TITLE: What is the support of a localised module?
QUESTION [5 upvotes]: Let $R$ be a noetherian commutative ring, and let $\mathfrak{m}$ be a maximal ideal of $R$. Let $M$ be a finitely-generated torsion $R_\mathfrak{m}$-module, considered as an $R$-module. Is it possible that $M_\mathfrak{p} \ne 0$ for some prime ideal $\mathfrak{p}$ not contained in $\mathfrak{m}$?
Intuitively/geometrically, this doesn't seem possible. What we have here is a coherent sheaf on $\operatorname{Spec} R_\mathfrak{m}$ and its direct image on $\operatorname{Spec} R$. What I want to argue is that for any point $\mathfrak{p}$ of $\operatorname{Spec} R$, $\mathfrak{p} \ne \mathfrak{m}$, there is an open neighbourhood $U$ of $\mathfrak{p}$ not containing $\mathfrak{m}$, and so therefore sections of $M$ over $U$ must be $0$... but this isn't quite true, since the image of $\operatorname{Spec} R_\mathfrak{m}$ is not just $\{ \mathfrak{m} \}$.
We know that $\mathfrak{p}$ is in the support of $M$ if and only if $\mathfrak{p}$ contains the annihilator of $M$. To show that $\mathfrak{p}$ is not in the support of $M$, we must exhibit an element of $R \setminus \mathfrak{p}$ that annihilates $M$. But why does such an element exist?

In case the answer to my very first question is "yes", let's add more hypotheses: suppose $R$ is a Dedekind domain. Then $\operatorname{Spec} R_\mathfrak{m}$ consists of an open point and a closed point, and the sections/stalk over the open point must be $0$ because $M$ is a torsion $R_\mathfrak{m}$-module. I feel convinced that this proves that $M_\mathfrak{p} = 0$ for all $\mathfrak{p} \ne \mathfrak{m}$... but how does this translate into algebra?

REPLY [2 votes]: In case the answer to my very first question is "yes", let's add more hypotheses: suppose $R$ is a Dedekind domain. Then $\operatorname{Spec} R_\mathfrak{m}$ consists of an open point and a closed point, and the sections/stalk over the open point must be $0$ because $M$ is a torsion $R_\mathfrak{m}$-module. I feel convinced that this proves that $M_\mathfrak{p} = 0$ for all $\mathfrak{p} \ne \mathfrak{m}$... but how does this translate into algebra?

Proof
Let $R$ be a Dedekind domain. Let $P$ be a maximal ideal of $R$ . Let M  be a finitely generated torsion $R_P$-module. Since $R_P$ is a principal ideal domain, $M$ is a finite product of cyclic torsion $R_P$-modules. Hence we can assume that $M$ is a finite product of modules of type $R_P/(P^n)R_P$. Since $R_P/(P^n)R_P$ is isomorphic to $R/P^n$, the support of $M$ as an $R$-module is {$P$}.