How badly can Krull's Hauptidealsatz fail for non-Noetherian rings? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T22:00:43Z http://mathoverflow.net/feeds/question/42510 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42510/how-badly-can-krulls-hauptidealsatz-fail-for-non-noetherian-rings How badly can Krull's Hauptidealsatz fail for non-Noetherian rings? Zev Chonoles 2010-10-17T15:15:04Z 2010-10-18T08:59:19Z <p>Krull's Hauptidealsatz (principal ideal theorem) says that for a Noetherian ring $R$ and any $r\in R$ which is not a unit or zero-divisor, all primes minimal over $(r)$ are of height 1. How badly can this fail if $R$ is a non-Noetherian ring? For example, if $R$ is non-Noetherian, is it possible for there to be a minimal prime over $(r)$ of infinite height?</p> http://mathoverflow.net/questions/42510/how-badly-can-krulls-hauptidealsatz-fail-for-non-noetherian-rings/42515#42515 Answer by Francesco Polizzi for How badly can Krull's Hauptidealsatz fail for non-Noetherian rings? Francesco Polizzi 2010-10-17T15:53:13Z 2010-10-17T15:53:13Z <p>I think that the answer is <strong>yes</strong>.</p> <p>Indeed, there are examples of integral domains $D$ such that every non-zero prime ideal of $D$ has infinite height.</p> <p>Look at the paper </p> <p>"Anti-archimedean rings and power series rings"</p> <p>D.D. Anderson; B.G. Kang; M H. Park</p> <p>Communications in Algebra, 1532-4125, Volume 26, Issue 10, 1998, Pages 3223 – 3238.</p> http://mathoverflow.net/questions/42510/how-badly-can-krulls-hauptidealsatz-fail-for-non-noetherian-rings/42601#42601 Answer by Hagen for How badly can Krull's Hauptidealsatz fail for non-Noetherian rings? Hagen 2010-10-18T08:59:19Z 2010-10-18T08:59:19Z <p>Valuation rings demonstrate quite clearly the failure of Krull's principal ideal theorem: take a valuation ring O of finite dimension. The prime ideals then form a chain </p> <p>$p_0:=0\subset p_1\subset\ldots\subset p_d$</p> <p>so that for every $i\in{1,\ldots ,d}$ there exists $r_i\in p_i\setminus p_{i-1}$. Obviously $p_i$ is a minimal prime over $r_iO$.</p> <p>For valuation domains of infinite dimension one has to consider the so-called limit-primes: a prime ideal $p$ of a commutative ring $R$ is called limit-prime if</p> <p>$p=\bigcup\limits_{q\in\mathrm{Spec} (R): q\subset p}q$.</p> <p>There exist valuation domains $O$ of infinite Krull dimension such that the maximal ideal $m$ of $O$ is no limit-prime. For example take a valuation ring such that the corresponding value group is</p> <p>$\mathbb{Z}\times\mathbb{Z}\times\ldots$ (countably many factors ordered lexigraphically).</p> <p>Then one can find $r\in m$ such that $m$ is minimal over $rO$.</p> <p>H</p>