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

$S\neq\varnothing$, 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 $A=\{a_{n}\mid n\mbox{ is a nonnegative integer}\}$. 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 $B=\{b_{m}\mid m\mbox{ is a nonnegative integer}\}$. 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 $T:=\{a_{i}\mid i\mbox{ is a nonnegative integer}\}$ is a subset of $S$ that is an $m$system. For if we pick any $a_{i}$ and $a_{j}$, if $i\leq j$, then $a_{i}$ is both the left and right sections of $a_{j}$, meaning that there are $r,s\in R$ such that $a_{j}=ra_{i}=a_{i}s$ (this can be easily proved inductively). As a result, $a_{i}(sy_{j})a_{j}=a_{j}y_{j}a_{j}\in S$, 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 $RI$ is an $n$system.
Proof.
Suppose $I$ is semiprime. Let $x\in RI$. Then $xRx\nsubseteq I$, which means there is an element $y\in R$ such that $xyx\notin I$. So $RI$ is an $n$system. Now suppose that $RI$ 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 $x\in RI$, then there is some $y\in R$ with $xyx\in RI$, contradicting condition on $x$. Therefore, $x\in I$, and $I$ is semiprime. ∎
Mathematics Subject Classification
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n system
If the empty set is considered an n system (by default) the characterization with n system is always true.