The $1$-dimensional Jacobian Conjecture over $\mathbb{Z}$-torsion free rings

Question

I am looking for further proofs, preferably in the literature, of the following result:

Proposition: Let $R$ be a unital commutative $\mathbb{Z}$-torsion free ring. If $f(x) \in R[x]$ with $f'(x) \in U(R[x])$, then there exists $g(x) \in R[x]$ such that $f(g(x)) = g(f(x)) = x$ in $R[x]$.

For example, $R = \mathbb{Z}$ is allowed (mixed characteristic) as well as $R = \mathbb{Z}[t]/(t^2)$, but $R = \mathbb{Z} \times \mathbb{Z} /(n)$ is not allowed.

I already have a proof of this fact which relies explicitly on a theorem of Comtet mentioned at least in MO415617, MO337330, and MO80828. So, if there is a proof in the literature based on a Comtet's type of result, I would like to know about this reference, but I am mostly interested in "new" proofs that do not invoke this type of result.

The $1$-dimensional Jacobian Problem in for $\mathbb{Z}$-torsion free rings as stated above should be well-known to experts (as a non-problem). But one issue when searching through the literature is that people are understandably mostly focused on the complex case in higher dimensions. Indeed the above result is completely trivial for any ring with $\operatorname{nil}(R) = 0$, not just $\mathbb{C}$.

EDIT: By Remy van Dobben de Bruyn's comment below, the name $\mathbb{Z}$-torsion free is a much more standard name than what I called strong characteristic 0 previously. Requiring only characteristic 0 is insufficient as shown by his example in the comments, but on the other hand requiring $R$ to be a $\mathbb{Q}$-algebra is a tad too strong as it excludes too many valid examples like the ones mentioned above. The point is that it suffices that multiplication by non-zero integers has a trivial kernel rather than non-zero integers being invertible in $R$.

Answer 1

Here is a straightforward commutative algebra argument. We have to show the following:

Lemma. Let $R$ be a $\mathbf Z$-torsion free ring and $f \colon R[x] \to R[x]$ an étale homomorphism of $R$-algebras. Then $f$ is an isomorphism.

Proof. As noted, this is trivial if $R$ is reduced: then $f'(x) \in R[x]^\times = R^\times$ is constant, so $\deg f = 1$ as $R$ is $\mathbf Z$-torsion free and $\tfrac{\partial}{\partial x}\sum_i a_ix^i = \sum_iia_ix^{i-1}$. We obtain the general case by reduction to the reduced case. Firstly, note that $f$ is defined and étale over some finitely generated subring $A \subseteq R$ (namely the subring generated by the coefficients of $f(x)$ and of $f'(x)^{-1}$), and the result for $f_A \colon A[x] \to A[x]$ implies the result for $f \colon R[x] \to R[x]$. Thus we may assume that $R$ is Noetherian. (Note that a subring of a $\mathbf Z$-torsion-free ring is still $\mathbf Z$-torsion-free.)

Then the radical $I = \operatorname{nil}(R)$ is finitely generated, so satisfies $I^n = 0$ for some $n \in \mathbf Z_{>0}$. Note that $R/I$ is again $\mathbf Z$-torsion free: if $x \in R$ and $m \in \mathbf Z_{>0}$ are such that $mx \in I$, then $m^n x^n = 0$ hence $x^n = 0$ since $R$ is $\mathbf Z$-torsion free. Write $\bar R$ for $R/I$, and $\bar f \colon \bar R[x] \to \bar R[x]$ for the reduction of $f$ modulo $I$.

Then $\bar f$ is still étale, hence an isomorphism since $R$ is reduced. By Nakayama's lemma [Tag 00DV(11)], we see that $f \colon R[x] \to R[x]$ is surjective since $\bar f$ is. It is also flat and finitely presented, hence isomorphic to the localisation $R[x] \to R[x]_e$ at an idempotent $e \in R[x]$ [Tag 00U8]. Since $\bar f$ is an isomorphism, we get $e \equiv 1 \pmod I$, i.e. $1-e$ is nilpotent. But $1-e$ is also idempotent, so $1-e=0$ and $e=1$. $\square$

In general, for thickenings $R \twoheadrightarrow R/I$ there is an equivalence of categories between étale $R$-algebras and étale $R/I$-algebras; see [Tag 0BQB] for the case of finite étale algebras, and [Tag 04DZ] for the general statement (in an even more general setting: thickenings are examples of universal homeomorphisms). So for these types of results, you usually only need to think about the reduced case.