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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.