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.


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 $k$.


[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,


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


Here are some comments:

  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 $X = \mathrm{Hom}_k(L, k_s)$.

Hope this helps!


[1] 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).

[2] 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

[3] This follows from the orbit-stabilizer theorem.


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