Theorem 9.22.4 (Fundamental theorem of infinite Galois theory). Let $L/K$ be a Galois extension. Let $G = \text{Gal}(L/K)$ be the Galois group viewed as a profinite topological group (Lemma 9.22.1). Then we have $K = L^ G$ and the map

\[ \{ \text{closed subgroups of }G\} \longrightarrow \{ \text{subextensions }K \subset M \subset L\} ,\quad H \longmapsto L^ H \]

is a bijection whose inverse maps $M$ to $\text{Gal}(L/M)$. The finite subextensions $M$ correspond exactly to the open subgroups $H \subset G$. The normal closed subgroups $H$ of $G$ correspond exactly to subextensions $M$ Galois over $K$.

**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 $M \setminus K$ is fixed by $\text{Gal}(L/K)$ as $E^{\text{Gal}(E/K)} = K$. This proves that $L^ G = K$.

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 $H$ containes $\text{Gal}(E/K)$ 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 $K$.
$\square$

## Comments (0)