Let $R$ be a commutative $\mathbb{Q}$-algebra which is not an integral domain,
for example: $R=\frac{\mathbb{Q}[t]}{(t^2-1)}$.


Let $k$ be an algebraically closed field of characteristic zero, and let $p,q \in k[x,y]$ such that $\operatorname{Jac}(p,q)=1$.


A [known result](https://eudml.org/doc/152524) by T. T. Moh says that if $(p,q)$ is a counterexample to the two-dimensional Jacobian Conjecture (namely, if $(x,y) \mapsto (p,q)$ is not an automorphism of $k[x,y]$), then $\deg(p) \geq 100$ or $\deg(q) \geq 100$.



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> Is this result still valid if we replace $k$ by $R$?
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**Remark:** See also [Proposition 1.1.12](https://books.google.co.il/books?id=wKJqqd8t8KUC&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false) and [this](https://mathoverflow.net/questions/315261/a-b-w-in-frac-mathbbqtt2-1x-y-operatornamejaca-b-1-o?noredirect=1&lq=1) related question.


Thank you very much!



The Jacobian Conjecture over a commutative $\mathbb{Q}$-algebra which is not an integral domain
The most general situation I can cover is the following:



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> **Proposition.** *Let $k$ be a field of characteristic $0$, and let $R$ be a $k$-algebra of finite type. If $f,g \in R[x,y]$ are polynomials such that $\det\operatorname{Jac}(f,g) = 1$ but $(f,g) \colon R[x,y] \to R[x,y]$ is not an isomorphism, then $\deg f \geq 100$ or $\deg g \geq 100$.*
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The proof relies on the following lemma:



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> **Lemma.** *Let $S$ be a scheme, let $X$ and $Y$ be flat $S$-schemes that are locally of finite presentation, and let $f \colon X \to Y$ be a morphism of $S$-schemes. If $f\_s \colon X\_s \to Y\_s$ is an isomorphism for all $s \in S$, then $f$ is an isomorphism.*
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*Proof.* The assumptions imply $f$ is flat by the fibrewise criterion of flatness [Tag [039E](https://stacks.math.columbia.edu/tag/039E)]. Hence, $f$ is universally open [Tag [01UA](https://stacks.math.columbia.edu/tag/01UA)]. Since each fibre $X\_y \to y$ is an isomorphism, we conclude that $f$ is a universal homeomorphism, so in particular $f$ is affine [Tag [04DE](https://stacks.math.columbia.edu/tag/04DE)]. It also follows that $f$ is universally closed, hence proper. Thus, $f$ is finite [Tag [01WN](https://stacks.math.columbia.edu/tag/01WN)], hence finite locally free of rank $1$. It is now an easy exercise (cf. e.g. [this blog post](https://lovelylittlelemmas.rjprojects.net/locally-free-algebras/) I wrote) that this forces $f$ to be an isomorphism. $\square$


**Remark.** Note that there are easy counterexamples without the flatness assumptions. For example, we can take $S = Y$ to be a nodal curve, and $X$ its normalisation minus one of the points above the node. This should convince you that the result is less trivial than it sounds.


*Proof of Proposition.* By the lemma, if $\phi = (f,g) \colon R[x,y] \to R[x,y]$ is not an isomorphism, then the same must be true for $\phi \otimes \kappa(\mathfrak p) \colon \kappa(\mathfrak p)[x,y] \to \kappa(\mathfrak p)[x,y]$ for some prime ideal $\mathfrak p \subseteq R$. Since the map $\kappa(\mathfrak p) \to \overline{\kappa(\mathfrak p)}$ is faithfully flat, we conclude that $\phi \otimes \overline{\kappa(\mathfrak p)}$ is also not an isomorphism.


But the condition $\det \operatorname{Jac}(f,g) = 1$ propagates to their images $\bar f, \bar g \in \overline{\kappa(\mathfrak p)}[x,y]$, so Moh's result shows that $\deg \bar f \geq 100$ or $\deg \bar g \geq 100$. This forces $\deg f \geq 100$ or $\deg g \geq 100$. $\square$


