CorollaryStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-11

How statement and proof provenance work

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.

For every prime $p$ and positive $n$ , $x^{n} - p$ is irreducible over $Q$

Statement

For every prime integer $p$ and every positive natural number $n$ , the polynomial $x^{n} - p$ is irreducible in $Q [x]$ .

Facts & Assumptions

Given: A prime integer $p$ and a natural number $n \geq 1$ .

[L1]

A primitive integer polynomial is irreducible over $Q$ when a prime divides every nonleading coefficient, does not divide the leading coefficient, and its square does not divide the constant coefficient (Eisenstein criterion over the integers).

[L2]

A prime integer satisfies $p > 1$ and has no positive divisor other than $1, p$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L3]

Nonzero integers cancel in products (The integers have no zero divisors; multiplicative cancellation).

[L4]

The only integer units are $1$ and $- 1$ ( $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ ).

Proof

technique · direct

1.1

The polynomial $x^{n} - p$ is primitive because its leading coefficient is the unit $1$ from [L4]; the prime $p$ divides every nonleading coefficient, including the zero intermediate coefficients, and does not divide $1$ .

givenL2L4

2.1

If $p^{2}$ divided $p$ , cancellation by the nonzero $p$ using [L3] would make $p$ a unit, contradicting [L2] and [L4]; thus Eisenstein's criterion [L1] applies and proves irreducibility.

step 1.1L1L2L3L4∎

Depends on

Used by

Eisenstein proves xⁿ-2 irreducible over ℚ for every positive n Example

Dependency tree · two levels

30 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, Irreducibility Tests in Q[T], Example 1.6 (standard reference, not scraped)

For every prime pp and positive nn, xn−pxn−p is irreducible over QQ