How badly can Krull's Hauptidealsatz fail for non-Noetherian rings?

Question

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?

Answer 1

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.

Answer 2

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