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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.

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  • $\begingroup$ You might wish to have a look at L.J.Ratliff Jr., 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
  • $\begingroup$ @JasonStarr - you're right. I think what I was thinking was that $\mathbb{q}_1$ lies over a prime that is strictly between $\mathfrak{p}_0$ and $\mathfrak{p}_2$. I'll edit the post. $\endgroup$ – benblumsmith Mar 13 '17 at 13:10
  • $\begingroup$ @JasonStarr - A propos of your correction, now it's not even completely clear to me in the geometric situation! Because the failure of $\mathfrak{q}_1$ to exist no longer implies that $\mathfrak{q}_0\subset\mathfrak{q}_2$ is saturated. $\endgroup$ – benblumsmith Mar 13 '17 at 13:36
  • $\begingroup$ @FredRohrer - Thank you, this looks extremely promising. $\endgroup$ – benblumsmith Mar 13 '17 at 13:45
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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 $y(y-1) \in B$.

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  • $\begingroup$ How do you check that $A$ is Noetherian? $\endgroup$ – R. van Dobben de Bruyn Dec 13 '19 at 19:04
  • $\begingroup$ $A$ is generated by $A'$ and $x$, where $A'$ is the subalgebra of $k[y] $ consisting of all polynomials $f$ such that $f(0) = f(1)$. $A'$ is finitely generated, since all subalgebras of $k[y] $ are. $\endgroup$ – auniket Dec 14 '19 at 2:09

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