# $m$-system

Let $R$ be a ring. A subset $S$ of $R$ is called an $m$-system if

• $S\neq\varnothing$, and

• for every two elements $x,\,y\in S$, there is an element $r\in R$ such that $xry\in S$.

$m$-Systems are a generalization of multiplicatively closet subsets in a ring. Indeed, every multiplicatively closed subset of $R$ is an $m$-system: any $x,y\in S$, then $xy\in S$, hence $xyy\in S$. However, the converse is not true. For example, the set

 $\{r^{n}\mid r\in R\mbox{ and }n\mbox{ is an odd positive integer}\}$

is an $m$-system, but not multiplicatively closed in general (unless, for example, if $r=1$).

Remarks. $m$-Systems and prime ideals of a ring are intimately related. Two basic relationships between the two notions are

1. 1.

An ideal $P$ in a ring $R$ is a prime ideal iff $R-P$ is an $m$-system.

###### Proof.

$P$ is prime iff $xRy\subseteq P$ implies $x$ or $y\in P$, iff $x,y\in R-P$ implies that there is $r\in R$ with $xry\notin P$ iff $R-P$ is an $m$-system. ∎

2. 2.

Given an $m$-system $S$ of $R$ and an ideal $I$ with $I\cap S=\varnothing$. Then there exists a prime ideal $P\subseteq R$ with the property that $P$ contains $I$ and $P\cap S=\varnothing$, and $P$ is the largest among all ideals with this property.

###### Proof.

Let $\mathcal{C}$ be the collection of all ideals containing $I$ and disjoint from $S$. First, $I\in\mathcal{C}$. Second, any chain $K$ of ideals in $\mathcal{C}$, its union $\bigcup K$ is also in $\mathcal{C}$. So Zorn’s lemma applies. Let $P$ be a maximal element in $\mathcal{C}$. We want to show that $P$ is prime. Suppose otherwise. In other words, $aRb\subseteq P$ with $a,b\notin P$. Then $\langle P,a\rangle$ and $\langle P,b\rangle$ both have non-empty intersections with $S$. Let

 $c=p+fag\in\langle P,a\rangle\cap S\quad\mbox{ and }\quad d=q+hbk\in\langle P,b% \rangle\cap S,$

where $p,q\in P$ and $f,g,h,k\in R$. Then there is $r\in R$ such that $crd\in S$. But this implies that

 $crd=(p+fag)r(q+hbk)=p(rq+rhbk)+(fagr)q+f\big{(}a(grh)b\big{)}k\in P$

as well, contradicting $P\cap S=\varnothing$. Therefore, $P$ is prime. ∎

$m$-Systems are also used to define the non-commutative version of the radical of an ideal of a ring.

Title $m$-system Msystem 2013-03-22 17:29:09 2013-03-22 17:29:09 CWoo (3771) CWoo (3771) 11 CWoo (3771) Definition msc 16U20 msc 13B30 m-system MultiplicativelyClosed NSystem PrimeIdeal