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.
$\operatorname{MaxSpec}(A)$ closed
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commutative-algebra
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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, – 2012-01-07
2 Answers
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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)
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0I don't have to assume that k is algebraically closed at all, even if I'm using the Nullstellensatz. Thanks again – 2012-01-07
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4This is a plagiarism of [this famo$u$s answer](http://math.stackexchange.com/a/74383/660) of Didier Piau. --- See [this Google Plus search](https://plus.google.com/u/0/s/%22Construct%20a%20function%20which%20is%20continuous%20in%22). – 2012-01-06