# Artin-Grothendieck vanishing, again

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 $X$ is an affinoid rigid space over a complete algebraically closed field, AND $X$ comes via base change from an affinoid defined over a discretely valued subfield, then $H^i(X,\mathbf{Z}/n)=0$ for all $i > \mathrm{dim}(X)$ and all $n$ 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 $\mathrm{Spa}A$ be an affinoid rigid space over a complete algebraically closed nonarchimedean field $K$. Set $\Lambda = \mathbf{Z}/n$ where $n$ is any integer prime to the residue characteristic. Then $R\Gamma(\mathrm{Spa}A,\Lambda) \in D^{\leq \mathrm{dim}A}(\Lambda)$.

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 $\mathrm{Spa}A$ 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 $R\Gamma(\mathrm{Spa}A,\Lambda) \in D^{\leq 1+\mathrm{dim}A}(\Lambda)$ in general.

For this we use induction on $\mathrm{dim}A$. I’ll assume for simplicity that $K$ has characteristic zero. Without loss of generality we can assume that $A$ is reduced. Then by excellence of affinoid algebras, we can pick some non-zero-divisor $f \in A$ such that $A[1/f]$ is regular. Fix a nonzero nonunit $\pi \in \mathcal{O}_K$, and for any $n \geq 1$ consider the rational subsets $U_n = \{ x\,with\,|f(x)| \geq |\pi|^n \}$ and $V_n = \{x\,with\,|f(x)| \leq |\pi|^n \}$ inside $\mathrm{Spa}A$. Set $W_n = U_n \cap V_n$, so we get a Mayer-Vietoras distinguished triangle

$R\Gamma(\mathrm{Spa}A,\Lambda) \to R\Gamma(U_n,\Lambda)\oplus R\Gamma(V_n,\Lambda) \to R\Gamma(W_n,\Lambda)\to$

for any $n$. Note that $U_n$ and $W_n$ are smooth affinoids, so their etale cohomology is concentrated in degrees $\leq \mathrm{dim}A$ by Step One. Therefore, truncating the above Mayer-Vietoras sequence we get a quasi-isomorphism

$\tau^{\geq \mathrm{dim}A+2}R\Gamma(\mathrm{Spa}A,\Lambda) \simeq \tau^{\geq \mathrm{dim}A+2}R\Gamma(V_n,\Lambda)$

for any $n$. Moreover, $\mathrm{Spa}(A/f) \sim \lim_{n} V_n$ 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 $n$, the previous quasi-isomorphism gives a quasi-isomorphism

$\tau^{\geq \mathrm{dim}A+2} R\Gamma(\mathrm{Spa}A,\Lambda) \simeq \tau^{\geq \mathrm{dim}A+2}R\Gamma(\mathrm{Spa}(A/f),\Lambda)$.

But now we win, because $A/f$ is an affinoid of dimension $\dim(A)-1$, so by the induction hypothesis its etale cohomology is concentrated in degrees $\leq \mathrm{dim}A$.

Step Three. Bootstrap from the result of Step Two by a trick. More precisely, let $X=\mathrm{Spa}A$ and $\Lambda=\mathbf{Z}/n$ be as in the statement of the main theorem. By Step Two, we just have to show that $H^{\mathrm{dim}+1}(X,\Lambda)=0$. By another application of Step Two, the complex $R\Gamma(X,\Lambda) \otimes_{\Lambda}^{\mathbf{L}} R\Gamma(X,\Lambda)$ has cohomology in degree $2\mathrm{dim}A+2$ given by $H^{\mathrm{dim}A+1}(X,\Lambda)^{\otimes 2}$, and its enough to show that the latter module is zero. But

$R\Gamma(X,\Lambda) \otimes_{\Lambda}^{\mathbf{L}} R\Gamma(X,\Lambda) \simeq R\Gamma(X \times X,\Lambda)$

by the Kunneth formula*, and $X \times X$ is a $2\mathrm{dim}A$-dimensional affinoid, so its cohomology is concentrated in degrees $\leq 2\mathrm{dim}A+1$ by yet another application of Step Two. This gives the result.

*The necessary result is that if $X$ and $Y$ are $K$-affinoid spaces, then $R\Gamma(X \times Y, \Lambda) \simeq R\Gamma(X,\Lambda) \otimes_{\Lambda}^{\mathbf{L}} R\Gamma(Y,\Lambda)$. 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 $X, Y$ have canonical adic compactifications $\overline{X},\overline{Y}$, and etale cohomology (with constant coefficients) doesn’t change if you replace $X$ or $Y$ by its compactification. But then $\overline{X}$ and $\overline{Y}$ are proper over $\mathrm{Spa}K$ (in the sense of Huber’s book), so it’s easy to show that

$R\Gamma(\overline{X} \times \overline{Y},\Lambda) \simeq R\Gamma(\overline{X},\Lambda) \otimes_{\Lambda}^{\mathbf{L}} R\Gamma(\overline{Y},\Lambda)$

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