Division rings arising from the endomorphism ring of a simple module I was reading about Schur's Lemma and I immediately asked myself: is every skew-field the endomorphism ring of some simple module?
The page on division rings on Wikipedia states the following:

In general, if R is a ring and S is a simple module over R, then, by
  Schur's lemma, the endomorphism ring of S is a division ring; every
  division ring arises in this fashion from some simple module.

How can the latter statement be proven?
Thank you
 A: If $D$ is a division ring, then it is naturally an $\operatorname{End}_D(D)$-module where multiplication is given by evaluation:
$$\phi .d := \phi(d).$$
Every element $d \in D$ defines a map "left multiplication by $d$"
$$m_d: D \to D, x \mapsto dx$$
which commutes with all $\phi \in \operatorname{End}_D(D)$, i.e. $m_d$ is an endomorphism of $D$ as an $\operatorname{End}_D(D)$-module. Now it easy to show that
\begin{align*}
\alpha: D &\longrightarrow \operatorname{End}_{\operatorname{End}_D(D)}(D),\\
d &\longmapsto m_d
\end{align*}
is an isomorphism of rings (the inverse sends $\theta$ to $\theta(1)$). So $D$ is the endomorphism ring of some module. We just have to show that $D$ is simple as an $\operatorname{End}_D(D)$-module. This is where we use that $D$ is a division ring. Suppose $E \subseteq D$ is a non-trivial $\operatorname{End}_D(D)$-submodule of $D$, i.e. $\phi(E) \subseteq E$ for all $\phi \in \operatorname{End}_D(D)$. Pick an $0 \neq e \in E$. Then for every $d \in D$ right multiplication by $e^{-1}d$ 
$$D \to D, x \mapsto xe^{-1}d$$
is an $D$-endomorphism of $D$ and its image contains $d = ee^{-1}d$. So in fact $E =D$.
A: Let $D$ be a division ring and consider $D$ as a right module over itself. Now consider the endomorphism ring of this module. If $s$ is some endomorphism, then $s(1) = a = a1$ for some $a\in D$ so the map $x \mapsto s(x) - ax$ is a morphism with non-trivial kernel, and hence $0$. This means that $s$ is simply multiplication by $a$ from the left, and we get that the endomorphism ring of $D$ as a module over itself is isomorphic to $D$.
