Let $R$ be a ring. A subset $S$ of $R$ is said to be an $n$ -system if

• $S\ne \varnothing$ , and
• for every $x\in S$ , there is an $r\in R$ , such that $xrx\in S$ .

$n$ -systems are a generalization of $m$ -systems in a ring. Every $m$ -system is an $n$ -system, but not conversely. For example, for any distinct $x,y\in R$ , inductively define the elements $$a_0=x,\ \mbox{ and }\ a_{i+1}=a_i y^i a_i\ \ \mbox{ for }i=0,1,2,\ldots.$$ Form the set $A=\lbrace a_n\mid n\mbox{ is a non-negative integer}\rbrace$ . In addition, inductively define $$b_0=y,\ \mbox{ and }\ b_{j+1}=b_j x^j b_j\ \ \mbox{ for }j=0,1,2\ldots,$$ and form $B=\lbrace b_m\mid m\mbox{ is a non-negative integer}\rbrace$ . Then both $A$ and $B$ are $m$ -systems (as well as $n$ -systems). Furthermore, $S=A\cup B$ is an $n$ -system which is not an $m$ -system.

The example above suggests that, given an $n$ -system $S$ and any $x\in S$ , we can ``construct'' an $m$ -system $T\subseteq S$ such that $x\in T$ . Start with $a_0=x$ , inductively define $a_{i+1}=a_iy_ia_i$ , where the existence of $y_i\in R$ such that $a_{i+1}\in S$ is guaranteed by the fact that $S$ is an $n$ -system. Then the collection $T:=\lbrace a_i\mid i\mbox{ is a non-negative integer}\rbrace$ is a subset of $S$ that is an $m$ -system. For if we pick any $a_i$ and $a_j$ , if $i\le j$ , then $a_i$ is both the left and right sections of $a_j$ , meaning that there are $r,s\in R$ such that $a_j=ra_i=a_is$ (this can be easily proved inductively). As a result, $a_i(sy_j)a_j=a_jy_ja_j\in S$ , and $a_j(y_jr)a_i=a_jy_ja_j\in S$ .

Remark $n$ -systems provide another characterization of a semiprime ideal: an ideal $I\subseteq R$ is semiprime iff $R-I$ is an $n$ -system.

Proof. Suppose $I$ is semiprime. Let $x\in R-I$ . Then $xRx\nsubseteq I$ , which means there is an element $y\in R$ such that $xyx\notin I$ . So $R-I$ is an $n$ -system. Now suppose that $R-I$ is an $n$ -system. Let $x\in R$ with the condition that $xRx\subseteq I$ . This means $xyx\in I$ for all $y\in R$ . If $x\in R-I$ , then there is some $y\in R$ with $xyx\in R-I$ , contradicting condition on $x$ . Therefore, $x\in I$ , and $I$ is semiprime.