Cohomology of rigid-analytic spaces

Question

Let $R$ be a complete discrete valuation ring and let $K$ be its field of fractions. Suppose $X$ is a smooth rigid-anaytic space over $K$. Often it is convenient to have a model of $X$ whose reduction has singularities which are as mild as possible--a semistable model. This amounts to having an admissible covering of $X$ by open affinoids $X_i$, each of which has good reduction, such that the reductions of any pair $X_i$ and $X_j$ meet transversally, if at all. (See the paper of Bosch/Lütkebohmert for definitions.) Let us assume such a semi-stable model of $X$ exists. Then the étale cohomology of $X$ can be computed from the combinatorics of the covering $X_i$, together with the étale cohomology of each $X_i$, via the weight spectral sequence of Rapaport-Zink.

Now suppose I have an open affinoid $Z\subset X$ which happens to have good reduction. My question is: Does there admit a semi-stable model of $X$ for which $Z$ belongs to the covering? Failing this, is there some sense one can make of my intuition that the cohomology of the reduction of $Z$ ought to contribute to the cohomology of $X$?

Feel free to edit/criticize my question to smithereens if you like.

Answer 1

Here's a first pass at your question; hopefully it will suggest something more definitive.

Let's imagine we were in the simplest case, where $X$ is a disk, with its smooth model being the formal affine line over $R$, and that $Z$ was the sub-disk of elements of absolute value less than or equal the absolute value of the uniformizer. Then we can find a semistable model in which $Z$ is one of the covering opens, by blowing up the formal affine line at the origin.

So in this test case, the answer seems to be yes .

Now in general, I think that Raynaud (and/or his collaborators or those who followed in his tradition) will say that the open immersion $Z \rightarrow X$ extends to an open immersion of formal models. So we can blow up the smooth model of $X$ and the smooth model of $Z$ so that the latter sits inside the former. What I'm not very certain about is how much you can control the nature of these blow-ups. (Presumably not at all in general, but you're starting in a fairly nice situation.)

Have you tried asking Brian Conrad yet?

Answer 2

In dimension 1, the answer should be yes because the semi-stable reduction is well understood. The only problem would be the difference between the rigid analytic reduction and the algebraic reduction as a formal covering defines a reduced analytic reduction but non necessarily reduced algebraic reduction (= special fiber of the formal scheme associated to the formal covering). The answer should be in Bosch-Lütkebohmert's paper on the stable reduction of curves and in Fresnel-van der Put's book (the second one).

In higher dimension, I am a litte skeptical. As Emerton pointed out, you could construct a model containning $\overline{Z}$ in its reduction at least birationally (result of Raynaud, explained in Melhmann's thesis in Münster). But it is not clear for me whether a semi-stable model dominating a given one model exist. Another difficulty in higher dimension is that the semi-stable reduction is not unique even in good reduction case (non-uniqueness of minimal birational model). However, you have two stable reductions whose irreducible components are all not uniruled, then there is a one-one birational correspondance between the components of two stable reductions (Abhyankar's lemma, I don't remember whether this requires desingularization).