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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17
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Separable degree is multiplicative in finite towers: [L:F]s=[L:K]s[K:F]s

Statement

For every finite tower FKL,

[L:F]s=[L:K]s[K:F]s.

Facts & Assumptions

Given: A finite tower FKL and an algebraic closure Ω/F.

[L1]

Separable degree counts embeddings into an algebraic closure (The separable degree [K:F]s as a count of embeddings into an algebraic closure).

[L2]

Restriction from F-embeddings of L to F-embeddings of K is surjective, and every fibre has cardinality [L:K]s (Restriction partitions embeddings in a finite tower into extension fibres).

Proof

technique · direct
1.1

By [L2], the finite set HomF(L,Ω) is the disjoint union of the restriction fibres indexed by HomF(K,Ω).

L2
2.1

There are [K:F]s fibres by [L1], and each has [L:K]s elements by [L2]. Counting the disjoint union gives the displayed product.

step 1.1L1L2algebra

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 35 results over 10 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources