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-set identities

Statement

Let $R$ be a commutative ring.

$V ((0)) = Spec (R)$ .
$V (R) = \emptyset$ .
For every family $(I_{λ})_{λ \in Λ}$ of ideals, $V ⁣ (\sum_{λ \in Λ} I_{λ}) = ⋂_{λ \in Λ} V (I_{λ}) .$
For every finite family $I_{1}, \dots, I_{n}$ of ideals with $n \geq 1$ , $V (I_{1} \dots I_{n}) = V (I_{1}) \cup \dots \cup V (I_{n}) .$

Facts & Assumptions

Given: A commutative ring $R$ .

[L1]

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

[L2]

Vanishing sets turn arbitrary sums into intersections (Vanishing sets of arbitrary sums).

[L3]

Vanishing sets turn finite products into unions (Vanishing sets of finite products).

Proof

technique · direct

1.1L1given

Every prime ideal contains $0$ , so $V ((0)) = Spec (R)$ . No prime ideal equals the whole ring, so no prime ideal contains $R$ ; hence $V (R) = \emptyset$ .

1.2L2L3

The arbitrary-sum identity is exactly [L2], and the finite-product identity is exactly [L3].

Vanishing-set identities

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources