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Let $R$ be an associative and non-unital ring. (Suppose that $R$ is $s$-unital, i.e. for each $x\in R$ there is $u,v\in R$ such that $ux=xv=x$.)

It is not difficult to show that if $R$ is a simple ring, then $Z(R)=\{ 0 \}$. Thus, non-unital simple rings are in some sense "extremely" non-commutative.

Are there any (common) examples of rings satisfying the following two conditions?

(1) $Z(R)=0$

and

(2) $R$ is non-simple.

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    $\begingroup$ If $R$ and $S$ are centerless then $R\times S$ is centerless and it is not simple. $\endgroup$
    – Fernando Muro
    Commented May 6, 2013 at 16:23
  • $\begingroup$ ... I forgot to say that $R$ and $S$ should be non-trivial. $\endgroup$
    – Fernando Muro
    Commented May 6, 2013 at 16:26
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    $\begingroup$ Thanks Fernando! That is a nice and elementary example. I was looking in the wrong place. $\endgroup$
    – Johan Öinert
    Commented May 6, 2013 at 19:21

1 Answer 1

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Take the semigroup ring $\mathbb{Z}S$ where $S=${ $a,b,c$} with multiplication $aS=bS=a, cS=c$. The elemwents $a$ and $c$ generate an ideal.

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