Lemma 10.62.4 (00L6)—The Stacks project

Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.62: Support and dimension of modules
Lemma 10.62.4 (cite)

Lemma 10.62.4. Let $R$ be a Noetherian ring. Let $I \subset R$ be an ideal. Let $M$ be a finite $R$-module. Then $I^ nM = 0$ for some $n \geq 0$ if and only if $\text{Supp}(M) \subset V(I)$.

Proof. Indeed, $I^ nM = 0$ is equivalent to $I^ n \subset \text{Ann}(M)$. Since $R$ is Noetherian, this is equivalent to $I \subset \sqrt{\text{Ann}(M)}$, see Lemma 10.32.5. This in turn is equivalent to $V(I) \supset V(\text{Ann}(M))$, see Lemma 10.17.2. By Lemma 10.40.5 this is equivalent to $V(I) \supset \text{Supp}(M)$. $\square$

Comments (2)

Comment #6774 by Yuto Masamura on November 29, 2021 at 03:03

This also follows from Lemma 10.40.5.

Indeed, $I^nM=0$ is equivalent to $I^n\subset\operatorname{Ann}(M)$ , hence to $I\subset\sqrt{\operatorname{Ann}(M)}$ since $R$ is Noetherian. This is equivalent to $V(I)\supset V(\operatorname{Ann}(M))$ , thus by Lemma 10.40.5 $V(I)\supset\operatorname{Supp}(M)$ .

Comment #6932 by Johan on January 20, 2022 at 16:16

OK, thanks and fixed here.

There are also:

3 comment(s) on Section 10.62: Support and dimension of modules

Comments (2)

Post a comment