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.

Depends on

Used by

Dependency tree · two levels

32 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Fields and Galois Theory, v5.10, Theorem 3.17 (standard reference, not scraped)
K. Conrad, The Galois Correspondence, Theorem 5.6 (standard reference, not scraped)