Degree Equals Galois Group Order - Imported Knowlpedia Content

Let $L/K$ be a finite field extension. Its Galois group is

\mathrm{Gal}(L/K)=\{\sigma:L\to L \text{ a field automorphism with }\sigma|_K=\mathrm{id}_K\},

i.e. the group of $K$ -automorphisms of $L$ .

Theorem (degree = group order for finite Galois extensions). If $L/K$ is a finite Galois extension, then

|\mathrm{Gal}(L/K)| = [L:K],

Remarks

A useful companion fact is the inequality valid for any finite extension:

|\mathrm{Aut}_K(L)| \le [L:K],

with equality if and only if $L/K$ is Galois (equivalently: finite, separable, and normal; see separable + normal = Galois).

Quadratic extensions. For $L=\mathbb{Q}(\sqrt2)$ over $K=\mathbb{Q}$ , we have $[L:K]=2$ . The two $K$ -automorphisms are $\sqrt2\mapsto \sqrt2 \quad\text{and}\quad \sqrt2\mapsto -\sqrt2,$ so $|\mathrm{Gal}(L/K)|=2=[L:K]$ .

A Galois splitting field of degree 6. Let $L$ be the splitting field over $\mathbb{Q}$ of $x^3-2$ . Then $L=\mathbb{Q}(\sqrt[3]{2},\zeta_3)$ is finite Galois over $\mathbb{Q}$ , and one finds $[L:\mathbb{Q}]=6 \quad\text{and}\quad \mathrm{Gal}(L/\mathbb{Q})\cong S_3,$ so $|\mathrm{Gal}(L/\mathbb{Q})|=6=[L:\mathbb{Q}]$ .

Contrast: a non-Galois finite extension. The simple extension $L=\mathbb{Q}(\sqrt[3]{2})$ has $[L:\mathbb{Q}]=3$ , but it is not normal (it does not contain the other complex cube roots), so it is not Galois. In fact, any $\mathbb{Q}$ -automorphism must send $\sqrt[3]{2}$ to another root of its minimal polynomial; only the real root lies in $L$ , so $\mathrm{Aut}_{\mathbb{Q}}(L)$ is trivial and $|\mathrm{Aut}_{\mathbb{Q}}(L)|=1<3$ .