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

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$Aut (K / F)$ is a group and $∣ Aut (K / F) ∣ \leq [K : F]_{s} \leq [K : F]$

Statement

For every field extension $K / F$ , composition makes $Aut (K / F)$ a group. If $K / F$ is finite, then

$∣ Aut (K / F) ∣ \leq [K : F]_{s} \leq [K : F] .$

In particular the relative automorphism group is finite.

Facts & Assumptions

Given: A field extension $K / F$ ; for the inequalities, a finite extension, an algebraic closure $Ω$ of $F$ , and the Axiom of Choice used by the embedding-extension theorem.

[F1]

An $F$ -automorphism of $K$ is an $F$ -isomorphism $K \to K$ (Relative field automorphisms and $Aut (K / F)$ ).

[A1]

Assuming the Axiom of Choice, an embedding $F \to Ω$ extends across the algebraic extension $K / F$ to an $F$ -embedding $τ : K \to Ω$ (Assuming Choice, a base-field embedding extends across every algebraic extension); the separable degree $[K : F]_{s}$ is the number of $F$ -embeddings $K \to Ω$ (The separable degree $[K : F]_{s}$ as a count of embeddings into an algebraic closure).

[L1]

For every finite field extension $K / F$ , one has $[K : F]_{s} \leq [K : F]$ (For a finite extension, $[K : F]_{s} \leq [K : F]$ ).

Proof

technique · direct

1.1

The identity map is an $F$ -automorphism, the composite of two $F$ -automorphisms is an $F$ -automorphism, and the inverse of an $F$ -isomorphism is again an $F$ -isomorphism fixing $F$ ; associativity is inherited from composition. Thus $Aut (K / F)$ is a group. Every such map fixes $0$ and $1$ , so no zero case is excluded.

F1algebra

1.2

For finite $K / F$ , choose $τ : K \to Ω$ as in [A1]. The map $σ \mapsto τ \circ σ$ sends $Aut (K / F)$ into the set of $F$ -embeddings $K \to Ω$ and is injective, because $τ \circ σ_{1} = τ \circ σ_{2}$ and injectivity of $τ$ imply $σ_{1} = σ_{2}$ .

A1F1choose

2.1

Step 1.2 and the definition of separable degree give $∣ Aut (K / F) ∣ \leq [K : F]_{s}$ , while [L1] gives $[K : F]_{s} \leq [K : F]$ . If $K = F$ , all three numbers are $1$ , so this includes the degree-one endpoint.

step 1.2L1∎

Depends on

Used by

For a finite extension, |Aut(K/F)| divides [K:F] Corollary
FALSE: every degree-n extension has exactly n automorphisms False statement
Artin's fixed-field theorem: [K:K^G]=|G| and Aut(K/K^G)=G Theorem
Equivalent characterizations of a finite Galois extension Theorem

Cited to discharge well-definedness by Relative field automorphisms and Aut(K/F).

Dependency tree · two levels

16 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, Chapter 3 (standard reference, not scraped)
K. Conrad, The Galois Correspondence, Section 4 (standard reference, not scraped)

Aut⁡(K/F)Aut(K/F) is a group and ∣Aut⁡(K/F)∣≤[K:F]s≤[K:F]∣Aut(K/F)∣≤[K:F]s​≤[K:F]