$\operatorname{Ann}_{R_\mathfrak{p}}(M_\mathfrak{p})=(\operatorname{Ann}_R(M))_\mathfrak{p}$

Question

Let $M$ be a $R$-module and $\mathfrak{p}$ be a prime ideal of $R$, where $R$ is a commutative ring with 1. Let $R_\mathfrak{p}$ denote $S^{-1}R$ and $M_\mathfrak{p}$ denote $S^{-1}M$, where $S=R\setminus \mathfrak{p}$. Then do we have the result $$\operatorname{Ann}_{R_\mathfrak{p}}(M_\mathfrak{p})=(\operatorname{Ann}_R(M))_\mathfrak{p}$$ According to what my professor said in class it seems that we need to require $M$ to be a finitely generated module. Otherwise the statement might not be true.
I actually showed this result without assuming $M$ to be finitely generated and thus it is very likely incorrect. Can someone show me where my mistakes are and the correct way to prove this result?
'$\subseteq$': Let $\frac{r}{s}\in \operatorname{Ann}_{R_\mathfrak{p}}(M_\mathfrak{p})$ with $r\in R,s\in R\setminus \mathfrak{p}$ and $\frac{r}{s} \cdot \frac{m}{t}=0$, $\forall \frac{m}{t}\in M_\mathfrak{p}$. Then in particular $\frac{r}{s}\cdot \frac{m}{1}=\frac{0}{1}$, $\forall m\in M$.
$\Rightarrow$ $\frac{rm}{s}=\frac{0}{1}$ $\Rightarrow$ $\forall m\in M$, $\exists a\in S$ with $arm=0$
$\Rightarrow$ $\forall m\in M$, $\exists a\in S$ with $\frac{ar}{as}\in \operatorname{Ann}_R(m)_\mathfrak{p}$
but $\frac{ar}{as}=\frac{r}{s}$ hence $\frac{r}{s}\in \operatorname{Ann}_R(m)_\mathfrak{p}$
'$\supseteq$': Let $\frac{r}{s}\in (\operatorname{Ann}_R(M))_\mathfrak{p}$. Then $rM=0$ and thus $\frac{r}{s}\cdot \frac{m}{t}=0$, $\forall \frac{m}{t}\in M_\mathfrak{p}$.