The saturation number of powers of graded ideals

TL;DR

This paper introduces the concept of saturation number for graded ideals in polynomial rings, analyzing its properties and behavior for various classes of ideals, including monomial, principal Borel, and squarefree Veronese ideals.

Contribution

It defines the saturation number for graded ideals and investigates its bounds and exact values for specific ideal classes, revealing linear and quasi-linear behaviors.

Findings

The saturation number is linearly bounded for all ideals.

For monomial ideals, the saturation number function is quasi-linear for large k.

Explicit formulas are provided for principal Borel and squarefree Veronese ideals.

Abstract

Let $S=K[x_1,\ldots,x_n]$ be the polynomial ring in $n$ variables over a field $K$ with maximal ideal $\frak{m}=(x_1,...,x_n)$, and let $I$ be a graded ideal of $S$. In this paper, we define the saturation number $\sat(I)$ of $I$ to be the smallest non-negative integer $k$ such that $I:\mm^{k+1}= I:\mm^k$. We show that $f(k)$ is linearly bounded, and that $f(k)$ is a quasi-linear function for $k\gg 0$, if $I$ is a monomial ideal. Furthermore, we show that $\sat(I^k)=k$ if $I$ is a principal Borel ideal and prove that $\sat(I_{d,n}^k) =\max\{l\:\; (kd-l)/(k-l) \leq n\},$ where $I_{d,n}$ is the squarefree Veronese ideal generated in degree $d$. \end{abstract}

Equations

$$I^{\mathrm{sat}}:=I:{\mathfrak{m}}^{\infty }=\bigcup _{k\ge 0}I:{\mathfrak{m}}^k$$

$$\alpha (M)=\min \{k:M_k\ne 0\}\text{and}\beta (M)=\max \{k:M_k\ne 0\},$$

$$\gamma (M)\le \sigma (M).$$

$$\gamma (M)-1=\gamma (M/0{:}_M\mathfrak{m})\le \sigma (M/0{:}_M\mathfrak{m})\le \sigma (M)-1,$$

$$\mathrm{sat}(I)\le \sigma (I^{\mathrm{sat}}/I).$$

$$\operatorname{sat}(I\cap{\mathfrak{m}}^{d})=\left\{\begin{array}[]{ll}0,&\text{if $d\leq\alpha(I)$,}\\ d-\alpha(I),&\text{if $d\geq\alpha(I)$.}\\ \end{array}\right.$$

$$\mathrm{sat}(I\cap {\mathfrak{m}}^d)=\mathrm{sat}((I\cap {\mathfrak{m}}^d):\mathfrak{m})+1=\mathrm{sat}(I\cap {\mathfrak{m}}^{d-1})+1=(d-1)-\alpha (I)+1=d-\alpha (I),$$

$$I^k=\bigcap _{j=1}^{d-1}(x_1,\dots ,x_{i_j}{)}^{kj}\cap {\mathfrak{m}}^{kd}.$$

$$B^k(u_{k,d,n}):\mathfrak{m}=B^{k-1}(u_{k-1,d-1,n})+B^k(u_{k,d,n}),$$

$$\mathrm{sat}(I_{d,n}^k)=\max \{l:(kd-l)/(k-l)\le n,l\le k\}.$$

$$\operatorname{sat}(I^{k})=\left\{\begin{array}[]{ll}k/2,&\text{if $k$ is even,}\\ (k-1)/2,&\text{if $k$ is odd}.\end{array}\right.$$

$$I=\bigcap _F{P}_F^{\tau (F)},$$

$$I_{3,2,\dots ,2}=\bigcap _{A,|A|=n-1}{P}_A\cap {\mathfrak{m}}^3.$$

Full text

This paper was done while the first author was visiting University of Kurdistan, Iran. He is thankful to the University for the hospitality.

Introduction

Throughout this paper, we assume that $S=K[x_{1},\ldots,x_{n}]$ is the polynomial ring in $n$ variables over a field $K$ with the unique graded maximal ideal $\mathfrak{m}=(x_{1},\ldots,x_{n})$, and that $I$ is a graded ideal of $S$. If $I$ is a monomial ideal, then we denote by $G(I)$ the unique minimal set of monomial generators of $I$.

