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 detect radicals

Statement

Assume the Axiom of Choice.

Let $R$ be a commutative ring and let $I, J ⊴ R$ be ideals. Then $V (I) = V (J) ⟺ \sqrt{I} = \sqrt{J} .$

Facts & Assumptions

Given: A commutative ring $R$ , ideals $I, J ⊴ R$ , and the Axiom of Choice.

[L1]

The radical of an ideal is the intersection of the prime ideals containing it (The radical of an ideal is the intersection of the prime ideals containing it).

[L2]

Radical is an idempotent ideal-valued operation (The radical of an ideal is an ideal).

Proof

technique · direct

1.1L1given

If $V (I) = V (J)$ , then the two ideals are contained in exactly the same prime ideals. Applying [L1] to both ideals shows that $\sqrt{I}$ and $\sqrt{J}$ are intersections over the same family of prime ideals, hence $\sqrt{I} = \sqrt{J}$ .

1.2L1L2given

Conversely, suppose $\sqrt{I} = \sqrt{J}$ . If $p \in V (I)$ , then $p$ contains $I$ , so [L1] gives $\sqrt{I} \subseteq p$ . Therefore $\sqrt{J} \subseteq p$ , hence $J \subseteq \sqrt{J} \subseteq p$ by [L2], and $p \in V (J)$ . The same argument with $I$ and $J$ reversed shows $V (J) \subseteq V (I)$ .

Vanishing sets detect radicals

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources