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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

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If $R$ is allowed to be non-commutatitive, then one can take $S$ as a right-$S$ module, which is simple, and the endomorphism ring is isomorphic to $S$ (acting by left multiplication), so this would be a trivial statement. I'm not sure that is intended though. – Marc van Leeuwen Feb 4 '13 at 17:46
No, the statement is slightly less trivial. We start with a division ring D, we want to find a simple module whose endomorphism ring is isomorphic to D. – user39280 Feb 4 '13 at 17:55
up vote 4 down vote accepted

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 $D$.

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Very clear answer. Thank you very much. – user39280 Feb 4 '13 at 17:35
I have a concern, is D, considered as a D-module, simple? – user39280 Feb 4 '13 at 17:52
Yes, as a $D$-submodule of $D$ is an ideal of $D$, the module is simple (any element is invertible from both sides, so $D$ has no non-trivial ideals). – Tobias Kildetoft Feb 4 '13 at 17:57
Be careful with non-commutativity issues! Left multiplication by $a$ is in general not a $D$-endomorphism of $D$. Right multiplication by $a$ is an endomorphism but the map $a \mapsto $ "right multiplication by $a$" is not a homomorphism of rings. – marlu Feb 4 '13 at 17:59
@marlu, the only problem here is that the endomorphism ring of $D$ as a left $D$-module is $D^{\mathrm{op}}$, so one can simply start with $D^{\mathrm{op}}$. – Mariano Suárez-Alvarez Feb 4 '13 at 18:05

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 $E =D$.

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