Exercise 5.5.F. on Ravi Vakil's Notes related to associated points Let $A$ be a Noetherian ring and $M$ a finitely generated $A$ module. In Ravi Vakil's notes he first states that the associated points of $M$ satisfy the following:
(A) The associated points of $M$ are precisely the generic points of irreducible components of the support of some element of $M$ on Spec $A$. 
I am working on the following exercise: Assuming (A) show that the associated points of $S^{-1}M$, where $S$ is a multiplicative subset of $A$, are precisely those associated points of $M$ that do not meet $S$.
I was able to show that: given $m/s \in S^{-1}M$ there is a bijective correspondence as sets between Supp $m/s$ and the the points on Supp $m$ that do not meet $S$.
From here I think I can use (A) somehow to complete the exercise, but 
I am having difficulties with it. I would appreciate any assistance! Thank you very much!
PS I asked this question also here working on this exercise. It might be helpful here, but I couldn't quite complete the exercise...
 A: $\DeclareMathOperator{\Supp}{Supp}$
$\DeclareMathOperator{\Spec}{Spec}$
$\DeclareMathOperator{\Ann}{Ann}$
First of all, note that the claim that $\Supp(\frac{m}{s})=\Supp(m)\cap\{\mathfrak{p} \in \Spec(A):\mathfrak{p}\cap S=\emptyset\}$ is false in general, which makes the proof a little harder. For a counter example, take $A$ an integral domain, $S$ a non-trivial multiplicatively closed subset and $m = 1$. Then $\frac{1}{1}$ is still supported everywhere, including primes meeting $S$. The problem is to show that the primes corresponding to generic points of $\Supp(\frac{m}{s})$ don't meet $S$, since it's just not true that any prime in this set doesn't meet $S$. This answer is long because it contains most of the thoughts I had when I was solving the problem. It is considerably shorter if you simply write out the concise argument and not all of the motivation included here.
To prove the claim, then:
Let's think for a moment about what $\Supp(m)$ is. Well, $\mathfrak{p} \not\in \Supp(m)$ means that $\exists r \not\in \mathfrak{p}$ such that $rm = 0$. In otherwords, some zerodivisor of $m$ lies outside of $\mathfrak{p}$. Thus $\mathfrak{p} \in \Supp(m)$ is equivalent to $\mathfrak{p}\supset\Ann_A(m)$, where $\Ann_A(m)$ is the set (in fact, an ideal) of zero-divisors of $m$. In yet more words, $\Supp(m) = V(\Ann_A(m))$. Then what is an irreducible component of this set? It is a maximal closed subset of the form $V(\mathfrak{p})$ for $\mathfrak{p}$ a prime. This is the same thing as closed subset of the form $V(\mathfrak{p})$ for $\mathfrak{p}$ a minimal prime lying over $\Ann_A(m)$.
Understanding $a)$ this characterisation of $\Supp(m)$ and $b)$ this characterisation of generic points of closed subsets in an affine scheme are, in my opinion, more important than doing the exercise itself.
With this in mind, we will see that proving the following will complete the exercise:

The minimal primes lying over $\Ann_A(\frac{m}{s})$ are precisely the minimal primes lying over $\Ann_A(m)$ that don't meet $S$.

We prove this in two parts:

i) The minimal primes lying over $\Ann_A(m)$ that don't meet $S$ are also minimal primes lying over $\Ann_A(\frac{m}{s})$.
ii) Any minimal prime lying over $\Ann_A(\frac{m}{s})$ is a minimal prime lying over $\Ann_A(m)$ that doesn't meet $S$.

