Vakil 5.5.F Associated primes/points behaves well with respect to localizing In the notes of Vakil's 'The rising sea', he attempted to explain the concepts of associated points geometrically. He discussed associated points of $M$ where $M$ is a finitely generated $A$ module (and $A$ is Noetherian ring). And he stated the important property (rather than defining it) first:

(A) The associated primes/points of $M$ are precisely the generic points of irreducible
components of the support of some element of $M$ (on Spec $A$).

And the exercise 5.5.F asks:

Show that the definition in (A) of associated primes/points behaves well with respect to localizing: if $S$ is a multiplicative subset of $A$, then the associated primes/points of $S^{-1}M$ are precisely those associated primes/points of $M$ that lie in Spec $S^{-1}A$, i.e., associated primes of $M$ that do not meet $S$.

This has being asked here (which is closed as a duplicate). After reading these answers I got a slightly different one following Vakil's approach, and wanted to share here. Any comments are appreciated.
 A: First suppose $p\in Supp\ m$ and $p\cap S=\emptyset$. Being in the support means that $m_p\ne 0$, that is
\begin{align*}
  rm\ne 0, \forall r\notin p.
\end{align*}
That is, $p$ contains the annihilator of $m$, $p\supset Ann_A(m)$. Now since $p$ is disjoint from $S$, for any $t\in S$ we have $rt\notin p$, hence $rtm\ne 0$. In other words
\begin{align}
    r t m\ne 0, \forall r\notin p, \forall t\in S,
\end{align}
which is nothing but the condition for $p\in Supp\frac{m}{s}$ (for any $s\in S$). This completes the first half.
On the other side, now suppose $p\in Supp \frac{m}{s}$ is an associated point of $S^{-1}M$. Let $t=1$ in the above equation we see that $rm\ne 0, \forall r\notin p$, which shows that $m_p\ne 0, p\in Supp\ m$, and similarly let $r=1$ we conclude that $Ann_A(m)\cap S=\emptyset$.
Then there must be a prime ideal $q\supset Ann_A(m)$ and being disjoint from $S$ (e.g. refer to here), so $q\in Supp\frac{m}{s}$ too.
But $p$ as an associated point, is minimal (among primes containing $Ann_A(m)$). Hence $p\subset q$ thus $p\cap S=\emptyset$ too.
