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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 Zariski-open subset is a union of distinguished opens

Statement

Let R be a commutative ring. If U⊆Spec⁡(R) is Zariski-open and U=Spec⁡(R)∖V(I) for an ideal I⊴R, then U=⋃f∈ID(f). In particular, every Zariski-open subset is a union of distinguished opens.

Facts & Assumptions

Given: A commutative ring R, an ideal I⊴R, and U=Spec⁡(R)∖V(I).

[L1]

D(f) is the set of prime ideals that do not contain f (Principal distinguished subsets of the prime spectrum).

Proof

technique · direct
1.1L1givenchoose

Let p∈U. Since p∉V(I), the ideal I is not contained in p. Choose f∈I∖p. Then p∈D(f) by [L1], so U⊆⋃f∈ID(f).

1.2L1given

Conversely, if p∈D(f) for some f∈I, then f∉p, so certainly I⊈p. Hence p∉V(I) and therefore p∈U. Thus ⋃f∈ID(f)⊆U.

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 prove the displayed equality, so every Zariski-open subset is a union of distinguished opens.

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