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 $T$?

In the analog commutative case, if $Jac(p,q)=1$, then $p$ and $q$ are algebraically independent, 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, (almost) all I can say now is that we can 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? There are more things to say about this non-commutative setting, for example, is an analog of Keller's theorem holds, namely, is it true that $D(T)=D(A_1)$ implies $T=A_1$? etc.