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.

A finite unit-ideal expression yields a finite distinguished-open subcover

Statement

Let $R$ be a commutative ring. If $1 = a_{1} f_{1} + \dots + a_{n} f_{n}$ for elements $a_{1}, \dots, a_{n}, f_{1}, \dots, f_{n} \in R$ , then $Spec (R) = D (f_{1}) \cup \dots \cup D (f_{n}) .$

Facts & Assumptions

Given: A commutative ring $R$ and an identity $1 = a_{1} f_{1} + \dots + a_{n} f_{n}$ in $R$ .

[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 $p \in Spec (R)$ . If $p$ were outside $D (f_{1}) \cup \dots \cup D (f_{n})$ , then [L1] would give $f_{i} \in p$ for every $i$ . Because $p$ is an ideal, it would then contain the sum $a_{1} f_{1} + \dots + a_{n} f_{n} = 1$ , impossible for a prime ideal.

L1givenalgebra

2.1

Therefore every prime ideal lies in at least one $D (f_{i})$ , so $Spec (R) = D (f_{1}) \cup \dots \cup D (f_{n})$ .

step 1.1

3.1

The displayed unit expression yields a finite distinguished-open subcover of the spectrum.

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

A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., Proposition (13.20) (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, v4.03, Proposition 14.4(c) (standard reference, not scraped)