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

Statement

Let $R$ be a commutative ring. If $U \subseteq Spec (R)$ is Zariski-open and $U = Spec (R) ∖ V (I)$ for an ideal $I ⊴ R$ , then $U = ⋃_{f \in I} D (f) .$ 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 $p \in U$ . Since $p \notin V (I)$ , the ideal $I$ is not contained in $p$ . Choose $f \in I ∖ p$ . Then $p \in D (f)$ by [L1], so $U \subseteq ⋃_{f \in I} D (f) .$

1.2L1given

Conversely, if $p \in D (f)$ for some $f \in I$ , then $f \notin p$ , so certainly $I ⊈ p$ . Hence $p \notin V (I)$ and therefore $p \in U$ . Thus $⋃_{f \in I} D (f) \subseteq U .$

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.

Every Zariski-open subset is a union of distinguished opens

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources