Lemma 10.62.2. Let $R$, $M$, $M_ i$, $\mathfrak p_ i$ as in Lemma 10.62.1. Then $\text{Supp}(M) = \bigcup V(\mathfrak p_ i)$ and in particular $\mathfrak p_ i \in \text{Supp}(M)$.

Comment #9033 by on

Just to save thinking time, here are the details: by induction in $n$. For $n=1$ there's only prime $\mathfrak{p}=\mathfrak{p}_1$, and we have $M\cong R/\mathfrak{p}$, whence $\operatorname{Supp}M=V(\mathfrak{p})$, by Lemma 10.40.5. Supposing the result true for $n-1$, we apply Lemma 10.40.9 and obtain $\operatorname{Supp}M=\operatorname{Supp}(M/M_{n-1})\cup\operatorname{Supp}M_{n-1}$. By the induction hypothesis, $\operatorname{Supp}M_{n-1}=\bigcup_{i=1}^{n-1}V(\mathfrak{p}_i)$; on the other hand, $\operatorname{Supp}(M/M_{n-1})=\operatorname{Supp}(R/\mathfrak{p}_n)=V(\mathfrak{p}_n)$.

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