Let $R$ be a ring such that $R$ is a simple $R$-module. Show that $R$ is a division ring. Let $R$ be a ring such that $R$ is a simple $R$-module. Show that $R$ is a division ring.
I have an idea for this but I would like to make sure it is correct. My idea is that $R$-submodules of $R$ are just the same as ideals in $R$. So if we take any non-zero element $r$ in $R$, then the ideal generated by $r$ must be the whole of $R$ (by simplicity) and so $r$ must be a unit and $R$ is a division ring. 
Thanks!
 A: Well, it's been a month, so here is the argument I had in mind. By assumption $R$ is simple as a left module over itself, hence every nonzero element $r \in R$ has a left inverse $s$, thus $sr = 1$. But $s$ also has a left inverse $t$. Hence $s$ has a left and right inverse, which must agree, and hence $r$ does too. 
A: Your approach is correct. An ideal is in particular a sub-module. So for a non-zero element $r$, the ideal generated by $r$ is all of $R$, so $r$ has an inverse. Now you do this for the left and from the right and then you a left and a right inverse. They are equal.
A: Perhaps you can argue as follows: by Schur's lemma, $End_R(R)\,$ is a division ring as $R$ is a simple $R-$module.
Now, let us define $\,\phi:End_R(R)\to R\,$ by $\,\phi(f):=f(1)\,$ . Show now 
1) $\phi\,$ is an $R-\,$module (left, right: as you wish) homomorphism
2) $\phi\,$ is bijective -- for the onto part you may want to check that $\,f_r: R\to R\,$ defined by $\,f_r(x):= xr\,$ is a (left; if you want right fix the definitions) $\,R\,$-endomorphism of $\,R\,$ . Injectivity is trivial -- .
