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

   $$c=p+fag\in ⟨P,a⟩\cap S\mathit{\hspace{1em}}\text{ and }\mathit{\hspace{1em}}d=q+hbk\in ⟨P,b⟩\cap S,$$

   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 non-commutative version of the radical of an ideal of a ring.