# Connection Between Finite Separable Extensions and Galois Group Actions

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 $$\text{open}^{\textbf{}}$$. 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 $$k_s^{\textbf{}}$$. 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 $$\operatorname{Hom}(L,k_s)_{\phi}\backslash \operatorname{Gal}(k)$$, $$\xi_*:\operatorname{Hom}(L,k_s)\xrightarrow{\cong} \operatorname{Hom}(L,k_s)_{\phi}\backslash \operatorname{Gal}(k)^{\textbf{}}.$$ 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.

 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?

 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).

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

1. Note that even though $$L$$ may not be a priori contained in $$k_s$$, it is nonetheless true that $$\phi(L) \subset k_s$$ (and $$\phi(L)$$ is isomorphic to $$L$$ since $$\phi$$ is injective, since $$L$$ is a field). Then the point is that the stabilizer is exactly isomorphic to $$\mathrm{Gal}(K/\phi(L))$$, which by Theorem 1.3.11 is a closed subgroup. But it's also a finite index subgroup, because $$L$$ is assumed to be a finite separable extension (one slightly complicated way to see this is to note that $$\phi(L)$$ is contained in some minimal finite Galois extension $$M$$ of $$k$$ such that $$M \subset k_s$$: this is called its Galois closure. Then you have a surjection of sets $$\mathrm{Gal}(M/k) \cong \mathrm{Gal}(k_s/k)/\mathrm{Gal}(k_s/M) \twoheadrightarrow \mathrm{Gal}(k_s/k)/\mathrm{Gal}(k_s/\phi(L))$$ and then note that $$\mathrm{Gal}(M/k)$$ is a finite group since $$M$$ is finite Galois. This implies that $$\mathrm{Gal}(k_s/\phi(L))$$ has finite index). Therefore, as you mentioned, Corollary 1.3.9 implies that $$\mathrm{Gal}(K/\phi(L))$$ is open, and you get continuity of the Galois action.
2. Note that a $$k$$-morphism $$\phi: L \to k_s$$ (which is always injective) preserves $$k$$ by definition. So if we take any polynomial $$f(x) = \sum c_nx^n$$ with $$c_n \in k$$ and pick $$\alpha \in L$$ such that $$f(\alpha) = 0$$, then this implies that $$f(\phi(\alpha)) = \sum c_n\phi(\alpha)^n = \phi(\sum c_n\alpha^n) = \phi(f(\alpha)) = \phi(0) = 0.$$
3. This is an application of the orbit-stabilizer theorem to a transitive group action: if a group $$G$$ acts on a set $$X$$ transitively (i.e. for every $$x,y \in X$$ there exists $$g \in G$$ such that $$g \cdot x = y$$) then there is a bijection between the set $$X$$ and $$G_x \backslash G$$ where $$G_x$$ is the stabilizer of any element in $$X$$ (it doesn't matter which element you pick, the stabilizers are all canonically isomorphic). The bijection is given by defining the map $$\rho_x: G \to X$$ as $$\rho(g) = g \cdot x$$, and then noting that the map acts as the identity on $$G_x$$. So now you just need to apply this formalism when $$G = \mathrm{Gal}(k)$$ and $$X = \mathrm{Hom}_k(L, k_s)$$.
 While $$L$$ may not be a subextension of $$k_s$$, $$\phi(L)$$ is. The stabilizer is, as you mentioned $$\mathrm{Gal}(k_s/\phi(L))$$. This has finite index because $$\phi(L)/k$$ is finite. Since it's not mentioned in theorem 1.3.11 let's prove this, as it doesn't seem to be mentioned in the theorem of Krull. Let $$L'$$ be the Galois closure of $$\phi(L)$$ inside $$k_s$$, then $$L'/k$$ is finite. We have that $$\mathrm{Gal}(k_s/k)/\mathrm{Gal}(k_s/L') \cong \mathrm{Gal}(L'/k)$$, so $$\mathrm{Gal}(k_s/L')$$ has finite index. Because $$\mathrm{Gal}(k_s/L') \subset \mathrm{Gal}(k_s/\phi(L))$$, it follows that $$\mathrm{Gal}(k_s/\phi(L))$$ has finite index as well. As you mentioned, a closed subgroup of finite index is open (this implication holds in any topological group).
 Let $$\varphi:A \to B$$ be a homomorphism of $$k$$-algebras and let $$f= \sum_{i=0}^n \lambda_ix^i\in k[x]$$ be a polynomial and suppose that $$f(a)=0$$ for some $$a \in A$$. Then we obtain using that $$f$$ is $$k$$-linear and a ring homomorphism: $$f(\varphi(a))=\sum_{i=0}^n \lambda_i \varphi(a)^i=\varphi(\sum_{i=0}^n \lambda_i a^i)=\varphi(f(a))=\varphi(0)=0$$. The proof is exactly the same as a for automorphism of field extensions