ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

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.

The element 1/p is integral over Z[1/p] but not over Z

Example

Let $p$ be a prime number. Inside $Q$ , the element $1 / p$ is integral over $Z [1 / p]$ but not integral over $Z$ .

Facts & Assumptions

Given: A prime number $p$ , the inclusion $Z \subseteq Q$ , and the localisation $Z [1 / p]$ .

[L1]

Integrality localises, and conversely denominators can be cleared after localisation (Integrality and integral closure commute with localisation).

Verification

technique · direct

1.1

The element $1 / p$ already belongs to the base ring $Z [1 / p]$ , so it satisfies the monic equation $T - 1 / p = 0$ there. Hence it is integral over $Z [1 / p]$ .

L1given

1.2

Suppose $1 / p$ were integral over $Z$ . Then there would be a monic equation $(1 / p)^{n} + a_{n - 1} (1 / p)^{n - 1} + \dots + a_{0} = 0$ with $a_{i} \in Z$ . Multiplying by $p^{n}$ gives $1 + a_{n - 1} p + \dots + a_{0} p^{n} = 0$ , impossible because the left side is congruent to $1$ modulo $p$ .

givenalgebra

2.1

Thus localisation makes $1 / p$ integral only after $p$ has become invertible in the base ring.

L1step 1.1step 1.2∎

Depends on

Integrality and integral closure commute with localisation

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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

J. S. Milne, A Primer of Commutative Algebra, v4.03, Proposition 6.7 (standard reference, not scraped)