ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06

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.

Index 2 obstructs reading factorisation from Z[sqrt 5] modulo 2

Example

For $A = Z [\sqrt{5}] ⊊ O_{K}$ with $K = Q (\sqrt{5})$ , reduction of the power polynomial $X^{2} - 5$ modulo $2$ gives $(X + 1)^{2}$ . But $O_{K} / 2 O_{K} ≅ F_{2} [X] / (X^{2} + X + 1)$ , which is a field. Thus the repeated factor in the nonmaximal power order is not a factorisation assertion about $O_{K}$ .

Facts & Assumptions

Given: $K = Q (\sqrt{5})$ , $A = Z [\sqrt{5}]$ , and $ω = (1 + \sqrt{5}) / 2$ .

[F1]

The order $A$ has index $2$ in $O_{K}$ (The nonmaximal quadratic order Z[sqrt 5] inside O_Q(sqrt 5)).

[F2]

The index-discriminant formula detects this nonmaximality (Order-index discriminant formula).

Verification

technique · direct

1.1

In $A / 2 A$ , the class of $\sqrt{5}$ satisfies $X^{2} - 5 \equiv X^{2} + 1 = (X + 1)^{2}$ , so $A / 2 A ≅ F_{2} [X] / ((X + 1)^{2})$ has a nonzero nilpotent.

F1givenalgebra

2.1

The element $ω$ satisfies $ω^{2} - ω - 1 = 0$ . Hence $O_{K} / 2 O_{K} ≅ F_{2} [X] / (X^{2} + X + 1)$ ; the polynomial has no root in $F_{2}$ , so this quotient is a field.

F1step 1.1algebra

3.1

The two quotient rings cannot agree, and [F2] identifies the reason as the index divisible by $2$ . Therefore reduction of the power polynomial in $A$ cannot by itself describe factorisation in the maximal order.

F2step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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

William Stein, Algebraic Number Theory, Section 6.1 (standard reference, not scraped)