Lemma 10.66.5 (0588)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.66: Weakly associated primes
Lemma 10.66.5 (cite)

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Every nonzero module has a weakly associated prime.

Lemma 10.66.5. Let $R$ be a ring. Let $M$ be an $R$-module. Then

\[ M = (0) \Leftrightarrow \text{WeakAss}(M) = \emptyset \]

Proof. If $M = (0)$ then $\text{WeakAss}(M) = \emptyset $ by definition. Conversely, suppose that $M \not= 0$. Pick a nonzero element $m \in M$. Write $I = \{ x \in R \mid xm = 0\} $ the annihilator of $m$. Then $R/I \subset M$. Hence $\text{WeakAss}(R/I) \subset \text{WeakAss}(M)$ by Lemma 10.66.4. But as $I \not= R$ we have $V(I) = \mathop{\mathrm{Spec}}(R/I)$ contains a minimal prime, see Lemmas 10.17.2 and 10.17.7, and we win. $\square$

Comments (1)

Comment #843 by Johan Commelin on July 24, 2014 at 18:38

Suggested slogan: A module is the zero module if and only if it has no weakly associated primes.

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