$s\in {\frak{p}}\cap S$ is nilpotent for $A_{\mathfrak{p}}/\operatorname{Ann}_A(\frac{m}{s})_{\mathfrak{p}}$ This question is motivated by About weakly associated primes
.
I have some detail in the solution that can not work out, I can rephrase the problem as :
Let $M$ be a $A$-module, we can taking localization at multiplicative set $S$, gets $S^{-1}M$ which can be treated as $A$-module.
Consider some $m/s\in S^{-1}M$, we have $\text{Ann}_A (m/s)$ which is an ideal of $A$, let $\text{Ann}_A(m/s)\subset{\frak{p}}$ be the minimal prime ideal containing annihilator.
I want to show that if $t\in S\cap \frak{p}$, then it's nilpotent element in:
$$A_{\mathfrak{p}}/(\operatorname{Ann}_A(\frac{m}{s}))_{\mathfrak{p}}$$
both solutions use this result, I have no idea how to prove it. Any help will be appreciated, thanks!
 A: The prime ideals of $A_\mathfrak{p}/(A_\mathfrak{p}\cdot\operatorname{Ann}_A(\frac{m}{s}))\cong\left(A/\operatorname{Ann}_A(\frac{m}{s})\right)_{\overline{\mathfrak{p}}}$ (with $\overline{\mathfrak{p}}=\mathfrak{p}/\operatorname{Ann}_A(\frac{m}{s})$) are in one-to-one correspondence with the prime ideals of $A_\mathfrak{p}$ containing $A_\mathfrak{p}\cdot\operatorname{Ann}_A(\frac{m}{s})$, which are in one-to-one correspondence with the prime ideals contained in $\mathfrak{p}$ containing $\operatorname{Ann}_A(\frac{m}{s})$. By construction, there is exactly one such ideal, namely $\mathfrak{p}$. Thus the only prime ideal of $A_{\mathfrak{p}}$ containing $A_\mathfrak{p}\cdot\operatorname{Ann}_A(\frac{m}{s})$ is $\mathfrak{p}A_\mathfrak{p}$, from which we conclude
$$
\sqrt{A_\mathfrak{p}\cdot\operatorname{Ann}_A(\frac{m}{s})}=\bigcap_{\substack{\mathfrak{q}\in\operatorname{Spec}A_{\mathfrak{p}}}\\A_\mathfrak{p}\cdot\operatorname{Ann}_A(\frac{m}{s})\subseteq\mathfrak{q}}\mathfrak{q}=\mathfrak{p}A_\mathfrak{p}.
$$
Therefore, if $t\in \mathfrak{p}$, then $t^n/1\in A_\mathfrak{p}\cdot\operatorname{Ann}_A(\frac{m}{s})$ for some $n$, and thus $t$ is nilpotent in $A_\mathfrak{p}/(A_\mathfrak{p}\cdot\operatorname{Ann}_A(\frac{m}{s}))\cong\left(A/\operatorname{Ann}_A(\frac{m}{s})\right)_{\overline{\mathfrak{p}}}$.