The ideal

[TABLE]

is called the saturation of $I$. Since $I\subseteq I:{\mathfrak{m}}\subseteq I:{\mathfrak{m}}^{2}\subseteq\ldots$ and $S$ is Noetherian, there exists an integer $k$ such that $I:{\mathfrak{m}}^{k+1}=I:{\mathfrak{m}}^{k}$. This implies that $\lambda(I^{\operatorname{sat}}/I)<\infty$. Here $\lambda(M)$ denotes the length of a module $M$.

In the first part of this paper we introduce $\operatorname{sat}(I)$ of $I$ which is defined to be the smallest non-negative integer $k$ such that $I:{\mathfrak{m}}^{k+1}=I:{\mathfrak{m}}^{k}$. We show that $\operatorname{sat}(I)\leq\operatorname{reg}(I)$ and obtain from this that $\operatorname{sat}(I^{k})$ is bounded by a linear function of $k$. It would be interesting to have this linear bound also for regular local rings. For monomial ideals one obtains an even better result. Indeed, we show that there exists a quasi-linear function $f$ such that $f(k)=\operatorname{sat}(I^{k})$ for $k\gg 0$ if all powers of $I$ have linear resolution. This is a consequence of the fact that $A=\bigoplus_{k\geq 0}(I^{k})^{\operatorname{sat}}$ is finitely generated when $I$ is a monomial ideal.

In the second part we study the saturation number of special classes of polymatroidal ideals. A polymatroidal ideal is a monomial ideal, generated in a single degree $d$ satisfying the following condition: for all monomials $u,v\in G(I)$ with $\deg_{x_{i}}(u)>\deg_{x_{i}}(v)$, there exists an index $j$ such that $\deg_{x_{j}}(v)>\deg_{x_{j}}(u)$ and $x_{j}(u/x_{i})\in I$ (see [8] or [9]). A squarefree polymatroidal ideal is called a matroidal ideal. Among the stable ideals, the principal Borel ideals are polymatroidal. The saturation number for this class of ideals behaves particularly nice. We show that $\operatorname{sat}(I^{k})=k$ for all $k$. The squarefree principal ideals are matroidal. Here the situation is more complicated. In this paper only consider the squarefree Veronese ideals. We denote by $I_{d,n}$ (see [8] or [13]) the squarefree Veronese ideal of degree $d$ in the variables $x_{1},...,x_{n}$, and show that $\operatorname{sat}(I_{d,n}^{k})=\max\{l\colon\;(kd-l)/(k-l)\leq n,\ l\leq k\}$ for all $k$. According to [15, Proposition 5] any polymatroidal ideal is of intersection type. This fact, allows us to compute the saturation number of a polymatroidal ideal in each concrete case. Moreover, it can be shown that if $I$ is any monomial ideal of intersection type with $\operatorname{Ass}(I)=\operatorname{Ass}^{\infty}(I)$, then $\operatorname{sat}(I^{k})=\operatorname{sat}(I)k$. Here $\operatorname{Ass}^{\infty}(I)=\bigcup_{k\geq 0}\operatorname{Ass}(I^{k})$ .

For unexplained notation or terminology, we refer the reader to [9] and [3]. Several explicit examples were performed with help of the computer algebra systems Macaulay2 [7] and CoCoA [1].

1. Upper bounds for the saturation number of an ideal and its powers

We start this section by the following definition.

Definition 1.1**.**

The number $\operatorname{sat}(I)=\min\{k\colon\;I:{\mathfrak{m}}^{k+1}=I:{\mathfrak{m}}^{k}\}$ is called the saturation number of $I$. **

It is clear that $\operatorname{sat}(I)=0$ if and only if $\operatorname{depth}(S/I)>0$. In particular, $\operatorname{sat}(I)=0$ if $I$ is a squarefree monomial ideal.

Let $M$ be a finitely generated graded $S$-module. We set

[TABLE]

if minimum and maximum exist and otherwise we set $-\infty$ and $\infty$ respectively. Note that if $M$ is a finite length graded $S$-module, then $\beta(M)$ is the largest degree of a socle element of $M$. For a finite length graded $S$-module $M$, we set $\sigma(M)=\beta(M)-\alpha(M)+1$ and $\gamma(M)=\min\{k\colon\;0:_{M}{\mathfrak{m}}^{k}=M\}$.

