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
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. $\square$
Your email address will not be published. Required fields are marked.
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: