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

Let $\mathfrak{p}_0\subset\mathfrak{p}_1\subset\mathfrak{p}_2$ be a saturated chain of primes in $A$ of length $2$.

Suppose $\mathfrak{q}_0,\mathfrak{q}_2$ lie over $\mathfrak{p}_0,\mathfrak{p}_2$, and $\mathfrak{q}_0\subset\mathfrak{q}_2$.

Is there necessarily a $\mathfrak{q}_1$ satisfying $\mathfrak{q}_0\subset\mathfrak{q}_1\subset\mathfrak{q}_2$ and lying over $\mathfrak{p}_1$?

It seems to me the answer is clearly *yes* if the rings $A,B$ are sufficiently geometric (**edit:** even this is no longer clear to me, see addendum), e.g. finitely generated algebras over an algebraically closed field, since in this case, if there is no $\mathfrak{q}_1$, then $\mathfrak{q}_0\subset\mathfrak{q}_2$ is saturated, and then $V(\mathfrak{q}_2)$ is a codimension one subvariety of $V(\mathfrak{q}_0)$, and $\operatorname{Spec}B\rightarrow\operatorname{Spec}A$ is a dimension-preserving map, so what is $V(\mathfrak{p}_1)$'s dimension?

But in general, it's not obvious to me. It seems to require that going-down holds in the integral extension of domains $A/\mathfrak{p}_0\subset B/\mathfrak{q}_0$, and because the going-down theorem requires an extra assumption of integral closure, shouldn't this fail sometimes?

So my question is:

How bad do $A,B$ have to be for $\mathfrak{q}_1$ to fail to exist? Can it happen for noetherian rings? Cohen-Macaulay rings? What's the "least pathological" example?

NB: This is crossposted from math.SE, where it hasn't gotten any answers after 2 weeks and a bounty. (**Edit:** it is no longer really the same question, see addendum.)

**Addendum:** A propos of an exchange in the comments with Jason Starr (partially deleted now), even in the geometric case it's no longer clear to me that the answer is yes. When I originally posted, I mistook the fact that a prime between $\mathfrak{q}_0$ and $\mathfrak{q}_2$ would necessarily lie over *some* prime strictly between $\mathfrak{p}_0$ and $\mathfrak{p}_2$ to imply it would lie over $\mathfrak{p}_1$. (Jason pointed out this mistake.) I corrected this by explicitly adding the requirement that $\mathfrak{q}_1$ lie over $\mathfrak{p}_1$, but now it is no longer true that the failure of $\mathfrak{q}_1$ to exist implies that $\mathfrak{q}_0\subset\mathfrak{q}_2$ is saturated. Thus, perhaps $V(\mathfrak{q}_2)$ is codimension 2 in $V(\mathfrak{q}_0)$, it's just that none of the codimension one subvarieties of $V(\mathfrak{q}_0)$ happen to lie over $V(\mathfrak{p}_1)$. It's not at all clear to me that this can't happen.

Fred Rohrer mentions a highly relevant paper by L.J. Ratliff in the comments. A cursory scan of this paper seems to indicate that the answer might be *no* most of the time. Ratliff states that the condition I'm asking for here is not even guaranteed if $A$ is a complete regular local ring (bottom of p. 778), though it's not completely clear from context whether he's requiring $B$ to be integral in this case. What's an example?

One last point is that these developments distinguish this question from the one I asked at math.SE, where I asked, "In what generality does it hold that if $\mathfrak{q}_0\subset\mathfrak{q}_2$ is saturated, and $\mathfrak{q}_i$ lies over $\mathfrak{p}_i$ in $A$, then $\mathfrak{p}_0\subset\mathfrak{p}_2$ is saturated?" This turns out (if I am thinking straight) to be the question addressed by Ratliff's paper, but I am asking for something stronger at present.

Going between rings and contractions of saturated chains of prime ideals,Rocky Mountain J. Math. 7 (1977), 777-787. (I have not studied this in detail myself.) $\endgroup$ – Fred Rohrer Mar 13 '17 at 12:46