Lemma 1.2**.**

Let $M\neq 0$ be a graded $S$-module of finite length. Then

[TABLE]

Proof.

We proceed by induction on $\gamma(M)$. The assertion is trivial if $\gamma(M)=1$. Let $\gamma(M)>1$. Then $\gamma(M/0:_{M}{\mathfrak{m}})=\gamma(M)-1$, $\beta(M/0:_{M}{\mathfrak{m}})\leq\beta(M)-1$ and $\alpha(M/0:_{M}{\mathfrak{m}})\geq\alpha(M)$. Thus, together with the induction hypothesis we obtain

[TABLE]

as desired.

Applied to the saturation of an ideal we obtain

Corollary 1.3**.**

Let $I\subset S$ be a graded ideal. Then

[TABLE]

In general the inequality $\operatorname{sat}(I)\leq\sigma(I^{\operatorname{sat}}/I)$ may be strict. The following example was communicated to us by Dancheng Lu and Lizhong Chu:

let $I=(xy,yz,zu,uv,vx^{2})^{5}$. Then $\operatorname{sat}(I)=2$ and $\gamma(I)\geq 4$.

We denote by $\operatorname{reg}(M)$ the Castelnuovo-Mumford regularity of a finitely generated graded $S$-module.

Corollary 1.4**.**

Let $I\subset S$ be a graded ideal. Then $\operatorname{sat}(I)\leq\operatorname{reg}(I).$

Proof.

It follows from Corollary 1.3 that $\operatorname{sat}(I)\leq\beta(I)+1$. Since $I^{\operatorname{sat}}/I$ is of finite length, we have $\beta(I)=\operatorname{reg}(I^{\operatorname{sat}}/I)$. By [4] and [6], $\operatorname{reg}(I^{\operatorname{sat}}/I)\leq\operatorname{reg}(S/I)=\operatorname{reg}(I)-1$. The desired conclusion follows.

The following result follows by Corollary 1.4 and [16] (see also  [5]).

Corollary 1.5**.**

Let $I\subset S$ be a graded ideal. Then there exists a linear function $f$ such that $\operatorname{sat}(I^{k})\leq f(k)$ for all $k\geq 0$.

Observe that $I=I^{\operatorname{sat}}\cap Q$ where $Q$ is an ${\mathfrak{m}}$-primary ideal. We call $I$ special if $Q$ is a power of ${\mathfrak{m}}$. In particular, $I^{\operatorname{sat}}\cap{\mathfrak{m}}^{d}$ is special for all $d$. In the sequel we will use

Proposition 1.6**.**

Let $I\subset S$ a graded ideal with $I=I^{\operatorname{sat}}$. Then

[TABLE]

Proof.

If $d\leq\alpha(I)$, then $I\cap{\mathfrak{m}}^{d}=I$, and since $I$ is saturated we have $\operatorname{sat}(I\cap{\mathfrak{m}}^{d})=0$. Now let $d\geq\alpha(I)$. We proceed by induction on $d-\alpha(I)$. If $d-\alpha(I)=0$, then the assertion is trivial. Now let $d-\alpha(I)>0$. We claim that $(I\cap{\mathfrak{m}}^{d}):{\mathfrak{m}}=I\cap{\mathfrak{m}}^{d-1}$. It is clear that $I\cap{\mathfrak{m}}^{d-1}\subset(I\cap{\mathfrak{m}}^{d}):{\mathfrak{m}}$. Conversely, let $x\in(I\cap{\mathfrak{m}}^{d}):{\mathfrak{m}}$. Then $x\in I:{\mathfrak{m}}=I$ since $I$ is saturated and $x\in{\mathfrak{m}}^{d}:{\mathfrak{m}}={\mathfrak{m}}^{d-1}$. Hence $x\in I\cap{\mathfrak{m}}^{d-1}$. Together with our induction hypothesis we obtain

[TABLE]

as desired.

