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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? – 2017-01-26
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    @Watson It is meant to denote the powerset – 2017-01-26

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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.

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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! – 2017-01-19
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    You can have other examples [here](http://math.stackexchange.com/questions/1954901) or on [this](https://ringtheory.herokuapp.com/) website (from [this](http://math.stackexchange.com/users/29335/rschwieb) user). – 2017-01-21
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    @watson Hi: thanks for the [suggestion!](http://ringtheory.herokuapp.com/rings/ring/61/) 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. – 2017-01-24
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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…] – 2017-01-24
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    @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 :) – 2017-01-24
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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. – 2017-01-24
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    @rschwieb : waouh! So great! Thank you very much. I think that "unique factorization domain" is not in the 1st column. – 2017-01-24
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    @Watson Properties that are (mostly) exclusively used with commutative rings [have a different web view](http://ringtheory.herokuapp.com/commrings/ring/61/) with a different collection of properties :) – 2017-01-24
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    @rschwieb : ah sorry, I didn't check. That's fine! Anyway, thank you very much for your work, dear rschwieb! – 2017-01-24
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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 $\geq 2$. – 2017-01-26