# 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.

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.

• Thanks for the answer. What is $U(T)$? Commented Aug 10, 2012 at 3:35
• The set of units (ie invertible elements) in $T$. And you're welcome. :)
– user5137
Commented Aug 10, 2012 at 3:36
• Wonderful example. Commented Aug 10, 2012 at 3:39