# A non-commutative analog of a result concerning a Jacobian pair

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, $$1=[A,B]=[\tilde{A}y,B]=\tilde{A}[y,B]+[\tilde{A},B]y$$.

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 $$\deg(w)=1$$.

• Perhaps I should post my above comment as an answer saying the following: The proof of Proposition 10.2.6 in A. van den Essen's book still holds if we replace $k[x,y]$ by $A_1(k)$, unless I am missing something. (I have checked all the arguments and they seem to still hold in the non-commutative case). – user237522 Jul 2 '19 at 21:17