# "Universal coefficent theorem" for pro-étale cohomology

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?

• 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. Jul 16 '20 at 0:58
• Would it be enough to assume that the structure morphism $\mathbb{Z}_\ell \rightarrow R$ is continuous? Jul 16 '20 at 1:26
• 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. Jul 16 '20 at 7:03
• @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. Jul 16 '20 at 13:51
• @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). Jul 16 '20 at 14:32

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 $$R\Gamma(X,-)$$).