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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?

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  • $\begingroup$ For $P\in R[x]$ arbitrary, take $A(x,y)=x^2$, $w(x,y)=2x^2P(x)$, $C(x,y)=2P(x)+xP'(x)$. $\endgroup$ – abx Nov 11 '18 at 15:45
  • $\begingroup$ 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). $\endgroup$ – user237522 Nov 11 '18 at 15:52
  • $\begingroup$ 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.) $\endgroup$ – Terry Tao Nov 11 '18 at 17:01
  • $\begingroup$ @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? $\endgroup$ – user237522 Nov 11 '18 at 19:45
  • $\begingroup$ (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. $\endgroup$ – user237522 Nov 11 '18 at 20:42

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