I was doing exercise 6.5 M (2022 version) in professor Vakil's FOAG. which asked me to show that :
Suppose $M$ is a finitely generated module over Noetherian $A$, and
$\mathfrak{p} \subset A$ is a prime ideal corresponding to an
irreducible component of $\operatorname{Supp} M \subset
\operatorname{Spec} A$. Show that $[\mathfrak{p}] \in$ Ass $M$.
There are two hints:
(Hint 1.) Suppose that $M$ is a finitely generated module over a
Noetherian ring $A$. $M$ has a (finite) filtration
$0=M_0 \subset M_1 \subset \cdots \subset M_n=M \quad$
where $M_{i+1} / M_i \cong A / p_i$ for some prime $p_i$.
(Hint 2.)We consider the short exact sequence $$ 0 \longrightarrow M^{\prime}
\longrightarrow M \longrightarrow M^{''} \longrightarrow 0 $$ of
A-modules. Suppose $\mathfrak{p} \in$ Ass $M^{\prime \prime}$, but
$\mathfrak{p} \notin \operatorname{Supp} M^{\prime}$ (a stronger
hypothesis than $\mathfrak{p} \notin$ Ass $M^{\prime}$ ). Then
$\mathfrak{p} \in$ Ass $M$.
Professor Vakil asked to use result in hint 1 ,2 to prove the exercise.
My attempt since $\text{Supp } M = V(\text{Ann } M)$ (as $M$ is finite generated). Therefore as $\frak{p}$ corresponds to the irreducible component means ${\frak{p}}$ is the minimal prime containing $\text{Ann } M$. Then I need to use the filtration in hint 1 and break it into a short exact sequence so that we can use hint 2, however, I don't know how to do it precisely.
Best Answer
Let's take a finite filtration of $M$ as in hint $(1)$: $$0 = M_0 \subseteq M_1 \subseteq \cdots \subseteq M_n = M$$ $$M_{i} /M_{i-1} \cong A/P_{i}$$
This filtration contains a lot of useful information! We'll make three easy observations:
Observation (1) $\bigcap_{j=1}^i P_j$ annihilates $M_i$. You can check this by an easy inductive argument, starting with the base case $M_1 \cong A/P_1$.
Observation(2) $\operatorname{Supp}(M) = \bigcup V(P_i)$. If $P$ is an element of the support, then there is an element $m$ which annihilator is contained in $P$. Let $m \in M_i$. Then we have seen in observation (1) that $\bigcap_{j=1}^i P_i$ annihilates $m$. So $\bigcap_{j=1}^i P_i \subseteq P$. By primeness, one of the $P_i$ must be contained in $P$, i.e. $P \in V(P_i)$ for some $i$. Conversely, suppose that $P$ contains one of the $P_i$. Then we have an exact sequence $0 \rightarrow M_{i-1} \rightarrow M_{i} \rightarrow A/P_i \rightarrow 0$. Localizing at $P$, $(A/P_i)_P \not= 0$ because $P_i \subseteq P$, and thus it is not possible that both $M_{{i-1}_P}$ and $M_{{i}_P}$ are zero. Thus $P$ is in the support of $M$.
Observation (3) If $P$ is a minimal element of the support of $M$, then $P = P_i$ for some $i$. This follows immediately from observation $(3)$.
Now let's put it all together with Hint (2). Let $P$ be a minimal element of the support. Then by observation (3), we can take $P = P_i$ for some $i$. Let's assume that $i$ is taken minimally. We have an exact sequence $0 \rightarrow M_{i-1} \rightarrow M_i \rightarrow A/P \rightarrow 0$. Of course, $P$ is the unique associated prime of $A/P$. By Hint (2), there are two cases to consider: (i) $P$ is an associated prime of $M_i$ or (ii) $P$ is in the support of $M_{i-1}$.
In case (i) $P$ is also an associated prime of $M$, and you are done. In case (ii), $P$ continues to be a minimal element of the support of $M$. Applying observation (3) to our filtration of $M_{i-1}$, we find that $P = P_j$ for some $j < i$, contrary to our previous assumption that $i$ was taken minimally. Therefore case (ii) is impossible.
I'd like to point out another method of proof, which personally I much prefer, and perhaps you will too because it carries on with your attempt to start with the fact that $\operatorname{Supp}(M) = V(\operatorname{Ann}(M))$.
The following fact is very useful in commutative algebra, and a good exercise to prove.
Fact Let $I$ be an ideal of a ring and $P$ a prime minimal over $I$. Then for each $a \in P$, there exists $n \in \mathbb{N}$ and $s \notin P$ such that $s a^n \in I$. If $P$ is finitely generated, then $n$ and $s$ can be taken such that $s P^n \subseteq I$.
Back to your problem, use the above fact to pick $n \in \mathbb{N}$ which is minimal with respect to the property that there exists an $s \notin P$ such that $s P^n M = 0$. Because $P$ is in the support of $M$, we know $n > 0$.
By our minimal choice of $n$, we know that $P$ is also in the support of $sP^{n-1} M$. Indeed, $sP^{n-1}M$ is a finite $A$-module, so if $P$ was not in its support, then we could find $s' \notin P$ with $s' P^{n-1}M = 0$.
Thus we can choose $0 \not= m \in sP^{n-1}$ such that $\operatorname{Ann}(m) \subseteq P$. Clearly also $Pm = 0$. Therefore $\operatorname{Ann}(m) = P$ and $P$ is an associated prime of $M$.