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?

EDIT: The answer is yes. See https://mathoverflow.net/questions/42510/how-badly-can-krulls-hauptidealsatz-fail-for-non-noetherian-rings

• you might want to "promote" this question to MO. It seems to be graduate/research level: e.g., I know a thing or two about commutative algebra but I don't know the answer to this offhand (I even looked in my own commutative algebra notes hopefully, but to no avail). – Pete L. Clark Oct 17 '10 at 12:32
• Done. Should I close here? – Zev Chonoles Oct 17 '10 at 15:27
• Told you you'd get an answer on MO. As for closing...I dunno what the convention is here. Maybe ask on meta? – Pete L. Clark Oct 18 '10 at 0:07

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