Let $A$ be a commutative ring with $1$. Suppose that $P \subseteq Q$ are prime ideals in $A$ and that $M$ is an $A$-module. Prove that the localization of the $A$-module $M_{Q}$ at $P$ is the localization $M_{P}$, i.e $(M_{Q})_{P} = M_{P}$.
Hint from the book: Use the fact that $S^{-1}A \otimes _{A} M \cong S^{-1}M$ as $S^{-1}A$ modules.
Here's what I have:
First set $S=A \setminus P$ and $T=A \setminus Q$, then by assumption $T \subset S$.
So using the hint:
$S^{-1}(T^{-1}M) \cong S^{-1}A \otimes_{A} T^{-1}M \cong S^{-1}A \otimes_{A} (T^{-1}A \otimes_{A} M) \cong (S^{-1}A \otimes_{A} T^{-1}A) \otimes_{A} M$
From here I'm stuck. Can you please help?