For monomial ideals, Corollary 1.5 can be improved as follows: a function $f\colon\;{\mathbb{Z}}\rightarrow{\mathbb{Z}}$ is called quasi-linear, if there exists an integer $d\geq 1$ and for $i=0,\ldots,d-1$, linear function $f_{i}(x)=a_{i}x+b_{i}$ with $a_{i},b_{i}\in{\mathbb{Q}}$ such that $f(k)=f_{i}(k)$ for $k\equiv i\operatorname{mod}d$.

Theorem 1.7**.**

Let $I\subset S$ be a monomial ideal. Then there exists a quasi-linear function $f$ such that $\sigma((I^{k})^{\operatorname{sat}}/I^{k})=f(k)$ for $k\gg 0$.

Proof.

We want to show that $f(k)=\beta((I^{k})^{\operatorname{sat}}/I^{k})-\alpha((I^{k})^{\operatorname{sat}}/I^{k})+1$ is quasi-linear. By Brodmann [2], $\operatorname{depth}S/I^{k}$ is constant for all $k\gg 0$. If $\operatorname{depth}S/I^{k}>0$ for all $k\gg 0$, then $f(k)=0$ for $k\gg 0$. Therefore, we may assume that there exists $k_{0}$ such that that $\operatorname{depth}S/I^{k}=0$ for $k\geq k_{0}$. Now let $k\geq k_{0}$, and let $\delta(I^{k})$ be the highest degree of a socle element of $(I^{k})^{\operatorname{sat}}/I^{k}$. Then $\delta(I^{k})=\beta((I^{k})^{\operatorname{sat}}/I^{k})$. On the other hand, $\delta(I^{k})=\max\{j\colon\beta_{n,j}(S/I^{k})=0\}-n=\operatorname{reg}_{n}(S/I^{k})$, where $\operatorname{reg}_{n}(M)$ denotes the $n$-regularity of a graded module, see [5]. It follows therefore from [5, Theorem 3.1] (see also [16]), that $\beta((I^{k})^{\operatorname{sat}}/I^{k})$ is a linear function for $k\gg k_{0}$. Thus it remains to be shown that $\alpha((I^{k})^{\operatorname{sat}}/I^{k})+1$ is a quasi-linear function.

Note that $\alpha((I^{k})^{\operatorname{sat}}/I^{k})$ is the least degree of a generator of $(I^{k})^{\operatorname{sat}}$. We denote this number by $a_{k}$, and have to show that the function $g(k)=a_{k}$ is quasi-linear for $k\gg 0$. In order to prove this we consider the graded $S$-algebra $A=\bigoplus_{k\geq 0}(I^{k})^{\operatorname{sat}}.$ Since $I$ is a monomial ideal, this $S$-algebra is finitely generated, see [10, Theorem 3.2]. Therefore, by [10, Theorem 2.1] there exists an integer $s$ such that $A^{(s)}=\bigoplus_{l\geq 0}A_{ls}$ is a standard graded $S$-algebra. Now let $k$ be any number $\geq s$, and let $k=ls+i$ with $0\leq i<s$. Then $A_{k}=(A_{s})^{l}A_{i}$. Thus $g(k)=a_{s}l+a_{i}$, and the desired conclusion follows.

2. The saturation number for polymatroidal ideals

As a first example of polymatroidal ideals we consider (squarefree) principal Borel ideals. Let $u=x_{1}^{a_{1}}\cdots x_{n}^{a_{n}}$ be a monomial in $S=K[x_{1},\ldots,x_{n}]$. We set $\deg_{x_{i}}(u)=a_{i}$ for $i=1,\ldots,n$. For a given integer $k\geq 1$, we let $I^{\leq k}$ be the ideal generated by all $u\in G(I)$ with $\deg_{x_{i}}(u)\leq k$ for $i=1,\ldots,n$. The ideal $I$ is squarefree if and only if $I=I^{\leq 1}$.

Definition 2.1**.**

Let $I\subset S=K[x_{1},\ldots,x_{n}]$ be a monomial ideal, and let $k\geq 1$ be an integer. Then $I$ is called $k$-strongly stable, if

1. (i)

$I=I^{\leq k}$; 2. (ii)

for all $u\in G(I)$ and all integers $1\leq i<j\leq n$ with $\deg_{x_{j}}(u)>0$ and $\deg_{x_{i}}(u)\leq k-1$ it follows that $x_{i}(u/x_{j})\in I$.

