Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: Literature-sourcedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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A degree-four polynomial can be reducible over Q\mathbb Q without having a rational root

Statement refuted

A polynomial over a field is irreducible whenever it has no root in that field.

Facts & Assumptions

Given: The polynomial f=(x2+1)(x2+2)=x4+3x2+2∈Q[x]f=(x^2+1)(x^2+2)=x^4+3x^2+2\in\mathbb Q[x].

[L1]

A root is equivalent to divisibility by the corresponding linear polynomial (Factor theorem over a commutative ring).

[L2]

The rational numbers form a field (The rationals form a field).

[L3]

The rational numbers form an ordered field, so squares are nonnegative and positive constants remain positive when added (The rationals form a totally ordered field).

Counterexample

technique · direct
1.1

The displayed equality is a factorization in the field polynomial ring [L2] into two positive-degree nonunits, so ff is reducible.

givenL2algebra
2.1

For every rational aa, [L3] gives a2+1>0a^2+1>0 and a2+2>0a^2+2>0, so f(a)>0f(a)>0 and ff has no rational root; by [L1] it has no rational linear factor, yet step 1.1 shows it reducible.

step 1.1L1L3algebra∎

Depends on

Used by

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Sources