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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.

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

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    $\begingroup$ 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

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