TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

For a finite extension, $[K : F]_{s} = [K_{s} : F]$

Statement

If $K / F$ is finite and $K_{s}$ is the separable closure of $F$ in $K$ , then

$[K : F]_{s} = [K_{s} : F] .$

Facts & Assumptions

Given: A finite extension $K / F$ and its relative separable closure $K_{s}$ .

[L1]

The field $K_{s}$ consists of the elements separable over $F$ (The separable closure of the base inside an algebraic extension).

[L2]

The extension $K / K_{s}$ is purely inseparable (An algebraic extension is purely inseparable over its separable closure).

[L3]

A finite purely inseparable extension has separable degree one (Pure inseparability and its conjugate, embedding, and separable-degree criteria).

[L4]

A finite separable extension has full separable degree (A finite extension is separable if and only if $[K : F]_{s} = [K : F]$ ).

[L5]

Separable degree is multiplicative in finite towers (Separable degree is multiplicative in finite towers: $[L : F]_{s} = [L : K]_{s} [K : F]_{s}$ ).

Proof

technique · direct

1.1L1L4

The finite extension $K_{s} / F$ is separable by [L1], so [L4] gives $[K_{s} : F]_{s} = [K_{s} : F]$ .

1.2L2L3

By [L2] and [L3], one has $[K : K_{s}]_{s} = 1$ .

2.1step 1.1step 1.2L5algebra∎

Multiplicativity [L5] in $F \subseteq K_{s} \subseteq K$ gives $[K : F]_{s} = [K : K_{s}]_{s} [K_{s} : F]_{s} = [K_{s} : F]$ . This includes $K_{s} = F$ .

For a finite extension, $[K : F]_{s} = [K_{s} : F]$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

For a finite extension, [K:F]s=[Ks:F][K:F]s​=[Ks​:F]

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

For a finite extension, $[K : F]_{s} = [K_{s} : F]$