The question is in the title, namely:

Suppose that $k$ is a field and $k^{sep}$ its absolute algebraic separable closure. Denote $\bar s :\mathrm{Spec}\, k^{sep} \rightarrow \mathrm{Spec}\, k$ the corresponding geometric point. Is $\pi_1^{et}(\mathrm{Spec}\, k, \bar s ) \simeq \mathrm{Gal}(k^{sep}/k)$ (as topological groups)?

This seems really as something that should hold, since this is a motivating example of étale fundamental groups. Let me describe where is my problem.

At the moment, I hold the following conflicting believes:

  1. $\pi_1^{et}(\mathrm{Spec}\, k, \bar s) \simeq \mathrm{Gal}(k^{sep}/k)$, as stated.

  2. $\pi_1^{et}(\mathrm{Spec}\, k, \bar s)$ is given as the group of automorphisms of the fibre functor $F_{\bar s}: \mathsf{FEt}_{\mathrm{Spec}\, k} \rightarrow \mathsf{Sets}, X \mapsto X_{\bar s}=X \times_k\mathrm{Spec}\,k^{sep}.$ Moreover (more importantly), $(\mathsf{FEt}_{\mathrm{Spec}\, k}, F_{\bar s})$ is a Galois category, e.g. as in Definition 52.3.6 on Stacksproject.

  3. There is an equivalence of Galois categories $(\mathsf{FEt}_{\mathrm{Spec}\, k}, F_{\bar s})$ and $(G\mathsf{-sets}, U)$, where $G=Aut(F_{\bar s}),$ $G\mathsf{-sets}$ denotes the cat. of all finite $G$-sets with continuous $G$-action and $U:G\mathsf{-sets}\rightarrow \mathsf{Sets}$ is the forgetful functor (cf. e.g. Proposition 52.3.10 on Stacksproject). In particular, $Aut(F_{\bar s})=Aut(U)$ as topological groups.
  4. Given a topological group $G$ and $U: G\mathsf{-sets} \rightarrow \mathsf{Sets}$ the forgetful functor, $Aut(U) \simeq \widehat G$, the profinite completion of $G$, as in Lemma 52.3.3 on Stacksproject.
  5. The group $\mathrm{Gal}(k^{sep}/k)$ is not necessarily its own profinite completion, because there are in general subgroups of finite index that are not open. I believe this is shown e.g. in Milne's field theory notes (chapter 7, Proposition 7.26) in the case $k=\mathbb{Q}$.

It seems to me that using 2.-4., one obtains that $$\pi_1^{et}(\mathrm{Spec}\, k, \bar s ) \simeq \widehat{\mathrm{Gal}(k^{sep}/k)},$$ which by 5. is in general not isomorphic to $\mathrm{Gal}(k^{sep}/k).$ That is in conflict with 1.

So more specifically:

Which one of the above statements is wrong? If neither one is, why is the conclusion wrong?

Thanks in advance for any help.


1 Answer 1


Your error is in the conclusion. In statement (4), $\widehat{G}$ is the profinite completion of $G$ as a topological group. That means you take the completion with respect to only continuous homorphisms from $G$ to finite groups, not all homomorphisms. Any profinite group with its profinite topology is its own profinite completion as a topological group.

  • $\begingroup$ Oh, I see! That is subtle. Thanks (again)! $\endgroup$ Sep 27, 2017 at 2:07

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