
Lemma 10.51.11. Let $R$ be a ring. Let $M$ be a finite length $R$-module. Choose any maximal chain of submodules

$0 = M_0 \subset M_1 \subset M_2 \subset \ldots \subset M_ n = M$

with $M_ i \not= M_{i-1}$, $i = 1, \ldots , n$. Then

1. $n = \text{length}_ R(M)$,

2. each $M_ i/M_{i-1}$ is simple,

3. each $M_ i/M_{i-1}$ is of the form $R/\mathfrak m_ i$ for some maximal ideal $\mathfrak m_ i$,

4. given a maximal ideal $\mathfrak m \subset R$ we have

$\# \{ i \mid \mathfrak m_ i = \mathfrak m\} = \text{length}_{R_{\mathfrak m}} (M_{\mathfrak m}).$

Proof. If $M_ i/M_{i-1}$ is not simple then we can refine the filtration and the filtration is not maximal. Thus we see that $M_ i/M_{i-1}$ is simple. By Lemma 10.51.10 the modules $M_ i/M_{i-1}$ have length $1$ and are of the form $R/\mathfrak m_ i$ for some maximal ideals $\mathfrak m_ i$. By additivity of length, Lemma 10.51.3, we see $n = \text{length}_ R(M)$. Since localization is exact, we see that

$0 = (M_0)_{\mathfrak m} \subset (M_1)_{\mathfrak m} \subset (M_2)_{\mathfrak m} \subset \ldots \subset (M_ n)_{\mathfrak m} = M_{\mathfrak m}$

is a filtration of $M_{\mathfrak m}$ with successive quotients $(M_ i/M_{i-1})_{\mathfrak m}$. Thus the last statement follows directly from the fact that given maximal ideals $\mathfrak m$, $\mathfrak m'$ of $R$ we have

$(R/\mathfrak m')_{\mathfrak m} \cong \left\{ \begin{matrix} 0 & \text{if } \mathfrak m \not= \mathfrak m', \\ R_{\mathfrak m}/\mathfrak m R_{\mathfrak m} & \text{if } \mathfrak m = \mathfrak m' \end{matrix} \right.$

This we leave to the reader. $\square$

Comment #3416 by Jonas Ehrhard on

I think it would improve readability, if we omit the variable $\mathcal{l}$ and just use $\text{lenght}_R(M)$, because $\mathcal{l}$ only appears twice.

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