$m$-systemMSystem2007-08-18 13:55:332008-05-24 13:00:36Definition\usepackage{amssymb,amscd}
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Let $R$ be a ring. A subset $S$ of $R$ is called an $m$-{\em system} if
\begin{itemize}
\item $S\ne \varnothing$, and
\item for every two elements $x,\,y\in S$, there is an element $r\in R$ such that $xry\in S$.
\end{itemize}
$m$-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 $$\lbrace r^n\mid r\in R \mbox{ and } n \mbox{ is an odd positive integer}\rbrace$$ is an $m$-system, but not multiplicatively closed in general (unless, for example, if $r=1$).
\textbf{Remarks}. $m$-Systems and prime ideals of a ring are intimately related. Two basic relationships between the two notions are
\begin{enumerate}
\item An ideal $P$ in a ring $R$ is a prime ideal iff $R-P$ is an $m$-system.
\begin{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.
\end{proof}
\item 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.
\begin{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.
\end{proof}
\end{enumerate}
$m$-Systems are also used to define the non-commutative version of the radical of an ideal of a ring.