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?
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NB: This is [crossposted from math.SE](https://math.stackexchange.com/questions/2160266/there-isnt-a-going-in-between-theorem-for-integral-ring-extensions-is-there), 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.



How bad does a ring have to be for a failure of "going-in-between"?
This is a partial answer, giving a two dimensional Noetherian counterexample. It starts from your observation that one has to consider the case where $A/\mathfrak{p}\_0$ is not integrally closed. Geometrically the construction is as follows: take a normal affine surface $Y$, a curve $C$ on $Y$, a point $P\_1 \in C$ and a point $P\_2 \in Y \setminus C$. Construct a non-normal surface $X$ by identifying $P\_1$ and $P\_2$; let $P'$ be the image of $P\_1$ and $P\_2$ on $X$ and $C'$ be the image of $C$ on $X$. Then the chains $X \supset C' \ni P'$ on $X$ and $Y \ni P\_2$ on $Y$ give a counterexample.


For an explicit example, take $Y = k^2$, where $k$ is any field, $C = x$-axis, $P\_1 = (0,0)$, $P\_2 = (0,1)$. So $B = k[x,y]$ and $A$ is the set of all polynomials $f \in B$ such that $f(0,0)= f(0,1)$. $B$ is integral over $A$, since e.g. $x \in B$ and $y(y-1) \in B$. 


