Lemma 10.52.4 (00IW)—The Stacks project

Lemma 10.52.4. Let $R$ be a local ring with maximal ideal $\mathfrak m$ . If $M$ is an $R$ -module and $\mathfrak m^ n M \not= 0$ for all $n \geq 0$ , then $\text{length}_ R(M) = \infty$ . In other words, if $M$ has finite length then $\mathfrak m^ nM = 0$ for some $n$ .

Proof. Assume $\mathfrak m^ n M \not= 0$ for all $n\geq 0$ . Choose $x \in M$ and $f_1, \ldots , f_ n \in \mathfrak m$ such that $f_1f_2 \ldots f_ n x \not= 0$ . The first $n$ steps in the filtration

$0 \subset R f_1 \ldots f_ n x \subset R f_1 \ldots f_{n - 1} x \subset \ldots \subset R x \subset M$

are distinct. For example, if $R f_1 x = R f_1 f_2 x$ , then $f_1 x = g f_1 f_2 x$ for some $g$ , hence $(1 - gf_2) f_1 x = 0$ hence $f_1 x = 0$ as $1 - gf_2$ is a unit which is a contradiction with the choice of $x$ and $f_1, \ldots , f_ n$ . Hence the length is infinite. $\square$

Comments (3)

Comment #7519 by Hao Peng on July 26, 2022 at 02:52

in fact the condition that M is a finite module is not needed here.

Comment #7651 by Stacks Project on August 14, 2022 at 13:09

Very good! Fixed here.

Comment #9926 by Alex on January 09, 2025 at 09:51

I think the indices on the products of the $f_i$ need to be reversed.

That is, we want to have $f_n \dots f_1 x \ne 0$ and the filtration should be $0 \subset R f_n \dots f_1 x \subset \cdots \subset R f_1 x \subset R x \subset M$ . Then if $R f_1 x = R f_2 f_1 x$ , we get $(1-gf_2)f_1 x$ (note no change here), which gives $f_1 x = 0$ , contradicting $f_n \dots f_1 x \ne 0$ .

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2 comment(s) on Section 10.52: Length

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