LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck 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.

A separating prime for an element outside a radical

Statement

Assume the Axiom of Choice.

Let $R$ be a commutative ring, let $I ⊴ R$ be an ideal, and let $f \in R$ . If $f \notin \sqrt{I}$ , then there exists a prime ideal $p$ of $R$ such that $I \subseteq p$ and $f \notin p$ .

Facts & Assumptions

Given: A commutative ring $R$ , an ideal $I ⊴ R$ , an element $f \notin \sqrt{I}$ , and the Axiom of Choice.

[L1]

An element belongs to $\sqrt{I}$ exactly when one of its positive powers lies in $I$ (The radical of an ideal).

[L2]

If every positive power of $f$ avoids $I$ , then some prime ideal contains $I$ but avoids $f$ (Separating an element from an ideal by a prime).

Proof

technique · direct

1.1L1given

Since $f \notin \sqrt{I}$ , [L1] says that $f^{n} \notin I$ for every integer $n \geq 1$ .

2.1L2step 1.1

Applying [L2] to step 1.1 yields a prime ideal $p$ with $I \subseteq p$ and $f \notin p$ .

3.1step 2.1∎

This prime separates $f$ from the radical of $I$ .

A separating prime for an element outside a radical

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources