CorollaryStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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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

1.1

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

givenalgebra

2.1

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 $f \notin p$ .

L1step 1.1

3.1

This is the required separating prime.

step 2.1∎

Depends on

A prime containing an ideal and avoiding a multiplicative set

Used by

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, A Primer of Commutative Algebra, v4.03, §2 Ideals (standard reference, not scraped)
A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., §2 Ideals (standard reference, not scraped)