Exercise $5.5.$M, Foundations of Algebraic Geometry by Ravi Vakil In the following, $A$ is a Noetherian ring.
The basic definition here is:

A prime $p\subsetneqq A$ is said to be associated to $M$ if $p$ is the annihilator of an element $m$ of $M$.

The exercise is:

$5.5.$M. EXERCISE.


(a) If $M$ is a finitely generated module over Noetherian $A$, show that M has a
filtration
$0 = M_0\subseteq M_1\subseteq ... \subseteq  M_n = M$
where $M_{i+1}/M_i\cong A/p_i$ for some prime ideal $p_i$.


(b) Show that the associated prime ideals of $M$ are among the $p_i$.


(c) Show that for each $i$, $\mathrm{Supp}(A/p_i)$ is contained in $\mathrm{Supp}(M)$, or equivalently, that every $p_i$ contains an associated prime. Hint: if $p_i$ does not contain an associated prime, then localize at $p_i$ to “make $M$ disappear”.

What I have done is (you can skip (a)(b) because most of my question is about (c)):
For (a): pick up a non-zero $m\in M$ with $Ann(m)\subsetneqq A$, since $A$ is Noetherian, $Ann(m)$ can be chosen as a maximal annihilator ideal(just take a maximal ideal of $\{Ann(n),Ann(m)\subseteq Ann(n)\}$as $Ann(m)$,by abuse of notation)which can be directly checked to be prime. There is $A/Ann(m)\cong (m)$, so just take $Ann(m)$ as $p_0$ and $(m)$ as $M_1$.Apply the above procedure for $M/(m)$ and so forth. Because $M$ is a Noetherian module, this procedure must end, it's done.
For (b): (a) can be rewritten as existing $m_1,...,m_n\in M$:
$1.$ $0 = M_0\subseteq (m_1)\subseteq (m_1,m_2)\subseteq...\subseteq (m_1,...,m_n) = M$
$2.$ $\forall x\in A\,(xm_i\in (m_1,..,m_{i-1})\Leftrightarrow x\in p_i)$
for a associated prime $p$, we choose a expression $Ann(a_1m_1+...+a_km_k)(a_k\neq 0)$ where $k$ is the smallest integer occuring among all possible expressions of $p$. It can be proved that $p=p_k$.
For (c): suppose (c) does not hold. If $M_{p_i}=0$, then exists $x\notin p_i$ annihilating $m_i$, which contradicts part (b) of my proof. Using injection ($M_{p_i}\rightarrow\prod(M_p)_{p_i},\,p$ associated primes) it suffices to prove $\forall$ associated $p,\,(M_p)_{p_i}=0$, but I have no clue to do this.
My questions are:
$1.$ am I correct for (a), (b)?
$2.$ why $M_{p_i}=0$ in part (c), assuming $p_i$ contains no associated prime? I know it's true because intuitively associated primes are generic points of $supp(m)$'s, but I should work this out just from the definition given above.
$3.$ I think what I have done for (a), (b) is not "elegant" enough. Is there a better way?
Any hint or reference will help.
Thanks :)
 A: I've also only just recently studied associated modules, so take what I say with a grain of salt. I'm putting it up here in case you find something useful.
Your (a) looks correct. If you want to make it marginally more "elegant," I don't think you even need to choose a specific element $m.$ If $M$ is nonzero, then $M$ has at least one associated prime $P,$ so there exists a submodule $N$ such that $N$ is isomorphic to $A/P.$ Then do the same for $M/N,$ and the induction must terminate because $A$ is Noetherian.
For (b), I'm not entirely convinced if we could choose such $m_1,\ldots,m_n.$ The way I would solve this problem is to consider the short exact sequence
$$0 \to M_i \to M_{i+1} \to A/p_i \to 0$$
of $A$-modules. Then we know that $\operatorname{Ass}(M_i) \subseteq \operatorname{Ass}(M_{i+1}) \subseteq \operatorname{Ass}(M_i) + \operatorname{Ass}(A/p_i),$ so the claim follows.
For (c), suppose as in the hint that $p_i$ does not contain an associated prime. Take any $m \neq 0.$ Then $\operatorname{ann}(m)$ is contained in some associated prime $q.$ By our assumption, $p_i$ does not contain $q.$ But this implies that $m=0$ in $M_{p_i}.$ Indeed, we can choose $a \in q\setminus p_i,$ and then we would have $am=0.$ But note that $m$ was chosen arbitrarily, so $M_{p_i} = 0.$ By (a), we know there exist submodules $M_i,M_{i+1}$ of $M$ such that $M_{i+1}/M_i \cong A/p_i.$ But when we localize both sides by $p_i,$ we get $0$ on the left hand side, but the right hand side equates to $(A/p_i)_{p_i} \neq 0.$
A: I do not have enough reputation yet to comment on the other answer, but it is not clear to me that choosing $a \in q \setminus p_i$ guarantees that $am = 0$, since the assumption is only that $\operatorname{Ann}(m) \subset q$.
I believe a more direct approach to answer your question (2) is as follows.
Taking the contrapositive of the statement, we have to prove that if $M_{\mathfrak{p}_i} \neq 0$ then $\mathfrak{p}_i$ contains an associated prime. We already showed (exercise 5.5.J) that $M_{\mathfrak{p}_i}$ has an associated prime $\mathfrak{q} = \operatorname{Ann}(m/s)$ for some $m \in M, s \in A \setminus \mathfrak{p}_i$, and one can check that $\mathfrak{q} = \operatorname{Ann}(m)$ is an associated prime of $M$. Since $m/s \neq 0$, we find that for all $r \in A \setminus \mathfrak{p}_i$, $rm \neq 0$, so it must be the case that $\mathfrak{q} \subset \mathfrak{p}_i$.
