TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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Every prime ideal of an Artinian ring is maximal

Statement

Let $R$ be a commutative Artinian ring and let $p$ be a prime ideal of $R$ . Then $p$ is maximal.

Facts & Assumptions

Given: A commutative Artinian ring $R$ and a prime ideal $p \subseteq R$ .

Proof

technique · direct

1.1

By Correspondence theorem: ideals of $R / I$ correspond to ideals of $R$ containing $I$ , ideals of $R / p$ correspond to ideals of $R$ containing $p$ . Hence every descending chain of ideals in $R / p$ lifts to a descending chain of ideals in $R$ , so $R / p$ is Artinian. Because $p$ is prime, $R / P$ is an integral domain if and only if $P$ is a prime ideal says that $R / p$ is an integral domain.

givenalgebra

2.1

The quotient $R / p$ is therefore an Artinian integral domain, so An Artinian integral domain is a field makes it a field. Then $R / M$ is a field if and only if $M$ is a maximal ideal forces $p$ to be maximal.

step 1.1givenalgebra

3.1

Hence every prime ideal of an Artinian ring is maximal.

step 2.1∎

Depends on

Used by

Dependency tree · two levels

17 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, Proposition 16.1 (standard reference, not scraped)
The Stacks Project, Section 10.53: Artinian rings (standard reference, not scraped)