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

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An algebraic extension that is a splitting field of a polynomial is normal

Statement

Let $E / F$ be algebraic. If $E$ is a splitting field over $F$ of a nonzero polynomial $f \in F [x]$ , then $E / F$ is normal.

Facts & Assumptions

Given: An algebraic extension $E / F$ that is a splitting field of $0 \neq f \in F [x]$ .

[F1]

Corresponding roots of a transported irreducible polynomial give an isomorphism between their simple adjunctions (A base-field isomorphism extends across simple adjunctions of corresponding roots of an irreducible polynomial).

[F2]

A base isomorphism extends to an isomorphism between splitting fields of corresponding polynomials (A base-field isomorphism extends to an isomorphism between splitting fields of corresponding polynomials).

[F3]

A splitting field is generated over its base by all roots of the polynomial (Polynomials that split and splitting fields of a polynomial or a family of polynomials).

[F4]

Normality requires every minimal polynomial of an element of $E$ to split over $E$ (A normal algebraic extension is one in which every minimal polynomial with a root in the extension splits there).

[F5]

Every nonzero polynomial over a field has a splitting field (Every nonzero polynomial over a field has a splitting field).

Proof

technique · direct conjugate argument

1.1

Fix $α \in E$ and let $m$ be its minimal polynomial over $F$ . By [F5], choose a splitting field $Ω / E$ of $m$ , and let $β$ be any root of $m$ in $Ω$ . By [F1], the identity on $F$ extends to an isomorphism $σ_{0} : F (α) \to F (β)$ sending $α$ to $β$ .

F1F5

2.1

The field $E$ is a splitting field of $f$ over $F (α)$ , because it splits $f$ and is generated by its roots. The field $E (β)$ is a splitting field of $f$ over $F (β)$ for the same reason. Since $σ_{0}$ fixes the coefficients of $f$ , [F2] extends it to an isomorphism $σ : E \to E (β)$ .

F2F3step 1.1

3.1

Every generator of $E$ over $F$ is a root of $f$ . The map $σ$ fixes $F$ and therefore carries each such generator to another root of $f$ , all of which already lie in $E$ . Hence $σ (E) \subseteq E$ . But $σ (E) = E (β)$ by surjectivity, so $β \in E$ .

F3step 2.1

4.1

Every root $β$ of the minimal polynomial $m$ lies in $E$ , so $m$ splits over $E$ . Since $α$ was arbitrary and $E / F$ is algebraic by hypothesis, [F4] proves normality.

F4step 1.1step 3.1∎

Depends on

Used by

An algebraic extension generated by elements whose minimal polynomials split in it is normal Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 42 results over 11 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 9.15.9 (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, Proposition 2.12 (standard reference, not scraped)