The Jacobian Conjecture over a commutative $\mathbb{Q}$-algebra which is not an integral domain 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 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 $\deg(q) \geq 100$.

Is this result still valid if we replace $k$ by $R$?

Remark: See also Proposition 1.1.12 and this related question.
Thank you very much!
 A: The most general situation I can cover is the following:

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 $\deg g \geq 100$.

The proof relies on the following lemma:

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.

Proof. The assumptions imply $f$ is flat by the fibrewise criterion of flatness [Tag 039E]. Hence, $f$ is universally open [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]. It also follows that $f$ is universally closed, hence proper. Thus, $f$ is finite [Tag 01WN], hence finite locally free of rank $1$. It is now an easy exercise (cf. e.g. this blog post 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 $\deg g \geq 100$. $\square$
A: The most general situation I can cover is of an integral domain of characteristic zero (this is why I have asked about a non-integral domain):

Let $k$ be a field of characteristic zero, and let $R$ be a $k$-algebra which is a (commutative) integral domain. If $f,g \in R[x,y]$ satisfy $\det \operatorname{Jac}(f,g)=1$ but $\phi: (x,y) \mapsto (f,g)$ is not an automorphism of $R[x,y]$, then $\deg(f) \geq 100$ or $\deg(g) \geq 100$.

Indeed, let $Q(R)$ be the field of fractions of $R$. $\phi$ is an $R$-algebra endomorphism of $R[x,y]$, and we can think of $\phi$ as a ${Q(R)}$-algebra endomorphism of $Q(R)[x,y]$, denote it by $\psi$.
We claim that $\psi$ is not an automorphism of $Q(R)[x,y]$; otherwise, if $\psi$ is an automorphism of $Q(R)[x,y]$, then by Lemma 1.1.8 we obtain that $\phi$ is an automorphism of $R[x,y]$, contrary to our assumption.
Since $\psi$ is not an automorphism of $Q(R)[x,y]$, by Moh's result we get that $\deg(f) \geq 100$ or $\deg(g) \geq 100$.
