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user237522
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Is $x \in A_1$ left algebraic over the subalgebra generated by $p$ and $q$, $[q,p]=1$?

Let $A_1:=A_1(x,y,k)$ be the first Weyl algebra over a field $k$ of characteristic zero, namely, the $k$-algebra generated by $x$ and $y$ with relation $yx-xy=1$. Let $f:(x,y) \mapsto (p,q)$ be a $k$-algebra homomorphism of $A_1$, so $[q,p]=1$. Denote the image of $A_1$ under $f$ by $T$ ($T=A_1(p,q,k)$ is the $k$-subalgebra of $A_1$ generated by $p$ and $q$).

Definition: Let $A \subseteq B$ be two arbitrary rings. We say that $b\in B$ is left algebraic over $A$ if there exists $a_m,a_{m-1},\ldots,a_1,a_0 \in A$ such that $a_mb^m+a_{m-1}b^{m-1}+\cdots+a_1b+a_0=0$.

Is $x$ left algebraic over $T$?

Several relevant remarks:

  • In the analog commutative case, if $Jac(p,q)=1$, then $p$ and $q$ are algebraically independent over $k$, so $k(p,q) \subseteq k(x,y)$ is a finite field extension, so $x$ is algebraic over $k(p,q)$ and then trivially $x$ is (left) algebraic over $k[p,q]$.

  • In the non-commutative setting, consider the division ring of fractions of $T$, denote it by $D(T)$, and the division ring of fractions of $A_1$, denote it by $D(A_1)$. Clearly, $D(T) \subseteq D(A_1)$, and $D(A_1)$ is a free $D(T)$-module, but I do not know if it is of finite rank or infinite rank.

  • If I am not wrong, if we will know that $x$ and $y$ are left algebraic over $T$ (= this is what I ask above, I do not know the answer yet), then $D(A_1)$ is of finite rank over $D(T)$. Indeed, $D(A_1)$ is finitely generated as a $D(T)$-algebra by $x$ and $y$, $x$ is left algebraic over $D(T)$ hence $D(T)(x)$ (= the $k$-algebra generated by $D(T)$ and $x$) is finitely generated as a left $D(T)$-module by $\{1,\ldots,x^{m-1}$ if $m$ is a degree of a polynomial over $D(T)$ having $x$ as a root, and $y$ is left algebraic over $D(T)(x)$ hence $A_1$ is finitely generated as a left $D(T)(x)$-module by some $\{1,\ldots,y^l\}$, so $D(A_1)$ is finitely generated as a left $D(T)$-module by $\{x^iy^j\}$, $1 \leq i \leq m$, $1 \leq j \leq l$.

  • There are more things to say about this non-commutative setting, for example, is an analog of Keller's theorem (birational case, Theorem 2.1 (b)) holds, namely, is it true that $D(T)=D(A_1)$ implies $T=A_1$? etc.

user237522
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