How badly can Krull's Hauptidealsatz fail for non-Noetherian rings? 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?
 A: 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 $r_i\in p_i\setminus p_{i-1}$. Obviously $p_i$ is a minimal prime over $r_iO$.
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
$p=\bigcup\limits_{q\in\mathrm{Spec} (R): q\subset p}q$.
There exist valuation domains $O$ of infinite Krull dimension such that the maximal ideal $m$ of $O$ is no limit-prime. 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
A: I think that the answer is yes.
Indeed, there are examples of integral domains $D$ such that every non-zero prime ideal of $D$ has infinite height.
Look at the paper 
"Anti-archimedean rings and power series rings"
D.D. Anderson; B.G. Kang; M H. Park
Communications in Algebra, 1532-4125, Volume 26, Issue 10, 1998, Pages 3223 – 3238.
