Lemma 10.52.4. Let $R$ be a local ring with maximal ideal $\mathfrak m$. Let $M$ be an $R$-module.

1. If $M$ is a finite module and $\mathfrak m^ n M \not= 0$ for all $n\geq 0$, then $\text{length}_ R(M) = \infty$.

2. 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 $f_1f_2 \ldots f_ n x \not= 0$. By Nakayama's Lemma 10.20.1 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. This can also be seen directly. 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, i.e., (1) holds. Combine (1) and Lemma 10.52.2 to see (2). $\square$

There are also:

• 2 comment(s) on Section 10.52: Length

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).