, $\omega(Q^{m})=\max_{k\in S_{m}}\{e(Q_{k})\}=\max_{k\in S_{m}}\{\sum\limits_{j=1}^{r}k_{j}a_{j}\}-r+1=(m-1)a_{s}+\omega(Q)$. 2. (2)

Fix $m\in\mathbb{N}$ and set $I_{1}=(x_{i_{1}}^{a_{1}},\cdot\cdot\cdot,x_{i_{r}}^{a_{r}})^{m},I_{2}=(x_{i_{1}},\cdot\cdot\cdot,x_{i_{s-1}},x_{i_{s}}^{ma_{s}},x_{i_{s+1}},\cdot\cdot\cdot,x_{i_{r}})$. It follows that $I_{1}\subseteq Q^{m}\subseteq I_{2}$ are $(x_{i_{1}},\cdot\cdot\cdot,x_{i_{r}})$-primary ideals, so we have

$ma_{s}=\omega(I_{2})\leq\omega(Q^{m})\leq\omega(I_{1})=(m-1)a_{s}+\sum_{j=1}^{r}a_{j}-r+1$ by Corollary 10, Lemma 4 and part 1 of this lemma. Therefore if $Q$ is $\omega$-linear, then $\omega(Q)=\displaystyle\lim_{m\to\infty}\cfrac{m\omega(Q)}{m}=\displaystyle\lim_{m\to\infty}\cfrac{\omega(Q^{m})}{m}=a_{s}$. Conversely, suppose that $\omega(Q)=a_{s}$ and fix $m\in\mathbb{N}$. Then since $x_{i_{s}}^{ma_{s}}\in G(Q^{m})$ we have $\omega(Q^{m})\geq ma_{s}=m\omega(Q)$ by Lemma 9. Hence $\omega(Q^{m})=m\omega(Q)$ by Corollary 7 and so $Q$ is $\omega$-linear.

∎

Corollary 8**.**

Let $I$ be an irreducible monomial ideal of $R=k[x_{1},\cdot\cdot\cdot,x_{n}]$ so that $I=(x_{i_{1}}^{a_{1}},\cdot\cdot\cdot,x_{i_{r}}^{a_{r}})$ for some $a_{1},\ldots,a_{n}\in\mathbb{N}$. Set $a_{s}=\max_{1\leq j\leq r}\{a_{j}\}$. Then the following are equivalent.

1. (1)

$I$* is $\omega$-linear.* 2. (2)

$\omega(I^{m})=m\omega(I)$* for some $m>1$.* 3. (3)

$\omega(I)=a_{s}$. 4. (4)

$a_{i}=1$* for each $i\neq s$. *

Proof.

$1\Rightarrow 2$: Obvious.

$2\Rightarrow 3$: Suppose that $\omega(I^{m})=m\omega(I)$ for some $m>1$. By Lemma 12.1 we have $\omega(I^{m})=(m-1)a_{s}+\omega(I)$. Hence $\omega(I)=a_{s}$.

$3\Leftrightarrow 4$: Immediate consequence of Lemma 4.

$3\Leftrightarrow 1$: Follows from Lemma 12.2.∎

Lemma 13**.**

Let $P$ be a monomial prime ideal of $R$. If $I,J$ are $P$-primary $\omega$-linear monomial ideals of $R$, then so is $I\cap J$.

Proof.

Without loss of generality we may assume that $\omega(I)\geq\omega(J)$. By Lemma 12.2, there is $j\in\{1,\cdot\cdot\cdot,r\}$ so that $x_{i_{j}}^{\omega(I)}\in G(I)$. There exists $a\in\mathbb{N}$ so $x_{i_{j}}^{a}\in G(J)$. Then again, by Lemma 12.2, $a\leq\omega(J)$. Now, $x_{i_{j}}^{\omega(I)}=lcm(x_{i_{j}}^{\omega(I)},x_{i_{j}}^{a})\in G(I\cap J)$. On the other hand, $\omega(I\cap J)=\omega(I)$ by Lemma 11. Hence $I\cap J$ is $\omega$-linear by Lemma 12.2. ∎

Given a monomial ideal $I$ of $R=k[x,y]$ we will write $I=(x^{a_{1}}y^{b_{1}},\ldots,x^{a_{r}}y^{b_{r}})$ where $\{a_{i}\}$ and $\{b_{i}\}$ are strictly decreasing and strictly increasing sequences of non-negative integers, respectively. Similarly, if $J$ is a monomial ideal of $R$ we write $J=(x^{c_{1}}y^{d_{1}},\ldots,x^{c_{s}}y^{d_{s}})$ where $\{c_{i}\}$ and $\{d_{i}\}$ are strictly decreasing and strictly increasing sequence of non-negative integers, respectively. Hence $b_{1}=a_{r}=0$ iff $I$ is $(x,y)$-primary, and $d_{1}=c_{s}=0$ iff $J$ is $(x,y)$-primary.

Lemma 14**.**

Let $R=k[x,y]$ and $I,J$ be $(x,y)$-primary monomial ideals with $\omega(I)\geq\omega(J)$. Then $\omega(IJ)\leq\omega(I)+\max\{c_{1},d_{s}\}$.

Proof.

We may assume that $c_{1}\geq d_{s}$. Then $e(I)=\omega(I)\geq\omega(J)\geq c_{1}$ by Lemma 9, so $(x,y)^{e(I)+c_{1}}=(x,y)^{e(I)}(x^{c_{1}},y^{c_{1}})=(\sqrt{I})^{e(I)}(x^{c_{1}},y^{c_{1}})\subseteq IJ$ are $(x,y)$-primary ideals. Therefore $\omega(IJ)\leq\omega((x,y)^{e(I)+c_{1}})=e(I)+c_{1}=\omega(I)+c_{1}$ by Lemma 10. ∎

We now classify $\omega$-linear monomial ideals $I$ in $R=k[x,y]$.

Lemma 15**.**

Let $R=k[x,y]$ and $I=(x^{a_{1}}y^{b_{1}},\cdot\cdot\cdot,x^{a_{r}}y^{b_{r}})$ be a monomial ideal of $R$. Then the following are equivalent.

1. (1)

$I$* is $\omega$-linear.* 2. (2)

$\omega(I^{n})=n\omega(I)$* for some $n>1$.* 3. (3)

$\omega(I)=\max\{a_{1}+b_{1},a_{r}+b_{r}\}$.

Proof.

Note that given $n\in\mathbb{N}$ and a monomial $f$ of $R$, by Lemma 6 we have

[TABLE]

Moreover, if $I$ is a principal ideal, then $I$ satisfies all of 1,2 and 3 by Corollary 3. Hence we may assume that $I$ is a $(x,y)$-primary monomial ideal of $R$. That is, $a_{r}=b_{1}=0$.

$1\Rightarrow 2$ is trivial.

$2\Rightarrow 3$: Suppose that $\omega(I^{n})=n\omega(I)$ for some $n>1$. Note that $\omega(I^{n-1})+\omega(I)\geq\omega(I^{n})=n\omega(I)$ by Lemma 11 and $\omega(I^{n-1})\leq(n-1)\omega(I)$ by Corollary 7, and thereby $\omega(I^{n-1})=(n-1)\omega(I)$. Hence we must have $\omega(I^{2})=2\omega(I)$. Since $\omega(I^{2})\leq\omega(I)+\max\{a_{1},b_{r}\}$ by Lemma 14, $\omega(I)=\omega(I^{2})-\omega(I)\leq\max\{a_{1},b_{r}\}$. On the other hand, $\omega(I)\geq\max\{a_{1},b_{r}\}$ by Lemma 9. Therefore $\omega(I)=\max\{a_{1},b_{r}\}$.

$3\Rightarrow 1$: Follows from Lemma 12.2. ∎

Lemma 16**.**

The set of monomial $\omega$-linear ideals of $R=k[x,y]$ is multiplicatively closed.

Proof.

