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

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 radical of a primary ideal is prime

Statement

Let $R$ be a commutative ring and let $Q ⊊ R$ be a primary ideal. Then $\sqrt{Q}$ is a prime ideal. In particular, $Q$ is $\sqrt{Q}$ -primary.

Facts & Assumptions

Given: A commutative ring $R$ and a primary ideal $Q ⊊ R$ .

[L1]

A primary ideal is a proper submodule whose quotient $R / Q$ has the property that every zero divisor acts nilpotently (Primary submodules and primary ideals).

[L2]

The radical $\sqrt{Q}$ consists of those $a \in R$ for which $a^{n} \in Q$ for some $n \geq 1$ (The radical of an ideal).

[L3]

A prime ideal is a proper ideal $P$ such that $a b \in P$ implies $a \in P$ or $b \in P$ (Prime ideals and maximal ideals in a commutative ring).

Proof

technique · direct

1.1

Let $a b \in \sqrt{Q}$ and assume $b \notin \sqrt{Q}$ . By [L2], some $n \geq 1$ has $(a b)^{n} \in Q$ , so in the quotient ring $A = R / Q$ one has $(\overset{ˉ}{a} \overset{ˉ}{b})^{n} = {\overset{ˉ}{a}}^{n} {\overset{ˉ}{b}}^{n} = 0.$ Because $b \notin \sqrt{Q}$ , the class $\overset{ˉ}{b}$ is not nilpotent in $A$ , so ${\overset{ˉ}{b}}^{n} \neq 0$ . Thus ${\overset{ˉ}{a}}^{n}$ kills the nonzero element ${\overset{ˉ}{b}}^{n}$ , which means that ${\overset{ˉ}{a}}^{n}$ is a zero divisor on $A$ .

L1L2givenalgebra

2.1

Since $Q$ is primary, [L1] makes every zero divisor on $A$ nilpotent. Therefore ${\overset{ˉ}{a}}^{n}$ is nilpotent, so $\overset{ˉ}{a}$ is nilpotent and hence $a \in \sqrt{Q}$ by [L2]. Thus $a b \in \sqrt{Q}$ and $b \notin \sqrt{Q}$ imply $a \in \sqrt{Q}$ , which is the primality condition from [L3].

L1L2L3step 1.1

3.1

Therefore $\sqrt{Q}$ is prime, and $Q$ is $\sqrt{Q}$ -primary by definition.

L1L2step 2.1∎

Depends on

Used by

A radical ideal in a Noetherian ring is the intersection of its minimal primes Corollary

Dependency tree · two levels

8 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 19.2 (standard reference, not scraped)
A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., Theorem (18.4) (standard reference, not scraped)