For any subset of associated points, is there a section whose support is the closure of that subset? (cf: Vakil 5.5.O)

Question

Let $A$ be a Noetherian ring and $M$ a finitely generated module. Vakil's Exercise 5.5.O says

Show that those subsets of Spec(A) which are the support of an elements of M are precisely those subsets which are the closure of a subset of the associated points.

I wonder if

for any subset of associated points, there is a section whose support is the closure of that subset.

So far, I am able to show

(1)for any associated point $p$, there is a section supported precisely on the closure of $p$.

(2) If $m ∈ M$, show that the support of $m$ is the closure of those associated points at which $m$ has nonzero germ.

The thing I am asking is more general than (2), I think.

Answer 1

We want to show that

for any subset of associated points, there is a section whose support is the closure of that subset.

Step 1: restricting to minimal elements does not change the closure.

As you have shown (I believe), if $\mathfrak{p}=\mathrm{ann}(m)$ is an associated prime, then $\mathrm{supp}(m)=\overline{\mathfrak{p}}$. Given any (necessarily finite) subset $\{\mathfrak{p}_1, \dots, \mathfrak{p}_r\}$ of $\mathrm{Ass}(M)$, say $\mathfrak{p}_i = \mathrm{ann}(m_i)$ for $1\leq i\leq r$, we claim that there is an element $m\in M$ whose support equals $\bigcup_{i=1}^r \overline{\mathfrak{p}_i}$. Note that if the minimal elements in $\{\mathfrak{p}_1, \dots, \mathfrak{p}_r\}$ are $\{\mathfrak{p}_1, \dots, \mathfrak{p}_s\}$, then $\bigcup_{i=1}^s \overline{\mathfrak{p}_i} = \bigcup_{i=1}^r \overline{\mathfrak{p}_i}$. So we may assume that each element of $\{\mathfrak{p}_1, \dots, \mathfrak{p}_r\}$ is minimal.

I claim that $m:=\sum_{i=1}^r m_i$ will be the desired element.

Step 2: inside the closure.

Indeed, at $\mathfrak{p}_i$ we have $m_i\neq 0\in M_{\mathfrak{p}_i}$. By minimality, for any other $j\neq i$, there exists $f_j \in \mathfrak{p}_j\setminus \mathfrak{p}_i$ and hence $m_j=0 \in M_{\mathfrak{p}_i}$. This shows that $m = m_i \neq 0 \in M_{\mathfrak{p}_i}$.

If $\mathfrak{q}\in \overline{\mathfrak{p}_i}$ for some $i$, then we have the localization homomorphism $M_{\mathfrak{q}} \to M_{\mathfrak{p}_i}$ under which $m\mapsto m\neq 0$, so $m\neq 0 \in M_{\mathfrak{q}}$.

Hence, $\mathrm{supp}(m) \supseteq \bigcup_i \overline{\mathfrak{p}_i}$.

Step 3: outside the closure.

Conversely, at any prime $\mathfrak{q} \notin \bigcup_i \overline{\mathfrak{p}_i}$, we have $\mathfrak{q} \not\supset \mathfrak{p}_i$ for all $i$. Then, there are $f_i\in \mathfrak{p}_i\setminus \mathfrak{q}$ for all $i$, so $m_i=0\in M_\mathfrak{q}$ for all $i$, which means that $m=0\in M_\mathfrak{q}$ and $\mathrm{supp}(m) = \bigcup_i \overline{\mathfrak{p}_i}$.