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
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
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