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LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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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.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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

Statement

Let R be a commutative ring, let U⊆Spec⁡(R) be Zariski-open, and let p∈U. Then there exists f∈R such that p∈D(f)⊆U.

Facts & Assumptions

Given: A commutative ring R, a Zariski-open set U⊆Spec⁡(R), and a point p∈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
1.1

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

L1givenchoose
2.1

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

step 1.1
3.1

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

step 2.1∎

Depends on

Used by

Dependency tree · two levels

4 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