# Example of a module whose support is not closed? [duplicate]

Possible Duplicate:
The support of a module is closed?

Is there a simple example of a module $M$ of a Noetherian commutative ring $R$ such that $\operatorname{Supp}(M)\subset\operatorname{Spec}(R)$ is not closed?

When typing this question, this answer popped up.

So I think we can take $R=\mathbb{Z}$, and let $M=\bigoplus_{\mathfrak{p}\in S}\mathbb{Z}/\mathfrak{p}$ for a nonclosed subset $S$ of $\operatorname{Spec}(\mathbb{Z})$.

Is there an actual explanation as to why the support of such $M$ is not closed in $\operatorname{Spec}(\mathbb{Z})$? I didn't gather one from the original answer.

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## marked as duplicate by Arturo Magidin, Asaf Karagila, Jennifer Dylan, Alex Becker, t.b.Aug 20 '12 at 14:08

Note that $(\mathbb{Z}/\mathfrak{p})_{\mathfrak{q}}=(0)$ whenever $\mathfrak{p}\neq \mathfrak{q}$, and $(\mathbb{Z}/\mathfrak{p})_{\mathfrak{p}} = \mathbb{Z}/\mathfrak{p}$. Moreover, localization commutes with direct sums, so for every $\mathfrak{q}$, $$M_\mathfrak{q} = \left(\bigoplus_{\mathfrak{p}\in S}\mathbb{Z}/\mathfrak{p}\right)_{\mathfrak{q}} = \bigoplus_{\mathfrak{p}\in S}(\mathbb{Z}/\mathfrak{p})_{\mathfrak{q}}.$$
So the support of $M$, that is, the set of primes $\mathfrak{q}$ such that $M_{\mathfrak{q}}\neq 0$ is precisely $S$, which by assumption is not closed.
This does work only if $S$ contains only maximal ideals. –  Niels Nov 27 '14 at 10:51