"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?
 A: 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 $R\Gamma(X,-)$).
