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Let $R$ be a finite ring. Is it possible that $R$ has an element $a\in R$ such that $a$ is a left divisor of zero and $a$ is not right divisor of zero?

Thanks.

rschwieb
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Richard
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  • Hi Richard: I guess you've tried something so far... can you describe what happened, briefly? It doesn't matter if it was a success or you got stuck. Adding these sorts of details will make your question more attractive to answer. – rschwieb Oct 09 '13 at 17:14
  • Thanks but I don't understand details. If there exist $b \neq 0$ s.t. $ba=0$ then $r\mapsto ra$ is not 1-1 and consequently not surjective. But why there is $c\neq 0$ s.t. $ac=0$? – Richard Oct 09 '13 at 17:29
  • Good start, Richard. For my hint, be sure to start by assuming $a$ isn't a right zero divisor. What can you tell me about the map? – rschwieb Oct 09 '13 at 17:38
  • (And do you want your ring to have identity?) – rschwieb Oct 09 '13 at 17:40
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    Dear Richard (and others), Regarding the existence of an identity, the argument here could be relevant. Regards, – Matt E Oct 09 '13 at 18:08
  • @rschwieb: Dear rschwieb, The argument linked in my previous comment could fit well with your suggested method to deal with the identity issue. Regards, – Matt E Oct 09 '13 at 18:09
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    If $a$ is not right divisor of zero then this map is bijection and if $R$ has units then there is $c$ such that $ca=1$. Hence if $a$ would be the left divisor, then for some $b \neq 0$ $ab=0$.Hence $b=(ca)b=c(ab)=0$ -a contradiction. – Richard Oct 09 '13 at 18:10
  • @Richard Great! :) – rschwieb Oct 09 '13 at 18:20
  • @MattE Ahh of course, some power of $a$ must be the identity map :) Good call! – rschwieb Oct 09 '13 at 18:24
  • @rschwieb: Dear rschwieb, Actually, I guess this doesn't work, as Boris Novikov's answer below shows. Regards, – Matt E Oct 09 '13 at 19:00
  • @MattE Hm, I thought I was convinced by your argument right away, and there are things to check in Boris's example... – rschwieb Oct 09 '13 at 19:14
  • @rschwieb: Dear rschwieb, Without $1$ in the ring, the map from $R$ to $End(R)$ by taking right multiplication (say) need not be injective (it won't be in Boris's example), and so just because something acts by right multiplication as the identity, doesn't make it an identity in the ring (and so it needn't act identically by left multiplication). I guess this means that commutativity is really needed in my argument from the other question. Regards, – Matt E Oct 09 '13 at 19:48
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    Yeah, you're right that since there are many things acting as right identity and none as the left identity, and that's where the argument breaks down :S Interesting. It's a rather dramatic example :) – rschwieb Oct 09 '13 at 19:54

2 Answers2

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Let $S$ be a finite semigroup of left zeroes ($ab=a$ for all $a,b\in S$), $F$ a finite field, $FS$ the semigroup ring of $S$ over $F$. Then $a(b-c)=0$, so every $a$ is a left divisor of zero but not a right divisor.

(Thanks @rschwieb for a remark)

Boris Novikov
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Hints (which come under the assumption that the ring has identity):

On a finite set $X$, a mapping from $X$ to $X$ is 1-1 iff it is onto.

Suppose $a$ is not a right divisor of zero.

The mapping $r\mapsto ra$ is a homomorphism of left $R$ modules from $R$ to $R$. By an elementary proposition, all left module homomorphisms from $R$ to $R$ have this form.

What do the hypotheses say about this map? Might it have an inverse?

rschwieb
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