Lemma 10.62.5. Let $R$, $M$, $M_ i$, $\mathfrak p_ i$ as in Lemma 10.62.1. The minimal elements of the set $\{ \mathfrak p_ i\}$ are the minimal elements of $\text{Supp}(M)$. The number of times a minimal prime $\mathfrak p$ occurs is

$\# \{ i \mid \mathfrak p_ i = \mathfrak p\} = \text{length}_{R_\mathfrak p} M_{\mathfrak p}.$

Proof. The first statement follows because $\text{Supp}(M) = \bigcup V(\mathfrak p_ i)$, see Lemma 10.62.2. Let $\mathfrak p \in \text{Supp}(M)$ be minimal. The support of $M_{\mathfrak p}$ is the set consisting of the maximal ideal $\mathfrak p R_{\mathfrak p}$. Hence by Lemma 10.62.3 the length of $M_{\mathfrak p}$ is finite and $> 0$. Next we note that $M_{\mathfrak p}$ has a filtration with subquotients $(R/\mathfrak p_ i)_{\mathfrak p} = R_{\mathfrak p}/{\mathfrak p_ i}R_{\mathfrak p}$. These are zero if $\mathfrak p_ i \not\subset \mathfrak p$ and equal to $\kappa (\mathfrak p)$ if $\mathfrak p_ i \subset \mathfrak p$ because by minimality of $\mathfrak p$ we have $\mathfrak p_ i = \mathfrak p$ in this case. The result follows since $\kappa (\mathfrak p)$ has length $1$. $\square$

Comment #2683 by Dario Weißmann on

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