The height of a principal prime ideal 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.
 A: 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 $$R=K[x,y/x,y/x^2,y/x^3,...],$$
i.e., 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 subset $S$ of $R$ containing all elements with nonzero constant term. It is clear that $S$ is a multiplicative subset of $R$, and let $T=R_S$ be the localization of $R$ at $S$--i.e., $$T=\left\{\frac{f(x,y)}{g(x,y)}\,\vert\,f(x,y)\in R, g(x,y)\in S\right\}.$$ 
It isn't 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 isomorphic to $\mathbb{Z}^n$ under the lexicographic ordering.
