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I was wondering for what kind of commutative rings we can always construct an infinite descending chain of distinct prime ideals?

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One example is $R[X_1,X_2,\ldots]$ for a domain $R$. There we have the infinite descending chain of prime ideals $I_1 \supsetneq I_2 \supsetneq \ldots $ with $I_k = (X_k, X_{k+1},\ldots)$. –  marlu Mar 14 '12 at 16:07
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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!

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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:

  1. 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.

  2. 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 ...

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Dear @Hagen, what you write is very interesting, but I don't quite understand the relationship between your assertions (I must admit that I only just had a quick look at them as of now). In particular : is the result in the first line proved in the next ones? Even if this sounds childish to you, I would be very grateful if you could take a few minutes to write your answer in a somewhat more structured format: Statement1: "blah" Reason : " blah" (or: "see following literature"), Statement 2: "blah" or "follows from statement 1", etc. But, let me say it again: your answer is very informative:+1 –  Georges Elencwajg Mar 14 '12 at 23:47
    
Dear Georges. Nothing childish about your critics (in fact your feedback in the past has always been valuable) - even more because I think I was too quick and things are not as simple as I thought they are. So I will probably correct or at least improve my post in the next few hours. Right now I am busy with something else. –  Hagen Mar 15 '12 at 8:39
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