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
Best Answer
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 $D$.