The ideal $I$ is called squarefree strongly stable, if it is $1$-strongly stable. $I$ is strongly stable, if $I$ is $k$-strongly stable for $k$ bigger than the maximal degree of a monomial in $G(I)$. In other words, if $u\in G(I)$ and $x_{j}$ divides $u$, then $x_{i}(u/x_{j})\in I$ for all $i\leq j$.

Let $u_{1},\ldots,u_{m}$ be monomials in $S$ with $\deg_{x_{i}}(u_{j})\leq k$ for $i=1,\ldots,n$ and $j=1,\ldots,m$. There exists a unique smallest $k$-strongly stable ideal containing $u_{1},\ldots,u_{m}$ which we denote by $B^{k}(u_{1},\ldots,u_{m})$. The monomials $u_{1},\ldots u_{m}$ are called Borel generators of $B^{k}(u_{1},\ldots,u_{m})$. If $I=B^{k}(u)$, then $I$ is called $k$-principal Borel, and $1$-principal Borel ideals are simple called principal Borel ideals.

Let $u,v$ be monomials of same degree and assume that $\deg_{x_{i}}(u)\leq k$ for $i=1,\ldots,n$. Then we write $v\prec_{k}u$ if $v\in B^{k}(u)$. This defines for each $d\geq 1$, a partial order on the set of monomials of degree $d$ whose exponents are bounded by $k$.

Theorem 2.2**.**

Let $u\in S$ be a monomial with $\deg_{x_{n}}(u)>0$, and let $I=B(u)$. Then $\operatorname{sat}(I^{k})=k$.

Proof.

It is observed in [8] that principal Borel ideals are polymatroidal. Therefore these ideals are of intersection type, as shown in [15, Proposition 5]. Let $u=x_{i_{1}}x_{i_{2}}\cdots x_{i_{d}}$ with $1\leq i_{1}\leq i_{2}\leq\cdots\leq i_{d}=n$. Then $I=\prod_{j=1}^{d}(x_{1},\ldots,x_{i_{j}})$, (see for example [12, Examples 2.8]).

We claim that $\operatorname{Ass}(I)=\{(x_{1},\ldots,x_{i_{1}}),(x_{1},\ldots,x_{i_{2}}),\ldots,(x_{1},\ldots,x_{i_{d}})\}$. We prove by induction on $d$. If $d=1$, then there is nothing to prove. Let $d\geq 2$ and the claim has been proved for fewer than $d$. It is clear that $\operatorname{Ass}(I)\setminus\{{\mathfrak{m}}\}=\cup_{j=2}^{d}\operatorname{Ass}(I:x_{i_{j}}^{\infty})$. Since $(I:x_{i_{j}}^{\infty})=\prod_{t=1}^{j-1}(x_{1},\ldots,x_{i_{t}})$, we can apply the inductive hypothesis to deduce our claim. Therefore, by applying [13, Corollary 4.10], we see that

[TABLE]

A monomial of least degree in $\bigcap_{j=1}^{d-1}(x_{1},\ldots,x_{i_{j}})^{kj}$ is $x_{1}^{k(d-1)}.$ Finally we apply Proposition 1.6 we get that $\operatorname{sat}(I)=kd-k(d-1)=k$, as desired.

Lemma 2.3**.**

Let $I$ be a polymatroidal ideal and $k\geq 1$ an integer. Then $I^{\leq k}$ is a polymatroidal ideal. In particular, if $u$ is $k$-bounded monomial. Then $B^{k}(u)$ is a polymatroidal ideal.

Proof.

We show that for $u,v\in G(I^{\leq k})$ the exchange property holds. Indeed, let $i$ be such that $\deg_{x_{i}}(u)>\deg_{x_{i}}(v)$, Since $I$ is polymatroidal, there exists $j$ such that $\deg_{x_{j}}(v)>\deg_{x_{j}}(u)$ and $x_{j}(u/x_{i})\in I$. Since $v\in G(I^{\leq k})$, it follows that $\deg_{x_{j}}(v)\leq k$, Therefore, $\deg_{x_{j}}(u)\leq k-1$. This implies that $x_{j}(u/x_{i})\in G(I^{\leq k})$.

Lemma 2.4**.**

Let $u\in S$ be a $k$-bounded monomial of degree $d$. Then $B^{k}(u):{\mathfrak{m}}=I+B^{k}(u)$, where $I=0$ or $I$ is a $(k-1)$-Borel ideal generated in degree $d-1$.

Proof.

We may assume that $\operatorname{depth}S/B^{k}(u)=0$. By Lemma 2.3, $B^{k}(u)$ is a polymatroidal ideal, and hence has a $d$-resolution, see [14, Lemma 1.3 ]. It follows that $B^{k}(u):{\mathfrak{m}}=I+B^{k}(u)$, where $I$ is generated in degree $d-1$. Let $v\in G(I)$. Then $x_{i}v\in G(B^{k}(u))$ for all $i$. Suppose $\deg_{x_{i}}(v)\geq k$ for some $i$. Then $\deg_{x_{i}}(x_{i}v)>k$, a contradiction. This shows that all generators of $I$ are $(k-1)$-bounded.

To complete the proof we must show that $I$ is $(k-1)$-Borel. Let $v\in G(I)$ with $x_{j}|v$, $i<j$ and $\deg_{x_{i}}(x_{i}v)\leq k-1$. Then we must show that $v_{0}=x_{i}(v/x_{j})\in G(I)$, that is, $x_{l}v_{0}\in B^{k}(u)$ for $l=1,\ldots,n$. Indeed, if $l=j$, then $x_{l}v_{0}=x_{i}v\in B^{k}(u)$. If $l\neq j$, then $x_{l}v_{0}=x_{i}(x_{l}v)/x_{j}\in B^{k}(u)$ because $x_{l}v\in B^{k}(u)$ and $B^{k}(u)$ is $k$-Borel.

The next result is taken from [11].

Lemma 2.5**.**

Let $u,v$ be $k$-bounded monomials of degree $d$. Let $u=x_{j_{1}}\cdots x_{j_{d}}$ with $j_{1}\leq j_{2}\leq\ldots\leq j_{d}$ and $v=x_{i_{1}}\cdots x_{i_{d}}$ with $i_{1}\leq i_{2}\leq\cdots\leq i_{d}$. Then $v\preceq_{k}u$ if and only if $i_{r}\leq j_{r}$ for $r=1,\ldots,d$.

Theorem 2.6**.**

Let as before $S=K[x_{1},\ldots,x_{n}]$ and $k$, $d$ be positive integers. Let $d=qk+r$ with integers $q,r\geq 0$ and $r<k$. We set $u_{k,d,n}=x_{n}^{k}x_{n-1}^{k}\cdots x_{n-q+1}^{k}x_{n-q}^{r}$. If $r=0$, then $u_{k,d,n}$ is defined in $S$ if and only if $\lfloor d/k\rfloor\leq n$ and if $r\neq 0$ then $u_{k,d,n}$ is defined in $S$ if and only if $\lfloor d/k\rfloor<n$, and we have

[TABLE]

if $u_{k-1,d-1,n}$ is defined in $S$. Otherwise, $B^{k}(u_{k,d,n}):{\mathfrak{m}}=B^{k}(u_{k,d,n})$.

Proof.

We may assume that $k<d$. Indeed,if $k\geq d$, then $u_{k,d,n}=x_{n}^{d}$ and $u_{k-1,d-1,n}=x_{n}^{d-1}$. In this case the assertion is obvious.

