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I'm trying to solve an exercise.

I should prove that if $R$ is a notherian ring and $\operatorname{Spec}(R)$ is discrete then $R$ is artinian.

I think it is enough to show that $\dim R=0$ since I can use that $R$ is artinian if and only if $R$ is noetherian and $\dim R=0$. Can you help me?

Thank you!

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3 Answers

You're on a good track, but try to think more about the discreteness. You want to say that there are no chains of prime ideals of length $1$; suppose that you had such a chain, say $\mathfrak p \subsetneqq \mathfrak q$. If $\operatorname{Spec}$ is discrete, then $\{\mathfrak p\}$ is a closed set. Is this possible? Recall what the closed subsets of $\operatorname{Spec}$ are defined to be in the Zariski topology.

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See Theorem 293 in $\S 13.3$ of these notes: among other things, it shows that the space $\operatorname{Spec} R$ is Hausdorff iff it is separated ("$T_1$") iff $R$ has dimension zero. Discrete spaces are Hausdorff, so there you go.

Note also that once you show that $R$ is Artinian, you will be able to conclude that $\operatorname{Spec} R$ is not only discrete but finite.

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those are very nice notes. Next time you upload an updated version of it to your site, could you please also include (in the PDF) the date when the document was last processed, so I know which version I have. –  Leon Lampret Oct 18 '11 at 16:38
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Suppose $\mathfrak{P}$ is prime, show that $\mathfrak{P}$ is maximal (Hint: $\mathfrak{P}$ is closed due to the topology on $\operatorname{Spec}(R)$).

What does that say about $\operatorname{Dim}(R)$?

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