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][1] (or [Lemma 1.14][2] or [Proposition 10.2.6][3]).

> **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][1] can be adjusted to the non-commutative case: 

**(i)** It is easy to see that [Lemma 1.3][1] 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,
$1=[A,B]=[\tilde{A}y,B]=\tilde{A}[y,B]+[\tilde{A},B]y$.

**Edit:** By considerations of $(1,1)$-degrees, we obtain that
$B$ is of the following form: $B=\lambda x + \mu +\tilde{B}y$,
for some $\tilde{B} \in A_1$, $\lambda \in k-\{0\}$, $\mu \in k$ (w.l.o.g. $\mu=0$). 

In other words, the apriori $B$ of the form $B=w+\tilde{B}y$ with $w \in k[x]$ and $\deg(w) \geq 1$, actually has $\deg(w)=1$.

Hopefully, this further information answers my question in the positive
(I have asked a separate [question][4] about this further information).



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I have asked this question [here][5].

Thank you very much!


  [1]: https://www.researchgate.net/publication/265368034_On_Appelgate-Onishi's_Lemmas
  [2]: https://link.springer.com/chapter/10.1007/978-94-015-8555-2_10
  [3]: https://www.springer.com/gp/book/9783764363505
  [4]: https://math.stackexchange.com/questions/3281015/is-a-specific-endomorphism-of-a-1-an-automorphism
  [5]: https://math.stackexchange.com/questions/3275302/a-non-commutative-analog-of-a-known-result-concerning-a-jacobian-pair