LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Primary submodules are exactly quotients with nilpotent zero divisors

Statement

Let $R$ be a commutative ring, let $M$ be a left $R$ -module, and let $Q ⊊ M$ be a proper submodule. Then $Q$ is primary if and only if the following classical condition holds:

for every $a \in R$ and every $m \in M$ , if $a m \in Q$ and $m \notin Q$ , then there exists $n \geq 1$ such that $a^{n} M \subseteq Q .$

Facts & Assumptions

Given: A commutative ring $R$ , a left $R$ -module $M$ , and a proper submodule $Q ⊊ M$ .

[L1]

A proper submodule $Q$ is primary exactly when every zero divisor on $M / Q$ acts nilpotently on $M / Q$ (Primary submodules and primary ideals).

[L2]

The quotient module $M / Q$ consists of cosets $m + Q$ (Quotient module $M / N$ with scalar multiplication on additive cosets).

Proof

technique · direct

1.1L1L2givenalgebra

Assume $Q$ is primary, and let $a m \in Q$ with $m \notin Q$ . Then $m + Q \neq 0$ in $M / Q$ by [L2], while $a (m + Q) = a m + Q = 0 + Q$ . Thus $a$ is a zero divisor on $M / Q$ . By [L1], some $n \geq 1$ satisfies $a^{n} (M / Q) = 0$ , which is equivalent to $a^{n} M \subseteq Q$ .

1.2L1L2given

Conversely, assume the displayed classical condition. Let $a$ be a zero divisor on $M / Q$ . Then there exists $m + Q \neq 0$ with $a (m + Q) = 0 + Q$ . By [L2], this means $m \notin Q$ and $a m \in Q$ . The hypothesis gives $n \geq 1$ with $a^{n} M \subseteq Q$ , equivalently $a^{n} (M / Q) = 0$ . Hence every zero divisor on $M / Q$ acts nilpotently, so $Q$ is primary by [L1].

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 prove the equivalence.

Depends on

Used by

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, Definition 19.1 and Lemma 19.5 (standard reference, not scraped)
A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., §18 (standard reference, not scraped)