$n$-system

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

• $S\ne \mathrm{\varnothing }$, 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 (http://planetmath.org/MSystem) 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

$$a_0=x,\text{ and }{a}_{i+1}=a_i{y}^i{a}_i\mathit{\hspace{1em}}\text{ for }i=0,1,2,\mathrm{\dots }.$$

Form the set $A=\{a_n\mid n\text{ is a non-negative integer}\}$. In addition, inductively define

$$b_0=y,\text{ and }{b}_{j+1}=b_j{x}^j{b}_j\mathit{\hspace{1em}}\text{ for }j=0,1,2\mathrm{\dots },$$

and form $B=\{b_m\mid m\text{ is a non-negative integer}\}$. 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 $a_{i+1}=a_i{y}_i{a}_i$, 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 $T:=\{a_i\mid i\text{ is a non-negative integer}\}$ 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 $a_j=r{a}_i=a_{i}s$ (this can be easily proved inductively). As a result, $a_i(s{y}_j)a_j=a_j{y}_j{a}_j\in S$, and $a_j(y_{j}r)a_i=a_j{y}_j{a}_j\in S$.

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 $xRx⊈I$, 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.∎