Let $I$ and $J$ be monomial $\omega$-linear ideals of $R$. By Lemma 6 we may assume that $I$ and $J$ are $(x,y)$-primary ideals of $R$. Then $\omega(I)=\max\{a_{1},b_{r}\}$, $\omega(J)=\max\{c_{1},d_{s}\}$ by Lemma 15. Now, $x^{a_{1}+c_{1}}$ and $y^{b_{r}+d_{s}}$ are elements of $G(IJ)$. Hence by Lemma 12.2 and Lemma 9, to show that $IJ$ is $\omega$-linear it suffices to show that $\omega(IJ)\leq\max\{a_{1}+c_{1},b_{r}+d_{s}\}$. Suppose that $\omega(I)=a_{1}$ and $\omega(J)=c_{1}$. Then all we have to show is $\omega(IJ)\leq a_{1}+c_{1}$, which follows from Lemma 11. The case when $\omega(I)=b_{r}$ and $\omega(J)=d_{s}$ can be derived in the exact same manner. Therefore, without loss of generality we may assume that $\omega(I)=a_{1}>b_{r}$ and $\omega(J)=d_{s}>c_{1}$. Observe now that $Ix^{c_{1}}+Jy^{b_{r}}$ is an $(x,y)$-primary ideal contained in $IJ$. Thus by Lemma 10 and Theorem 2 we have

[TABLE]

∎

Recall that given an ideal $I$ of a commutative ring $R$, an element $f\in R$ is said to be integral over $I$ if there is some $k\in\mathbb{N}$ and $c_{i}\in I^{i}$ for each $i\in\{1,\ldots,k\}$ so that

[TABLE]

The set of elements of $R$ integral over $I$ is called the integral closure of $I$ and denoted by $\overline{I}$. $I$ is said to be integrally closed if $I=\overline{I}$.

Corollary 9**.**

Every integrally closed monomial ideal of $R=k[x,y]$ is $\omega$-linear.

Proof.

Let $I$ be an integrally closed monomial ideal of $R$. It is well known that $R$ is an integrally closed domain (i.e., $R$ is an integral domain that contains every nonzero element of the quotient field of $R$ that is integral over $R$), and that each principal ideal of $R$ is integrally closed, and the product of an integrally closed ideal of $R$ and a nonzero element of $R$ yields another integrally closed ideal of $R$. Hence by Lemma 6 we may assume that $I$ is $(x,y)$-primary. Now by [14, Proposition 2.6] there are monomial ideals $I_{1}=(\{x^{r-i}y^{b_{i}}\}_{i=0}^{r})$ and $I_{2}=(\{x^{a_{i}}y^{i}\}_{i=0}^{r})$ of $R$ with $0=b_{0}<b_{1}<\cdot\cdot\cdot<b_{r}$ and $a_{0}>a_{1}>\cdot\cdot\cdot>a_{r}=0$ so $I=I_{1}I_{2}$. Thus by Lemma 16, it suffices to show that $I_{1}$ and $I_{2}$ are $\omega$-linear. By Theorem 2, $\omega(I_{1})=\max_{0\leq i\leq r-1}\{c_{i}\}$, where $c_{i}=r-i+b_{i+1}-1$ for each $i\in\{0,1,\cdot\cdot\cdot,r-1\}$. Since $c_{i+1}-c_{i}=b_{i+1}-(b_{i}+1)\geq 0$ for each $i\in\{0,1,\cdot\cdot\cdot,r-1\}$, we have $\omega(I_{1})=c_{r-1}=b_{r}=\max\{r,b_{r}\}$ and $I_{1}$ is $\omega$-linear by Lemma 15. The proof that $I_{2}$ is $\omega$-linear follows similarly. ∎

Remark 1**.**

1. (1)

