LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 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.

The closure of a prime is its vanishing set

Statement

Assume the Axiom of Choice.

Let $R$ be a commutative ring and let $p \in Spec (R)$ . Then $\overline{{p}} = V (p) .$

Facts & Assumptions

Given: A commutative ring $R$ , a prime ideal $p \subseteq R$ , and the Axiom of Choice.

[L1]

Every Zariski-closed subset has a unique radical defining ideal (Every Zariski-closed subset has a unique radical defining ideal).

[L2]

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

Proof

technique · direct

1.1L2given

Since $p \supseteq p$ , the point $p$ lies in $V (p)$ by [L2]. Because $V (p)$ is closed, the closure $\overline{{p}}$ is contained in $V (p)$ .

1.2L1L2

Let $Z$ be a closed subset containing $p$ . By [L1], write $Z = V (I)$ for its radical defining ideal $I$ . Since $p \in Z$ , fact [L2] gives $I \subseteq p$ . Therefore every prime ideal containing $p$ also contains $I$ , so $V (p) \subseteq V (I) = Z$ .

2.1step 1.1step 1.2∎

Step 1.2 shows that every closed set containing $p$ also contains $V (p)$ . Hence $V (p)$ is the smallest closed set containing $p$ , that is, $\overline{{p}} = V (p)$ .

The closure of a prime is its vanishing set

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources