# What is the etale cohomology of G_m on Spec Z?

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.

• 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. – Meric Feb 2 at 20:55
• @Meric This is an answer (not a comment). – Martin Brandenburg Feb 3 at 0:24
• What is a non-trivial etale cover of Spec(Z)? Asking for a friend. – Yosemite Stan Feb 5 at 4:26
• 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.) – SebastianJB Feb 6 at 13:47
• 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. – Meric Feb 8 at 18:31