Prove that if $V$ is a simple R-module, then $End_R(V)$ is a division ring.
$proof:$
Let $\gamma \in end_R(V)$. Then $\gamma: V \rightarrow V$ is injective because $ker(\gamma) \leq V$ and so $ker(\gamma)=0$.
Thus $\gamma^{-1} : \gamma(V) \rightarrow V$ is a r-module homomorphism such that $\gamma^{-1} \gamma(v) = v$ $\forall v \in V$. Also, we can extend $\gamma^{-1}$ to all of $V$ by maknig $\gamma^{-1}(v)=0$ $\forall v \in V - \gamma(V)$
Therefore, the extension of $\gamma^{-1}$ from $\gamma(V)$ to all of $V$ is an element of $End_R(V)$ and is a left inverse for $\gamma$. Thus $End_R(V)$ is a division ring.
We don't need to do any extending of $\gamma^{-1}$ from $\gamma(V)$ to $V$ since $\gamma(V)=V$. The image of a homomorphism is a submodule, so if $V$ is simple and $\gamma$ is not the zero homomorphism then $\gamma(V)=V$, so $\gamma$ is surjective and we are done. Other than that the proof is good.