PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck 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.

A normal extension generated by finitely many elements is the splitting field of the product of their minimal polynomials

Statement

Let $E / F$ be normal and suppose $E = F (α_{1}, \dots, α_{m})$ for some $m \in N$ . If $p_{j}$ is the minimal polynomial of $α_{j}$ over $F$ , then $E$ is a splitting field over $F$ of $p_{1} p_{2} \dots p_{m} .$ For $m = 0$ , the product is $1$ and the assertion reads $E = F$ .

Facts & Assumptions

Given: A normal extension $E / F$ with the displayed finite generating family.

[F1]

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

[F2]

Each $α_{j}$ is a root of its minimal polynomial $p_{j}$ (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).

[F3]

A splitting field is generated over $F$ by all roots of a polynomial that splits there (Polynomials that split and splitting fields of a polynomial or a family of polynomials).

Proof

technique · direct

1.1F3

If $m = 0$ , the generating hypothesis says $E = F$ , and the nonzero constant $1$ has empty root set, so [F3] makes $F$ its splitting field.

1.2F1F3

Suppose $m > 0$ . By [F1], every $p_{j}$ splits over $E$ , so their product splits over $E$ . Let $K$ be the subfield of $E$ generated over $F$ by all roots of that product. Then $K \subseteq E$ .

2.1F2F3step 1.2∎

Each generator $α_{j}$ is one of those roots by [F2], so $E = F (α_{1}, \dots, α_{m}) \subseteq K$ . Thus $E = K$ , and [F3] says exactly that $E$ is the splitting field of the product.

Depends on

Dependency tree · 2 levels

Sources

J. S. Milne, Fields and Galois Theory, Proposition 2.12