I apologize if this is straightforward, but I can't seem to find a reference: What is $H^1_{et}(\mathrm{Spec}(\mathbb{Z}), \mathbf{G}_m)$? It shouldn't be as simple as noting that the etale fundamental group of Spec Z is trivial, since Spec Z does admit non-trivial etale covers, just not finite ones.
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11$\begingroup$ It's much simpler: $H^1_{et}(X,\mathbf{G}_m)$ is always the Picard group $\mathrm{Pic}(X)$ for any scheme $X$. If $X = \mathrm{Spec}(R)$ for a PID R (e.g., $R = \mathbf{Z}$), then this group is trivial by algebra. $\endgroup$– MericCommented Feb 2, 2020 at 20:55
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5$\begingroup$ @Meric This is an answer (not a comment). $\endgroup$– Martin BrandenburgCommented Feb 3, 2020 at 0:24
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1$\begingroup$ What is a non-trivial etale cover of Spec(Z)? Asking for a friend. $\endgroup$– Yosemite StanCommented Feb 5, 2020 at 4:26
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$\begingroup$ The map $\mathrm{Spec}\, \mathbb{Z}[\sqrt{-1}, \frac{1}{2}] \times \mathbb{Z}[\frac{1 + \sqrt{5}}{2}, \frac{1}{5}] \to \mathrm{Spec} \, \mathbb{Z}$ works. Extensions of number rings are ramified along the vanishing of their discriminants, so if we delete that vanishing, we have an etale map. Doing this with a pair of number rings with relatively prime discriminants gives us a surjective etale map. (My algebraic number theory is rusty, so I'm using the table on en.wikipedia.org/wiki/… to build my specific example.) $\endgroup$– SebastianJBCommented Feb 6, 2020 at 13:47
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1$\begingroup$ You could also just look at Zariski covers, like $\mathbf{Z} \to \mathbf{Z}[1/2] \times \mathbf{Z}[1/3]$. These covers are enough to compute (via Cech cohomology) the relevant cohomology group anyways. $\endgroup$– MericCommented Feb 8, 2020 at 18:31
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