If $A$ is an arbitrary commutative ring, is $\operatorname{MaxSpec}(A)$ closed as a subset of $\operatorname{Spec}(A)$? I wanted to think of a counterexample, but so far without success. I tried to consider generic points, but if $\operatorname{MaxSpec}$ is a proper subset of $\operatorname{Spec}$, then it cannot contain one, so this approach won't work.
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2Dear Steffi, How many rings can you write down in which MaxSpec$(A)$ is a closed subset of Spec$(A)$? Regards, – Matt E Jan 07 '12 at 00:06
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$$\mathbb Z$$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$
Pierre-Yves Gaillard
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4This is a plagiarism of this famous answer of Didier Piau. --- See this Google Plus search. – Pierre-Yves Gaillard Jan 06 '12 at 20:33
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If the maximal spectrum is a closed subset, then there is an ideal $I\subseteq A$ such that the maximal ideals of $A$ are precisely the prime ideals which contain $I$. Can you use this observation to construct an example of an $A$ where this does not hold?
(An easy way to make use of this is to look for a ring such that the intersection of the maximal ideals is zero—that is, with trivial Jacobson radical—so that the choice of $I$ is thereby severly limited)
Mariano Suárez-Álvarez
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Thanks for both answers, but I can only choose one :-) I thought of k[X] where k is algebraically closed. This is a Jacobson ring and hence the Jacobson radical equals the nilradical which is zero because k[X] is an integral domain. Then $I$ must be the zero ideal which is generic. I think that "k is algebraically closed" can be dropped so that I don't have to use the Nullstellensatz. Thanks!! – Steffi Jan 07 '12 at 14:48
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I don't have to assume that k is algebraically closed at all, even if I'm using the Nullstellensatz. Thanks again – Steffi Jan 07 '12 at 15:03