Even if $I$ and $J$ are $\omega$-linear monomial primary ideals such that $\sqrt{I}=\sqrt{J}$, we may have $\omega(I\cap J)<\omega(IJ)<\omega(I)+\omega(J)$. Indeed, set $R=k[x,y]$, $I=(x^{3},xy,y^{2})$ and $J=(x^{2},xy,y^{3})$. Then both $I$ and $J$ are $\omega$-linear $(x,y)$-primary ideals of $R$. However, $IJ=(x^{5},x^{3}y,x^{2}y^{2},xy^{3},y^{5})$, so $\omega(IJ)=5<6=\omega(I)+\omega(J)$. On the other hand, $\omega(I\cap J)=\max\{\omega(I),\omega(J)\}=3$ by Corollary 2. 2. (2)

Not every $\omega$-linear monomial ideal of $R=k[x,y]$ is integrally closed. For example, set $I=(x^{3},xy^{2},y^{4})$. Then $\omega(I)=4$ by Theorem 2, and $I$ is $\omega$-linear by Lemma 15. However, $(x^{2}y)^{2}=x^{3}(xy^{2})\in I^{2}$ and $x^{2}y\not\in I$. Thus $I$ is not integrally closed (**[10*, Theorem 1.4.2]**). *

So far, we have considered only primary $\omega$-linear monomial ideals, and most of the proof is solely based on the fact that $e(I)=\omega(I)$ when $I$ is a primary ideal. We now show that there exists a class of (integrally closed) nonprimary $\omega$-linear monomial ideals. In fact, some of the squarefree monomial ideals are $\omega$-linear, as we will see in the next two lemmas.

Recall that a graph $G$ consists of a set of vertices $V=\{v_{1},...,v_{n}\}$ and a set of edges $E\subseteq\{v_{i}v_{j}|v_{i},v_{j}\in V\}$, and is called bipartite if there exists two disjoint subsets $U_{1},U_{2}$ of $V$ such that $E\subseteq\{v_{i}v_{j}\mid v_{i}\in U_{1},v_{j}\in U_{2}\}$. The edge ideal of $G$ is defined to be the ideal $I=(\{x_{i}x_{j}|v_{i}v_{j}\in E\})$ of $R=k[x_{1},...,x_{d}]$, where $k$ be a field and $d$ is the number of vertices of $G$. Given a graph $G=(V,E)$, a subset $W$ of $V$ is said to be a vertex cover if given $v_{i}v_{j}\in E$, either $v_{i}\in W$ or $v_{j}\in W$. A vertex cover $W$ of $G$ is said to be a minimal vertex cover if each proper subset of $W$ is not a vertex cover of $G$.

If $I$ is an edge ideal of a graph, then it is a squarefree monomial ideal and a monomial prime ideal $P$ is a minimal ideal of $I$ if and only if the set of vertices that corresponds to $P$ is a minimal vertex cover. Also, a graph is bipartite if and only if it has no cycle of odd length as its subgraph.

Our first example of a nonprimary $\omega$-linear ideal is the edge ideal of a bipartite graph.

Lemma 17**.**

Let $R=k[x_{1},\cdot\cdot\cdot,x_{n}]$. If $I$ is an ideal of $R$ that is also the edge ideal of a bipartite graph $G$, then $I$ is $\omega$-linear.

Proof.

Let $I$ be an edge ideal of a graph $G$ and let $P_{1},\cdot\cdot\cdot,P_{r}$ be the set of (incomparable) minimal prime ideals of $I$. Recall that a graph $G$ is bipartite if and only if

[TABLE]

for each $m\in\mathbb{N}$ ([15, Theorem 5.9]). Hence if $G$ is bipartite, then by Lemma 8, $\omega(I^{m})=\sum\limits_{i=1}^{r}e(P_{i}^{m})=\sum\limits_{i=1}^{r}m=mr$ for each $m\in\mathbb{N}$. Therefore the conclusion follows. ∎

There are nonbipartite graphs whose edge ideals are $\omega$-linear.

Lemma 18**.**

Let $R=k[x_{1},\cdot\cdot\cdot,x_{n}]$. Let $I=(x_{1}x_{2},x_{2}x_{3},\cdot\cdot\cdot,x_{n-1}x_{n},x_{n}x_{1})$ (that is, $I$ is the edge ideal of a cycle graph of length $n$). Then $I$ is $\omega$-linear.

