# The groups $H^i(k,\mathbb{Z})$ for $i=1,2$

This question is related to my post Interpretation of some maps involving cohomology groups.

$$C$$ 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 $$H^1(k,\mathbb{Z}^n) = 0$$?

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, we have $$H^1(\Gamma _k,\mathbb{Z}[\Gamma _k/H])=H^1(H,\mathbb{Z})=0$$.

• 1) By the isomorphism $H^1(k,\mathbb{Z}/n) \cong \mathrm{Hom}(\Gamma_k,\mathbb{Z}/n)$, it seems like $\mathbb{Z}/n$ also has trivial Galois action right? Feb 28, 2021 at 11:17
• 2) So such an $H$ would be of the form $\mathrm{Gal}(L/k)$ where $L$ is a finite extension of $k$, i.e., $H$ is a finite subgroup of $\Gamma_k$ and therefore $H^1(H,\mathbb{Z}) = 0$. Does $n$ here play a part in the number of such $H$ in the direct sum? Feb 28, 2021 at 11:25
• 1) Yes, the exact sequence is of trivial $\Gamma _k$-modules. $\qquad\qquad\quad$ 2) Well, each summand of $\Bbb{Z}^n$ as a $\Gamma _k$-module corresponds to an orbit of the $n$ points under $\Gamma _k$, so $n=\sum (\Gamma _k:H)$ over the $H$ which appear.
– abx
Feb 28, 2021 at 12:50
• Then something seems off with my understanding, because $(\Gamma_k:H)$ is not finite... If $H$ is an infinite subgroup, then how is $H^1(H,\mathbb{Z}) = 0$? Feb 28, 2021 at 15:43
• Yes, $(\Gamma _k:H)$ is finite: $H$ is the stabilizer in $\Gamma _k$ of one of the points at infinity.
– abx
Feb 28, 2021 at 16:00