There exists an element $x$ of $M$ such that $W^{-1}R\dfrac{x}{1}=W^{-1}Rz$ and $\operatorname{ann}_R x=\operatorname{ann}_{W^{-1}R} z\cap R$

Question

Let $R$ be a Noetherian ring, $W$ a multiplicative subset of $R$, $M$ an $R$-module, and $z$ an element of $W^{-1}M$. Show that there exists an element $x$ of $M$ so that $W^{-1}R\dfrac{x}{1}=W^{-1}Rz$ and $\operatorname{ann}_R x=\operatorname{ann}_{W^{-1}R} z\cap R$ (i.e. $Rx\hookrightarrow W^{-1}R\dfrac{x}{1}=W^{-1}Rz$).

For the first part: Let $z\in W^{-1}M$ and $x\in M$. Well then $\dfrac{x}{1}\in W^{-1}M$. Since $z$ is an arbitrary element of $W^{-1}M$ and $R$ is Noetherian then $W^{-1}R\dfrac{x}{1}=W^{-1}Rz$. Is this correct?

Also how do I begin the following part? I know that $\operatorname{ann}_R(M)=\{a\in R \ | \ aM=0\}$. But how do I use that when I have intersections?