Dixmier's lemma: endomorphisms of a simple module over a countable-dimensional algebra

By [F1], $D$ is a division ring. The map $\mathbb{C}\to D$, $\lambda \mapsto \lambda {\mathrm{id}}_M$, is an injective unital ring homomorphism onto a central copy of $\mathbb{C}$ by [F2], so $D$ is a $\mathbb{C}$-division algebra containing $\mathbb{C}$ in its center.

Let $V$ be a $\mathbb{C}$-vector space spanned by a sequence $b_1,b_2,\dots$ and let $L\subseteq V$ be linearly independent. Then $L$ is at most countable. Indeed, put $V_k=\mathrm{span}(b_1,\dots ,b_k)$; the recursive rule that retains $b_i$ exactly when $b_i\mathrm{\notin }\mathrm{span}(\text{retained }{b}_j,j<i)$ is a definition by recursion on $i$ and is a canonical ordered basis $(c_1,\dots ,c_{d_k})$ of $V_k$ with $d_k\le k$: retained elements are independent and each $b_i$ lies in the span of the retained elements up to $i$. Hence every $v\in V_k$ has a unique coordinate vector in ${\mathbb{C}}^{d_k}$. Order $\mathbb{C}$ by real part and then imaginary part, and ${\mathbb{C}}^{d_k}$ lexicographically. Since $V={\bigcup }_k{V}_k$, for $v\in L$ the set of $k$ with $v\in V_k$ is nonempty and $n(v):=\min \{k:v\in V_k\}$ is well defined; let $L_k=\{v\in L:n(v)=k\}$. Each $L_k$ is linearly independent in $V_k$, hence finite of size at most $d_k\le k$ by [F6], so the set $\{w\in L_k:w<v\}$ is finite. Now $f(v):=(n(v),\mathrm{\#}\{w\in L_{n(v)}:w<v\})$ lies in $\mathbb{N}\times \mathbb{N}$ and is injective: if $n(v)=n(w)$ and the two counts agree, then the coordinate vectors of $v$ and $w$ are not distinct, because if one preceded the other lexicographically the corresponding counts would differ; totality of the lexicographic order gives $v=w$. Thus $L$ injects into the at most countable set $\mathbb{N}\times \mathbb{N}$ and is at most countable.

For $\mathfrak{g}$ finite-dimensional, $U(\mathfrak{g})$ has a countable $\mathbb{C}$-basis by [F7].

Since $M$ is a quotient of $A$ as a $\mathbb{C}$-vector space — the map $a\mapsto av$ is a surjective $\mathbb{C}$-linear map for any $0\mathrm{\ne }v\in M$ by [F3] — it is spanned by the images of a countable basis of $A$. Applying step 1.2 with this spanning sequence, every linearly independent subset of $M$ is at most countable.

Suppose, for contradiction, that $D\mathrm{\ne }\mathbb{C}$ and choose $x\in D\setminus \mathbb{C}$; then $x$ is transcendental over $\mathbb{C}$. For otherwise $p(x)=0$ for a nonzero $p\in \mathbb{C}[t]$; by [F4] write $p=c{\prod }_i(t-a_i)$ with $c\mathrm{\ne }0$, so $c{\prod }_i(x-a_i)=0$ and one factor $x-a_i$ is zero, giving $x=a_i\in \mathbb{C}$, a contradiction.

Let $x$ be transcendental as in step 2.2. Evaluation $\mathbb{C}[t]\to D$, $p\mapsto p(x)$, is an injective unital homomorphism whose image is commutative because $x$ commutes with the central copy of $\mathbb{C}$. Every nonzero $p$ has $p(x)\mathrm{\ne }0$ and $D$ is a division ring, so $p(x)$ is a unit; the formula $f\mathrm{/}g\mapsto f(x)g(x{)}^{-1}$ is therefore well defined — two representations of the same fraction cross-multiply, and multiplying the resulting identity by the inverses gives equality — and defines an injective field homomorphism $\mathbb{C}(t)\to D$ by [F5]. In particular, for $a\in \mathbb{C}$ the elements $u_a:=(x-a{)}^{-1}$ exist in $D$. They are $\mathbb{C}$-linearly independent: if ${\sum }_i{\lambda }_i{u}_{a_i}=0$ with distinct $a_i$, multiplying by ${\prod }_i(x-a_i)$ gives $p(x)=0$ for $p(t)={\sum }_i{\lambda }_i{\prod }_{j\mathrm{\ne }i}(t-a_j)$; injectivity of $p\mapsto p(x)$ forces $p=0$, and evaluating at $t=a_j$ gives ${\lambda }_j{\prod }_{i\mathrm{\ne }j}(a_j-a_i)=0$, so ${\lambda }_j=0$ because the factors $a_j-a_i$ are nonzero.

Fix $0\mathrm{\ne }v\in M$. The evaluation map ${\mathrm{ev}}_v\text{\hspace{0.17em}⁣}:D\to M$, $T\mapsto Tv$, is $\mathbb{C}$-linear and injective: if $Tv=0$ then $T(av)=a(Tv)=0$ for every $a\in A$ because $T$ is $A$-linear, and $Av=M$ by [F3], so $T=0$. Hence the vectors $({\mathrm{ev}}_v(u_a){)}_{a\in \mathbb{C}}=((x-a{)}^{-1}v{)}_{a\in \mathbb{C}}$ form a $\mathbb{C}$-linearly independent subset $S\subseteq M$ by step 3.1. The assignment $a\mapsto (x-a{)}^{-1}v$ is a bijection $\mathbb{C}\to S$ — injective because its values are linearly independent — and $\mathbb{C}$ is uncountable by [F6], so $S$ is uncountable.

Step 2.1 makes every linearly independent subset of $M$, in particular $S$, at most countable, while step 4.1 makes $S$ uncountable; this contradiction forces $D=\mathbb{C}$, so every $A$-endomorphism of $M$ is a scalar. Consequently, for $z\in Z(A)$ the map ${\mu }_z\text{\hspace{0.17em}⁣}:M\to M$, $m\mapsto zm$, is $A$-linear because ${\mu }_z(am)=z(am)=(az)m=a(zm)$ for $a\in A$, so ${\mu }_z=\chi (z){\mathrm{id}}_M$ for some $\chi (z)\in \mathbb{C}$; the assignment $\chi \text{\hspace{0.17em}⁣}:Z(A)\to \mathbb{C}$ is a unital algebra homomorphism since ${\mu }_{z{z}^{\mathrm{\prime }}}={\mu }_z\circ {\mu }_{z^{\mathrm{\prime }}}$ and ${\mu }_1={\mathrm{id}}_M$ with $M\mathrm{\ne }0$. By step 1.3 the hypotheses hold for $A=U(\mathfrak{g})$ with $\mathfrak{g}$ finite-dimensional.