[[Schur's lemma]]
# Dixmier's lemma

Let $A$ be a [[K-monoid]] over $\mathbb{K}$ and $V$ be a [[Simple module|simple]] $A$-[[Module over a unital associative algebra|module]].
If the [[cardinality]] $\abs{\mathbb{K}} > \dim_{\mathbb{K}} V$, then every $A$-module endomorphism $\vartheta \in \lMod{A}(V,V)$ is an [[algebraic element]] over $\mathbb{K}$.[^1969] #m/thm/module 

  [^1969]: 1969\. [[Sources/@quillenEndomorphismRingSimple1969|On the endomorphism ring of a simple module over an enveloping algebra]], p. 171

> [!check]- Proof
> By [[Schur's lemma]], $\lMod A(V,V)$ is a [[division algebra]] over $\mathbb{K}$.
> Suppose $\vartheta \in \lMod A (V,V)$ is [[Algebraic element|transcendental]] over $\mathbb{K}$, i.e. $p(\vartheta) = 0$ iff $p = 0$ for $p(x) \in \mathbb{K}[x]$.
> The division algebra generated by $\vartheta$ is then
> $$
> \begin{align*}
> \mathbb{K}(\vartheta) &= \left\{  \frac{p(\vartheta)}{q(\vartheta)} : p(x), q(x) \in \mathbb{K}[x], q \neq 0  \right\} \\
> &= \{ f(\vartheta) : f(x) \in \mathbb{K}(x) \}
> \end{align*}
> $$
> where $\mathbb{K}(x)$ is the [[field of rational functions]] for $\mathbb{K}$,
> and we have a straightforward isomorphism of division $\mathbb{K}$-algebras $\mathbb{K}(x) \cong \mathbb{K}(\vartheta)$.
> By [[Lower bound on the dimension of the field of rational functions]] we have the inequality
> $$
> \begin{align*}
> \abs{\mathbb{K}} \leq \dim_{\mathbb{K}} \mathbb{K}(x) = \dim_{\mathbb{K}} \mathbb{K}(\vartheta)
> \end{align*}
> $$
> Since $V$ is a [[vector space]] over $\mathbb{K}(\vartheta)$ with scalar multiplication given by the action of $\vartheta$, we have
> $$
> \begin{align*}
> \dim_{\mathbb{K}} \mathbb{K}(\vartheta) \leq \dim_{\mathbb{K}} V
> \end{align*}
> $$
> and thus
> $$
> \begin{align*}
> \abs{\mathbb{K}} \leq \dim_{\mathbb{K}} V
> \end{align*}
> $$
> a contradiction. <span class="QED"/>

#
---
#state/tidy | #lang/en | #SemBr