Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17
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An algebraic extension is purely inseparable over its separable closure

Statement

Let K/F be algebraic and let Ks be the separable closure of F in K. Then K/Ks is purely inseparable.

Facts & Assumptions

Given: An algebraic extension K/F, its separable closure Ks, and an element α∈K.

[L1]

The field Ks consists exactly of the elements of K separable over F (The separable closure of the base inside an algebraic extension).

[L2]

In characteristic p, the minimal polynomial of α has a unique form g(xpe) with g irreducible and separable (In characteristic p, every irreducible polynomial is uniquely g(xpe) with g irreducible and separable).

[L3]

Pure inseparability is the elementwise p-power condition, with only the trivial case in characteristic zero (Purely inseparable algebraic extensions).

[L5]

In a perfect field, every nonconstant irreducible polynomial is separable (Perfect fields: every irreducible polynomial is separable).

Proof

technique · direct
1.1L1L2algebra

Suppose char⁡F=p>0 and write the minimal polynomial of α as g(xpe) using [L2]. Then g(αpe)=0, and the minimal polynomial over F of αpe divides the separable polynomial g, so αpe is separable over F and belongs to Ks by [L1].

2.1step 1.1L3

Thus every element of K has a suitable p-power in Ks, so [L3] makes K/Ks purely inseparable.

3.1L1L3L4L5∎

In characteristic zero [L4] and [L5] make every irreducible polynomial separable, so Ks=K by [L1]; the extension K/Ks is trivial and purely inseparable by [L3].

Depends on

Used by

Dependency tree · two levels

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Sources