LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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.

Every point of a Zariski-open set has a distinguished-open neighbourhood inside it

Statement

Let $R$ be a commutative ring, let $U \subseteq Spec (R)$ be Zariski-open, and let $p \in U$ . Then there exists $f \in R$ such that $p \in D (f) \subseteq U .$

Facts & Assumptions

Given: A commutative ring $R$ , a Zariski-open set $U \subseteq Spec (R)$ , and a point $p \in U$ .

[L1]

Every Zariski-open subset is a union of distinguished opens (Every Zariski-open subset is a union of distinguished opens).

Proof

technique · direct

1.1L1givenchoose

By [L1], the open set $U$ is a union of distinguished opens. Since $p \in U$ , there exists $f \in R$ with $p \in D (f)$ and $D (f) \subseteq U$ .

2.1step 1.1

The chosen $D (f)$ is therefore a distinguished-open neighbourhood of $p$ contained in $U$ .

3.1step 2.1∎

Hence every point of a Zariski-open set has a distinguished-open neighbourhood inside that open set.

Every point of a Zariski-open set has a distinguished-open neighbourhood inside it

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources