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

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.

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 $$\mathrm{supp}(m) = \bigcup_i \overline{\mathfrak{p}_i}$$.