In algebraic topology, for any space with finite homology type, the universal coefficient theorem states that for any abelian group $G$, we have $$H^n(X,G)\cong \left( H^n(X,\mathbb{Z})\otimes G\right)\oplus \text{Tor}_1(H^{n+1}(X,\mathbb{Z}),G).$$ My question is whether the analogous statement is true for the pro-étale cohomology, namely if $R$ is a $\mathbb{Z}_\ell$-algebra, do we have $$H^n_{proét}(X,\underline{R})\cong \left(H^n_{proét}(X,\underline{\mathbb{Z}_\ell})\otimes R\right)\oplus \text{Tor}_1(H_{proét}^{n+1}(X,\underline{\mathbb{Z}_\ell}),R)$$ for a sufficiently nice scheme? I'm mostly interested in the case of a smooth, projective scheme over some algebraically closed field (possibly of positive characteristic). Also, would this decomposition respect that Galois action on the cohomology?
-
$\begingroup$ Note: $\mathbf Z_\ell$ needs to be endowed with a topology for it to give the 'right' cohomology in the pro-étale site. So there should probably be a natural topology on $R$ as well (somehow compatible with the one on $\mathbf Z_\ell$) for this to have any chance of success. $\endgroup$– R. van Dobben de BruynCommented Jul 16, 2020 at 0:58
-
$\begingroup$ Would it be enough to assume that the structure morphism $\mathbb{Z}_\ell \rightarrow R$ is continuous? $\endgroup$– curious math guyCommented Jul 16, 2020 at 1:26
-
$\begingroup$ I do not think it is reasonable to consider any topology on $R$: the point of this formula is to compare the (derived) tensor product with $R$ within the pro-étale topos with the (derived) tensor product externally (i.e. within the ordinary derived category of $\mathbf{Z}_\ell$-modules), where everything is discrete. $\endgroup$– D.-C. CisinskiCommented Jul 16, 2020 at 7:03
-
$\begingroup$ @Denis-CharlesCisinski I don't understand what you mean. If you take $R = \mathbf Z_\ell$ with the discrete topology (i.e. the true constant sheaf), then you should get something very different than the usual $\mathbf Z_\ell$ cohomology. $\endgroup$– R. van Dobben de BruynCommented Jul 16, 2020 at 13:51
-
$\begingroup$ @R.vanDobbendeBruyn I mean $R$ should be discrete, not $\mathbf{Z}_\ell$. This is what happen when you go from $\mathbf{Z}_\ell$-linear coefficients to $\mathbf{Q}_\ell$-linear ones: $\mathbf{Q}_\ell=\mathbf{Z}_\ell\otimes\mathbf{Q}$ with $\mathbf{Q}$ discrete. My point is that, in the "universal coefficients theorem", there is a tensor/tor with cohomology groups considered as discrete objects, so that $R$ has to be discrete. My answer below explains how to proceed (I underline the continuous $\mathbf{Z}_\ell$ to differientiate from the discrete one). $\endgroup$– D.-C. CisinskiCommented Jul 16, 2020 at 14:32
1 Answer
The only condition you need on $X$ for such a formula to hold is that it is coherent (=quasi-compact and quasi-separated).
Let $R$ be a discrete ${\mathbf{Z}_\ell}$-module. We consider the sheaf $\underline{\mathbf{Z}}_\ell$ on the pro-étale site defined as the limit of the constant sheaves $\mathbf{Z}/\ell^i\mathbf{Z}$. This is an algebra on the constant sheaf associated to the discrete ring $\mathbf{Z}_\ell$. Hence we may define $\underline{R}$ as the tensor product of $R$ with $\underline{\mathbf{Z}}_\ell$ (exercise: it is in fact the derived tensor product, since $\underline{\mathbf{Z}}_\ell$ has no $\ell$-torsion stalkwise; this is where pecularities of the pro-étale site have a role to play, in a proof which is otherwise very formal). Let $R\Gamma(X,-)$ denote the derived global sections on the pro-étale topos of $X$. To prove that the canonical map $$R\Gamma(X,\underline{R})\leftarrow R\Gamma(X,\underline{\mathbf{Z}}_\ell)\otimes^L_{\mathbf{Z}_\ell}R$$ is invertible in the derived category of $\mathbf{Z}_\ell$-modules, assuming that $X$ is coherent, we may assume that $R$ is of finite type because $R\Gamma(X,-)$ commutes with filtered colimits (this is precisely what coherence is good for). If $R=S\oplus T$, it is sufficient to prove it for $S$ and $T$ separately. Hence, without loss of generality, we may assume that $R=\mathbf{Z}_\ell$, in which case this is trivial, or that $R=\mathbf{Z}/\ell^i\mathbf{Z}$, in which case this is always true as well (taking the cone of the multiplication by $\ell^i$ commutes with $R\Gamma(X,-)$).
