TITLE: Nagata's bizzare examples
QUESTION [6 upvotes]: Hi,
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

REPLY [9 votes]: 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.