LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck 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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A distinguished-open cover of the spectrum forces the covering ideal to be the unit ideal

Statement

Assume the Axiom of Choice.

Let $R$ be a commutative ring and let $(f_{λ})_{λ \in Λ}$ be a family of elements of $R$ such that $Spec (R) = ⋃_{λ \in Λ} D (f_{λ}) .$ Then the ideal generated by the family ${f_{λ}}_{λ \in Λ}$ is the unit ideal $R$ .

Facts & Assumptions

Given: A commutative ring $R$ , a family $(f_{λ})_{λ \in Λ}$ in $R$ , and the Axiom of Choice.

[L1]

In a nonzero commutative ring, every proper ideal is contained in a maximal ideal (In a nonzero commutative ring, every proper ideal is contained in a maximal ideal).

Proof

technique · direct

1.1L1givenchoose

Let $J$ be the ideal generated by ${f_{λ}}_{λ \in Λ}$ . If $J \neq R$ , then $J$ is a proper ideal, so $R$ is nonzero and [L1] gives a maximal ideal $m$ containing $J$ .

2.1step 1.1given

For every $λ$ , one has $f_{λ} \in J \subseteq m$ , so $m \notin D (f_{λ})$ . Hence $m \notin ⋃_{λ \in Λ} D (f_{λ})$ , contradicting the assumed cover of $Spec (R)$ .

3.1step 2.1∎

The contradiction shows that $J$ cannot be proper. Therefore $J = R$ .

A distinguished-open cover of the spectrum forces the covering ideal to be the unit ideal

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources