The Stacks project

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

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

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

Consider the canonical map

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. $\square$

Proof. By Lemma 9.15.7 given $\tau : L \to L$ in $\text{Gal}(L/K)$ the restriction $\tau |_ M : M \to M$ is an element of $\text{Gal}(M/K)$. This defines the homomorphism $c$. Continuity follows from the universal property of the topology: the action

is continuous as $M \subset L$ and the action $\text{Gal}(L/K) \times L \to L$ is continuous. Hence continuity of $c$ by part (1) of Lemma 9.22.1. Lemma 9.15.7 also shows that the map is surjective. $\square$

Proof. Every subfield of $L$ containing $K$ is separable over $K$ (follows immediately from the definition). Let $S \subset L$ be a finite subset. Then $K(S)/K$ is finite and there exists a tower $L/E/K(S)/K$ such that $E/K$ is finite Galois, see Lemma 9.16.5. Hence $E = L_\lambda $ for some $\lambda \in \Lambda $. This certainly implies the set $\Lambda $ is not empty. Also, given $\lambda _1, \lambda _2 \in \Lambda $ we can write $L_{\lambda _ i} = K(S_ i)$ for finite sets $S_1, S_2 \subset L$ (Lemma 9.7.5). Then there exists a $\lambda \in \Lambda $ such that $K(S_1 \cup S_2) \subset L_\lambda $. Hence $\lambda \geq \lambda _1, \lambda _2$ and $\Lambda $ is directed (Categories, Definition 4.21.4). Finally, since every element in $L$ is contained in $L_\lambda $ for some $\lambda \in \Lambda $, it follows from the description of filtered colimits in Categories, Section 4.19 that $\mathop{\mathrm{colim}}\nolimits L_\lambda = L$.

If $\lambda \geq \lambda '$ in $\Lambda $, then we obtain a canonical surjective map $G_\lambda \to G_{\lambda '}$, $\sigma \mapsto \sigma |_{L_{\lambda '}}$ by Lemma 9.21.8. Thus we get an inverse system of finite groups with surjective transition maps.

Recall that $G = \text{Aut}(L/K)$. By Lemma 9.22.2 the restriction $\sigma |_{L_\lambda }$ of a $\sigma \in G$ to $L_\lambda $ is an element of $G_\lambda $. Moreover, this procedure gives a continuous surjection $G \to G_\lambda $. Since the transition mappings in the inverse system of $G_\lambda $ are given by restriction also, it is clear that we obtain a canonical continuous map

Continuity by definition of limits in the category of topological groups; recall that these limits commute with the forgetful functor to the categories of sets and topological spaces by Topology, Lemma 5.30.3. On the other hand, since $L = \mathop{\mathrm{colim}}\nolimits L_\lambda $ it is clear that any element of the inverse limit (viewed as a set) defines an automorphism of $L$. Thus the map is bijective. Since the topology on both sides is profinite, and since a bijective continuous map of profinite spaces is a homeomorphism (Topology, Lemma 5.17.8), the proof is complete. $\square$

Proof. We will use the result of finite Galois theory (Theorem 9.21.7) without further mention. Let $S \subset L$ be a finite subset. There exists a tower $L/E/K$ such that $K(S) \subset E$ and such that $E/K$ is finite Galois, see Lemma 9.16.5. In other words, we see that $L/K$ is the union of its finite Galois subextensions. For such an $E$, by Lemma 9.22.2 the map $\text{Gal}(L/K) \to \text{Gal}(E/K)$ is surjective and continuous, i.e., the kernel is open because the topology on $\text{Gal}(E/K)$ is discrete. In particular we see that no element of $L \setminus K$ is fixed by $\text{Gal}(L/K)$ as $E^{\text{Gal}(E/K)} = K$. This proves that $L^ G = K$.

By Lemma 9.21.4 given a subextension $L/M/K$ the extension $L/M$ is Galois. It is immediate from the definition of the topology on $G$ that the subgroup $\text{Gal}(L/M)$ is closed. By the above applied to $L/M$ we see that $L^{\text{Gal}(L/M)} = M$.

Conversely, let $H \subset G$ be a closed subgroup. We claim that $H = \text{Gal}(L/L^ H)$. The inclusion $H \subset \text{Gal}(L/L^ H)$ is clear. Suppose that $g \in \text{Gal}(L/L^ H)$. Let $S \subset L$ be a finite subset. We will show that the open neighbourhood $U_ S(g) = \{ g' \in G \mid g'(s) = g(s)\} $ of $g$ meets $H$. This implies that $g \in H$ because $H$ is closed. Let $L/E/K$ be a finite Galois subextension containing $K(S)$ as in the first paragraph of the proof and consider the homomorphism $c : \text{Gal}(L/K) \to \text{Gal}(E/K)$. Then $L^ H \cap E = E^{c(H)}$. Since $g$ fixes $L^ H$ it fixes $E^{c(H)}$ and hence $c(g) \in c(H)$ by finite Galois theory. Pick $h \in H$ with $c(h) = c(g)$. Then $h \in U_ S(g)$ as desired.

At this point we have established the correspondence between closed subgroups and subextensions.

Assume $H \subset G$ is open. Arguing as above we find that $H$ contains $\text{Gal}(L/E)$ for some large enough finite Galois subextension $E$ and we find that $L^ H$ is contained in $E$ whence finite over $K$. Conversely, if $M$ is a finite subextension, then $M$ is generated by a finite subset $S$ and the corresponding subgroup is the open subset $U_ S(e)$ where $e \in G$ is the neutral element.

Assume that $K \subset M \subset L$ with $M/K$ Galois. By Lemma 9.22.2 there is a surjective continuous homomorphism of Galois groups $\text{Gal}(L/K) \to \text{Gal}(M/K)$ whose kernel is $\text{Gal}(L/M)$. Thus $\text{Gal}(L/M)$ is a normal closed subgroup.

Finally, assume $N \subset G$ is normal and closed. For any $L/E/K$ as in the first paragraph of the proof, the image $c(N) \subset \text{Gal}(E/K)$ is a normal subgroup. Hence $L^ N = \bigcup E^{c(N)}$ is a union of Galois extensions of $K$ (by finite Galois theory) whence Galois over $K$. $\square$

Proof. This is a reformulation of Lemma 9.22.2. $\square$