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 be a field and a -algebra (yes unital). Let be a finite dimensional simple -module and (which is a division algebra by Schur’s Lemma). Then the canonical homomorphism from to is surjective.

**Proof**

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

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

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

If is not nilpotent, then the minimal polynomial of factors as where . In particular it has two coprime factors, so by the Chinese remainder theorem has a non-trivial idempotent. In particular, contains a non-trivial idempotent.

If on the other hand is nilpotent, then we can find with , and . Since is irreducible, there exists with . Then and . So is not nilpotent, and we can run the argument of the previous paragraph again to conclude that contains a non-trivial idempotent.

Let be this non-trivial idempotent. We will use the fact that is an irreducible -module. This is part of a general fact, that is a bijection betwen simple -modules with and simple -modules. Furthermore, from the nature of the constructions of both directions of this bijection, we can conclude that

Now we can perform an induction on dimension. The inductive hypothesis tells us that and are both surjective. To finish, given the symmetry between and , it suffices to show that given any -hyperplane with and any -line , there is a nonzero element in with kernel and image . (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 allows us to find a nonzero element with . Pick nonzero and . Since is simple there exists with . Then does the trick and our proof is complete.