There isn't a "going in-between" theorem for integral ring extensions, is there?

Question

Let $A\subset B$ be an integral extension of commutative, unital rings.

We have the well-known "incomparability", "lying-over", and "going-up" theorems (5.9-5.11 in Atiyah-Macdonald, or see here). Lying-over asserts that every prime of $A$ has a prime of $B$ that contracts to it ("lies over" it). Incomparability asserts that two distinct primes of $B$ that lie over the same prime of $A$ are incomparable i.e. neither contains the other. Going-up asserts that if there is a chain of primes in $A$ and you fix a prime of $B$ lying over the minimal prime in this chain, then there exists a chain of primes in $B$ of the same length, each lying over the corresponding prime of $A$, with your fixed prime as the minimal member.

We also have the "going-down theorem" (Atiyah-Macdonald 5.16 or same link as before), but only if $A$ and $B$ are domains and $A$ is integrally closed. This is like the going-up theorem except you fix a prime of $B$ lying over the maximal prime of the chain in $A$, and extend downward instead of upward.

I just realized I may have been subtly assuming the existence of a "going-in-between theorem":

If $\mathfrak{q}_1\subset\mathfrak{q}_2$ are primes of $B$ and $\mathfrak{p}_1\subset\mathfrak{p}_2$ are their intersections with $A$, then incomparability implies that if $\mathfrak{p}_2$ covers $\mathfrak{p}_1$ (i.e. there are no primes properly between them), then the same is true about $\mathfrak{q}_2$ and $\mathfrak{q}_1$.

I thought that the converse was true as well, i.e. that if $\mathfrak{q}_2$ covers $\mathfrak{q}_1$, then $\mathfrak{p}_2$ covers $\mathfrak{p}_1$.

But I realize now that I believed this due to a faulty application of the going-up theorem. If there is a prime of $A$ between $\mathfrak{p}_1$ and $\mathfrak{p}_2$, going-up asserts there will be a prime of $B$ lying over it and containing $\mathfrak{q}_1$, but there is no guarantee that it will be inside $\mathfrak{q}_2$.

Going-up can guarantee it will be contained in some prime of $B$ lying over $\mathfrak{p}_2$, but not necessarily $\mathfrak{q}_2$.

Does it ever happen in an integral ring extension $A\subset B$ that a covering relation $\mathfrak{q}_1\subset\mathfrak{q}_2$ in $B$ intersects with $A$ to yield primes $\mathfrak{p}_1$ and $\mathfrak{p}_2$ with a properly intermediate prime $\mathfrak{p}_1\subset\mathfrak{p}^\star\subset \mathfrak{p}_2$? What's an example?

If it does happen, then how bad does a ring have to be for it to happen? Can it happen in a noetherian ring? Or is it ruled out somehow by the Krull height theorem? If it can happen in a noetherian ring, how about a catenary ring? (Obviously not a local catenary ring.) How about a Cohen-Macaulay ring?

Aside: This question is motivated by a technical point in an argument I gave in this question.

Second aside: It seems clear to me on geometric grounds that the proposed $\mathfrak{p}^\star$ cannot exist if $A$ and $B$ are affine coordinate rings over a field: for in this setting, if $\mathfrak{q}_1$ is covered by $\mathfrak{q}_2$, then $V(\mathfrak{q}_1)$ is an irreducible variety, and $V(\mathfrak{q}_2)$ is a codimension one irreducible subvariety, and $V(\mathfrak{p}_1)$ and $V(\mathfrak{p}_2)$ are their images under a dimension-preserving map, in which case what is $V(\mathfrak{p}^\star)$'s dimension? But this logic bundles a lot of nuanced dimension theory into a black box. This question is an attempt to open the box and examine the contents.

Answer 1

I'm surprised this never got properly answered! I am teaching commutative algebra this semester and just came across this same subtlety (in working out the argument for the catenary property of finitely-generated algebras over a field).

It sounds like there might be counterexamples in general. For finitely-generated algebras, though, the proof is actually quite short and direct, so I thought I would post it here. (Ratliff's paper contains results and assertions that imply it, using a bit more machinery and general theory.)

Proposition 1. Let $\phi : A \to B$ be an integral morphism of finitely-generated algebras over a field. Let $\mathfrak{q}_1 \subsetneq \mathfrak{q}_2$ be a strict minimal inclusion of prime ideals in $B$ and let $\mathfrak{p}_i := \phi^{-1}(\mathfrak{q}_i)$. Then $\mathfrak{p}_1 \subsetneq \mathfrak{p}_2$ is a strict, minimal inclusion.

Proof. Strictness follows by the incomparability property. By quotienting by $\mathfrak{q}_1$ and $\mathfrak{p}_1$ we can assume $\mathfrak{q}_1 = \mathfrak{p}_1 = 0$. Now $A \subseteq B$ is an integral extension of finitely-generated domains over a field, $\operatorname{ht} \mathfrak{q}_2 = 1$ and we want to show $\operatorname{ht} \mathfrak{p}_2 = 1$. Note that the usual way to get equality of heights is by the Going-Down property, which we may not have here. Fortunately:

Proposition 2. Let $A \subseteq B$ be an integral extension of finitely-generated domains over a field. Let $\mathfrak{q} \subseteq B$ be a prime ideal. Then $\operatorname{ht} \mathfrak{q} = \operatorname{ht} \mathfrak{q} \cap A.$

Proof. Let $k[x_1, \ldots, x_n] \subset A$ be a Noether normalization. Then we have $$\operatorname{ht} \mathfrak{q} \leq \operatorname{ht} \mathfrak{q} \cap A \leq \operatorname{ht} \mathfrak{q} \cap k[x_1, \ldots, x_n] = \operatorname{ht} \mathfrak{q}.$$ The two inequalities follow from the incomparability property. The equality is from the Going-Down property from $B$ to $k[x_1, \ldots, x_n]$. //

So, given $\mathfrak{p}_1 \subsetneq \mathfrak{p}^\star \subsetneq \mathfrak{p}_2$, we do indeed find some $\mathfrak{q}_1 \subsetneq \mathfrak{q}^\star \subsetneq \mathfrak{q}_2$. Note however that we have not asserted that $\mathfrak{q}^\star$ lies over $\mathfrak{p}^\star$.