Let $X$ be a quasi-projective variety over a $p$-adic field. Denote by $X^{an}$ its rigid analytification. Does $\pi_1^{et}(X)=\pi_1^{et}(X^{an})$?


1 Answer 1


Yes, if $X$ is proper (by GAGA). Otherwise, this is false for covers of degree divisible by $p$. See Example 51 in Ducros's survey "Étale Cohomology of Schemes and Analytic Spaces", though unfortunately he does not give a reference.

For an example of this failure in equal characteristic $p$, see section 7.4 in de Jong–van der Put "Etale cohomology of rigid analytic spaces" (Doc. Math. 1995). They construct an example of a $\mathbf{Z}/p$-covering of $\mathbf{A}^{1, \rm an}$ which is not the analytification of a finite etale covering of the line.

In mixed characteristic, almost the same argument should work if we replace Artin-Schreier coverings with Kummer coverings of degree $p$ (i.e. replace equations $T^p-T=f$ with $T^p=f$), but I didn't check the details.

  • 4
    $\begingroup$ Piotr, Thanks! I agree with you about the proper and equal characteristic case. In mixed characteristic case, I found the answer right now. The result is also ture (of course, here the fundamental group means finite etale ones). see Theorem 3.1 in scribd.com/document/14391207/… $\endgroup$
    – Yang
    Jan 20 at 15:09
  • 1
    $\begingroup$ Oops! Indeed, my answer was too hasty. I will leave it here for now and edit or delete later. $\endgroup$ Jan 20 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.