I am trying to understand the proof of Lemma 1.5.1 given in "Galois Groups and Fundamental Groups" by Tamás Szamuely.
Let $k$ be a base field and $k_s\subset \bar{k}$ a separable and algebraic closure, and let $\operatorname{Gal}(k):=\operatorname{Gal}(k_s/k),$ the absolute galois group. Let us furthermore define $\mathbf{FinSep}_k$ be the category of finite separable extensions of $k$ and $\operatorname{Gal}(k)-\mathbf{FinSetCon}$ be the category of finite sets with continuous and transitive left $\operatorname{Gal}(G)$-action. I want to understand the construction of the functor $$\Phi:\mathbf{FinSep}_k\to \operatorname{Gal}(k)-\mathbf{FinSetCon}.$$
So, I have to figure out where to send objects.
Let $L\in \mathbf{FinSep}_k$, and make the following assignment $L\mapsto \operatorname{Hom}_k(L,k_s)$, which is the set of $k$-algebra homomorphisms. We have a natural action by the Galois group, on the set, by
$$\varphi:\operatorname{Gal}(k)\times \operatorname{Hom}(L,k_s)\to \operatorname{Hom}(L,k_s)\\(\sigma,\phi)\mapsto \sigma\circ\phi.$$
Group action axioms.
The axioms of a group action is checked as follows: $e(x)=x$ is an element of $\operatorname{Gal}(k)$, and serves as the identity transformation in the definition of group actions: $e\circ\phi=\phi$. That $\varphi(\sigma\circ \tau,\phi)=\varphi(\sigma,\varphi(\tau,\phi))$ follows since composition of functions are associative.
Transformation is continuous.
$\operatorname{Hom}_k(L,k_s)$ is equipped with the discrete topology and so, the action is continuous if and only if the stabilizer $\operatorname{Hom}_k(L,k_s)_{\phi}$ is open for all $\phi\in \operatorname{Hom}_k(L,k_s)$. The stabilizer consists of all $\sigma\in\operatorname{Gal}(k)$, such that $\sigma\circ\phi=\phi.$ That is, all elements fixing $\phi(L)$.
By using Theorem 1.3.11, one can conclude that $\operatorname{Hom}(L,k_s)_{\phi}$ is $\text{open}^{\textbf{[1]}}$. If that is the case, we have that $\operatorname{Gal}(k)$ acts continuously.
Transitivity.
By the primitive element theorem $L$ is generated by a primitive element $\alpha\in k$, with minimal polynomial, $f$, say. An element $\phi\in\operatorname{Hom}_k(L,k_s)$ is given by mapping $\alpha$ to a root of $f$ in $k_s^{\textbf{[2]}}$. The Galois group permutes roots transitively, which shows that the Galois group gives a transitive action.
Next, they claim that the above argument shows that the map $\xi:\sigma\circ \phi\mapsto \sigma\operatorname{Hom}(L,k_s)_{\phi}$, induces an isomorphism of $\operatorname{Hom}(L,k_s)$ and the left coset space $\operatorname{Hom}(L,k_s)_{\phi}\backslash \operatorname{Gal}(k)$, $$\xi_*:\operatorname{Hom}(L,k_s)\xrightarrow{\cong} \operatorname{Hom}(L,k_s)_{\phi}\backslash \operatorname{Gal}(k)^{\textbf{[3]}}.$$
If $U$ is normal, we obtain the quotient $\operatorname{Gal}(k)/\operatorname{Hom}(L,k_s)_{\phi}$. By Theorem 1.3.11, this arises if and only if $L$ is Galois over $k$.
Questions.
[1] I'm not really sure how I apply Theorem 1.3.11 to conclude that the stabilizer is open. First things first, in this section, we do not assume $L$ is a subextension of $k_s$. But in Theorem 1.3.11, we do need to assume it, at least that is how I understand it (which is probably wrong). Even if I understood that it is possible to apply it, I do not see how.
Theorem 1.3.11 tells us what the closed subgroups looks like. There is also Corollary 1.3.9 in the book which tells me that the open subgroups of a profinite group are precisely the closed subgroups of finite index – which might be something I could use to conclude what we want.
I'm just confused about The Krull Theorem, and how I might apply it, do you have any idea?
[2] I haven't been doing Galois Theory for a while, so I am a little rusty. But I know that the Galois group sends root of the minimal polynomial to roots of the same minimal polynomial. But they are claiming that an arbitrary $k$-morphism, $\phi:L\to k_s$ does so too.
Why does it do that, or which result might be applied to conclude that? (This is probably obvious, and I am sorry for that).
[3] Between which objects does the map $\sigma\circ\phi\mapsto \sigma U$ operate? I'm stuck on the word "induce", which map does it induce from? Is it the following map we are considering
$$
\xi:\operatorname{Hom}(L,k_s)\to \{(\sigma\operatorname{Hom}(L,k_s)_{\phi})\}_{\sigma\in\operatorname{Gal}(k)},
$$
and from that one define the map $\sigma\circ \phi\mapsto \sigma\operatorname{Hom}(L,k_s)_{\phi}$?
When we are talking about isomorphisms now, I guess we are thinking about everything as $\operatorname{Gal}(k)$-sets? Maybe I shouldn't say $\sigma$ is in the Galois group in the above map? Maybe $\sigma$ is in the set $\operatorname{Hom}(L,k_s)$? Do I fix any of $\sigma$ or $\phi$ when I define the above map?
Sorry, that was a lot of questions. But I am a little bit confused about this construction. There are quite a few things to keep track of, I think.
I would be just as happy if you took your time to answer just one question, as answering all of the questions! If anything is unclear, I am happy to edit.
Best wishes,
Joel
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 $$\operatorname{Gal}(L|k)\cong \operatorname{Gal}(K|k)/\operatorname{Gal}(K|L).$$
Best Answer
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Hope this helps!