# Constructing Finite Separable Field Extensions From Sets Equipped With a Continuous Galois Group Action

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 $$\Psi$$.

## Questions

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,

Joel

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 $$\operatorname{Gal}(L|k)\cong \operatorname{Gal}(K|k)/\operatorname{Gal}(K|L).$$

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 $$G_k/H\cong Hom_k(L,k_s)$$.

• 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?
– Joel
Nov 15, 2020 at 10:16
• Given your action $H$ is the stabilizer of any element. For a transitive action this will determinate $L=k_s^H$ up to isomorphism. 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.

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 $$k$$.