This question is related to my post [Interpretation of some maps involving cohomology groups](https://mathoverflow.net/questions/385059/interpretation-of-some-maps-involving-cohomology-groups?noredirect=1#comment980961_385059).


$C$ is a smooth geometrically integral affine curve over a number field $k$, and $C\_1$ is its smooth completion. We focus on the case where $C\_1$ differs from $C$ by a point (e.g., $C$ can be an elliptic curve). Denote the absolute Galois group of $k$ by $\Gamma\_k$.


By a comment by @abx, we obtain an exact sequence $$1 \rightarrow \bar{k}^\* \rightarrow \bar{k}[C]^\* \rightarrow \mathbb{Z} \rightarrow 0.$$


We have only one copy of $\mathbb{Z}$ since only one point is missing on $C$.


The long cohomology sequence associated to this exact sequence gives us
$$H^1(k,\mathbb{Z}) \rightarrow \mathrm{Br}(k) \rightarrow H^2(k,\bar{k}[C]^\*) \rightarrow H^2(k, \mathbb{Z}).$$


In this case the second map is injective. To see why, recall that the cohomology of $\mathbb{Z}$ arises from the exact sequence
$$0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Q} \rightarrow \mathbb{Q}/\mathbb{Z} \rightarrow 1$$ which has trivial $\Gamma\_k$-action. Therefore $H^1(k,\mathbb{Z}) = \mathrm{Hom}(\Gamma\_k,\mathbb{Z})$. This group is trivial since $\mathbb{Z}$ is torsion-free and thus for each finite extension $L$ of $k$ contained in $\bar{k}$, the group $\mathrm{Hom}(\mathrm{Gal}(L/k),\mathbb{Z})$ is zero.


To interpret the elements in $H^2(k,\mathbb{Z})$, we use the fact that $\mathbb{Z}$ has trivial Galois action and so any class of 2-cocycle $f \in H^2(k,\mathbb{Z})$ is of the form $\Gamma\_k \times \Gamma\_k \rightarrow \mathbb{Z}$ satisfying


$$f(g,hk)+f(h,k) = f(gh,k)+f(g,h).$$


This doesn't seem very helpful in understanding the group, unlike the 1-cocycle case above. However, there are some well-known cases. For example, if $K$ is a local field and $L/K$ is a finite unramified extension with Galois group $G$, or when $G$ is the profinite completion of $\mathbb{Z}$, then
$$H^2(G,\mathbb{Z}) = H^1(G,\mathbb{Q}/\mathbb{Z}) = \mathbb{Q}/\mathbb{Z}.$$


**Question 1.** Is this also true when $G = \Gamma\_k$, i.e., is $H^2(k,\mathbb{Z})=\mathbb{Q}/\mathbb{Z}$?


**Question 2.** For the general case where $C\_1\backslash C$ consists of $n$ points, do we still have $H^1(k,\mathbb{Z}^n) = 0$?



The groups $H^i(k,\mathbb{Z})$ for $i=1,2$
**Question 1**: No. By the exact sequence $0\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}/n\rightarrow 0$, the $n$-torsion of $H^2(k,\mathbb{Z})$ is $H^1(k, \mathbb{Z}/n)\cong \operatorname{Hom}(\Gamma \_k,\mathbb{Z}/n) $, a huge group.


**Question 2**: I think so. Indeed the $\Gamma \_k$-module $\mathbb{Z}^n$ is a permutation module, i.e. a direct sum of modules $\mathbb{Z}[\Gamma \_k/H]$, with $H$ a subgroup of $\Gamma \_k$. By the [Shapiro lemma](https://en.wikipedia.org/wiki/Shapiro%27s_lemma#Statement_for_group_cohomology), we have $H^1(\Gamma \_k,\mathbb{Z}[\Gamma \_k/H])=H^1(H,\mathbb{Z})=0$.


