# Does every regular Noetherian domain have finite Krull dimension?

Background: A Noetherian ring is said to be regular if its localizations at all prime (or maximal) ideals are regular local rings. Without this assumption, there are counter-examples.

Thanks.

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All the examples of regular Noetherian domains are coordinate rings of some nice (finite-dimensional) variety, in which case the result is obvious. –  Alex Becker Jan 23 '13 at 7:46
@Alex: how about the ring $\mathbb Z$? (Not that it is infinite-dimensional, but I would not say "all teh examples":) –  Serge Lvovski Jan 23 '13 at 7:52
@Serge Whoops, I meant to say "I tend to think of". –  Alex Becker Jan 23 '13 at 9:08
It is not clear to me "without" which "assumption" do you mean? Do you mean without noetherian assumption? As far as I know, the definition of a regular local ring includes noetherian. Are you asking for a non-noetherian ring whose local rings are regular (thus noetherian) local rings? –  Mahdi Majidi-Zolbanin Jan 23 '13 at 14:01
No, I meant that without the regularity assumption, there are examples of noetherian domains of infinite Krull dimension. –  shenghao Feb 18 '13 at 13:18

No. An example is given in K. Fujita, Infinite dimensional Noetherian Hilbert domains, Hiroshima Math. J. 5 (1975), 181-185.

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How about the example of Nagata given in Exercise 4 of Chapter 11 of Atiyah-MacDonald? –  user28172 Jan 23 '13 at 8:19
@nosr: cf. Neil's answer and my comment on it. –  Fred Rohrer Jan 23 '13 at 14:20

Here's my favorite example (is this the one nosr refers to in his comment? I'm pretty sure it's also due to Nagata). Let $k$ be a field and $A=k[x_1, x_2, x_3, \ldots]$ be polynomial ring in countably many variables. Let $P_1 := (x_1)$, $P_2 := (x_2, x_3)$, $P_3 := (x_4, x_5, x_6)$, and in general $P_n$ is generated by the "next" $n$ variables. That is, $P_n := \left(x_{{n \choose 2} + 1}, x_{{n \choose 2} + 2}, \ldots, x_{{n+1} \choose 2}\right)$. Let $W := A \setminus \bigcup_{n=1}^\infty P_n$, and let $R := W^{-1}A$. Then every prime ideal of $R$ is in some $P_nR$, each of which is a maximal ideal of $R$, and $R_{P_n R} \cong k[y_1, \dotsc, y_n]_{(y_1, \dotsc, y_n)}$ is certainly a regular local ring. Hence $R$ is a regular Noetherian ring. But as it has essentially polynomial rings of every dimension as localizations, $R$ has infinite dimension.

On the other hand, every Noetherian local ring has finite Krull dimension.

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This is indeed (a special case of) the original Example 1 in the Appendix of Nagata's Local rings. Your last sentence might be confusing, as it remains true without the word "regular". –  Fred Rohrer Jan 23 '13 at 14:19
Good point, Fred. I have edited it accordingly. –  Neil Epstein Jan 23 '13 at 14:34
"Hence $R$ is a regular Noetherian ring" - Well, noetherian cannot be checked stalkwise, and here it really requires some work. –  Martin Brandenburg Jan 23 '13 at 14:54
@Martin: A slick way to handle the noetherian property in this example is to apply a nice result of Cohen: a commutative ring is noetherian if every prime ideal is finitely generated (see Theorem 3.4 in Matsumura's "Commutative Ring Theory" for a clever elementary proof). –  user28172 Jan 23 '13 at 15:01
Nagata's Lemma E1.1 (that yields noetherianness of the example in question) is similarly elementary as Matsumura's proof (or the original proof) of Cohen's Theorem, but it finishes the whole thing. –  Fred Rohrer Jan 23 '13 at 15:50