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This question asks the following: "Suppose $k$ is a field of characteristic zero and $P$ and $Q$ are commuting elements of the first Weyl algebra. Is it true that $P$ and $Q$ are polynomials in some third element $H$?"

It was answered in the negative (the counterexample seems to be due to Dixmier and Makar-Limanov).

Now replace the field of characteristic zero $k$ by a non-normal integral domain of characteristic zero $R$, for example $R=\mathbb{C}[t^2,t^3]$.

Is it possible to adjust $P$ and $Q$ to some $\tilde{P}$ and $\tilde{Q}$ satisfying the following three conditions:

(i) $[\tilde{P},\tilde{Q}]=0$.

(ii) $\tilde{P}$ and $\tilde{Q}$ are not polynomials in some third element $H$.

(iii) $\tilde{P}$ has a 'Dixmier mate', namely, there exists $B$ in the first Weyl algebra over $R$ such that $[\tilde{P},B]=1$.

Remarks:

(1) The analogous question in $R[x,y]$ can be found here.

(2) In contrast, in $k[x,y]$, it is true that if $\operatorname{Jac}(p,q)=0$, then $p=u(h)$ and $q=v(h)$, for some $h \in k[x,y]$, $u(T),v(T) \in k[T]$; see, for example, Corollary 1.3 in Nagata's paper ($k$ is an algebraically closed field of characteristic zero) or this question ($k$ is an arbitrary field). $k$ can be replaced by any normal integral domain of characteristic zero, see this answer.

Any hints and comments are welcome!

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  • $\begingroup$ If we allow $H$ to equal $\tilde{P}$ or $\tilde{Q}$, then the answer is negative, due to arxiv.org/abs/0912.5202. Indeed, apply Theorem 2.11 to (i)+(iii) and get that $\tilde{Q} \in k[\tilde{P}]$. Therefore, (ii) is not satisfied (just take $H=\tilde{P}$). $\endgroup$
    – user237522
    Dec 14, 2018 at 10:13

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