MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $f:X_0\rightarrow X$ is a closed immersion of locally noetherian schemes such that the topological spaces of $X_0$ and $X$ are identical (or, more generally, if $f$ is a universal homeomorphism), then it is known (see e.g. SGA1, Thm I.8.2, or, more generally, SGA1, Thm. IX.4.10) that pullback along $f$ induces an equivalence between the categories of etale $X$-schemes and etale $X_0$-schemes.

The main property of étale morphisms is the "infinitesimal lifting criterion", after which the definition of formally étale morphisms is modeled.

Is a "topological invariance result" as above also true for the categories of formally étale $X_0$- and $X$-schemes?

share|cite|improve this question
I thought etale and formally etale are the same? At least that's what remark I.3.22 in Milne's book seems to say – Moshe May 4 '12 at 17:09
Not precisely: etale=formally etale + locally of finite presentation. – Lars May 4 '12 at 18:15
I see, thanks... – Moshe May 4 '12 at 20:47
It would be great if you add what you've tried so far. At first sight, this seems to be just the definition of formally etale. If not, please explain. – Martin Brandenburg May 5 '12 at 12:57
Martin, that's what I thought at first. I agree that the fully faithfulness follows more or less directly from the definition of a formally etale morphism. The essential surjectivity (for $X_0\rightarrow X$ a nilp. thickening) is usually proven using the reduction to "standard etale" morphisms, which to my knowledge does not work for formally etale morphisms. The more general case ($X_0\rightarrow X$ a universal homeomorphism) is an application of faithfully flat descent (that's why it is in Exp. IX of SGA1). – Lars May 5 '12 at 15:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.