Proof.

Since a cycle of even length is bipartite, by Lemma 17 we may assume that $n=2l+1$ for some $l\in\mathbb{N}$. Fix $m\in\mathbb{N}$. $I$ is a squarefree monomial ideal, so $I=P_{1}\cap\cdot\cdot\cdot\cap P_{r}$ where $P_{1},\cdot\cdot\cdot P_{r}$ are the minimal prime ideals of $I$ ([10, Lemma 1.3.5]). Thus by Lemma 8 we have $\omega(I)=\sum_{i=1}^{r}e(P_{i})=r$, and we only need to show that $\omega(I^{m})=mr$. Note that since $I$ is an edge ideal of a cycle of length $2l+1$, Ass$(R/I^{m})=\{P_{1},\cdot\cdot\cdot,P_{r}\}$ if $m\leq l$ and Ass$(R/I^{m})=\{P_{1},\cdot\cdot\cdot,P_{r},\mathfrak{m}\}$ if $m>l$ ([5, Lemma 3.1]). Hence if $m\leq l$, then $I^{m}=\cap_{i=1}^{r}P_{i}^{m}$ and $\omega(I^{m})=\sum\limits_{i=1}^{r}e(P_{i}^{m})=mr$ by Lemma 8, so we are done. Assume that $m>l$. Then $I^{m}=(\cap_{i=1}^{r}P_{i}^{m})\cap Q$ is the canonical primary decomposition of $I^{m}$, where $Q$ is an $\mathfrak{m}$-primary monomial ideal of $R$ ([10, Proposition 1.4.4]). Now, $Q=(x_{1}^{a_{1}},\cdot\cdot\cdot,x_{n}^{a_{n}},f_{1},\cdot\cdot\cdot,f_{t})$ for some $a_{i}\in\mathbb{N}$ and monomials $f_{i}$. Since $I$ is a squarefree monomial ideal and $Q$ is a primary component of $I^{m}$, we must have $a_{i}\leq m$ for each $i\in\{1,\cdot\cdot\cdot,n\}$ and then $e(Q)\leq e((x_{1}^{a_{1}},\cdot\cdot\cdot,x_{n}^{a_{n}}))\leq mn-n+1\leq mr$ by Lemma 4 and since $n\leq r$. It follows that $\omega(I^{m})=\max\{\sum\limits_{i=1}^{r}e(P_{i}^{m}),e(Q)\}=\max\{mr,e(Q)\}=mr$ by Lemma 7 and Lemma 8. ∎

We were unable to show whether every edge ideal is $\omega$-linear, but we do have the following.

Lemma 19**.**

Let $I$ be a square-free monomial ideal. Then $\omega(I^{m})\geq m\omega(I)$ for each $m\in\mathbb{N}$.

Proof.

Let $P_{1},\cdot\cdot\cdot,P_{r}$ be minimal prime ideals of $I$. Then $I=\cap_{i=1}^{r}P_{i}$ and $\omega(I)=r$ by Lemma 8. Set $f_{i}=\sum\limits_{x_{j}\in G(P_{i})}x_{j}$ for each $i\in\{1,\cdot\cdot\cdot,r\}$. Then $f:=\prod_{i=1}^{r}f_{i}\in\prod_{i=1}^{r}P_{i}\subseteq I$, so $f^{m}\in I^{m}$. However, $\cfrac{f^{m}}{f_{i}}\not\in P_{i}^{m}$, so $\cfrac{f^{m}}{f_{i}}\not\in I^{m}$([10, Proposition 1.4.4]). Thus $I^{m}$ is not $(mr-1)$-absorbing and $\omega(I^{m})\geq m\omega(I)$. ∎

We close the section with the following question: Is every integrally closed monomial ideal $\omega$-linear? Integrally closed monomial ideals considered in this note (monomial ideals in $R=k[x,y]$, irreducible monomial ideals, or edge ideal of bipartite graphs) were all $\omega$-linear. Note also that if this question has an affirmative answer, then it follows that every edge ideal is $\omega$-linear.