October | 2023 | C'est un blog

I’ve never been at ease with the Jacobson density theorem, every proof that I’d seen felt somewhat weird and never had any stickiness in my mind. So I came up with my own proof, I hope you like it.

The Theorem (Jacobson Density Theorem}
Let $k$ be a field and $A$ a $k$ -algebra (yes unital). Let $S$ be a finite dimensional simple $A$ -module and $D=\operatorname{End}_A(S)$ (which is a division algebra by Schur’s Lemma). Then the canonical homomorphism from $A$ to $\operatorname{End}_D(S)$ is surjective.

Proof
First, without loss of generality (replacing $A$ by a quotient), we can assume that $S$ is faithful.

Let us first deal with the case when $A$ is a division algebra. Then $S=A$ . And $D=A$ as well, acting by right multiplication. And we are done.

Since $A$ is not a division algebra it has a non-zero element $a$ which is not invertible.

If $a$ is not nilpotent, then the minimal polynomial $m(x)\in k[x]$ of $a$ factors as $x^np(x)$ where $n,\deg(p(x))>0$ . In particular it has two coprime factors, so by the Chinese remainder theorem $k[x]/(m(x))$ has a non-trivial idempotent. In particular, $A$ contains a non-trivial idempotent.

If on the other hand $a$ is nilpotent, then we can find $v,w\in V$ with $av=0$ , $aw=v$ and $v\neq 0$ . Since $V$ is irreducible, there exists $b\in A$ with $bv=w$ . Then $(ba)v=0$ and $(ba)w=w$ . So $ba$ is not nilpotent, and we can run the argument of the previous paragraph again to conclude that $A$ contains a non-trivial idempotent.

Let $e$ be this non-trivial idempotent. We will use the fact that $eS$ is an irreducible $eAe$ -module. This is part of a general fact, that $S\mapsto eS$ is a bijection betwen simple $A$ -modules $S$ with $eS\neq 0$ and simple $eAe$ -modules. Furthermore, from the nature of the constructions of both directions of this bijection, we can conclude that

$\operatorname{End}_A(S)\cong \operatorname{End}_{eAe}(eS).$

Now we can perform an induction on dimension. The inductive hypothesis tells us that $eAe \to \operatorname{End}_{D}(eS)$ and $(1-e)A(1-e)\to \operatorname{End}_{D}((1-e)S)$ are both surjective. To finish, given the symmetry between $e$ and $1-e$ , it suffices to show that given any $D$ -hyperplane $H$ with $eS\subset H$ and any $D$ -line $\ell\subset eS$ , there is a nonzero element in $A$ with kernel $H$ and image $\ell$ . (I thought about this step myself in terms of matrices, but actually writing it up like that feels a little gauche).

The inductive hypothesis applied to $(1-e)A(1-e)$ allows us to find a nonzero element $c\in A$ with $ker(c)=H$ . Pick nonzero $v\in \im(c)$ and $w\in \ell$ . Since $S$ is simple there exists $d\in A$ with $dv=w$ . Then $dc$ does the trick and our proof is complete.

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