Let $R$ be a ring (say noetherian of finite Krull dimension, possibly with additional hypotheses) henselian along the ideal $(p)$, and let $\hat{R}$ be the $p$-adic completion. Is it true that the étale cohomology of $R[1/p]$ and $\hat{R}[1/p]$ with mod $p$ coefficients coincide? I believe that this should be true (for $K$-theoretic reasons), but I was wondering if there is a direct argument.

  • $\begingroup$ @BenLim. Presumably when the OP writes "$p$", the OP intends the prime integer $p$. Thus, the fraction field $K$ is characteristic $0$, and every field extension of $K$ is separable. $\endgroup$ – Jason Starr Mar 10 '18 at 9:25

TL;DR: Your expectation is right. In fact, there is a third object to compare with $R[1/p]$ and $\hat R[1/p]$, the affinoid rigid space ${\rm Spf}(\hat R)^{\rm rig}$. The cohomology comparison is given by the Gabber-Fujiwara theorem: see Corollary 6.6.4 in [1].

Let us consider the following general setup: let $(A, I)$ be a henselian couple with $A$ noetherian (i.e. $A$ is henselian along $I$), and let $\hat A$ be the $I$-adic completion. We set $$X={\rm Spec}\, A, \quad \hat X = {\rm Spec}\, \hat A,$$ $$U= X - V(I), \quad \hat U = \hat X - V(\hat I) \quad \text{ where }\hat I = I\cdot \hat A.$$ Let $\varepsilon\colon \hat U\rightarrow U$ be the canonical map.

The following is a classical theorem of Elkik.

Theorem 1 (Cor. p. 579 in [2]). The restriction functor $$ \varepsilon^*\colon(\text{finite etale covers of }U)\to (\text{finite etale covers of }\hat U) $$ is an equivalence of categories.

Equivalently, for any finite group $G$, the restriction map $H^1(U, G)\to H^1(\hat U, G)$ is bijective, or $\pi_1(\hat U)\to \pi_1(U)$ is an isomorphism.

The Gabber-Fujiwara theorem (see [1], Cor. 6.6.3) is a far-reaching generalization of this, treating the higher cohomology of arbitrary torsion etale sheaves. Actually, the theorem compares the cohomology of $U$ to the cohomology of the rigid space $U^{\rm rig} = {\rm Spf}(\hat A)^{\rm rig}$. Whatever this object is, it is clear from the notation that $U^{\rm rig}$ depends only on $(\hat A, \hat I)$, and in particular we get the following as a corollary (see [1], Cor.6.6.4):

Theorem 2 (The formal base change theorem). Let $\mathscr{F}$ be an etale sheaf of sets (resp. torsion groups, resp. torsion abelian groups) on $U$. Then the pull-back map $$ \varepsilon^*\colon H^q(U, \mathscr{F})\longrightarrow H^q(\hat U, \varepsilon^* \mathscr{F}) $$ is an isomorphism for $q=0$ (resp. for $q\leq 1$, resp. for $q\geq 0$).

In particular, at least in good cases (finite cohomological dimension, geometrically unibranch) the Artin-Mazur etale homotopy types of $U$ and $\hat U$ are equivalent. (Disclaimer: as written, [1] treats only abelian cohomology, but I'm sure the nonabelian statements can be easily deduced from Elkik).

Addendum. A forthcoming book by Abbes (a sequel to [3]) is expected to contain a new proof of the Gabber-Fujiwara theorem, based on Gabber's affine analog of the proper base change theorem.

Curiosity. In fact in the $p$-adic situation you describe, both cohomology groups agree with the cohomology of the fundamental group: Theorem 6.7 in here states that for every noetherian $\mathbf{Z}_{(p)}$-algebra $A$ such that $(A, pA)$ is a henselian pair, the scheme $X={\rm Spec}\, A[1/p]$ is a $K(\pi, 1)$. For $\mathbf{F}_p$-coefficients, this can be deduced (via Gabber-Fujiwara) from earlier work of Scholze (Theorem 4.9 in [4]).

[1] Fujiwara, K.: Theory of tubular neighborhood in étale topology. Duke Math. J. 80 (1), 15–57 (1995).

[2] Renee Elkik, Solutions d’equations a coefficients dans un anneau henselien, Ann. Sci. Ecole Norm. Sup. (4) 6 (1973), 553–603 (1974). MR 0345966 (49 #10692)

[3] Abbes, A.: Éléments de géométrie rigide. Volume I. Progress in Mathematics, vol. 286, Birkhäuser/Springer Basel AG, Basel (2010).

[4] Scholze, P.: $p$-adic Hodge theory for rigid analytic varieties. Forum Math. Pi 1, e177 (2013)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.