Krull's Hauptidealsatz (principal ideal theorem) says that for a Noetherian ring $R$ and any $r\in R$ which is not a unit or zerodivisor, all primes minimal over $(r)$ are of height 1. How badly can this fail if $R$ is a nonNoetherian ring? For example, if $R$ is nonNoetherian, is it possible for there to be a minimal prime over $(r)$ of infinite height?
2 Answers
I think that the answer is yes.
Indeed, there are examples of integral domains $D$ such that every nonzero prime ideal of $D$ has infinite height.
Look at the paper
"Antiarchimedean rings and power series rings"
D.D. Anderson; B.G. Kang; M H. Park
Communications in Algebra, 15324125, Volume 26, Issue 10, 1998, Pages 3223 – 3238.

1$\begingroup$ Thanks! Amazingly, your answer also takes care of my motivation for this question. I had been wondering how a ring might have every prime be of the same height, and it's clear that the only options are 0 and $\infty$; but since there must always be minimal primes, $\infty$ is technically impossible, but we can (WLOG) let 0 be a prime of height 0 and see if we can get the rest to be the same. So I wanted to check that even primes we'd normally expect to be small (e.g. minimal primes over a principal ideal) could still be of infinite height. $\endgroup$ Oct 17, 2010 at 16:26
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
$p_0:=0\subset p_1\subset\ldots\subset p_d$
so that for every $i\in\{1,\ldots ,d\}$ there exists $r_i\in p_i\setminus p_{i1}$. Obviously $p_i$ is a minimal prime over $r_iO$.
For valuation domains of infinite dimension one has to consider the socalled limitprimes: a prime ideal $p$ of a commutative ring $R$ is called limitprime if
$p=\bigcup\limits_{q\in\mathrm{Spec} (R): q\subset p}q$.
There exist valuation domains $O$ of infinite Krull dimension such that the maximal ideal $m$ of $O$ is no limitprime. For example take a valuation ring such that the corresponding value group is
$\mathbb{Z}\times\mathbb{Z}\times\ldots$ (countably many factors ordered lexigraphically).
Then one can find $r\in m$ such that $m$ is minimal over $rO$.
H

$\begingroup$ +1 Thanks for your enlightening examples. What would be the best place to learn more about notnecessarilyNoetherian valuation rings? $\endgroup$ Oct 20, 2010 at 1:40

1$\begingroup$ Valuation domains are frequently studied via the associated Krull valuations, so that any book on valuation theory can be used. Personally I learned a lot from Otto Endler's Valuation Theory and ZariskiSamuel, Commutative Algebra Vol. 2. Moreover Robert Gilmer's book on Multiplicative Ideal Theory and FuchsSalce, Modules over valuation domains. H $\endgroup$– HagenOct 20, 2010 at 8:08