What is an example of a Noetherian Semi-local ring with an infinite number of prime ideals?

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.

• What does the notation $P$ in $P(|Specm(R)|)$ mean, actually? – Watson Jan 26 '17 at 13:32
• @Watson It is meant to denote the powerset – Pax Kivimae Jan 26 '17 at 22:28

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 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
• @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 ringtheory.herokuapp.com – for instance there is an amazing example of Noetherian ring with infinite Krull dimension…] – Watson Jan 24 '17 at 19:43