CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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An irreducible polynomial over $R$ has degree $1$ or an even degree

Statement

If $f \in R [x]$ is irreducible, then $\deg f = 1$ or $\deg f$ is even.

Facts & Assumptions

Given: An irreducible polynomial $f \in R [x]$ .

[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 $a \in R$ , and a polynomial $p \in R [x]$ , one has $p (a) = 0$ if and only if $x - a$ divides $p$ (Factor theorem over a commutative ring).

Proof

technique · direct

1.1

Suppose $\deg f$ is odd. Then [L1] gives $a \in R$ with $f (a) = 0$ .

givenL1

2.1

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 $\deg f = 1$ .

step 1.1L2algebra

3.1

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.

step 2.1algebra∎

Depends on

Used by

The complex numbers are algebraically closed Theorem

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Keith Conrad, Applications of Galois Theory, Theorem 2.1 (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, v5.10, Chapter 5 (standard reference, not scraped)

An irreducible polynomial over RR has degree 11 or an even degree