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


1 Answer 1


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

  • $\begingroup$ 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? $\endgroup$ Feb 28, 2021 at 11:17
  • $\begingroup$ 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? $\endgroup$ Feb 28, 2021 at 11:25
  • $\begingroup$ 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. $\endgroup$
    – abx
    Feb 28, 2021 at 12:50
  • $\begingroup$ Then something seems off with my understanding, because $(\Gamma_k:H)$ is not finite... If $H$ is an infinite subgroup, then how is $H^1(H,\mathbb{Z}) = 0$? $\endgroup$ Feb 28, 2021 at 15:43
  • $\begingroup$ Yes, $(\Gamma _k:H)$ is finite: $H$ is the stabilizer in $\Gamma _k$ of one of the points at infinity. $\endgroup$
    – abx
    Feb 28, 2021 at 16:00

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