Let $k$ be a field of characteristic zero and let $E=E(x,y) \in k[x,y]$. Define $t_x(E)$ to be the maximum among $0$ and the $x$-degree of $E(x,0)$. Similarly, define $t_y(E)$ to be the maximum among $0$ and the $y$-degree of $E(0,y)$.

The following nice result appears in several places, see for example,Proposition 2.1 (or Lemma 1.14 or Proposition 10.2.6).

Nice result: Let $A,B \in k[x,y]$ satisfy $\operatorname{Jac}(A,B) \in k-\{0\}$ (such $A,B$ is called a Jacobian pair). Assume that the $(1,1)$-degree of $A$, $\deg(A)$, is $>1$ and the $(1,1)$-degree of $B$, $\deg(B)$, is $>1$. Then the numbers $t_x(A),t_y(A),t_x(B),t_y(B)$ are all positive.

My question: Is the same result holds in the first Weyl algebra over $k$, $A_1(k)$? where instead of the Jacobian we take the commutator.

My answer: Of course, we must first define $t_x(A),t_y(A),t_x(B),t_y(B)$ in $A_1(k)$; it seems to me that the same definition holds for $A_1(k)$, or am I missing something? Perhaps it is not possible to consider $E(x,0)$, where $E \in A_1(k)$?

If I am not wrong, the proof of Proposition 2.1 can be adjusted to the non-commutative case:

(i) It is easy to see that Lemma 1.3 has a non-commutative analog.

(ii) Replacing the Jacobian by the commutator yields a similar result (use $[ab,c]=a[b,c]+[a,c]b$), and then the same conclusion follows.

One has to be careful what exactly is the similar result, since, for example, $[y^3,B]=y^2c+ycy+cy^2$, where $c:=[y,B]$. Then $[y^3,B]=3cy^2+\epsilon$, where $\epsilon \in A_1$ has degree $<\deg(c)+2$. We can consider the highest $(0,1)$-degree terms.

Indeed, suppose that $t_x(A)=0$. Then $A$ is divisible by $y$, so $A=\tilde{A}y^t$, where $t \geq 1$ and $\tilde{A}$ is an element of $A_1$ not divisible by $y$.

Actually, immediately $t=1$, because Lemma 1.3 (= the commutative and its non-commutative analog) says that if $(1,0) \notin \operatorname{Supp}(A)$, then $(0,1) \in \operatorname{Supp}(A)$.

We have, $1=[A,B]=[\tilde{A}y^t,B]=\tilde{A}[y^t,B]+[\tilde{A},B]y^t =\tilde{A}(t[y,B]y^{t-1}+\epsilon)+[\tilde{A},B]y^t =\tilde{A}t[y,B]y^{t-1}+\tilde{A}\epsilon+[\tilde{A},B]y^t$.

One cannot just take $y=0$ and get a contradiction, I guess (we are working in $A_1$ not in $k[x,y]$). So we should probably reach the conclusion by considerations of degrees.

I have asked this question here.

Thank you very much!