due to Nagata and his clever and bizzare examples I'm unsure in this:

1) Is there a regular ring of infinite Krull dimension?

2) Is it true that: Regular ring of finite Krull dimension = Commutative Noetherian ring of finite global dimension? (answer is yes if R is local by Serre's Theorem)

Regards, David


1 Answer 1


For 1), Nagata's counterexample goes as follows:

Let $\mathbb{N}=A_1\cup A_2 \cup \cdots$ be a partition such that $|A_i|<|A_{i+1}|$. Take $S=k[x_1,x_2,\ldots]$, prime ideals $I_i=\langle x_i | i\in A_i \rangle$ and consider the localization $S[U^{-1}]$ in the multiplcative set $U=S \setminus \bigcup_i I_i$. Then the height of each of the $I_i$ is $|A_i|$ and so $S[U^{-1}]$ has infinite Krull dimension. It not so hard to show that the ring is regular, but a more surprising point is that it is also Noetherian.

For 2) I believe the answer is yes by this reference.

  • $\begingroup$ And why is this noetherian? $\endgroup$
    – David
    Jan 11, 2011 at 18:03
  • $\begingroup$ The ring is in fact noetherian, but showing this is non-trivial (for a reference, see Nagata's original paper). I edited my answer above. $\endgroup$
    – J.C. Ottem
    Jan 11, 2011 at 18:32

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