We first show that $u_{k-1,d-1,n}\in B^{k}(u_{k,d,n}):{\mathfrak{m}}$. For this it suffices to show that $x_{n}u_{k-1,d-1,n}\in B^{k}(u_{k,d,n})$, because $B^{k}(u_{k,d,n})$ is $k$-Borel. Indeed, let $u_{k,d,n}=x_{i_{1}}x_{i_{2}}\cdots x_{i_{d}}$ with $i_{1}\leq i_{2}\leq\ldots\leq i_{d}=n$ and $x_{n}u_{k-1,d-1,n}=x_{j_{1}}x_{j_{2}}\cdots x_{j_{d}}$ with $j_{1}\leq j_{2}\leq\ldots\leq j_{d}=n$. By Lemma 2.5 we must show that $j_{l}\leq i_{l}$ for $l=1,\ldots,d$. Figure 1 illustrates this comparison. The integers $i_{l}$ and $j_{l}$ are labeled from right to left. Then for any $l$ with $1\leq l\leq d$ the boxes with the same $x$ coordinate give us the value of $i_{l}$ and $j_{l}$. For example, for $l_{1}$ in Figure 1 we obtain $i_{l_{1}}=n-q+1$ and $j_{l_{1}}=n-q^{\prime}+1$, and for $l_{2}$ we obtain $i_{l_{2}}=n-2$ and $j_{l_{2}}=n-3$. From the equations $d=qk+r$ and $d-1=q^{\prime}(k-1)+r^{\prime}$ with $0\leq r<k$ and $0\leq r^{\prime}<k-1$, it follows that $q\leq q^{\prime}$ because $d/k\leq(d-1)/(k-1)$. Therefore Figure 1 shows that the desired inequalities hold.

Now since $u_{k-1,d-1,n}\in B^{k}(u_{k,d,n}):{\mathfrak{m}}$, and since by Lemma 2.4 the monomials of degree $d-1$ in $B^{k}(u_{k,d,n}):{\mathfrak{m}}$ generate $(k-1)$-Borel ideal, we see that $B^{k-1}(u_{k-1,d-1,n})\subset B^{k}(u_{k,d,n}):{\mathfrak{m}}$.

Conversely, note that with respect to $\preceq_{k-1}$, the monomial $u_{k-1,d-1,n}$ is the unique largest $(k-1)$-bounded monomial of degree $d-1$. Therefore, $B^{k}(u_{k,d,n}):{\mathfrak{m}}\subseteq B^{k-1}(u_{k-1,d-1,n})+B^{k}(u_{k,d,n})$, because Lemma 2.4 implies that the monomials $B^{k}(u_{k,d,n}):{\mathfrak{m}}$ which do not belong to $B^{k}(u_{k,d,n})$ are $(k-1)$-bounded of degree $d-1$.

As an immediate consequence of Theorem 2.6 we obtain

Corollary 2.7**.**

Given integers $k,d,n,$ with $1\leq d\leq n$ and $k\geq 1$. Then

[TABLE]

Example 2.8**.**

Let $I=(x_{1}x_{2},x_{1}x_{3},x_{2}x_{3})$. Then $\operatorname{sat}(I)=0$ and for $k\geq 2$ we have

[TABLE]

In fact, $\operatorname{sat}(I)=0$ because $I$ is squarefree. Now let $k\geq 2$. If $k=2a$, then $(kd-l)/(k-l)=(4a-l)/(2a-l)\leq 3$ if and only if $l\leq a$. Therefore, $a=\max\{l\colon\;(kd-l)/(k-l)\leq n,\ l\leq k\}$ and $\operatorname{sat}(I^{k})=k/2$. Finally, if $k=2a+1$, then $(kd-l)/(k-l)=(4a+2-l)/(2a+1-l)\leq 3$ if and only if $l\leq a$. Therefore, $a=\max\{l\colon\;(kd-l)/(k-l)\leq n,\ l\leq k\}$ and $\operatorname{sat}(I^{k})=(k-1)/2$. **

Now let $I$ be any polymatroidal ideal. In [15, Proposition 5] it is shown that $I$ is of intersection type, which means that $I$ is the intersection of powers of monomial prime ideals. In other words, there exists monomial prime ideals $P_{1},\ldots,P_{r}$ and positive integers $a_{i}$, and $d\geq 0$ such that $I=P_{1}^{a_{1}}\cap\ldots\cap P_{r}^{a_{r}}\cap{\mathfrak{m}}^{d}$. Notice that $I^{\operatorname{sat}}=P_{1}^{a_{1}}\cap\ldots\cap P_{r}^{a_{r}}$, so that $I=I^{\operatorname{sat}}\cap{\mathfrak{m}}^{d}$.

