This question is a follow-up to the following question:

Connection Between Finite Separable Extensions and Galois Group Actions

I would like to understand Theorem 1.5.2 in the book "Galois groups and Fundamental Groups" by Tamás Szamuely.

Definitions and Previous Work

Let $k$ be a field and $k_s$ a separable and algebraic closure. Let furthermore $\mathbf{FinSep}_k$ be the category of finite separable extensions of $k$. Let furthermore $\operatorname{Gal}(k):=\operatorname{Gal}(k_s|k)$, be the absolute Galois group. We define $\operatorname{Gal}(k)-\mathbf{FinSetCont}$ denote the finite sets equipped with a continuous group action by the absolute Galois group.

I would like to show that we have an anti-equivalence $$\mathbf{FinSep}_k\simeq \operatorname{Gal}(k)-\mathbf{FinSetCont}.$$ I have shown that we have a functor $$\Phi:\mathbf{FinSep}_k\to \operatorname{Gal}(k)-\mathbf{FinSetCont},$$ by assigning $$L\mapsto \operatorname{Hom}(L,k_s)\\ \Big(M\xrightarrow{f} L\Big )\mapsto\phi\circ f,$$ where $\phi:L\to k_s$.

Constructing Inverse

In the book, they show that the functor is essentially surjective and fully faithful.

However, by doing it like that, it doesn't seem to give me a clue about how to map a finite set equipped with a continuous Galois action to a a finite separable extension of $k$ (I may be wrong).

I would like to have a functor $$\Psi:\operatorname{Gal}(k)-\mathbf{FinSetCont}\to \mathbf{FinSep}_k,$$ and then show that it is an inverse to $\Phi$.

I think it should be possible to do by the Fundamental Theorem of Galois - I am not sure though.

If $S$ is a set equipped with a continuous action by $\operatorname{Gal}(k)$, where should I send it? I would like to have a map like the following one, I think $$\operatorname{Gal}(k_s|L)\mapsto L=k_s^{\operatorname{Gal}(k_s|L)}.$$

But in order to do this, it feels like I need to recover the Galois group from the category of $\operatorname{Gal}(k)-\mathbf{Sets}$.

I did some work on the étale fundamental group of varieties defined over the complex numbers earlier. I showed the connection to the classical fundamental group by taking the profinite completion. In order to do this connection, I showed the following:

Let $\widehat{\mathbf{G}}$ be the profinite completion of a group $\mathbf{G}$, considered as groupoids, and define the category $\mathbf{\widehat{G}-FinSets}$ to be the category of functors $q:\mathbf{\widehat{G}}\to\mathbf{FinSets}$, from the profinite completion to the category of finite sets. Then I showed that $$\operatorname{Aut}(q)\cong\mathbf{\widehat{G}}.$$ Maybe I can use this somehow? Maybe a variation of it? Notice that I didn't use anything about continuity in this construction, I think.

I talked with a professor earlier about this. He said that it may be good to understand the following equivalence of categories: $$ \mathbf{\widehat{G}-FinSetCont}\simeq \mathbf{G-FinSet}. $$

I am not sure how to use that though. Or well, he didn't mention in which direction the above result might be good, and I don't think I used it to construct $\Phi$. So I suppose he meant I might need it to construct $\Psi$.


1. Can I use $\operatorname{Aut}(q)$ somehow?

2. Or should I prove the equivalence the professor recommended: $\mathbf{\widehat{G}-FinSetCont}\simeq \mathbf{G-FinSet}$? Maybe this lets me use the classical fundamental theorem of Galois somehow?

Or maybe use a combination of 1. and 2.?

Do you have any idea where I should send the objects and morphisms to define $\Psi$, and what I need to do so? I really have no idea, and I would be really happy if you could help me to figure this out.

Best wishes,


I always try to mention some useful results, this time I wasn't sure if I should mention the fundamental theorem of Galois for finite extension or the Krull variant - but I did choose to mention the latter in the end.

Theorem 1.3.11 (Krull) Let $L$ be a subextension of the Galois extension $K|k$. Then $\operatorname{Gal}(K|L)$ is a closed subgroup of $\operatorname{Gal}(K|k)$. Moreover, the maps $$ L\mapsto H:=\operatorname{Gal}(K|L)\qquad \text{ and }\qquad H\mapsto L:=K^H $$ yield an inclusion-reversing bijection between subfields $K\supset L\supset k$ and closed subgroups $H\subset G$. A subextension $L|k$ is Galois over $k$ if and only if $\operatorname{Gal}(K|L)$ is normal in $\operatorname{Gal}(K|k)$; In this case there is a natural isomorphism $$\operatorname{Gal}(L|k)\cong \operatorname{Gal}(K|k)/\operatorname{Gal}(K|L).$$


2 Answers 2


For $L/k$ finite separable then $G_k$ acts on $Hom_k(L,k_s)$ and $L =k_s^H$ where $H$ is the stabilizer of $Id$. Since the action is transitive then the other stabilizers are conjugate to $H$ and $k_s^{\sigma H \sigma^{-1}} = \sigma(L)$. Thus the correspondence is between isomorphism classes of finite separable extensions of $k$ and transitive continuous actions of $G_k$ on finite sets.

For a class of separable extension given by $L/k$ there is only one finite continuous transitive action $\rho$, up to a permutation of the finite set. This is because $\rho$ defines an action of $G_k$ on $G_k/H\cong Hom_k(L,k_s)$.

  • $\begingroup$ Hi reuns, thanks for your answer! I'm sorry, but I don't quite understand what this tells me about the construction of $\Psi$ (I'm sure this is telling me that, it's just me who doesn't understand). When I construct $\Psi$, I need an arbitrary $\operatorname{Gal}(k)$-set, don't I? If this is the case, I cannot use $\operatorname{Hom}_k(L,k_s)$, or am I mistaken? $\endgroup$
    – Joel
    Nov 15, 2020 at 10:16
  • 1
    $\begingroup$ Given your action $H$ is the stabilizer of any element. For a transitive action this will determinate $L=k_s^H$ up to isomorphism. $\endgroup$
    – reuns
    Nov 15, 2020 at 10:39

It is not true that there is an anti-equivalence between the category of finite separable extensions of $k$ and the category of continuous actions of the profinite completion of $\operatorname{Gal}(k)$ on finite sets. There is certainly a contravariant functor $$\Phi:\mathbf{FinSep}_k\to \operatorname{Gal}(k)-\mathbf{FinSetCont}$$ but there is no contravariant functor $\Psi$ serving as an inverse of $\Phi$.

The fundamental theorem of Grothendieck Galois theory says that there is a anti-equivalence between the category of commutative separable algebras over $k$ and the category of continuous actions of the profinite completion of $\operatorname{Gal}(k)$ on finite sets.

A commutative separable algebra over $k$ is more general than a finite separable extension of $k$. It's the same as a finite product of finite separable extensions of $k$. The finite separable extensions of $k$ correspond only to the transitive continuous actions of the profinite completion of $\operatorname{Gal}(k)$ on finite sets.

This article explains what's going on:

It shows how to construct, from any continuous action of the profinite completion of $\operatorname{Gal}(k)$ on a finite set, a commutative separable algebra over $k$.


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