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Sсhur's lemma states:

Lemma. Let $R$ be a finite-dimensional algebra over an algebraically closed field $K$ and let $M$ be a simple $R$-module. Then $\textrm{End}_R(M)\cong K.$

Proof. Since $M$ is finitely generated, it's clearly finite-dimensional over $K.$ Consider $\Psi\in\textrm{End}_R(M).$ It is a $K$-linear operator. Let $\alpha$ be an eigenvalue of $\Psi.$ Then $\Psi-\alpha\cdot\textrm{Id}$ has got no inverse. Hence, $\Psi=\alpha\cdot\textrm{Id}.$

I want to prove the following:

Proposition. Let $$ be a $\mathbb C$- algebra and let $$ be a simple $$-module with a countable basis (over $\mathbb C$). Then $\textrm{End}_R(M)\cong \mathbb C.$

Here one cannot refer to the finiteness of the dimension. How to get around this difficulty?

Marius S.L.
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Maxim Nikitin
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    You want Dixmier's lemma, see this post (it's stated for group representations but the argument works fine for modules): https://math.stackexchange.com/questions/4551157/representations-of-abelian-groups/4551288#4551288 – Qiaochu Yuan Jan 02 '23 at 21:12
  • @QiaochuYuan Maybe even vote to close as duplicate? – Pedro Jan 04 '23 at 00:12

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