Lemma 10.66.9. Let $R$ be a ring. Let $M$ be an $R$-module. Let $\mathfrak p$ be a prime ideal of $R$ which is finitely generated. Then

$\mathfrak p \in \text{Ass}(M) \Leftrightarrow \mathfrak p \in \text{WeakAss}(M).$

In particular, if $R$ is Noetherian, then $\text{Ass}(M) = \text{WeakAss}(M)$.

Proof. Write $\mathfrak p = (g_1, \ldots , g_ n)$ for some $g_ i \in R$. It is enough the prove the implication “$\Leftarrow$” as the other implication holds in general, see Lemma 10.66.6. Assume $\mathfrak p \in \text{WeakAss}(M)$. By Lemma 10.66.2 there exists an element $m \in M_{\mathfrak p}$ such that $I = \{ x \in R_{\mathfrak p} \mid xm = 0\}$ has radical $\mathfrak pR_{\mathfrak p}$. Hence for each $i$ there exists a smallest $e_ i > 0$ such that $g_ i^{e_ i}m = 0$ in $M_{\mathfrak p}$. If $e_ i > 1$ for some $i$, then we can replace $m$ by $g_ i^{e_ i - 1} m \not= 0$ and decrease $\sum e_ i$. Hence we may assume that the annihilator of $m \in M_{\mathfrak p}$ is $(g_1, \ldots , g_ n)R_{\mathfrak p} = \mathfrak p R_{\mathfrak p}$. By Lemma 10.63.15 we see that $\mathfrak p \in \text{Ass}(M)$. $\square$

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