# Solutions to $w_x=CA_x$, $w_y=CA_y$ other than $w=f(A)$ and $C=f'(A)$?

Let $$R \in \{\mathbb{C}[t^2,t^3], \mathbb{Z}[\sqrt{5}]\}$$, and let $$A=A(x,y) \in R[x,y]$$ with $$\deg(A) \geq 1$$ (total degree).

I wish to prove or find a counterexample to the following claim:

If there exist $$w=w(x,y),C=C(x,y) \in R[x,y]$$ such that $$w_x=CA_x$$ and $$w_y=CA_y$$ (the subscripts denote partial derivatives), then necessarily $$w=f(A)$$ and $$C=f'(A)$$, for some $$f \in R[T]$$.

I have asked the above question here, but have not received an answer; hopefully, my question is appropriate for MO.

Actually, my 'original' question is as follows:

Suppose that $$A,B,w \in R[x,y]$$ satisfy two conditions: (i) $$\operatorname{Jac}(A,B)=1$$. (ii) $$\operatorname{Jac}(A,w)=0$$. Is it true that $$w \in R[A]$$?

Perhaps, if there exists a counterexample to my original question, then it is easier to find a counterexample in the first Weyl algebra than in $$R[x,y]$$, where the Jacobian is replaced by the commutator? See this question.

New edit 2: If I am not missing something, the following is a counterexample to my original question: $$R=\frac{\mathbb{C}[a,b,c]}{\langle a^2-1, bc \rangle}$$, $$A=(\bar{a}+i\bar{b})x+\bar{b}y$$, $$B=\bar{b}x+(\bar{a}-i\bar{b})y$$, $$w=\bar{c}x$$; see this for more elaboration. Please, what do you think?

Any hints and comments are welcome!

New edit: A counterexample if $$R$$ is not a $$\mathbb{Q}$$-algebra: Adjusting Warning 1.1.17, we obtain: Let $$R:=\frac{\mathbb{Z}[t]}{(2t)}$$, $$R[x,y]$$, $$A=x-\bar{t}x^2$$, $$B=Y$$, $$w=x$$. We have:

(i) $$\operatorname{Jac}(A,B)=1$$. (ii) $$\operatorname{Jac}(A,w)=(1-2\bar{t}x)0-01=10-01=0$$. (iii) $$w=X \notin R[x-\bar{t}x^2]$$ from considerations of degrees.

Motivation: The motivation to my questions is a result of Cheng-Mckay-Wang that showed that the answer to my 'commutative' question is positive for $$R=\mathbb{C}$$, and a result by J.A. Guccione, J.J. Guccione, and C. Valqui that showed that the answer to my 'non-commutative' question is positive for $$R=k$$, a field of characteristic zero.

Then it is possible to show (I can add a complete explanation, if one will ask me to) that the answer to my question is still positive if we replace $$R=\mathbb{C}$$ by any field of characteristic zero.

Moreover, the answer to my question is still positive if we replace $$R=\mathbb{C}$$ by any normal integral domain of characteristic zero; this follows quite immediately from this answer (I can add a complete explanation, if one will ask me to). See also this answer.

Therefore, we are left with the case that $$R$$ is a non-normal integral domain of characteristic zero, for example: $$R \in \{\mathbb{C}[t^2,t^3], \mathbb{Z}[\sqrt{5}]\}$$.

We have: (i) $$A_xB_y-A_yB_x=1$$ and (ii) $$A_xw_y-A_yw_x=0$$. Then, $$A_xw_y=A_yw_x$$, so $$\frac{w_y}{A_y}=\frac{w_x}{A_x}=:C$$ for some $$C \in R(x,y)$$; hence, $$w_x=CA_x$$ and $$w_y=CA_y$$. Multiply (i) by $$C$$ and get that $$R[x,y] \ni w_xB_y-w_yB_x=CA_xB_y-CA_yB_x=C$$.

Next, as was calculated here (my above mentioned MSE question), $$C_yA_x=C_xA_y$$, namely, $$\operatorname{Jac}(A,C)=0$$. Observe that $$w_x=CA_x$$ (or $$w_y=CA_y$$) implies that $$\deg(w)=\deg(C)+\deg(A)$$, so $$\deg(C) < \deg(w)$$. By induction on the degree of an element $$e \in R[x,y]$$ in the 'centralizer of $$A$$' (= $$\operatorname{Jac}(A,e)=0$$) we obtain that $$C \in R[A]$$.

Then, if I am not wrong, from $$C \in R[A]$$, $$w_x=CA_x$$ and $$w_y=CA_y$$, it follows that $$w \in R[A]$$: Write $$C=g(A)$$ for some $$g \in R[T]$$, so $$w_x=g(A)A_x=\frac{\partial}{\partial{x}}f(A)$$, where $$f'(T)=g(T)$$. Then, $$w=f(A)+H(y)$$. Then $$CA_y=w_y=g(A)A_y+H'(y)=CA_y+H'(y)$$, so $$H'(y)=0$$, namely, $$H(y)=r \in R$$, concluding that indeed $$w=f(A)+r \in R[A]$$.

The base case of the induction is $$e=\lambda x+\mu y+\nu$$; w.l.o.g $$\mu=0$$ so $$e=\lambda x+\nu$$. Then we have, $$\operatorname{Jac}(A,e)=-A_y \lambda=0$$, so $$A \in R[x]$$. It is easy to see that $$A \in R[x]$$ which has a Jacobian mate (= $$\operatorname{Jac}(A,B)=1$$) must be of degree one: $$A=\delta x + \epsilon$$, and then indeed $$e=\lambda x+\nu \in R[x]=R[A]$$.

It seems that the above arguments are valid for any integral domain $$R$$ of characteristic zero, normal or not; am I right?

• For $P\in R[x]$ arbitrary, take $A(x,y)=x^2$, $w(x,y)=2x^2P(x)$, $C(x,y)=2P(x)+xP'(x)$. – abx Nov 11 '18 at 15:45
• Thank you! Truly, I meant that $A$ has a Jacobian mate ($x^2$ does not have a Jacobian mate). But it is my 'fault' that I have not mentioned this explicitly. (In some calculations I did in order to fotmulate this question, I have assumed that $A$ has a Jacobian mate). – user237522 Nov 11 '18 at 15:52
• By applying $\partial_y$ to $\omega_x = C A_x$ and $\partial_x$ to $\omega_y = C A_y$ we get $C_y A_x = C_x A_y$. If $A_x,A_y$ have no common factor, this implies that $C_x = D A_y$ and $C_y = D A_x$ for some $D$, and then you can establish the claim by induction on degree. (At least when $R$ is a field of characteristic zero. There could be divisibility obstructions in the integral domains you have.) – Terry Tao Nov 11 '18 at 17:01
• @TerryTao, thank you very much! (I have mentioned that $C_yA_x=C_xA_y$ in math.stackexchange.com/questions/2993124/…, but did not know how exactly this may help). In my case, indeed $A_x$ and $A_y$ have no common factor (since there exists $B \in R[x,y]$ such that $A_xB_y-A_yB_x=1$). However, I wish that $R$ will be a non-normal integral domain, not a field; is there a counterexample in that case? – user237522 Nov 11 '18 at 19:45
• (I guess that it was meant that: $C_x=DA_x$ and $C_y=DA_y$). Now, perhaps it would be easier to find a counterexample to the non-commutative analogous question, in the first Weyl algebra; I will soon ask this in a separate question. – user237522 Nov 11 '18 at 20:42