It's well known that under some "weak" hypothesis, such as finitely generation, the support of an A-module is closed in Spec(A). It is true also in the most general case?
The support of a module is closed?
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commutative-algebra
1 Answers
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Pick a non-closed subset $S$ of $\operatorname{Spec}\mathbb Z$, and let $\displaystyle M=\bigoplus_{\mathfrak p\in S}\:\mathbb Z/{\mathfrak p}$.
NB: Editing this answer required solving a captcha.
(Thanks go to Akhil for catching a stupid mistake)
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0Very nice. I suppose by $\mathbb{Z}_p$ you mean the ring of $p$-adic integers. :) – 2010-12-09
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0Do you mean $\mathbb{Z}/p$? The support of the localization $\mathbb{Z}_{(p)}$ is the entire space. – 2010-12-09
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0Nice but doesn't seem quite correct to me! What about $S=\{ (0) \}$ ? The generic point is non-closed and you get with your formula $M=\mathbb Z$, whose support is the whole of $\mathbb Z$ obviously, hence closed. To correct this $S$ must contain maximal ideals only. – 2014-11-27