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?
The support of a module is closed?
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$\begingroup$
commutative-algebra
modules
maximal-and-prime-ideals
1 Answers
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Pick a non-closed subset $S$ of $\operatorname{Spec}\mathbb Z$, and let $\displaystyle M=\bigoplus_{\mathfrak p\in S}\:\mathbb Z/{\mathfrak p}$.
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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