Lemma 10.66.2. Let $R$ be a ring. Let $M$ be an $R$-module. Let $\mathfrak p$ be a prime of $R$. The following are equivalent:

1. $\mathfrak p$ is weakly associated to $M$,

2. $\mathfrak pR_{\mathfrak p}$ is weakly associated to $M_{\mathfrak p}$, and

3. $M_{\mathfrak p}$ contains an element whose annihilator has radical equal to $\mathfrak pR_{\mathfrak p}$.

Proof. Assume (1). Then there exists an element $m \in M$ such that $\mathfrak p$ is minimal among the primes containing the annihilator $I = \{ x \in R \mid xm = 0\}$ of $m$. As localization is exact, the annihilator of $m$ in $M_{\mathfrak p}$ is $I_{\mathfrak p}$. Hence $\mathfrak pR_{\mathfrak p}$ is a minimal prime of $R_{\mathfrak p}$ containing the annihilator $I_{\mathfrak p}$ of $m$ in $M_{\mathfrak p}$. This implies (2) holds, and also (3) as it implies that $\sqrt{I_{\mathfrak p}} = \mathfrak pR_{\mathfrak p}$.

Applying the implication (1) $\Rightarrow$ (3) to $M_{\mathfrak p}$ over $R_{\mathfrak p}$ we see that (2) $\Rightarrow$ (3).

Finally, assume (3). This means there exists an element $m/f \in M_{\mathfrak p}$ whose annihilator has radical equal to $\mathfrak pR_{\mathfrak p}$. Then the annihilator $I = \{ x \in R \mid xm = 0\}$ of $m$ in $M$ is such that $\sqrt{I_{\mathfrak p}} = \mathfrak pR_{\mathfrak p}$. Clearly this means that $\mathfrak p$ contains $I$ and is minimal among the primes containing $I$, i.e., (1) holds. $\square$

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