TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck 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 $K / F$ .

[L1]

Every finite separable extension is simple (A finite extension generated by elements all but possibly one of which are separable is simple).

[L2]

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

[L3]

Separable degree is at most ordinary degree for every finite extension (For a finite extension, $[K : F]_{s} \leq [K : F]$ ).

[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 $m_{α}$ ).

[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

1.1L1L4L6

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

1.2L2L3L5

Conversely, assume $[K : F]_{s} = [K : F]$ and fix $α \in K$ . 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 $a c = b d$ , while [L3] gives $a \leq b$ and $c \leq d$ .

2.1step 1.2L4algebra

The inequalities give $a c \leq b c \leq b d$ ; 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.

3.1step 2.1L6

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

4.1step 1.1step 3.1∎

Steps 1.1 and 3.1 establish the biconditional.

Depends on

Used by

Dependency tree · two levels

19 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

P. L. Clark, Field Theory, Chapters 4 and 5 (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, Chapters 3 and 5 (standard reference, not scraped)

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