The support of a module is closed?

Question

It's well known that under some "weak" hypothesis, such as finitely generated, the support of an A-module is closed in Spec(A). It is true also in the most general case?

Answer 1

Pick a non-closed subset $S$ of $\operatorname{Spec}\mathbb Z$, and let $\displaystyle M=\bigoplus_{\mathfrak p\in S}\:\mathbb Z/{\mathfrak p}$.

Answer 2

As @Niels said, the answer above is not quite correct. Here is a modified version.

Let $\mathcal{P}$ be an infinite set of prime numbers of $\mathbb{Z}$ such that there are infinitely many prime numbers of $\mathbb{Z}$ that are not contained in $\mathcal{P}$. Then the support of the $\mathbb{Z}$-module $M=\oplus_{p\in\mathcal{P}}\mathbb{Z}/p\mathbb{Z}$ is equal to $\mathcal{P}$.

Proof

Claim: $M_\mathfrak{q}\neq 0\iff q\in\mathcal{P}$ where $(q)=\mathfrak{q}$. If $q\in\mathcal{P}$, then $\mathbb{Z}/q\mathbb{Z}$ survives under the localization at $\mathfrak{q}$; If $q\notin\mathcal{P}$, then the localization inverts the annihilators of $\mathbb{Z}/p\mathbb{Z}$ and thus $M=0$. $\text{Spec}(\mathbb{Z})=\{(0)\}\cup\{(p)\vert p\ \text{is prime}\}$. Since $\mathcal{P}$ is infinite, $\mathcal{P}$ is not closed. Also, it is not open since any nonempty open subset of $\text{Spec}(\mathbb{Z})$ must contain $(0)$ and all but finitely many primes. Thus, we have constructed a $\mathbb{Z}$-module $M$ such that the support of $M$ is neither closed nor open.