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Let $R$ be a ring with right identity $e$. If for each $a \in R\setminus \{0\}$ there exists $x \in R$ such that $xa = e$. Show that $R$ is a division ring.

I can show this result for finite ring $R$, but finiteness is not given. Firstly I think considering $(ax - e)a$ would be enough to prove this, but now I realize only right identity condition is given.

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  • $\begingroup$ I would give that as definition for division ring. What do you mean then? $\endgroup$ – Blumer May 6 '17 at 19:17
  • $\begingroup$ To show division ring I have to show $ax = xa = e.$ But from above how can I show $ax = e?$ $\endgroup$ – Rwitam Jana May 6 '17 at 19:22
  • $\begingroup$ Ok, you are right. Sorry. $\endgroup$ – Blumer May 6 '17 at 19:24
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Of course if $R=\{0\}$ everything is trivial (and $R$ isn't really a division ring) so we suppose this is not the case, and $e\neq 0$.

Suppose $ex-x\neq 0$ for some $x$. Then there exists $y$, such that $y(ex-x)=e$, but $y(ex-x)=yex-yx=yx-yx=0$, not $e$. By this contradiction, $ex=x$ for all $x$. So $e$ is a two-sided identity.

The rest follows the trivial routine: if $x\neq 0$, then $\exists y\neq 0$ such that $yx=e$. Then $\exists z$ such that $zy=e$. But $x=(zy)x=z(yx)=z$, so in fact $xy=e$ as well. All elements are invertible.

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