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CorollaryStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-13
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  • 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.
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Every nonconstant polynomial over a field has a root in some field extension

Statement

Every nonconstant polynomial f∈F[x] has a root in some field extension of F.

Facts & Assumptions

Given: A field F and a nonconstant polynomial f∈F[x].

[F1]

Every nonzero nonunit polynomial over a field is a finite product of irreducible polynomials (Every nonzero nonunit polynomial over a field factors into irreducible polynomials).

[F2]

If p is monic irreducible, then F[x]/(p) is a field extension in which x+(p) is a root of p (F[x]/(p) for monic irreducible p is a field extension containing the root x+(p) with unique reduced representatives).

Proof

technique · direct
1.1F1

A nonconstant polynomial is nonzero and not a unit, so [F1] supplies an irreducible factor q of f.

2.1step 1.1algebra

Divide q by its nonzero leading coefficient to obtain a monic irreducible factor p; (p)=(q) and still p∣f.

3.1F2step 2.1

By [F2], K=F[x]/(p) is a field extension and a=x+(p) satisfies p(a)=0.

4.1step 2.1step 3.1algebra∎

Since f=ph for some h∈F[x], evaluation gives f(a)=p(a)h(a)=0.

Depends on

Dependency tree · 2 levels

Sources