CorollaryStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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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.

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Separating an element from an ideal by a prime

Statement

Assume the Axiom of Choice.

Let $R$ be a commutative ring, let $I ⊴ R$ be an ideal, and let $f \in R$ . If $f^{n} \notin I$ for every integer $n \geq 1$ , 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 $f \in R$ whose positive powers all avoid $I$ , and the Axiom of Choice.

[L1]

If an ideal is disjoint from a multiplicative set, then some prime ideal contains it and stays disjoint from that multiplicative set (A prime containing an ideal and avoiding a multiplicative set).

Proof

technique · direct

1.1givenalgebra

The set $S = {1, f, f^{2}, \dots}$ is multiplicative, and the hypothesis says exactly that $I \cap S = \emptyset$ .

2.1L1step 1.1

Applying [L1] to the ideal $I$ and the multiplicative set $S$ yields a prime ideal $p$ with $I \subseteq p$ and $p \cap S = \emptyset$ . In particular $f \notin p$ .

3.1step 2.1∎

This is the required separating prime.

Depends on

A prime containing an ideal and avoiding a multiplicative set

Separating an element from an ideal by a prime

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources