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Let $R$ be a regular local ring (I am particularly interested in the case when $R$ is the local ring of a point on a smooth scheme of finite type over a field). Let $G$ be the etale fundamental group of $Spec R$, i.e., the profinite group classifying the proper etale morphisms into $Spec R$.

Then to any finite abelian group with a $G$-module structure one can assign an etale sheaf on $Spec R$; this is a kind of inverse image functor. Conversely, to any etale sheaf of $\mathbb Z/m$-modules on $Spec R$ one can assign a $G$-module over $\mathbb Z/m$; this is a kind of direct image.

What can be said about these two functors? Is the direct image exact in this case? If I start with a $G$-module, take its inverse image to $Spec R$, and then take the direct image, do I get the same $G$-module that I started with?

Perhaps, a more concrete question: do the etale cohomology of $Spec R$ coincide with the profinite group cohomology of $G$? Say, with coefficients in a finite module over $G$, or even with finite constant coefficients?

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  • $\begingroup$ Does Appendix A of Freitag-Kiehl "etale cohomology and the Weil conjecture", or Milne's "etale cohomology", answer your questions? I am thinking that $R$ being a regular local ring might be a red herring. The constructions you allude to above give an equiv of abelian cats between lcc sheaves of ab gps on $X$ and continuous $\pi_1(X)$-modules for $X$ an arbitrary connected Noetherian (and perhaps even that isn't necessary but I've never thought about the non-Noeth case) scheme. $\endgroup$
    – Kevin Buzzard
    Commented Nov 3, 2010 at 14:13
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    $\begingroup$ I think the answer to your concrete question will be 'yes' only if you assume the ring to be henselian (which is unlikely for the cases you seem to require). Otherwise, the topology of finite etale covers (i.e. the one corresponding to the site of $G$-sets) is a lot coarser than the topology of general etale covers, and the direct image functor from the latter to the former can have non-trivial cohomology. $\endgroup$
    – Keerthi Madapusi
    Commented Nov 3, 2010 at 15:39
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    $\begingroup$ Perhaps to summarize the discussion: finite G-modules will embed fully faithfully into etale sheaves on Spec(R), the essential image being those etale sheaves which are locally finite constant. These are not all as one can have, e.g., the pushforward of the constant sheaf from the complement of the closed point, which will be constructible but not constant. For your concrete question I will disagree with the previous commenters and say that my guess is "yes": the etale cohomology of Spec(R) with coefficients in such a locally finite constant sheaf will equal the cohomology of the [cont'd] $\endgroup$
    – Dustin Clausen
    Commented Nov 3, 2010 at 15:57
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    $\begingroup$ corresponding G-module because, by Artin's theorem on good covers, such a Spec(R) will be a K(pi,1) (at least in the geometric case). $\endgroup$
    – Dustin Clausen
    Commented Nov 3, 2010 at 15:57
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    $\begingroup$ My understanding of Clausen's comments is limited; I certainly cannot add anything to them. However, I think I now see that the functor $F_*$ is indeed not exact in the nonhenselian case. It suffices to take $R$ to be the local ring of a point on a curve over an algebraically closed field of characteristic zero. Then $\mathbb{Z}/n\to i_*\mathbb{Z}/n$ is an epimorphism of etale sheaves. The functor $F_*$ transforms it to the monomorphism $\mathbb{Z}/n\to \mathbb{Z}/n(G)$, where $\mathbb{Z}/n(G)$ is the group of locally constant $\mathbb{Z}/n$-valued functions on $G=\pi_1^{et}(R)$. $\endgroup$
    – Leonid Positselski
    Commented Nov 4, 2010 at 23:07

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