# The division ring of fractions of the first Weyl algebra and its subrings

The first Weyl algebra, $A_1(k)= k\langle x,y | yx-xy=1\rangle$, where $k$ is a field of characteristic zero, is known to be a simple Noetherian ring, hence it has a (left) division ring of fractions (see this question), denote it by $D_{x,y}$.

Let $f:A_1(k) \to A_1(k)$ be a $k$-algebra endomorphism (by simplicity, it is necessarily injective. In 1968, Dixmier conjectured that $f$ must be surjective). Denote $p:=f(x)$ and $q:=f(y)$. The image of $A_1(k)$ under $f$, $k\langle p,q | qp-pq=1\rangle$, is isomorphic to $A_1(k)$, so it also has its own (left) division ring of fractions, denote it by $D_{p,q}$. Clearly, $D_{p,q} \subseteq D_{x,y}$.

Please, I have two questions:

(1) Is it true that every element of $D_{x,y}$ is of the form $b^{-1}a$, where $0 \neq b,a \in A_1(k)$? I think that the answer is trivially yes.

$D_{x,y}$ is a free left $D_{p,q}$-module, and $D_{p,q}$ has the IBN property, so the rank of $D_{p,q} \subseteq D_{x,y}$ is well-defined (in analogy to the dimension of a field extension); see this question.

(2) Is the rank of $D_{p,q} \subseteq D_{x,y}$ finite or infinite?

If I am not wrong, the rank is necessarily $\leq \aleph_0$ (since probably a subset of $x^iy^j$ can serve as a basis?). In analogy: If $f:k[x,y] \to k[x,y]$ has an invertible Jacobian, then Keller's theorem (Theorem 2.1, birational case) says that if $k(p,q)=k(x,y)$ then $k[p,q]=k[x,y]$, namely such $f$ is surjective, hence an automorphism (surjective iff bijective, in a Noetherian ring). In the commutative case, $k(p,q) \subseteq k(x,y)$ is finite dimensional.

My suspecture: Similarly to Keller's theorem for polynomial rings, here we also have: If $D_{p,q}=D_{x,y}$, then $f$ is surjective (so the Dixmier Conjecture is true if the division rings are equal).

I wonder if it is less difficult to show an equality of the division rings, $D_{p,q}=D_{x,y}$, than to show an equality of the fields of fractions, $k(p,q)=k(x,y)$.

• As a rule of thumb, things tend to get harder, not easier, when commutativity is removed. – David Hill Sep 14 '17 at 20:30