TheoremStatement: 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.

Eisenstein criterion over the integers

Statement

Let $f = a_{n} x^{n} + \dots + a_{0} \in Z [x]$ be primitive with $n \geq 1$ . If there is a prime $p$ such that

$p ∤ a_{n}, p ∣ a_{i} for every i < n, p^{2} ∤ a_{0},$

then $f$ is irreducible in $Q [x]$ .

Facts & Assumptions

Given: A primitive polynomial $f = a_{n} x^{n} + \dots + a_{0}$ and a prime $p$ satisfying the displayed divisibility conditions.

[L1]

A rational factorization of a primitive integer polynomial clears to a primitive integer factorization (Gauss lemma: primitive factorisations over $Q$ can be cleared to primitive factorisations over $Z$ ).

[L2]

Polynomial rings over fields are unique factorisation domains, hence domains (For every field $F$ , $F [x]$ is a unique factorisation domain).

[L3]

Reduction of coefficients modulo $p$ is a polynomial-ring homomorphism (Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism).

[L4]

The ring $Z / p$ is the quotient $Z / p Z$ (For every $n \in N$ , the congruence-class ring $Z / n$ is the quotient ring $Z / n Z$ ).

[L5]

A prime $p$ is greater than $1$ and its only positive divisors are $1, p$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L6]

The ring $Z / p$ is a field (For every prime $p$ , the two operations on $Z / p$ make it a field).

Proof

technique · contradiction

1.1

Suppose for contradiction that $f$ is reducible over $Q$ ; [L1] gives $f = G H$ with primitive integer polynomials $G, H$ of positive degree.

assume-contragivenL1

2.1

By [L3] and [L4], reduction modulo $p$ gives $\overset{ˉ}{G} \overset{ˉ}{H} = {\overset{ˉ}{a}}_{n} x^{n}$ in the domain $(Z / p) [x]$ of [L6] and [L2]. Because $p ∤ a_{n}$ , the leading coefficients of $G$ and $H$ both survive reduction: their product is $a_{n}$ , so neither is divisible by $p$ . Thus $\deg \overset{ˉ}{G} = \deg G > 0$ and $\deg \overset{ˉ}{H} = \deg H > 0$ . Comparing the least nonzero terms in the product ${\overset{ˉ}{a}}_{n} x^{n}$ now shows that both reductions are monomials of positive degree, so the constant coefficients of $G$ and $H$ are divisible by $p$ .

step 1.1L2L3L4L5L6algebra

3.1

The constant coefficient $a_{0}$ is the product of those two constant coefficients, so step 2.1 gives $p^{2} ∣ a_{0}$ , contradicting the hypothesis; hence $f$ is irreducible over $Q$ .

step 2.1L5discharge-contradiction∎

Depends on

Used by

For every prime p and positive n, xⁿ-p is irreducible over ℚ Corollary
An algebraic extension need not be finite Counterexample
An annihilating polynomial need not be minimal: √2 is a root of both x²-2 and x⁴-4 Counterexample
ℚ(³√2)/ℚ is separable and nonnormal with trivial automorphism group Counterexample
The real cube root of two is algebraic but not algebraically constructible Counterexample
Frobenius in a small cyclotomic field Example
Nonabelian frobenius conjugacy class Example
ℚ(√2,√3) has four embeddings into ℚ̄ Example
ℚ(√2)/ℚ has degree two Example
ℚ(√2)≅ℚ[x]/(x²-2) with basis 1,√2 Example
ℚ(³√2) has three embeddings into ℚ̄ but only one ℚ-automorphism Example
ℚ(³√2)/ℚ has degree three Example
The full S₃ correspondence for the splitting field of x³-2 Example
The minimal polynomial of √2+√3 over ℚ is x⁴-4x²+1 Example
The splitting field of x²-2 over ℚ is ℚ(√2), with roots ±√2 Example
The splitting field of x³-2 over ℚ is ℚ(³√2,ω) with ω=(-1+i√3)/2 Example
The ten-field D₄ correspondence for the splitting field of x⁴-2 Example
Translation turns x⁴+1 into an Eisenstein polynomial Example
x³-2 has discriminant -108 and Galois group S₃ over ℚ Example
x⁴+x³+x²+x+1 has Galois group C₄ over ℚ Example
x⁵-6x+3 over ℚ is not solvable by radicals Example
Φ₇(t+1) is Eisenstein at seven Example
FALSE: a polynomial solvable by radicals must have abelian Galois group False statement
FALSE: every irreducible quintic over ℚ is insoluble by radicals False statement
FALSE: the isomorphism between two splitting fields that fixes the base field is unique False statement
Φ_pʳ(t)=∑_k<pt^kpʳ⁻¹, and Φ_pʳ(t+1) is Eisenstein at p Proposition

Dependency tree · two levels

44 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

Thomas W. Judson, Abstract Algebra: Theory and Applications, Theorem 17.17 (standard reference, not scraped)
Keith Conrad, Irreducibility Tests in Q[T], Appendix A.4 (standard reference, not scraped)