This question comes from the naive belief that $|Spec(R)|\subset P(|Specm(R)|)$, which I now know is only true if $R$ is a Jacobson ring, which lead me to believe that Semilocal rings are characterized by having a finite number of prime ideals, rather than just maximal. I now believe this is false, but I cannot come up with a counterexample. Any help would be greatly appreciated.
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What does the notation $P$ in $P(|Specm(R)|)$ mean, actually? – Watson Jan 26 '17 at 13:32
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1@Watson It is meant to denote the powerset – Pax Jan 26 '17 at 22:28
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You can try the local ring $\Bbb Q[X,Y]_{(X,Y)}$, which is noetherian (as localization of a noetherian ring — $\Bbb Q[X,Y]$ is noetherian by the Hilbert basis theorem). The ideals $(X+aY)$ in $\Bbb Q[X,Y]_{(X,Y)}$ are prime and pairwise distinct, so that the spectrum of $\Bbb Q[X,Y]_{(X,Y)}$ is infinite.
Watson
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I can't believe I didn't even check if for local rings. I think I might be a bit off today. Thank you so much! – Pax Jan 19 '17 at 20:44
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@watson Hi: thanks for the suggestion! When I checked, I did not see another such ring in storage, so it is valuable indeed. I'd like to help flesh it out more, but I need a second opinion. It is UFD that is not a PID, is that correct? It seems like $(X,Y)$ should not be principal, but I just want to run it by another person. – rschwieb Jan 24 '17 at 19:26
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@rschwieb : you're welcome :-). I agree that, as a localization of a UFD, it is a UFD, and it has dimension $2$, so it is not a PID. [By the way, it would be so great to have some property giving the Krull dimension of the rings at http://ringtheory.herokuapp.com/ – for instance there is an amazing example of Noetherian ring with infinite Krull dimension…] – Watson Jan 24 '17 at 19:43
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1@Watson I will think about adding it as something visible for commutative rings. As a substitute, I have been inserting it into the ring's extra info. But it would be nice to be able to search on Krull dimension for commutative rings :) – rschwieb Jan 24 '17 at 19:54
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@Watson The ring is much more complete now, if you revisit the link. You might be able to tell me what further can be said about the commutative-properties. – rschwieb Jan 24 '17 at 20:09
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@rschwieb : waouh! So great! Thank you very much. I think that "unique factorization domain" is not in the 1st column. – Watson Jan 24 '17 at 20:11
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@Watson Properties that are (mostly) exclusively used with commutative rings have a different web view with a different collection of properties :) – rschwieb Jan 24 '17 at 20:22
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@rschwieb : ah sorry, I didn't check. That's fine! Anyway, thank you very much for your work, dear rschwieb! – Watson Jan 24 '17 at 20:28
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As for local rings, dimension 0 implies that Spec has only one element ; dimension 1 implies that Spec has at most two elements. So we try to focus on rings with Krull dimension $\geq 2$. – Watson Jan 26 '17 at 11:12