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CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27
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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.

An irreducible polynomial over R has degree 1 or an even degree

Statement

If f∈R[x] is irreducible, then deg⁡f=1 or deg⁡f is even.

Facts & Assumptions

Given: An irreducible polynomial f∈R[x].

[L1]

Every odd-degree real polynomial has a real root (Every odd-degree real polynomial has a real root).

[L2]

For a commutative ring R, an element a∈R, and a polynomial p∈R[x], one has p(a)=0 if and only if x−a divides p (Factor theorem over a commutative ring).

Proof

technique · direct
1.1givenL1

Suppose deg⁡f is odd. Then [L1] gives a∈R with f(a)=0.

2.1step 1.1L2algebra

By [L2], the linear polynomial x−a divides f. Since f is irreducible and x−a is nonconstant, this forces f to be associated to x−a, so deg⁡f=1.

3.1step 2.1algebra∎

Therefore an irreducible real polynomial can have odd degree only in the linear case. If it is not linear, its degree is not odd and hence is even.

Depends on

Dependency tree · 2 levels

Sources