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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.

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  • $\begingroup$ In view of the example given by user52824 below, I wish to emphasize my comment 1: I am in fact interested in the case where $f$ is quasicompact. $\endgroup$ – Matthieu Romagny Nov 3 '14 at 8:50
  • $\begingroup$ ... and $S$ is noetherian. $\endgroup$ – Matthieu Romagny Nov 3 '14 at 8:57
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[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)).

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  • $\begingroup$ Why it is true that $f$ is an iso on local rings at points of $Y$? $\endgroup$ – Matthieu Romagny Nov 3 '14 at 9:51
  • $\begingroup$ @MatthieuRomagny: All local rings on $S$, and hence on $Y$, are $\mathbf{F}_2$ since everything in $A$ is idempotent. $\endgroup$ – user27920 Nov 3 '14 at 11:00
  • $\begingroup$ OK, I get it. Thank you for this nice contribution. The particular situation that I have in mind has additional features (like quasicompactness) that your example doesn't, but it helped me anyway to understand things better. Thanks again! (And if you happen to have ideas in the quasicompact case...) $\endgroup$ – Matthieu Romagny Nov 3 '14 at 22:40
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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.

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  • $\begingroup$ Vlad the Impaler? Is that Halloween humor? $\endgroup$ – Jason Starr Nov 5 '14 at 2:37
  • $\begingroup$ Vlad, very nice use of the classic punctured spectrum dimension induction trick. $\endgroup$ – user27920 Nov 5 '14 at 7:46
  • $\begingroup$ Two questions: 1) I'd be happy to have details on the proof that $\mathfrak{m}'=\mathfrak{m}B$. 2) For formal glueing in the very last step, the quoted lemma from the Stacks Project requires to know that $B$ is a finite $A$-module, how do you deal with this issue? $\endgroup$ – Matthieu Romagny Nov 8 '14 at 15:51
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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.

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