I was wondering for what kind of commutative rings we can always construct an infinite descending chain of distinct prime ideals?
When do infinite descending chains of prime ideals exist in commutative rings?
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2One 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 $I_k = (X_k, X_{k+1},\ldots)$. – 2012-03-14
2 Answers
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!
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 ...
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0Dear $G$eorges. Nothing childish $a$bout 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. – 2012-03-15