Proof explanation for $S^{-1}(\operatorname{Ann}_{A}(M)) = \operatorname{Ann}_{S^{-1} A}(S^{-1}M)$ The proposition and the proof are below, my problem is the last line can't understand how it gets from $\operatorname{Ann}(S^{-1} M)$ to $S^{-1} \mathfrak{a}$. Please explain it a little more ?

Proposition 3.14. Let $M$ be a finitely generated $A$-module, $S$ a multiplicatively closed subset of $A$. Then $S^{-1}(\operatorname{Ann}(M)) = \operatorname{Ann}(S^{-1}M)$.
Proof: If this is true for two $A$-modules, $M$,$N$, it is true for $M + N$:
  \begin{align*}
     S^{-1}(\operatorname{Ann}(M+N))
  &= \text{$S^{-1}(\operatorname{Ann}(M) \cap \operatorname{Ann}(N))$ by (2.2)}
  \\
  &= \text{$S^{-1}(\operatorname{Ann}(M)) \cap S^{-1}(\operatorname{Ann}(N))$ by (3.4)}
  \\
  &= \text{$\operatorname{Ann}(S^{-1}M) \cap \operatorname{Ann}(S^{-1}N)$ by hypothesis}
  \\
  &= \operatorname{Ann}(S^{-1}M + S^{-1}N)
   = \operatorname{Ann}(S^{-1}(M+N)).
\end{align*}
  Hence it is enough to prove (3.14) for $M$ generated by a single element:
  then $M \cong A/\mathfrak{a}$ (as $A$-module), where $\mathfrak{a} = \operatorname{Ann}(M)$;
  $S^{-1}M \cong (S^{-1} A)/(S^{-1} \mathfrak{a})$ by (3.4), so that $\operatorname{Ann}(S^{-1}M) = S^{-1} \mathfrak{a} = S^{-1}(\operatorname{Ann}(M))$.


Since $S^{-1}M \cong (S^{-1}A)/(S^{-1}\mathfrak{a})$ where $\mathfrak{a} = Ann(M)$, it follows that $Ann(S^{-1}M) = S^{-1}\mathfrak{a}$ from the following fact:
If $R$ is any ring and $I$ is an ideal of $R$, then $Ann(R/I) = I$. Indeed, if $x \in Ann(R/I)$, then $xr \in I$ for all $r \in R$ and so in particular $x = x.1 \in I$. The converse that every element of $I$ annihilates $R/I$ is obvious.
