PropositionStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

If $K / F$ is normal and $F \subseteq E \subseteq K$ , then $K / E$ is normal

Statement

If $K / F$ is a normal algebraic extension and $F \subseteq E \subseteq K$ , then $K / E$ is a normal algebraic extension.

Facts & Assumptions

Given: A normal algebraic extension $K / F$ and an intermediate field $E$ .

[F1]

Normality means that the minimal polynomial over the base of every element of the extension splits there (A normal algebraic extension is one in which every minimal polynomial with a root in the extension splits there).

[F2]

The minimal polynomial divides every base-field polynomial that vanishes at the element (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).

[F3]

For every field $L$ , the polynomial ring $L [x]$ is a unique factorisation domain (For every field $F$ , $F [x]$ is a unique factorisation domain).

Proof

technique · direct

1.1

Every element of $K$ is algebraic over $F$ , hence also algebraic over $E$ because the same polynomial lies in $E [x]$ . Thus $K / E$ is algebraic.

1.2

Fix $α \in K$ . Let $m_{F} \in F [x]$ and $m_{E} \in E [x]$ be its minimal polynomials. By [F2], $m_{E}$ divides $m_{F}$ in $E [x]$ .

2.1

Normality of $K / F$ makes $m_{F}$ split over $K$ . In the unique factorisation domain $K [x]$ from [F3], every divisor of that product of linear factors is itself a product of linear factors, so $m_{E}$ splits over $K$ . Since $α$ was arbitrary, [F1] makes $K / E$ normal.

F1F3step 1.2∎

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: 28 results over 7 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

The Stacks Project, Lemma 9.15.6 (standard reference, not scraped)

If K/FK/F is normal and F⊆E⊆KF⊆E⊆K, then K/EK/E is normal