Galois Extensions and the Galois Group

Galois theory establishes a profound correspondence between field extensions and groups, translating field-theoretic problems into group-theoretic ones.

The Galois Group

Definition8.1Galois extension and Galois group

A field extension $K/F$ is Galois if it is algebraic, normal, and separable. The Galois group $\mathrm{Gal}(K/F)$ is the group of all $F$ -automorphisms of $K$ (automorphisms fixing $F$ pointwise), with composition as the group operation.

For a finite Galois extension: $|\mathrm{Gal}(K/F)| = [K:F]$ .

ExampleGalois groups of small extensions

$\mathrm{Gal}(\mathbb{Q}(\sqrt{2})/\mathbb{Q}) = \{1, \sigma\}$ where $\sigma(\sqrt{2}) = -\sqrt{2}$ . Isomorphic to $\mathbb{Z}/2\mathbb{Z}$ .
$\mathrm{Gal}(\mathbb{Q}(\sqrt[3]{2}, \omega)/\mathbb{Q}) \cong S_3$ (symmetric group on 3 letters), acting on the three roots of $x^3 - 2$ .
$\mathrm{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p) = \langle \mathrm{Frob} \rangle \cong \mathbb{Z}/n\mathbb{Z}$ where $\mathrm{Frob}(x) = x^p$ .
$\mathrm{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \cong (\mathbb{Z}/n\mathbb{Z})^\times$ (units modulo $n$ ).

Fixed Fields

Definition8.2Fixed field

For a group $G$ of automorphisms of a field $K$ , the fixed field is $K^G = \{a \in K : \sigma(a) = a \text{ for all } \sigma \in G\}$ . This is always a subfield of $K$ .

Theorem8.1Artin's theorem

If $G$ is a finite group of automorphisms of a field $K$ , then $K/K^G$ is a Galois extension with $\mathrm{Gal}(K/K^G) = G$ and $[K:K^G] = |G|$ .

Characterizations

Theorem8.2Equivalent conditions for Galois

For a finite extension $K/F$ , the following are equivalent:

$K/F$ is Galois.
$K$ is the splitting field of a separable polynomial over $F$ .
$|\mathrm{Aut}(K/F)| = [K:F]$ .
$K^{\mathrm{Aut}(K/F)} = F$ (the fixed field equals the base field).

RemarkNon-Galois extensions

$\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ is not Galois: $|\mathrm{Aut}(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q})| = 1$ (the only $\mathbb{Q}$ -automorphism maps $\sqrt[3]{2}$ to a real cube root of 2, of which $\sqrt[3]{2}$ is the only one), while $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}] = 3$ . The extension is not normal.