Theorem 9.21.7 (Fundamental theorem of Galois theory). Let $L/K$ be a finite Galois extension with Galois group $G$. Then we have $K = L^ G$ and the map

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

is a bijection whose inverse maps $M$ to $\text{Gal}(L/M)$. The normal subgroups $H$ of $G$ correspond exactly to those subextensions $M$ with $M/K$ Galois.

**Proof.**
By Lemma 9.21.4 given a subextension $L/M/K$ the extension $L/M$ is Galois. Of course $L/M$ is also finite (Lemma 9.7.3). Thus $|\text{Gal}(L/M)| = [L : M]$ by Lemma 9.21.2. Conversely, if $H \subset G$ is a finite subgroup, then $[L : L^ H] = |H|$ by Lemma 9.21.6. It follows formally from these two observations that we obtain a bijective correspondence as in the theorem.

If $H \subset G$ is normal, then $L^ H$ is fixed by the action of $G$ and we obtain a canonical map $G/H \to \text{Aut}(L^ H/K)$. This map has to be injective as $\text{Gal}(L/L^ H) = H$. Hence $|G/H| = [L^ H : K]$ and $L^ H$ is Galois by Lemma 9.21.2.

Conversely, assume that $K \subset M \subset L$ with $M/K$ Galois. By Lemma 9.15.7 we see that every element $\tau \in \text{Gal}(L/K)$ induces an element $\tau |_ M \in \text{Gal}(M/K)$. This induces a homomorphism of Galois groups $\text{Gal}(L/K) \to \text{Gal}(M/K)$ whose kernel is $H$. Thus $H$ is a normal subgroup.
$\square$

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