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 be a commutative ring. If is Zariski-open and for an ideal , then In particular, every Zariski-open subset is a union of distinguished opens.
Facts & Assumptions
Given: A commutative ring , an ideal , and .
is the set of prime ideals that do not contain (Principal distinguished subsets of the prime spectrum).
Proof
Let . Since , the ideal is not contained in . Choose . Then by [L1], so
Conversely, if for some , then , so certainly . Hence and therefore . Thus
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
- J. S. Milne, A Primer of Commutative Algebra, v4.03, §14 (standard reference, not scraped)
- The Stacks Project, Section 10.21: Open and closed subsets of spectra (standard reference, not scraped)