Lemma 9.22.1. Let $E/F$ be a Galois extension. Endow $\text{Gal}(E/F)$ with the coarsest topology such that

$\text{Gal}(E/F) \times E \longrightarrow E$

is continuous when $E$ is given the discrete topology. Then

1. for any topological space $X$ and map $X \to \text{Aut}(E/F)$ such that the action $X \times E \to E$ is continuous the induced map $X \to \text{Gal}(E/F)$ is continuous,

2. this topology turns $\text{Gal}(E/F)$ into a profinite topological group.

Proof. Throughout this proof we think of $E$ as a discrete topological space. Recall that the compact open topology on the set of self maps $\text{Map}(E, E)$ is the universal topology such that the action $\text{Map}(E, E) \times E \to E$ is continuous. See Topology, Example 5.30.2 for a precise statement. The topology of the lemma on $\text{Gal}(E/F)$ is the induced topology coming from the injective map $\text{Gal}(E/F) \to \text{Map}(E, E)$. Hence the universal property (1) follows from the corresponding universal property of the compact open topology. Since the set of invertible self maps $\text{Aut}(E)$ endowed with the compact open topology forms a topological group, see Topology, Example 5.30.2, and since $\text{Gal}(E/F) = \text{Aut}(E/F) \to \text{Map}(E, E)$ factors through $\text{Aut}(E)$ we obtain a topological group. In other words, we are using the injection

$\text{Gal}(E/F) \subset \text{Aut}(E)$

to endow $\text{Gal}(E/F)$ with the induced structure of a topological group (see Topology, Section 5.30) and by construction this is the coarsest structure of a topological group such that the action $\text{Gal}(E/F) \times E \to E$ is continuous.

To show that $\text{Gal}(E/F)$ is profinite we argue as follows (our argument is necessarily nonstandard because we have defined the topology before showing that the Galois group is an inverse limit of finite groups). By Topology, Lemma 5.30.4 it suffices to show that the underlying topological space of $\text{Gal}(E/F)$ is profinite. For any subset $S \subset E$ consider the set

$G(S) = \{ f : S \to E \mid \begin{matrix} f(\alpha )\text{ is a root of the minimal polynomial} \\ \text{of }\alpha \text{ over }F\text{ for all }\alpha \in S \end{matrix} \}$

Since a polynomial has only a finite number of roots we see that $G(S)$ is finite for all $S \subset E$ finite. If $S \subset S'$ then restriction gives a map $G(S') \to G(S)$. Also, observe that if $\alpha \in S \cap F$ and $f \in G(S)$, then $f(\alpha ) = \alpha$ because the minimal polynomial is linear in this case. Consider the profinite topological space

$G = \mathop{\mathrm{lim}}\nolimits _{S \subset E\text{ finite}} G(S)$

Consider the canonical map

$c : \text{Gal}(E/F) \longrightarrow G,\quad \sigma \longmapsto (\sigma |_ S : S \to E)_ S$

This is injective and unwinding the definitions the reader sees the topology on $\text{Gal}(E/F)$ as defined above is the induced topology from $G$. An element $(f_ S) \in G$ is in the image of $c$ exactly if (A) $f_ S(\alpha ) + f_ S(\beta ) = f_ S(\alpha + \beta )$ and (M) $f_ S(\alpha )f_ S(\beta ) = f_ S(\alpha \beta )$ whenever this makes sense (i.e., $\alpha , \beta , \alpha + \beta , \alpha \beta \in S$). Namely, this means $\mathop{\mathrm{lim}}\nolimits f_ S : E \to E$ will be an $F$-algebra map and hence an automorphism by Lemma 9.8.11. The conditions (A) and (M) for a given triple $(S, \alpha , \beta )$ define a closed subset of $G$ and hence $\text{Gal}(E/F)$ is homeomorphic to a closed subset of a profinite space and therefore profinite itself. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).