# 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 $$R$$?

Thank you very much!

• Note that your ring $R$ is isomorphic to $\mathbb Q \times \mathbb Q$ by the Chinese remainder theorem, so the Jacobian conjecture over $R$ is just 'two copies of the Jacobian conjecture over $\mathbb Q$'. Since the degree of a polynomial over $R$ is just the maximum of the degrees on the two components, the result by Moh immediately carries over to $R$. – R. van Dobben de Bruyn Nov 18 '18 at 0:27
• @R.vanDobbendeBruyn, thank you very much! You can write your comment as an answer, if you like. Please two questions: (1) Moh's result is over any field of characteristic zero, not necessarily algebraically closed, right? (In Moh's paper the base field is algebraically closed, but in other papers dealing with such results it is not assumed to be algebraically closed). (2) What if we replace $R$ by an arbitrary $\mathbb{Q}$-algebra which is not an integral domain? – user237522 Nov 18 '18 at 0:49

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 $$\deg g \geq 100$$. $$\square$$

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 $$\deg(g) \geq 100$$.