CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-17

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The separable degree of $F (α) / F$ is the number of distinct roots of $m_{α}$

Statement

If $α$ is algebraic over $F$ , then $[F (α) : F]_{s}$ equals the number of distinct roots of the minimal polynomial $m_{α}$ in any algebraic closure of $F$ .

Facts & Assumptions

Given: An algebraic element $α$ over $F$ and an algebraic closure $Ω / F$ .

[L1]

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

[L2]

Such embeddings of $F (α)$ correspond bijectively to the distinct roots of $m_{α}$ ( $F$ -embeddings of $F (α)$ into an algebraically closed field correspond to the distinct roots of $m_{α}$ ).

[L3]

The embedding count is independent of the chosen algebraic closure (The separable degree is independent of the chosen algebraic closure).

Proof

technique · direct

1.1

By [L2], the embedding set counted in [L1] is in bijection with the distinct-root set of $m_{α}$ in $Ω$ .

L1L2

2.1

Taking finite cardinalities gives the assertion, and [L3] removes dependence on $Ω$ .

step 1.1L3∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 38 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

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)

The separable degree of F(α)/FF(α)/F is the number of distinct roots of mαmα​