The Jacobson Density Theorem

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.

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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