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CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17
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For a finite extension, [K:F]s≤[K:F]

Statement

For every finite field extension K/F, one has [K:F]s≤[K:F].

Facts & Assumptions

Given: A finite extension K/F.

[L2]

Embeddings of a simple algebraic extension correspond to the distinct roots of its minimal polynomial (F-embeddings of F(α) into an algebraically closed field correspond to the distinct roots of mα).

[L3]

Ordinary extension degrees multiply in finite towers (Tower law for finite extensions: [L:F]=[L:K][K:F]).

[L4]

A finite extension has a finite basis over its base (The degree [K:F]=dim⁡FK of a finite field extension).

Proof

technique · direct
1.1L4choose

Choose a finite F-basis of K by [L4]; its elements generate K, so adjoining them successively gives a finite tower of simple extensions from F to K.

1.2L2

At each simple step, [L2] counts embeddings by distinct roots of a minimal polynomial, so its separable degree is at most the degree of that polynomial, which is the ordinary degree of the step.

2.1step 1.1step 1.2L1L3algebra∎

Multiplying the inequalities in step 1.2 and using [L1] and [L3] for the two tower products gives [K:F]s≤[K:F].

Depends on

Used by

Dependency tree · two levels

12 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