Let $R$ be a unital ring, and consider $R$ to be a left $R$-module.

Show that $R$ is a division ring if and only if every nonzero left $R$-module contains a submodule isomorphic to $R$ (as modules).

I proved one direction, but am unsure of the other:

$(\implies)$ Assume $R$ is a division ring. Let $M$ be a nonzero left $R$-module. Take nonzero $x\in M$ and define $\phi:R\to M$ by $\phi(r)=rx$. Note that $\phi$ is a module homomorphism and $\ker\phi=\{0\}$. Thus $R\cong R/\ker\phi\cong\text{Im}\phi$, where $\text{Im}\phi\leq M$.

$(\impliedby)$ Here I was provided a hint: Let $I\neq R$ be a left ideal of $R$, then $R/I$ is a nonzero left $R$-module. By hypothesis, $R/I$ contains a submodule isomorphic to $R$.

I suppose that I should prove that $I=0$, so that $R$ has only two left ideals, and thus a division ring, however I am stuck on this currently.

Thanks for any help.

Update: I also know this result, in case it may be useful: $R$ is a division ring iff $R$ is simple as an $R$-module.


Pick $I$ to be a maximal left ideal, and I think you will rapidly see your way to the end.

  • 1
    $\begingroup$ Thanks. If $I$ is a maximal left ideal, then $R/I$ is a simple module, so that $R\cong R/I$ is a simple module, then by the fact "R is a division ring iff R is simple as an R-module.", we can conclude. $\endgroup$ – yoyostein Nov 26 '16 at 8:02

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.