Lemma 10.52.11. 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
$n = \text{length}_ R(M)$,
each $M_ i/M_{i-1}$ is simple,
each $M_ i/M_{i-1}$ is of the form $R/\mathfrak m_ i$ for some maximal ideal $\mathfrak m_ i$,
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 10.52.10 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 10.52.3, 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.
$\square$
Comments (2)
Comment #3416 by Jonas Ehrhard on
Comment #3478 by Johan on
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