TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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The fundamental theorem of finite Galois theory

Statement

Let $K / F$ be finite Galois and let $G = Gal (K / F)$ . The assignments $H \mapsto K^{H}$ and $E \mapsto Gal (K / E)$ are mutually inverse inclusion-reversing bijections between subgroups $H \leq G$ and intermediate fields $F \subseteq E \subseteq K$ . Moreover,

$[K : K^{H}] = ∣ H ∣ and [K^{H} : F] = [G : H] .$

Facts & Assumptions

Given: A finite Galois extension $K / F$ , its finite group $G$ , the fact that $K / E$ is finite Galois for every intermediate field $E$ (A finite Galois extension is Galois over every intermediate field), the tower law (Tower law for finite extensions: $[L : F] = [L : K] [K : F]$ ), and the finite-group formula $∣ G ∣ = ∣ H ∣ [G : H]$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L1]

If $H$ is a finite group of automorphisms of $K$ , then $[K : K^{H}] = ∣ H ∣$ and $Aut (K / K^{H}) = H$ (Artin's fixed-field theorem: $[K : K^{G}] = ∣ G ∣$ and $Aut (K / K^{G}) = G$ ).

[L2]

For a finite extension $L / E$ with $G = Aut (L / E)$ , being Galois, being the splitting field of a separable polynomial, $∣ G ∣ = [L : E]$ , and $L^{G} = E$ are equivalent (Equivalent characterizations of a finite Galois extension).

Proof

technique · direct

1.1L1

For the subgroup-to-field-to-subgroup direction, Artin applied to $H$ gives $Gal (K / K^{H}) = Aut (K / K^{H}) = H$ . This includes $H = {1}$ , whose fixed field is $K$ , and $H = G$ .

1.2L1L2given

For the field-to-subgroup-to-field direction, put $H = Gal (K / E)$ . Since $K / E$ is finite Galois, [L2] gives $∣ H ∣ = [K : E]$ . Artin gives $[K : K^{H}] = ∣ H ∣ = [K : E]$ , while $E \subseteq K^{H}$ ; the tower law forces $E = K^{H}$ . This includes $E = K$ and $E = F$ .

2.1step 1.1step 1.2L1L2given∎

If $H_{1} \subseteq H_{2}$ , then every element fixed by $H_{2}$ is fixed by $H_{1}$ , so $K^{H_{2}} \subseteq K^{H_{1}}$ ; the reverse map is likewise inclusion-reversing. Artin gives $[K : K^{H}] = ∣ H ∣$ , while [L2] applied to $K / F$ gives $[K : F] = ∣ G ∣$ , so the tower and Lagrange formulas give $[K^{H} : F] = ∣ G ∣ / ∣ H ∣ = [G : H]$ . Together with steps 1.1 and 1.2 these statements prove the claimed bijections and degree formulas.

The fundamental theorem of finite Galois theory

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources