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, 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][1], where it hasn't gotten any answers after 2 weeks and a bounty.

  [1]: http://math.stackexchange.com/questions/2160266/there-isnt-a-going-in-between-theorem-for-integral-ring-extensions-is-there