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A formal consequence of Krull's principal ideal theorem is the following:

If $A$ is a Noetherian ring, and $I$ is an ideal generated by $r$ elements, then any prime ideal which is minimal among those that contain $I$ has height at most $r$.

This statement implies that for Noetherian rings, principal prime ideals have height at most $1$.

My question is if this is true for any ring, i.e., is a principal prime ideal of a ring always of height at most $1$?

The above question is clearly true if the following statement is true: If every maximal ideal of a ring is finitely generated, then the ring is Noetherian. (Note that this statement is true if we replace maximal ideals by prime ideals)

However, I am not sure if this latter assertion is true, although I do not have a counterexample for it.

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This latter assertion isn't true as you can see here math.stackexchange.com/questions/183199/… –  user26857 Oct 18 '12 at 18:38

1 Answer 1

up vote 7 down vote accepted

Actually, no. In any valuation domain, the prime ideals are linearly ordered by inclusion, so there exists at most one nonzero principal prime ideal.

In particular, let $K$ be any field, and let $$R=K[x,y/x,y/x^2,y/x^3,...],$$

ie elements of $R$ are "polynomials" in $x$ and $y$ over $K$, except you can divide $y$ by $x$ as many times as you like.

Consider the following subset $S$ of $R$ containing all elements with nonzero constant term. It is clear that $S$ is a multiplicative subset of $R$, so let $T=R_S$ be the localization of $R$ at $S$--ie $$T=\left\{\frac{f(x,y)}{g(x,y)}\,\vert\,f(x,y)\in R, g(x,y)\in S\right\}.$$

It isn't too difficult to show that any nonzero element of $T$ is of the form $ux^n y^m$ where $u\in U(T)$, $m\geq 0$, and if $m=0$ then $n\geq 0$ (basically, take an arbitrary nonzero element of $T$ and factor out all of the $x$'s and $y$'s that you can).

So, we have the following chain of prime ideals in $T$ (and, in fact, these are all the prime ideals of $T$): $$0\subsetneq (y,y/x,y/x^2,y/x^3,\cdots)\subsetneq xT$$

You can generalize this to a chain of length $n$ by taking a valuation domain with value group order isomorphic to $\mathbb{Z}^n$ under the lexicographic ordering.

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Thanks for the answer. What is $U(T)$? –  Rankeya Aug 10 '12 at 3:35
    
The set of units (ie invertible elements) in $T$. And you're welcome. :) –  user5137 Aug 10 '12 at 3:36
2  
Wonderful example. –  Rankeya Aug 10 '12 at 3:39

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