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

•
$S\ne \mathrm{\varnothing}$, 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
$$\{{r}^{n}\mid r\in R\text{and}n\text{is an odd positive integer}\}$$ 
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.
An ideal $P$ in a ring $R$ is a prime ideal iff $RP$ is an $m$system.
Proof.
$P$ is prime iff $xRy\subseteq P$ implies $x$ or $y\in P$, iff $x,y\in RP$ implies that there is $r\in R$ with $xry\notin P$ iff $RP$ is an $m$system. ∎

2.
Given an $m$system $S$ of $R$ and an ideal $I$ with $I\cap S=\mathrm{\varnothing}$. Then there exists a prime ideal $P\subseteq R$ with the property that $P$ contains $I$ and $P\cap S=\mathrm{\varnothing}$, 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 $\u27e8P,a\u27e9$ and $\u27e8P,b\u27e9$ both have nonempty intersections^{} with $S$. Let
$$c=p+fag\in \u27e8P,a\u27e9\cap S\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}d=q+hbk\in \u27e8P,b\u27e9\cap S,$$ 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
$$crd=(p+fag)r(q+hbk)=p(rq+rhbk)+(fagr)q+f\left(a(grh)b\right)k\in P$$ as well, contradicting $P\cap S=\mathrm{\varnothing}$. Therefore, $P$ is prime. ∎
$m$Systems are also used to define the noncommutative version of the radical^{} of an ideal of a ring.
Title  $m$system 

Canonical name  Msystem 
Date of creation  20130322 17:29:09 
Last modified on  20130322 17:29:09 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  11 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 16U20 
Classification  msc 13B30 
Synonym  msystem 
Related topic  MultiplicativelyClosed 
Related topic  NSystem 
Related topic  PrimeIdeal 