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

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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.

Kronecker's one-root step: adjoining a root removes a linear factor and lowers the remaining degree

Statement

Let $F$ be a field and let $f \in F [x]$ have degree $n \geq 1$ . There is a root $α$ in a field extension of $F$ such that, for $K = F (α)$ , there is a polynomial $g \in K [x]$ satisfying $f (α) = 0, f = (x - α) g, \deg g = n - 1.$ If a field extension $L / K$ splits $g$ , then $f$ splits over $L$ .

Facts & Assumptions

Given: A field $F$ and a polynomial $f \in F [x]$ of degree $n \geq 1$ .

[F1]

Every nonconstant polynomial over a field has a root in some field extension (Every nonconstant polynomial over a field has a root in some field extension).

[F2]

For a polynomial over a commutative ring, $f (α) = 0$ if and only if $x - α$ divides $f$ (Factor theorem over a commutative ring).

[F3]

Every field is an integral domain, and over an integral domain degrees of nonzero polynomial products add (Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring, Over an integral domain, degrees add under multiplication of nonzero polynomials).

[F4]

A polynomial splits when it is a nonzero scalar times a product of linear factors (Polynomials that split and splitting fields of a polynomial or a family of polynomials).

Proof

technique · constructive

1.1

Since $n \geq 1$ , the polynomial is nonconstant. By [F1], choose an extension $H / F$ and a root $α \in H$ , and let $K = F (α) \subseteq H$ .

F1construct

2.1

By [F2], $f = (x - α) g$ for some $g \in K [x]$ . Since $f$ is nonzero, so is $g$ ; also $x - α$ is nonzero. Thus [F3] gives $n = 1 + \deg g$ and hence $\deg g = n - 1$ .

F2F3step 1.1algebra

3.1

If $g$ splits over $L$ , adjoining the factor $x - α$ to its linear factorisation gives a linear factorisation of $f$ over $L$ . This also covers $n = 1$ , when $g$ is a nonzero constant and its factor product is empty.

F4step 2.1discharge-construct∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 43 results over 16 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

T. Judson, Abstract Algebra: Theory and Applications, Section 21.2 (standard reference, not scraped)