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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-17
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A finite extension is separable if and only if [K:F]s=[K:F]

Statement

A finite extension K/F is separable if and only if [K:F]s=[K:F].

Facts & Assumptions

Given: A finite extension K/F.

[L3]

Separable degree is at most ordinary degree for every finite extension (For a finite extension, [K:F]s[K:F]).

[L4]

For a simple extension, separable degree is the number of distinct roots of the minimal polynomial (The separable degree of F(α)/F is the number of distinct roots of mα).

[L5]

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

[L6]

An extension is separable when every element has separable minimal polynomial over the base (Separable algebraic elements and separable extensions).

Proof

technique · direct
1.1

If K/F is separable, [L1] gives K=F(α). The polynomial mα is separable, so its number of distinct roots equals its degree; [L4] therefore gives [K:F]s=[K:F].

L1L4L6
1.2

Conversely, assume [K:F]s=[K:F] and fix αK. Put a=[F(α):F]s, b=[F(α):F], c=[K:F(α)]s, and d=[K:F(α)]. Then [L2] and [L5] give ac=bd, while [L3] gives ab and cd.

L2L3L5
2.1

The inequalities give acbcbd; equality of the endpoints and positivity of extension degrees force a=b. By [L4], the minimal polynomial of α therefore has as many distinct roots as its degree and is separable.

step 1.2L4algebra
3.1

Since α was arbitrary, every element of K is separable over F, so [L6] makes K/F separable. This proves the reverse implication.

step 2.1L6
4.1

Steps 1.1 and 3.1 establish the biconditional.

step 1.1step 3.1

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 47 results over 11 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