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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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An extension that is both separable and purely inseparable is trivial

Statement

If an algebraic extension $K / F$ is both separable and purely inseparable, then $K = F$ .

Facts & Assumptions

Given: An algebraic extension $K / F$ that is separable and purely inseparable, and an element $α \in K$ .

[L1]

Separability makes the minimal polynomial of every element separable (Separable algebraic elements and separable extensions).

[L2]

Pure inseparability makes every element have exactly one distinct conjugate over the base (Pure inseparability and its conjugate, embedding, and separable-degree criteria).

[L3]

Every algebraic element has a monic irreducible minimal polynomial over the base (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).

Proof

technique · direct

1.1

Let $m_{α}$ be the minimal polynomial from [L3]. It is separable by [L1], so all of its roots are distinct, but [L2] says it has only one distinct root. Hence $\deg m_{α} = 1$ .

L1L2L3

2.1

A degree-one minimal polynomial puts $α$ in $F$ . Since $α \in K$ was arbitrary, $K = F$ .

step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 42 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)