Question

Can anyone prove that a Weyl Algebra is not isomorphic to a matrix ring over a division ring?

Answer 1

Notation: The Weyl algebra is $$k[x_1, x_2, \ldots, x_n, \partial_1, \partial_2, \ldots, \partial_n]$$ with the obvious relations.

The Weyl algebra doesn't contain any division rings larger than $k$, and it is infinite dimensional over $k$. So, assuming you don't allow infinite matrices, that's a proof.

How to see that it doesn't contain any division ring larger than $k$? I just need to show that any nonconstant differential operator is not invertible. One way to see this is to notice that multiplying differential operators multiplies symbols, and the symbol of a nonconstant differential operator is a nonconstant polynomial.

Answer 2

A different proof would be to show that a Weyl algebra is not semisimple, that is, that it is not a direct sum of simple submodules as a left module over itself. However, note that there is an infinite descending chain of left submodules of a Weyl algebra given by $A_n\supseteq A_nd\supseteq A_nd^2\supseteq A_nd^3\supseteq...$ where $d$ is any non-invertible element. A direct sum of a finite number of simple modules can't have an infinite descending chain of submodules. Then, by the converse of Artin-Wedderburn, $A_n$ is not a direct sum of matrix algebras over a divsion ring.

Of course, showing this sequence of submodules never stabilizes can be done by looking at the associated graded algebra, and noting that the $\overline{A_nd^n}$ are always distinct there. However, then this answer starts getting closer to David's answer, so maybe this wasn't a truly different proof.

Answer 3

I would only like to add a simple proof that the Weyl algebra doesn't even HAVE any (non-trivial) finite-dimensional representations. Already in the case n=1, consider the relations

Now suppose you had a finite dimensional representation, and take the trace of both sides of the above.

It implies that the identity acts as 0 so the whole representation does.