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Added assumptions to the original question to fit with the actual situation I'm interested in.
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Matthieu Romagny
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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, and universally schematically dominant, iswith $f$$S$ noetherian, anis $f$ 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.

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, universally bijective, and universally schematically dominant, 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...

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.

Source Link
Matthieu Romagny
  • 4.5k
  • 1
  • 31
  • 37

Is this formally étale morphism of schemes an isomorphism?

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, universally bijective, and universally schematically dominant, 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...