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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 USpec(R) is Zariski-open and U=Spec(R)V(I) for an ideal IR, then U=fID(f). In particular, every Zariski-open subset is a union of distinguished opens.

Facts & Assumptions

Given: A commutative ring R, an ideal IR, 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
1.1

Let pU. Since pV(I), the ideal I is not contained in p. Choose fIp. Then pD(f) by [L1], so UfID(f).

L1givenchoose
1.2

Conversely, if pD(f) for some fI, then fp, so certainly Ip. Hence pV(I) and therefore pU. Thus fID(f)U.

L1given
2.1

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

step 1.1step 1.2

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