Note that $i)$ and $ii)$ are actually stronger than what is being asked for, minimal primes lying over annihilators are referred to as weakly associated primes and can be defined over non-Noethrain affine schemes and neither our proof nor the discussion above use that $A$ was Noetherian.
The proof of $i)$ is not so hard, if $\mathfrak{p}$ is any prime lying over $\Ann_A(m)$ that doesn't meet $S$, suppose $\frac{m}{s}$ wasn't supported at $\mathfrak{p}$. Then there exists $t \in S, r \not\in \mathfrak{p}$ such that $trm = 0$. In particular, $tr \in \Ann_A(m) \subset \mathfrak{p}$. But then $\mathfrak{p}$ is prime, so since $r \not\in \mathfrak{p}, t \in \mathfrak{p}\cap S$, a contradiction. Thus $\mathfrak{p}$ is in the support of $\frac{m}{s}$ (i.e. lies over $\Ann_A(\frac{m}{s})$) and it must be minimal amongst primes that do so since it was minimal over $\Ann_A(m) \subset \Ann_A(\frac{m}{s})$.
The proof of $ii)$ is slightly more involved, here's a slightly modified version of an answer I gave to a different question, which was inspired by the answer of zcn to the same question.
Suppose $\mathfrak{p}$ is minimal over $\operatorname{Ann}_A(\frac{m}{s})$. Now suppose $t \in \mathfrak{p}\cap S$. Then since $\mathfrak{p}_\mathfrak{p}$ is the unique minimal prime lying over $\operatorname{Ann}_A(\frac{m}{s})_{\mathfrak{p}}$ in $A_{\mathfrak{p}}$, $t$ is nilpotent in $A_{\mathfrak{p}}/\operatorname{Ann}_A(\frac{m}{s})_{\mathfrak{p}}\cong (A/\operatorname{Ann}_A(\frac{m}{s}))_{\mathfrak{p}}$, that is to say there exists some $x \in A\setminus \mathfrak{p}$ with $t^nx \in \operatorname{Ann}_A(\frac{m}{s})$. But then $x \in \operatorname{Ann}_A(\frac{m}{s}) \subset \mathfrak{p}$, a contradiction.
There are a few ways to see that $x$ annihilates $\frac{m}{s}$ if $t^nx$ does, if you don't see it immediately (I didn't!). One is to actually write down from definitions what it means for an element of $A$ to annihilate an element of $S^{-1}M$ and observe that for $a \in A, s \in S$, $a$ annihilates an element iff $sa$ does, or you can just write $\frac{m}{s} = \frac{t^nm}{t^ns}$ and then use that $x(\frac{m}{s}) = x\frac{t^nm}{t^ns}= \frac{xt^n}{s}\frac{1}{t^n} = 0$. The first method is perhaps more enlightening, and can be thought of as being a consequence of the fact that $s$ "acts as a unit on $M$".
A: At the risk of making mistakes here, I shall write my solutions with a disclaimer that it might not be correct.
As you have observed, you have shown that $\text{supp}(\frac{m}{s})=\text{supp}(m)\cap \{{\frak{p}}\in\text{Spec}{A}:\ {\frak{p}}\cap S=\emptyset\}$
Now, this a closed set in $\text{Spec}S^{-1}A$, so we let it be $V_{S^{-1}A}(I)$ for some $I$ ideal in $S^{-1}A$. Similarly, we let $V_{A}(J)$ to be $\text{supp}(m)$ for some $J\subset A$ an ideal.
Claim: $V_{S^{-1}A}(S^{-1}J)=V_{S^{-1}A}(I)$
Proof If $\frak{p}$ is not in the RHS, then $m/s$ is zero in $(S^{-1}A)_{\frak{p}}$. This means that there exists $f/t\notin \frak{p}$ such that $(fm)/(st)=0$ in $S^{-1}A$. Hence there exists $s'\in S$ such that $s'fm=0\in M$. Hence, if we let $\phi:A\rightarrow S^{-1}A$ be the obvious localisation map, then $m=0$ in $A_{\phi^{-1}(\frak{p})}$. Hence $\phi^{-1}({\frak{p}})\notin \text{supp}(m)$, and so $\phi^{-1}({\frak{p}})$ does not contain $J$. Hence ${\frak{p}}=S^{-1}\phi^{-1}(\frak{p})$ does not contain $S^{-1}J$.
Conversely, if $\frak{p}$ does not lie in the LHS, then $\phi^{-1}(\frak{p})$ does not contain $J$. So it does not lie in the support of $m$ and hence $m=0$ in $A_{\phi^{-1}(\frak{p})}$. Hence there exists $t\notin\phi^{-1}(\frak{p})$ such that $tm=0$ in $M$. Since $(S^{-1}A)_{\frak{p}}=A_{\phi^{-1}(\frak{p})}$, hence $m/s=tm/ts=0$ in $(S^{-1}A)_{\frak{p}}$, and so $\frak{p}$ does not lie in the support of $m/s$ and hence it does not contain $I$. 

An irreducible component of $V_{S^{-1}A}(I)$ corresponds to a minimal prime ideal $\frak{m}$ that contains $I$. Similarly an irreducible component of $V_{A}(J)$ corresponds to a minimal prime in $A$ that contains $J$.
Claim Let $\phi:A\rightarrow S^{-1}A$ be the localisation map. I claim that this map induces a bijection between minimal primes containing $I$ in $S^{-1}A$ and minimal primes in $A$ that contains $J$.
The prime $\frak{m}$ being a minimal element on the LHS is an associated point of $m/s$.
On the other hand, via $\phi:A\rightarrow S^{-1}A$ the localisation map, we let ${\frak{m}}'=\phi^{-1}(\frak{m})$. It is a prime ideal that avoids $S$. Hence it remains to show that (i) it lies in the support of $m$ and (ii) it is a minimal prime amongst all those that contains $J$ (because there is a dictionary between irreducible components and minimal primes).
To prove (i), suppose $m=0$ in $A_{\frak{m'}}$, then there exists $f\notin\frak{m}'$ such that $fm=0\in M$. So $fm=0\in (S^{-1}A)_{\frak{m}}$, and hence it is done.
To prove (ii), suppose there is another $\frak{n}\subset \frak{m'}$ that contains $J$. Then by the claim, $S^{-1}\frak{n}\subset \frak{m}$ and both contains $I$. By minimality of $\frak{m}$, $S^{-1}\frak{n}=\frak{m}.$ Hence $\frak{n}=\frak{m'}$.
So $\frak{m}'$ is an associated prime of $m$ as advertised.
