Is this formally étale morphism of schemes an isomorphism?

Question

In the last days I came to consider the following question which I'd be happy to see answered by the affirmative:

if $f:X\to S$ is a morphism of schemes which is formally étale, quasicompact, universally bijective, universally schematically dominant, with $S$ noetherian, is $f$ an isomorphism?

Comments :

1. you may assume $X$ and $S$ affine if you like,
2. I can show that it is enough to show that $f$ is a monomorphism, i.e. the diagonal morphism $\Delta:X\to X\times_SX$ is an isomorphism. Under the assumptions, $\Delta$ is a closed bijective immersion defined by an ideal $I$ such that $I^2=I$. If $I$ were nilpotent this would imply $I=0$, but in general...

EDIT. In fact my comment no 2 assumes that $f$ is quasicompact (an assumption which was absent from the first version of the question). Here is how it goes. If $f$ is quasicompact and universally schematically dominant, it follows from results of Olivier and Mesablishvili (see Mesablishvili, More on descent theory for schemes, Georgian Math. J. 2004; treated also in the Stacks Project here) that $f$ is an effective epimorphism. It is a categorical fact that an effective epi which is a mono is iso. This explains comment no 2. Since I am pleased to assume that $f$ is quasicompact and $S$ is noetherian, I modified the question accordingly.

Answer 1

[I originally gave what I thought to be a counterexample in the affine case, but I realized it violates universal schematic dominance, so below I give a non-qc counterexample that was originally a comment.]

Let $A = \mathbf{F}_2^I$ be a product of copies of $\mathbf{F}_2$ indexed by an infinite set $I$, and let $S = {\rm{Spec}}(A)$. Observe that every local ring on $S$ is $\mathbf{F}_2$ since every element of $A$ is idempotent. Let $X$ be the disjoint union of the evident collection of clopen points (indexed by $I$) and the reduced structure $Y$ on the closed complement of their union. Let $f:X \rightarrow S$ be the natural map. This is a universal bijection (built from the stratification by $Y$ and $S-Y$) and an isomorphism on local rings, so faithfully flat. In particular, it is universally schematically dominant. It is not even qc (since $I$ is infinite), so not an isomorphism. The interesting thing is that it is formally etale. This amounts to showing $f|_Y:Y \rightarrow S$ is formally etale.

More generally, if $A$ is any ring and $J$ is an ideal such that $J^2 = J$ (such as $A$ as above and $J$ the ideal of $Y$ in $S$) then $A \rightarrow A/J$ is formally etale (that being a non-flat map when $J \ne 0$ and ${\rm{Spec}}(A/J)$ is not open in Spec($A$), such as happens above, so perhaps slightly surprising at first sight). This is the standard counterexample to EGA 0$_{\rm{IV}}$ 19.10.3(i) and to EGA IV$_4$ 18.4.6(i) (whose "proof" ends by invoking EGA 0$_{\rm{IV}}$, 19.10.3(i)).

Answer 2

Assume $S = Spec(A)$ and $X = Spec(B)$. So $A$ is a Noetherian ring. Translating the conditions we have a formally \'etale ring map $A \to B$ such that for all $A$-algebras $A'$ the ring map $A' \to A' \otimes_A B$ is injective and bijective on spectra. We have to show that $A = B$.

Let $\mathfrak m$ be a maximal ideal. As $A/\mathfrak m \to B/\mathfrak mB$ is universally bijective we see that $B/\mathfrak m B$ has a unique prime ideal $\mathfrak m'$ and $A/\mathfrak m \subset B/\mathfrak m'$ is purely inseparable. As $A/\mathfrak m \to B/\mathfrak m B$ is formally \'etale we see that $\mathfrak m' = \mathfrak m B$ and that $A/\mathfrak m \subset B/\mathfrak m'$ is separated. Thus $A/\mathfrak m = B/\mathfrak m'$.

If for all maximal ideal $\mathfrak m$ of $A$ the map $A_\mathfrak m \to B_\mathfrak m$ is an isomorphism, then so is $A \to B$. Hence we may assume $A$ is local. In particular $\dim(A) < \infty$.

Assume $A$ is local with maximal ideal $\mathfrak m$ and $A$ of dimension $d$. Induction on $d$.

If $d = 0$ then $\mathfrak m^n = 0$ for some $n \geq 1$. Above we have seen that $A/\mathfrak m = B/\mathfrak m B$ and hence $\mathfrak m^i /\mathfrak m^{i + 1} \to \mathfrak m^i B/\mathfrak m^{i + 1} B$ is surjective for all $i$. Hence $A \to B$ is surjective. Since $A \to B$ is formally \'etale we get that it is an isomorphism.

If $d > 0$, pick $f \in \mathfrak m$ such that $\dim(R/fR) < d$ and $\dim(R_f) < d$. Then by induction on $d$ we see that $A_f \to B_f$ is an isomorphism and that $A/fA \to B/fB$ is an isomorphism. Using that $A \to B$ is formally \'etale we see that $A \to B$ induces an isomorphism $A^\wedge \to B^\wedge$ of $f$-adic completions (argument similar to above). Then it follows from formal glueing that $A = B$, see Tag 05ET.

Answer 3

I think that without additional finiteness assumptions, the answer is no in general. Here is a counterexample, in characteristic $p>0$.

Let $R$ be the quotient of the polynomial ring $\mathbb{F}_p[x_1,x_2,x_3,\dots]$ by the ideal generated by the elements $x_n-x_{2n}x_{2n+1}$ for $n\ge 1$. Let $I=(x_1,x_2,x_3,\dots)\subset R$. By the construction of $R$ we see that all the generators of $I$ lie in $I^2$, hence $I=I^2$.

Let $R^{\text{p}}=\varinjlim (R\stackrel{F}\to R\stackrel{F}\to \dots)$ be the perfection of $R$, let $J=IR^{\text{p}}$, and $A=R^{\text{p}}/J$. Since $R^{\text{p}}$ is perfect, the map $\mathbb{F}_p\to R^{\text{p}}$ is formally étale. Since $J=J^2$ in $R^{\text{p}}$, the quotient map $R^{\text{p}}\to A$ is formally étale. By composition the map $\mathbb{F}_p \to A$ is formally étale. Then $A=R^{\text{p}}/J$ is a local, nonreduced, zero-dimensional ring with residue field $\mathbb{F}_p$, which is formally étale over $\mathbb{F}_p$. Thus the map $\text{Spec}(A)\to \text{Spec}(\mathbb{F}_p)$ has all the required properties and is not an isomorphism.