most recent 30 from http://mathoverflow.net 2013-05-26T06:02:04Z http://mathoverflow.net/feeds/user/2223 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21067/noetherian-rings-of-infinite-krull-dimension moby 2010-04-12T03:22:25Z 2011-03-28T15:33:08Z

<p>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. </p> <p>The only example I know of is the one due to Nagata [1962]: 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.</p> <p>The question is: what are other examples of Noetherian rings of infinite Krull dimension?</p>

http://mathoverflow.net/questions/7389/what-are-the-most-overloaded-words-in-mathematics/7565#7565 moby 2009-12-02T06:38:43Z 2009-12-02T06:38:43Z

<p>Canonical... would be a canonical example. I guess.</p>

http://mathoverflow.net/questions/21067/noetherian-rings-of-infinite-krull-dimension moby 2010-04-12T05:23:55Z 2010-04-12T05:23:55Z

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?