Alphabeta Math
LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27
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.

Vanishing sets of finite products

Statement

Let R be a commutative ring, and let I1,…,In⊴R with n≥1. Then V(I1⋯In)=V(I1)∪⋯∪V(In).

Facts & Assumptions

Given: A commutative ring R, ideals I1,…,In⊴R, and an integer n≥1.

[L1]

V(K) is the set of prime ideals containing K (The prime spectrum and vanishing sets).

[L2]

A prime ideal contains a factor whenever it contains a product (Prime ideals and maximal ideals in a commutative ring).

Proof

technique · direct
1.1L1givenalgebra

If p∈V(Ij) for some j, then Ij⊆p. Every element of I1⋯In lies in Ij, so I1⋯In⊆p, and therefore p∈V(I1⋯In). This proves V(I1)∪⋯∪V(In)⊆V(I1⋯In).

1.2L1L2choosealgebra

Conversely, let p∈V(I1⋯In) and suppose that no Ij is contained in p. For each j, choose aj∈Ij∖p. Then a1⋯an∈I1⋯In⊆p, so repeated use of [L2] forces some aj∈p, a contradiction. Hence Ij⊆p for some j, and p∈V(Ij).

2.1step 1.1step 1.2∎

The two inclusions prove V(I1⋯In)=V(I1)∪⋯∪V(In).

Depends on

Used by

Dependency tree · two levels

7 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