LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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.

r one s two integral element membership

Statement

A commutative Noetherian $(S_{2})$ domain whose height-one localizations are DVRs is integrally closed.

Facts & Assumptions

Given: The objects and hypotheses in the statement. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

[F1]

r one s two intersection of height one localisations: If $R$ is a commutative Noetherian domain satisfying $(S_{2})$ , then inside its fraction field $K$ one has $R = ⋂_{ht p = 1} R_{p}$ . For a field the empty intersection is interpreted as $K = R$ .

[F2]

Valuation rings are integrally closed: Every valuation ring is an integrally closed domain.

Proof

1.1

Let $u \in Frac R$ satisfy a monic equation over $R$ . At each height-one prime the same equation is monic over $R_{p}$ . A DVR is a valuation ring, hence integrally closed, so $u \in R_{p}$ .

F2given

2.1

The height-one intersection theorem now gives $u \in R$ . If there are no height-one primes, its empty-intersection convention says $R$ is already the fraction field; the conclusion remains valid. Since $u$ was arbitrary, $R$ is integrally closed.

F1step 1.1∎

Depends on

Used by

serre normality criterion two directions Corollary

Dependency tree · two levels

12 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

Proposition 8.41 final proof paragraph, p.58 (standard reference, not scraped)