When do infinite descending chains of prime ideals exist in commutative rings? I was wondering for what kind of commutative rings we can always construct an infinite descending chain of distinct prime ideals? 
 A: I doubt that there is a crisp definitive answer to your question, but here are a few implications and, maybe more importantly, non-implications.
For the sake of concision, let us call strongly infinite dimensional a ring in which  there  exists an infinite strictly decreasing sequence of prime ideals  $\;\mathfrak p_0 \supsetneq \mathfrak p_1 \supsetneq \mathfrak p_2\supsetneq...$  
a) If a ring is  strongly infinite dimensional, it is infinite dimensional.
This is obvious.  
b) If a ring is noetherian it cannot  be strongly  infinite dimensional.
Indeed, by definition every ideal (prime or not) in a noetherian ring is generated by a finite number $r$ of elements.
And by  a generalization of Krull's Hauptidealsatz,  if a prime ideal $\mathfrak p$ is generated by $r$ elements, then $height(\mathfrak p)\leq r$.
Hence no prime ideal $\mathfrak p$ can be the beginning of an infinite  strictly descending chain of prime ideals.
c) If a ring is infinite dimensional, it needn't be strongly  infinite dimensional.
Indeed there exist examples (due to Nagata) of infinite dimensional noetherian rings, and we have just seen in b) that a noetherian ring cannot be  strongly  infinite dimensional.   
d) If a  ring is non-noetherian it needn't be strongly  infinite  dimensional.
For example consider a field $k$ and the non-noetherian ring $k[X_i\mid i\in \mathbb N]/\langle X_i\cdot X_j\mid i,j\in \mathbb N\rangle $.
This ring  is non-noetherian but since its only prime ideal is $\mathfrak m=\langle \bar X_i\mid i\in \mathbb N\rangle $, there is no risk that it could be strongly  infinite  dimensional!
A: This is a correction of my previous post.
The statement "For a local ring $O$ to be infinite dimensional and to be strongly infinite dimensional are equivalent." seems to be wrong.
What remains of my previous post seems to be this:


*

*Let $R$ be a strictly infinite dimensional ring, and let $S$ be an extension ring of $R$ such that for $S/R$ the Lying-Over-Theorem and the Going-Down-Theorem hold. Then $S$ is strictly infinite dimensional. In particular this holds if $R$ is an integrally closed domain, $S$ is a domain, and $S/R$ is an integral extension.
For $R$ we can for example take a polynomial ring in infinitely many variables over a field.

*Kang (Proc. AMS 126 (3), 1998) has shown that for any infinite chain of primes in a domain $R$, there exists a valuation ring $O\supseteq R$ of the field of fractions $K$ of $R$ which contains a chain of primes lying over the given one. We can conclude that if $R$ is a strictly infinite dimensional domain, then there exists a strictly infinite dimensional valuation domain $O$ of the field of fractions $K$ of $R$. General valuation theory then yields that the transcendence degree of the extension $K/k$, where $k$ is the prime field of $K$, is infinite. Consequently $R$ contains a polynomial ring in infinitely many variables over $k$ or over $\mathbb{Z}$. This last conclusion is rather weak, because we only use the fact that $O$ has infinite dimension, not the particular structure of the spectrum. There seems to be room for improvement here ...
