1
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.