CorollaryStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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A finite module that vanishes at a prime vanishes on some principal neighbourhood of that prime

Statement

Let $M$ be a finitely generated left $R$ -module and let $p$ be a prime ideal of $R$ . If $M_{p} = 0$ , then there exists $s \notin p$ such that the localisation of $M$ at the multiplicative set $S_{s} : = {1, s, s^{2}, \dots}$ is zero.

Facts & Assumptions

Given: A commutative ring $R$ , a finitely generated left $R$ -module $M$ , and a prime ideal $p$ with $M_{p} = 0$ .

[L1]

For a finite module, the support is the set of primes containing its annihilator (For a finite module, support is the set of primes containing the annihilator).

[L2]

The support condition $p \in {Supp}_{R} (M)$ means $M_{p} \neq 0$ (Support of a module).

[L3]

In a localisation, each denominator becomes a unit (The localisation relation is an equivalence relation and fraction arithmetic is well defined).

Proof

technique · direct

1.1L1L2choose

Since $M_{p} = 0$ , [L2] says $p \notin {Supp}_{R} (M)$ . By [L1], ${Ann}_{R} (M) ⊈ p$ , so choose $s \in {Ann}_{R} (M) ∖ p$ .

2.1step 1.1L3algebra∎

Let $S_{s} = {1, s, s^{2}, \dots}$ . In the localisation $S_{s}^{- 1} M$ , the element $s / 1 \in S_{s}^{- 1} R$ is a unit by [L3], and it annihilates every element because $s$ annihilates all of $M$ . Therefore $S_{s}^{- 1} M = 0$ .

A finite module that vanishes at a prime vanishes on some principal neighbourhood of that prime

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources