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Here's my dilemma: I'm trying to do some work with the pro-étale topology and would like to compute very basic instances of pro-étale cohomology to see what I can extract from a pro-étale Kummer exact sequence (I'd also like to get my hands dirty to get started with the pro-étale topology). The classical Kummer exact sequence is given at finite level by $$1\longrightarrow \mathcal{O^*}(X)/\mathcal{O^*}(X)^n\longrightarrow H^1(X,\mu_n)\longrightarrow Pic_n(X)\longrightarrow 1$$ for $X$ defined over an algebraically closed field of characteristic zero. This follows from the long exact sequence in cohomology. For the pro-étale setting, $H^1$ should still be equivalent to line bundles/$\mathbb{G}_m-$torsors (I don't see how Tag 03P7 wouldn't also hold for the pro-étale site): so $H^1_{proét}(X)=Pic(X)$. But alas, I'm not sure if $H^2_{\text{proét}}(X,\mathbb{G}_m)=0$. I'm also wondering, what do we know about $H^1_{proet}(X,\mathbb{Z})$? I know this vanishes in etale cohomology whenever the etale fundamental group is profinite (always) (this follows from the equivalence of $H^1$ and continuous homs). So when $X$ is geometrically unibranch, $H^1_{proet}(X,\mathbb{Z})=0$ because in that case etale and pro-etale $\pi_1$ coincide. But I have no idea what holds in general. Maybe more generally, is there some relationship between $H_{et}$ and $H_{proet}$ that we can exploit to compute one or the other?

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  • $\begingroup$ I think that $H^1_{\text{proét}}(X,\mathbf Z)$ should be nonzero when $X$ is a nodal cubic, since (the pro-discrete completion of?) its pro-étale fundamental group is the discrete group $\mathbf Z$ (by Lemma 7.4.6 in Bhatt–Scholze). $\endgroup$ Commented Apr 13 at 20:37
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    $\begingroup$ There is a fully faithful embedding of etale sheaves into pro-etale sheaves, and this preserves cohomology, see Proposition 5.2.6 (2) of The pro-etale topology for schemes. This applies in particular to $\mathbb Z$ or $\mathbb G_m$, so their pro-etale cohomology agrees with the usual etale cohomology. The pro-etale $\pi_1$ is a red herring for these questions. $\endgroup$ Commented Apr 15 at 11:13

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