Corollary 1.6 implies that if $a$ the least degree of a generator of $P_{1}^{a_{1}}\cap\ldots\cap P_{r}^{a_{r}}$, and $d\geq a$, then $\operatorname{sat}(I)=d-a$.

Example 2.9**.**

Let $I=(x_{1},x_{2})\cap(x_{2},x_{3})\cap(x_{2},x_{4})\cap(x_{3},x_{4})\cap(x_{1},x_{2},x_{3},x_{4})^{5}$. An element of least degree in $(x_{1},x_{2})\cap(x_{2},x_{3})\cap(x_{2},x_{4})\cap(x_{3},x_{4})$ is $x_{2}x_{3}$. Therefore, $\operatorname{sat}(I)=5-2=3$.**

By applying monomial localization one obtains

Corollary 2.10**.**

Let $I$ be a monomial ideal with the property that $\operatorname{Ass}(I)=\operatorname{Ass}^{\infty}(I)$ and that all powers of $I$ are of intersection type. Then $\operatorname{sat}(I^{k})=\operatorname{sat}(I)k$ for all $k$.

Since power of polymatroidal ideals are again polymatroidal we have

Corollary 2.11**.**

Let $I$ be a polymatroidal ideal with $\operatorname{Ass}(I)=\operatorname{Ass}^{\infty}(I)$. Then $\operatorname{sat}(I^{k})=\operatorname{sat}(I)k$ for all $k$.

Example 2.12**.**

Let $I$ be a transversal polymatroid. In other words, $I$ is a product of monomial prime ideals. Then $\operatorname{sat}(I^{k})=\operatorname{sat}(I)k$. This follows from Corollary 2.11 because $\operatorname{Ass}(I)=\operatorname{Ass}^{\infty}(I)$, (see[13, Corollary 4.6]).**

In order to compute $\operatorname{sat}(I)$ of a polymatroidal ideal we have to determine its presentation as an intersection of powers of monomial prime ideals, as described in [15]: let ${\mathcal{P}}$ be a discrete polymatroid on the ground set $[n]$ of rank $d$ with rank function $\rho$, see [8]. The complementary rank function $\tau\colon 2^{[n]}\rightarrow{\mathbb{Z}}_{+}$ is given by $\tau(F)=d-\rho([n]\setminus F)$ for all $F\in 2^{[n]}$.

A subset $F\subset[n]$ is called $\tau$-closed, if $\tau(G)<\tau(F)$ for any proper subset $G$ of $F$, and $F$ is called $\tau$-separable if there exist non-empty subsets $G$ and $H$ of $F$ with $G\cap H=\emptyset$ and $G\cup H=F$ such that $\tau(G)+\tau(H)=\tau(F)$. If $F$ is not $\tau$-separable, then it is called $\tau$-inseparable.

With this information the intersection presentation of polymatroidal ideal is given as follows:

Theorem 2.13** (Theorem 12, [15]).**

Let $I$ be a polymatroidal ideal associated with the discrete polymatroid $\mathcal{P}$ with complementary rank function $\tau$. Then

[TABLE]

where the intersection is taken over all $F\subset[n]$ which are $\tau$-closed and $\tau$-inseparable.

Example 2.14**.**

Consider the polymatroidal ideal $I_{3,2,\ldots,2}\subset K[x_{1},\ldots,x_{n}]$. Its rank function is given by $\rho(\emptyset)=0$, $\rho(A)=2$ if $|A|=1$ and $\rho(A)=3$ if $|A|\geq 2$. Therefore, $\tau(A)=0$ if $|A|\leq n-2$, $\tau(A)=1$ if $|A|=n-1$ and $\tau(A)=3$ if $|A|=n$. It follows that $A$ is $\tau$-closed and $\tau$-inseparable, if and only if $|A|=n-1$ or $|A|=n$. Thus Theorem 2.13 implies that

[TABLE]

Since $x_{1}x_{2}\in\bigcap_{A,|A|=n-1}P_{A}$, Corollary 1.6 implies that $\operatorname{sat}(I_{3,2,\ldots,2})=1$. **