The last few days I read up on étale morphisms of rings / affine schemes and the étale fundamental group. I have two closely connected questions.

  1. For $K$ a field a ring-homomorphism $K \rightarrow A$ is étale if and only if $A\cong L_1 \times ... \times L_r$ as $K$-algebras, where all $L_i \mid K$ are finite separable field extensions. It is often stated that the étale fundamental group of a field is it's absolute Galois-group $\operatorname{Gal}(K^s,K)$, where $K^s$ is the separable closure. I don't quite see how we can go from a product of finite separable extensions to single finite separable extensions, ie. why $$\pi_1^\text{et}(K) = \lim \operatorname{Aut}_K(L_1 \times \dots \times L_r) \;\;\text{and}\;\; \operatorname{Gal}(K^s,K) = \lim \operatorname{Aut}_K(L)$$ are supposed to give the same result, especially, since a product of finite separable extensions may not admit a morphism monomorphism into a common separable extension. Edit:There exists morphisms though.

  2. Why is a finite field $F$ considered to be an étale Eilenberg-Maclane space? When the issues in 1 are resolved, I see why $\pi_1^\text{et}(F) = \widehat{\Bbb Z}$, but just for spaces this shouldn't be the only requirement. I tried to look for more "evidence", but it seems like higher étale fundamental groups are quite hard to define. So is there some other naive reason, one could say that?

As always, thank you for your time and efforts!

  • 3
    $\begingroup$ I don't know enough about etale homotopy to resolve part 2, but a finite product of finite separable extensions always admits a morphism in to a common separable extension. For each $L_i$, one can find a map $L_i\to K^{sep}$; now consider the subfield of $K^{sep}$ generated by all the images. This will be a separable finitely generated algebraic extension of $K$ and therefore finite separable. $\endgroup$
    – KReiser
    Oct 26, 2023 at 17:27
  • $\begingroup$ You are absolutely right. I was thinking along these lines until I noted that a product has the universal property in the wrong direction and that there certainly cannot be an embedding… But we only need a morphism! $\endgroup$ Oct 26, 2023 at 18:10


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