Example of an $R$-module $M$ that is not simple but $\mathrm{End}_R(M)$ is a field. I came across Schur's theorem:

If an $R$-module $M$ is simple, then $\mathrm{End}_R(M)$ is a skew field. ($M$ is simple if the only submodule are $M$ and $0$, and $M \neq 0$.)

I heard that it doesn't have an reciproc. Can you guys give me an example of an $R$-module $M$ that is not simple but $\mathrm{End}_R(M)$ is a field, and explain way? I've struggle a bit. 
 A: Let $K$ be some field and consider the two matrices
$$
 A =
\begin{pmatrix}
 1 & 1 \\
 0 & 0
\end{pmatrix}
\quad\text{and}\quad
B =
\begin{pmatrix}
 0 & 0 \\
 0 & 1
\end{pmatrix}
$$
and let $R = K\langle X,Y \rangle$ be the $K$-subalgebra of $\mathrm{M}_2(K)$ generated by these two matrices. Consider $M = K^2$ as an $R$-module via the usual matrix multiplication.
Because $R$ is a unital $K$-algebra every $R$-endomorphism of $M$ must be $K$-linear. Each $K$-linear endomorphism $f$ of $M = K^2$ is given by multiplication with a $2 \times 2$-matrix $F$, and that $f$ is already an $R$-module homomorphism is equivalent to $f$ commuting with the action of both $A$ and $B$ on $K^2$, i.e. $F$ commuting with $A$ and $B$.
Now an easy calculation shows that every matrix which commutes with both $A$ and $B$ is already a scalar multiple of the identity matrix. Thus $\mathrm{End}_R(M) = K$ consists of multiplication with scalars. But $M$ is not simple, because $\langle e_1 \rangle$ is a one-dimensional submodule.

In the case of modules over a $K$-algebra we case see why the converse of Schur does not necessarily hold: If $V$ is a module over a $K$-algebra $A$, then every $a \in A$ gives an endomorphism $f_a \colon V \to V$, $v \mapsto a.v$. For a linear map $g \colon V \to V$ to be an $A$-module homomorphism we need $g$ to respect certain subspaces of $V$. If for example $v \in V$ is an eigenvector of $f_a$ to some eigenvector $\lambda$ then the same goes for $g(a)$. So $g$ must respect the eigenspaces of the endomorphisms $f_a$.
Observations like this can be used to easily construct modules with one-dimensional $A$-endomorphism rings, i.e. every $A$-endomorphism is given by multiplication with a scalar from $K$. But as we have seen above this does not suffice to show that the module is simple.
(If a remember correcty Verma modules of semisimple, complex (finite dimensional?) Lie algebras are a nice example of this, because the heighest weight vectors must be invariant under module homomorphisms, but they already generate the whole module.)

A interesting observation is that if $M$ is an $R$-module such that $\mathrm{End}_R(M)$ is a skew field, then $M$ is indecomposable: If we have $M = M_1 \oplus M_2$ for some submodules $M_1$ and $M_2$ then we have a projection $p \colon M \to M$, i.e. a module homomorphism with $p^2 = p$, such that $\ker p = M_1$ and $\mathrm{im} \ p = M_2$. But because $\mathrm{End}_R(M)$ is a skew field it follows from $p^2 = p$ that $p = 0$ or $p = 1$, so either $M_1 = M$ or $M_2 = M$.
