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


$\begingroup$ The field of constants. I have edited the answer; see if that is clearer. $\endgroup$ – David E Speyer Nov 8 '09 at 13:49

1$\begingroup$ I guess you could also use the fact (which follows easily by looking at highest order terms) that the Weyl algebra is a domain and not a division ring. So it can't be a matrix ring over a division ring. $\endgroup$ – GS Jan 14 '10 at 10:40
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 noninvertible 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 ArtinWedderburn, $A_n$ 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.

$\begingroup$ How does d^n being distinct show that And^n is distinct? $\endgroup$ – Casebash Nov 8 '09 at 5:18




I would only like to add a simple proof that the Weyl algebra doesn't even HAVE any (nontrivial) finitedimensional representations. Already in the case n=1, consider the relations $$[\partial_x,x]=1.$$ 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.

$\begingroup$ I was going to say this, but I wasn't sure if it still works when the underlying ring is noncommutative. $\endgroup$ – Qiaochu Yuan Nov 9 '09 at 15:01

$\begingroup$ Which ring are you calling underlying? The base field? This is part of the standard proof that W_n is simple. In fact not only can it not have finite dimensional representations, but it's smallest representations have socalled Gelfand Kirillov dimension n, meaning that they are infinitedimensional, and graded, and the dimension of the kth piece is on the order of k^{n1}. These are called holonomic modules. The prototype example is C[x_1,...x_n] with its natural action. The dimension of the kth graded part is \choose{k+n1}{n1}, which is approximately k^{n1} as k>\infty. $\endgroup$ – David Jordan Nov 9 '09 at 16:05

1$\begingroup$ The question is about matrix rings over division rings, not over fields. Unless you're just stating a weaker result? $\endgroup$ – Qiaochu Yuan Nov 9 '09 at 23:13

1$\begingroup$ ...and, moreover, the division rings in questions need not be finite dimensional algebras over the basefield of $A_n$. $\endgroup$ – Mariano SuárezÁlvarez Jun 26 '11 at 17:03
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 $y_1x_1x_1y_11,\ldots,y_nx_nx_ny_n1$ and $x_i x_jx_jx_i, y_i y_j  y_j y_i$ 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 $y_1x_1x_1y_1=1$, the image of which, under $\phi$, is equal to
$$ BAAB=\phi(y_1x_1x_1y_1)=\phi(1)=I. $$
Hence the matrices $A$ and $B$ have to satisfy $BAAB=I$. Taking the trace of the left hand side of this equality yields
$$ tr(BAAB)=tr(BA)tr(AB)=tr(AB)tr(AB)=0 $$
whereas the trace of the right hand side is equal to $tr(I)=m$. This is a contradiction.
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 wellknown fact:
The identity element of a unital Banach algebra can not be a commutator,
i.e. $abba\neq 1$ 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.

1$\begingroup$ I think by "unital ring" you mean "unital commutative (associative) ring". Otherwise, there are obviously ring homomorphisms. Similarly tr(AB) = tr(BA) requires commutativity (it has counterexamples in every noncommutative ring). Unfortunately, the original question specifically allows matrix rings over noncommutative rings, so this doesn't answer the question. $\endgroup$ – Jack Schmidt Jun 21 '11 at 20:27

$\begingroup$ Jack, you are of course right. Thanks for pointing this out. I have tried to make my "answer" a bit more clear now. $\endgroup$ – Johan Öinert Jun 26 '11 at 16:16