Edit: Badam Baplan gives a remarkable example below, which answers the question completely. I would still appreciate any other examples that others may come up with; in particular, is an example still possible if we ask for $R$ to be Noetherian?
Let $R$ be a (commutative, unital) ring and $I<R$ an ideal. Recall that we define $$\operatorname{ht}I=\min\{\operatorname{ht}P:P\in\operatorname{Spec}R,P\geqslant I\},$$ where $\operatorname{ht}P$ is defined in the usual way for a prime ideal $P$. Is there a nice example of an integral domain $R$ and an ideal $I$ such that (i) $I$ contains no non-zero prime ideals, and (ii) $\operatorname{ht}I>1$? Note: by Krull's height theorem, if $R$ is Noetherian then $I$ cannot be principal.
If we drop the condition that $R$ is an integral domain, then we can get examples where $\operatorname{ht}I$ is arbitrarily large. For instance, let $F$ be your favorite field, and define $A$ to be the polynomial ring $F[x_1,\dots,x_n]$. Choose any non-zero ideal $M\leqslant A$, with some $o\in M\setminus\{0\}$ of minimal degree. (For later, note that $o\notin x_1M+\dots+x_nM$; call this fact $(\star)$.) Now, consider the ring structure on the product $R:=A\times M$ defined by
- $(a,m)+(b,n)=(a+b,m+n)$
- $(a,m)\cdot(b,n)=(ab,an+bm)$
for all $a,b\in A$ and $m,n\in M$. Note that every element $(0,m)$ is nilpotent, so any prime ideal of $R$ must contain $0\times M$. Conversely, if $P$ is a prime ideal of $A$, then $P\times M$ is a prime ideal of $R$. With this in mind, consider the ideal $I<R$ generated by the elements $(x_i,0)$ for $1\leqslant i\leqslant n$. First note that $I$ contains no prime ideals; indeed, one can explicitly compute $I$ to be the product $$(x_1A+\dots+x_nA)\times(x_1M+\dots +x_nM),$$ where both direct factors are considered as ideals of $A$. In particular, by fact $(\star)$, we have $(0,o)\notin I$, and so (by the remark above, since $(0,o)$ is nilpotent), $I$ cannot contain any prime ideal.
On the other hand, consider the prime ideal $P:=\langle x_1,\dots,x_n\rangle\times M>I$. This is the unique minimal prime lying above $I$, since $P=I+(0\times M)$ and every prime must contain $0\times M$. However, we have $\operatorname{ht}P=n$, since $$0\times M<\langle x_1\rangle\times M<\langle x_1,x_2\rangle\times M<\dots<P$$ is a strictly ascending chain of primes below $P$. Thus, since $P$ is the unique minimal prime above $I$, we have $\operatorname{ht}I=n$.
One can extend this example in the natural way to find an example where $\operatorname{ht}I=\infty$. So this resolves the question completely if we do not require $R$ to be an integral domain. However, I'm struggling to find an example where $R$ is an integral domain. Can anyone think of a nice example?
I tried attempting to consider the example above as a quotient of an appropriate polynomial ring, but didn't make any progress. My thinking was that we could perhaps find a desirable prime lying below the kernel of the projection map, and then quotient by that to find a domain with some similar behavior. However, this doesn't seem to work out very nicely.
As a smaller follow-up question, what is the geometric significance (if any) of such a ring?
Best Answer
One of my favorite examples of this kind of wonky behavior originates from Krull's (false) conjecture that every completely integrally closed domain is the intersection of its one-dimensional valuation overrings. Nakayama disproved this conjecture in On Krull's conjecture concerning completely integrally closed integrity domains, I and II. He did this by constructing an integral domain $D$ which possesses the following properties:
So for Nakayama's purposes every valuation overring of $D$ is a localization of $D$ at a prime ideal and thus has infinite dimension.
For the purposes of your question, every nonzero principal ideal of $D$ has infinite height, but a principal ideal of $D$ never contains a prime ideal.
Indeed, for any Archimedean domain (i.e. $\bigcap d^nD = 0$ for all $d \in D$) it is is impossible for a prime to be properly contained in a principal ideal. Consequently a principal prime in an Archimedean domain necessarily has height $1$. Hence in an Archimedean domain with no finite height nonzero primes, it is impossible for a nonzero prime to be contained in a principal ideal.
(Updated to mention low-dimensional Noetherian examples) In the 1970's there was focused interest in the possible poset structure of the prime ideals of a Noetherian ring. One of the motivating questions was a specialization of your own: do finite intersections of maximal ideals contain prime ideals?
Among the advances in this period was a construction of Heitmann, which takes in a number of compatible 'special' Noetherian domains $D_\alpha$ with prime posets $P_\alpha$ and spits out a Noetherian domain having (up to order isomorphism) as its prime poset $\coprod P_\alpha$, the disjoint union of the $P_\alpha$ glued together at $(0)$.
This construction gives you a way to find $n$-dimensional Noetherian domain counterexamples to your question. For example, you can apply it to a couple of $n$-dimensional rings $D_1, D_2$. Pick a maximal ideal $M_i$ coming from the $P_i$ component. Clearly $M_1 \cap M_2$ is height $n$. But $M_1 \cap M_2$ cannot contain a nonzero prime, by disjointness of the $P_i$.
You can check out Heitmann's paper Prime ideal posets in Noetherian rings for details of the construction.