A few years ago I started thinking about whether there was a natural rigid analytic version of the Artin-Grothendieck vanishing theorem. Last summer this grew into an obsession, and I managed to prove some general results. In particular, I showed that if is an affinoid rigid space over a complete algebraically closed field, AND comes via base change from an affinoid defined over a discretely valued subfield, then for all and all prime to the residue characteristic. I also proved a similar result with a non-constant coefficient sheaf, assuming moreover that the base field is of characteristic zero. This all got written up here.

Now, the hypothesis of definability over a discretely valued field is clearly stupid and shouldn’t be there, but I wasn’t able to remove it. So I was extremely happy this morning when Akhil Mathew and Bhargav Bhatt sent me an expanded version of their paper on arc-descent, in which they give a beautiful proof of rigid analytic Artin-Grothendieck vanishing without any superfluous assumptions. Their arguments are phrased in terms of algebraic geometry, rather than rigid analysis; in this post I want to recast (mostly for my own benefit I guess) the essential point of their argument in rigid analytic language.

The key is to prove the following.

**Theorem (Bhatt-Mathew).** *Let be an affinoid rigid space over a complete algebraically closed nonarchimedean field . Set where is any integer prime to the residue characteristic. Then .*

This implies the characteristic zero case of Conjecture 1.2 in my paper.

The proof proceeds in three steps.

**Step One: **Treat the case where is smooth. This was already done by Berkovich in the 90’s and I’ll take it for granted, although BM give their own nice argument for it. (Both arguments eventually appeal to the classical Artin-Grothendieck vanishing theorem.)

**Step Two: **Prove the weaker statement that in general.

For this we use induction on . I’ll assume for simplicity that has characteristic zero. Without loss of generality we can assume that is reduced. Then by excellence of affinoid algebras, we can pick some non-zero-divisor such that is regular. Fix a nonzero nonunit , and for any consider the rational subsets and inside . Set , so we get a Mayer-Vietoras distinguished triangle

for any . Note that and are smooth affinoids, so their etale cohomology is concentrated in degrees by Step One. Therefore, truncating the above Mayer-Vietoras sequence we get a quasi-isomorphism

for any . Moreover, in the sense of adic spaces, which implies that the etale cohomology of the left-hand side is the colimit of the etale cohomologies of the right-hand sides. Therefore, passing to the colimit over , the previous quasi-isomorphism gives a quasi-isomorphism

.

But now we win, because is an affinoid of dimension , so by the induction hypothesis its etale cohomology is concentrated in degrees .

**Step Three. **Bootstrap from the result of Step Two by a trick. More precisely, let and be as in the statement of the main theorem. By Step Two, we just have to show that . By another application of Step Two, the complex has cohomology in degree given by , and its enough to show that the latter module is zero. But

by the Kunneth formula*, and is a -dimensional affinoid, so its cohomology is concentrated in degrees by yet another application of Step Two. This gives the result.

*The necessary result is that if and are -affinoid spaces, then . I’m not sure if this is in the literature; Bhargav and Akhil prove (an algebraic form of) it in their paper. However, it is easy to deduce this from the results in Huber’s book. The point is that have canonical adic compactifications , and etale cohomology (with constant coefficients) doesn’t change if you replace or by its compactification. But then and are proper over (in the sense of Huber’s book), so it’s easy to show that

by the usual combination of proper base change and the projection formula.