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.

Answer 4

This is not an answer to the original question. However, it is related and I think that it is worth mentioning.

Assuming that ring morphisms take identity elements to identity elements, we can show the following for the $n$:th Weyl algebra, with very basic methods.

Choose an arbitrary positive integer $n$ and put $A_n := \mathbb{C}\langle x_1,\ldots,x_n,y_1,\ldots,y_n \rangle / I$ where $I$ is the ideal generated by the elements $y_1x_1-x_1y_1-1,\ldots,y_nx_n-x_ny_n-1$ and $x_i x_j-x_jx_i, y_i y_j - y_j y_i$ for $i,j \in 1,\ldots,n$.

Claim:
There does not exist a positive integer $m$ and an associative, commutative and unital ring $R$ such that there is a ring morphism

$$ \phi : A_n \to M_m(R).$$

Proof:
Seeking for a contradiction, suppose that there is some $m$ and some associative, commutative and unital ring $R$ such that $\phi$ exists. Denote the images in $M_m(R)$ of $x_1$ respectively $y_1$, under $\phi$, by $A:=\phi(x_1)$ respectively $B:=\phi(y_1)$. The image of $1$ will be the identity matrix $I$.

Consider the element $y_1x_1-x_1y_1=1$, the image of which, under $\phi$, is equal to

$$ BA-AB=\phi(y_1x_1-x_1y_1)=\phi(1)=I. $$

Hence the matrices $A$ and $B$ have to satisfy $BA-AB=I$. Taking the trace of the left hand side of this equality yields

$$ tr(BA-AB)=tr(BA)-tr(AB)=tr(AB)-tr(AB)=0 $$

whereas the trace of the right hand side is equal to $tr(I)=m$. This is a contradiction.

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Corollary of the above proof:
The same claim holds if we replace $M_m(R)$ by any unital Banach algebra.

This is easily seen by using the following well-known fact:
The identity element of a unital Banach algebra can not be a commutator, i.e. $ab-ba\neq 1$ for any elements $a,b$ of the Banach algebra.

This applies to the case $M_m(R)$ with $R=\mathbb{C}$, because $M_m(\mathbb{C})$ is a unital C*-algebra and in particular a unital Banach algebra. So in this particular case, this gives us an alternative way of making the desired conclusion without using the trace.