TITLE: Endomorphism ring is local
QUESTION [5 upvotes]: Let $V$ be a module and $S \subset V$ a simple submodule such that $S$ is contained in every non-zero submodule of $V$. Moreover, assume that every endomorphism of $S$ occurs as the restriction of an endomorphism of $V$. 
Prove that:
(i) The endomorphism ring $R:= \text{End}(V)$ is local.
(ii) $\text{End}(R) / J(\text{End}(R)) \cong \text{End}(S)$, where $J( \cdot)$ is the Jacobson radical.
Definition: Local ring is a ring with $1 \neq 0 $ and if $x$  is an element of the ring, then $x$ is invertible or $1-x$ is invertible.
It is the first time I study module theory, so any help would be really appreciated.
P.S. It is not for homowork, it is just an exercise from some notes that I study.

REPLY [4 votes]: We might get the idea to send $f:V\to V$ to $f|_S:S\to V$. Actually we can say more: if $f(S)\neq 0$, then $S\subseteq f(S)$, but since $f(S)$ is a nonzero image of a simple module, it is also simple, so $f(S)=S$. If $f(S)=\{0\}$ then $f(S)\subseteq S$ trivially. Thus $f|_S:S\to S$. Is it a ring homomorphism? Yes: check it out. 
The assignment $f\mapsto f|_S$ is a ring homomorphism  $\phi:\text{End}(V)\to \text{End}(S)$. By the hypothesis, $\phi$ is surjective. Since $V$ is simple, $\text{End}(S)$ is a division ring, and that means $\ker(\phi)$ is a maximal left ideal that is also two-sided.
Let's examine $\ker(\phi)$ further. By definition, it is $\{f\in \text{End}(V)\mid f(S)=\{0\}\}=\{f\in \text{End}(V)\mid S\subseteq\ker(f)\}$. But remembering that $S$ is contained in all nonzero left ideals, this means that the kernel is finally $\{f\in \text{End}(V)\mid \ker(f)\neq\{0\}\}$. 
This says that the ideal is exactly the set of nonunits of $\text{End}(V)$, which is enough to conclude it is local already. But if you really need to use the $x,1-x$ criterion, you still can. If $f$ is not invertible, then $S\subseteq \ker(f)$. For any $s\in S$ then, $(1-f)(s)=s$, showing that $S$ is not in the kernel of $1-f$, and therefore the kernel of $1-f$ must be $0$, so $1-f$ is invertible.
So in summary, we've established that $\text{End}(V)$ is a local ring whose kernel consists of all noninvertible transformations of $V$. That kernel is necessarily the Jacobson radical, so part (ii) is done.