CounterexampleConstruction: Literature-sourcedVerification: Literature-sourcedprecheck passaudited 2026-08-11

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A degree-four polynomial can be reducible over $Q$ 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 $f = (x^{2} + 1) (x^{2} + 2) = x^{4} + 3 x^{2} + 2 \in Q [x]$ .

[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 $f$ is reducible.

givenL2algebra

2.1

For every rational $a$ , [L3] gives $a^{2} + 1 > 0$ and $a^{2} + 2 > 0$ , so $f (a) > 0$ and $f$ 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

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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

Neil Donaldson, Math 120B Notes, example after Theorem 23.8 (standard reference, not scraped)

A degree-four polynomial can be reducible over QQ without having a rational root