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.
Best Answer
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.