Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-11
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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.
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These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Translation turns x4+1 into an Eisenstein polynomial

Example

The polynomial x4+1 is irreducible over Q, although Eisenstein's criterion does not apply to it before a translation.

Facts & Assumptions

Given: The polynomial f=x4+1∈Q[x] and the substitution τ(f)(x)=f(x+1).

[L1]

The universal property makes substitution x↦x+1 a ring homomorphism; substitution x↦x−1 is its inverse (Universal property of R[x]: a coefficient homomorphism and the image of x determine a unique ring homomorphism).

[L2]

Eisenstein's criterion proves a primitive integer polynomial irreducible over Q from the stated prime-divisibility conditions (Eisenstein criterion over the integers).

[L3]

An integer is prime when it exceeds 1 and has no positive divisors other than 1 and itself (Prime and composite integers: p is prime when p>1 and its only positive divisors are 1 and p).

[L4]

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

Verification

technique · direct
1.1

No prime can make f=x4+1 Eisenstein before translation, because the criterion would require that prime to divide the constant coefficient 1. A direct divisor check using [L3] shows that 2 is prime. Binomial expansion gives τ(f)=x4+4x3+6x2+4x+2. It is monic and hence primitive, and it is Eisenstein at 2: every nonleading coefficient is even, the leading coefficient is odd, and 4 does not divide the constant coefficient 2. Thus [L2] makes τ(f) irreducible.

givenL2L3algebra
2.1

Over the field [L4], [L1] makes τ an automorphism; a nontrivial factorization of f would map to one of τ(f), so irreducibility of τ(f) implies irreducibility of f.

step 1.1L1L4∎

Depends on

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Sources