Lemma 10.157.2. Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module. The following are equivalent:

1. $M$ has no embedded associated prime, and

2. $M$ has property $(S_1)$.

Proof. Let $\mathfrak p$ be an embedded associated prime of $M$. Then there exists another associated prime $\mathfrak q$ of $M$ such that $\mathfrak p \supset \mathfrak q$. In particular this implies that $\dim (\text{Supp}(M_{\mathfrak p})) \geq 1$ (since $\mathfrak q$ is in the support as well). On the other hand $\mathfrak pR_{\mathfrak p}$ is associated to $M_{\mathfrak p}$ (Lemma 10.63.15) and hence $\text{depth}(M_{\mathfrak p}) = 0$ (see Lemma 10.63.18). In other words $(S_1)$ does not hold. Conversely, if $(S_1)$ does not hold then there exists a prime $\mathfrak p$ such that $\dim (\text{Supp}(M_{\mathfrak p})) \geq 1$ and $\text{depth}(M_{\mathfrak p}) = 0$. Since $\text{depth}(M_{\mathfrak p}) = 0$, we see that $\mathfrak p \in \text{Ass}(M)$ by the two Lemmas 10.63.15 and 10.63.18. Since $\dim (\text{Supp}(M_{\mathfrak p})) \geq 1$, there is a prime $\mathfrak q \in \text{Supp}(M)$ with $\mathfrak q \subset \mathfrak p$, $\mathfrak q \not= \mathfrak p$. We can take such a $\mathfrak q$ that is minimal in $\text{Supp}(M)$. Then by Proposition 10.63.6 we have $\mathfrak q \in \text{Ass}(M)$ and hence $\mathfrak p$ is an embedded associated prime. $\square$

Comment #1164 by Hu Fei on

''Conversely, if $(S_1)$ does not'' should add ''hold''

Comment #7078 by Yuto Masamura on

I suggest a proof of the omitted part in the last statement. I think we also need Proposition 10.63.6 in addition to the lemmas cited above.

Since $\operatorname{depth}(M_{\mathfrak p})=0$, we see that $\mathfrak p\in\operatorname{Ass}(M)$ by the two lemmas (Lemmas 10.63.15, 10.63.18). Since $\dim(\operatorname{Supp}(M_{\mathfrak p}))\ge1$, there is a prime $\mathfrak q\in\operatorname{Supp}(M)$ with $\mathfrak q\subsetneq\mathfrak p$. We can take such a $\mathfrak q$ that is minimal in $\operatorname{Supp}(M)$. Then by Proposition 10.63.6 we have $\mathfrak q\in\operatorname{Ass}(M)$ and hence $\mathfrak p$ is an embedded associated prime.

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