Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

(I don't mind seeing a completely different example either, I just figured I'd ask about this one since it's already here.)

share|cite|improve this question

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

P.S. With your reputation, you should be able to leave a comment in that answer. You could have followed up with a comment asking why the support is not closed, instead of asking a brand new question. (Clicking on Mariano's name would have revealed he is still active, so he would have seen your comment). – Arturo Magidin Feb 20 '12 at 7:44
up vote 4 down vote accepted

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.

share|cite|improve this answer
Thanks for clearing it up. – Buble Feb 21 '12 at 9:11
This does work only if $S$ contains only maximal ideals. – Niels Nov 27 '14 at 10:51

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