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

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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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A finite intersection of primary submodules with one radical is primary

Statement

Let $R$ be a Noetherian commutative ring, let $M$ be a finitely generated left $R$ -module, and let $r \geq 1$ . Let $Q_{1}, \dots, Q_{r} ⊊ M$ be $p$ -primary submodules. Then $Q = Q_{1} \cap \dots \cap Q_{r}$ is also $p$ -primary.

Facts & Assumptions

Given: A Noetherian commutative ring $R$ , a finitely generated left $R$ -module $M$ , an integer $r \geq 1$ , a prime ideal $p$ , and $p$ -primary submodules $Q_{1}, \dots, Q_{r} ⊊ M$ .

[L1]

A proper submodule is primary exactly when $a m \in Q^{'}$ and $m \notin Q^{'}$ imply $a^{n} M^{'} \subseteq Q^{'}$ for some $n \geq 1$ (Primary submodules are exactly quotients with nilpotent zero divisors).

[L2]

A primary submodule $Q^{'}$ is $p$ -primary when $\sqrt{{Ann}_{R} (M^{'} / Q^{'})} = p$ (Primary submodules and primary ideals).

Proof

technique · direct

1.1L1L2givenchoosealgebra

Because $r \geq 1$ , one has $Q \subseteq Q_{1} ⊊ M$ , so $Q$ is proper. Suppose $a m \in Q$ and $m \notin Q$ . Choose $i$ with $m \notin Q_{i}$ . Since $Q_{i}$ is primary, [L1] gives $e \geq 1$ with $a^{e} M \subseteq Q_{i}$ , so $a \in \sqrt{{Ann}_{R} (M / Q_{i})} = p$ by [L2].

2.1L1L2step 1.1choosealgebra

For each $j$ , step 1.1 gives $a \in p = \sqrt{{Ann}_{R} (M / Q_{j})}$ , so choose $e_{j} \geq 1$ with $a^{e_{j}} M \subseteq Q_{j}$ . The nonempty finite list has a maximum $E$ , and then $a^{E} M \subseteq Q_{j}$ for every $j$ , hence $a^{E} M \subseteq Q$ . By [L1], the proper submodule $Q$ is primary.

3.1L2step 2.1algebra∎

If $a \in p$ , the same finite-maximum argument as in step 2.1 gives a power of $a$ in ${Ann}_{R} (M / Q)$ , so $p \subseteq \sqrt{{Ann}_{R} (M / Q)}$ . Conversely, if a power of $a$ annihilates $M / Q$ , it also annihilates every $M / Q_{j}$ because $Q \subseteq Q_{j}$ , so $a \in \sqrt{{Ann}_{R} (M / Q_{j})} = p$ . Thus $\sqrt{{Ann}_{R} (M / Q)} = p$ , and [L2] makes $Q$ $p$ -primary.

A finite intersection of primary submodules with one radical is primary

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources