# Noetherian rings of infinite Krull dimension?

Since Noetherian rings satisfy the ascending chain condition, every such ring must contain infinitely many chains of prime ideals s.t. the heights of these chains are unbounded.

The only example I know of is the one due to Nagata : we take a polynomial ring in infinitely many variables over a field, and consider the infinite collection of prime ideals formed by disjoint subsets of the variables. Then we localise the ring by the complement of the union of these prime ideals. With a little work, we can show that by appropriate choice of the subsets, the localised ring will be Noetherian and of infinite Krull dimension. Eisenbud (ex. 9.6) provides a good walkthrough.

The question is: what are other examples of Noetherian rings of infinite Krull dimension?

• The fact that Nagata had to come up with a fairly involved construction means that they probably don't come up in practice, so would you like us to look up other artificial examples? – Harry Gindi Apr 12 '10 at 5:05
• Such Noetherian rings are pathological by nature, so artificial examples are probably the way to go... working first from geometry, then to the related algebraic treatment? – moby Apr 12 '10 at 5:23
• The question could be improved by giving it more focus: one could presumably modify Nagata's construction in various small ways -- e.g. by starting with $\mathbb{Z}$ instead of a field -- but you are probably not interested in such examples. So, what kind of features are you looking for in other examples? E.g. "Does there exist a Noetherian ring of infinite Krull dimension such that...X?" By filling in X, you ask a binary question, which all of a sudden mathematicians are interested in answering. Just asking "What's out there?" doesn't have the same appeal. – Pete L. Clark Apr 12 '10 at 7:20
• The entire point of coming up with pathological examples is to show that certain conditions are necessary. Presumably the only motivation to come up with more pathological examples to illustrate necessity of a specific condition is if they are simpler to construct, or if you've come up with some sort of "construction scheme" that gives you a whole class of pathological examples. – Harry Gindi Apr 12 '10 at 9:14
• @Harry: Oftentimes one wants to prove a result for all noetherian rings and the infinite dimensional case is a sticking point. For example Lemma 3.1.5 in Brian Conrad's notes on Grothendieck duality has a very nice (and fairly involved) proof due to Gabber. The lemma is almost trivial when the ring has finite Krull dimension. So having examples at hand could be very helpful. – Jesse Burke Apr 12 '10 at 23:33

## 2 Answers

If you check the book "Krull Dimension" Memoirs of the American Mathematical Society 133 1973, you will find examples of commutative Noetherian integral domains of arbitrary infinite ordinal Krull dimension

• The complete reference: R. Gordon and J. C. Robson, Krull dimension, Memoirs Amer. Math. Soc, No. 133 (1973). – YCor Jan 12 '13 at 20:17

As you did not ask your ring to be commutative, you can probably take differential operators on your Nagata's example. You may want to look at the 1982 paper by Goodearl and Warfield where they construct a commutative ring and its differential operator ring, both of which have infinite Krull dimensions.