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Let $k$ be a number field and denote by $H^i(k,-)$ the Galois cohomology functor $H^i(\mathrm{Gal}(\bar{k}/k),-)$. Let $X$ be a smooth geometrically integral curve over $k$. One can easily show that the map $H^1(k,\mathrm{Pic}^0(X_{\bar{k}})) \rightarrow H^1(k,\mathrm{Pic}(X_{\bar{k}}))$ is surjective. Indeed, applying the functor $H^1(k,-)$ to the exact sequence $$0 \rightarrow \mathrm{Pic}^0(X_{\bar{k}}) \rightarrow \mathrm{Pic}(X_{\bar{k}}) \rightarrow \mathrm{NS}(X_{\bar{k}}) \rightarrow 0,$$ we get the exact sequence $$H^1(k,\mathrm{Pic}^0(X_{\bar{k}})) \rightarrow H^1(k,\mathrm{Pic}(X_{\bar{k}})) \rightarrow H^1(k,\mathbb{Z}).$$ The last term is zero because $\mathbb{Z}$ has trivial Galois action and so any 1-cocycle $f \in H^1(k,\mathbb{Z})$ is simply an element of $\mathrm{Hom}(\mathrm{Gal}(\bar{k}/k),\mathbb{Z})$. But $\mathbb{Z}$ is torsion-free and thus all such maps are zero maps.

Question 0. Here I did not assume that $X$ is projective, but would all of the above still hold without this assumption? For example, I've never seen the Neron-Severi group of an affine curve discussed in any literature.

Now moving on, we have an exact sequence of Galois modules $$0 \rightarrow \mathrm{Pic}(X_{\bar{k}})_{\mathrm{tor}} \rightarrow \mathrm{Pic}(X_{\bar{k}}) \rightarrow \mathrm{Pic}(X_{\bar{k}})_{\mathrm{free}} \rightarrow 0$$ where $\mathrm{Pic}(X_{\bar{k}})_\mathrm{tor}$ denotes the maximal torsion subgroup of $\mathrm{Pic}(X_{\bar{k}})$ and $\mathrm{Pic}(X_{\bar{k}})_\mathrm{free} = \mathrm{Pic}(X_{\bar{k}})/\mathrm{Pic}(X_{\bar{k}})_\mathrm{tor}$, the maximal free quotient.

Question 1. In this case $\mathrm{Pic}(X_{\bar{k}})_\mathrm{free}$ certainly does not have trivial Galois action, so we cannot reduce 1-cocycles to group homomorphisms $\mathrm{Gal}(\bar{k}/k) \rightarrow \mathrm{Pic}(X_{\bar{k}})_\mathrm{free}$. Therefore how do we go about computing $H^1(k,\mathrm{Pic}(X_{\bar{k}})_\mathrm{free})$?

I have thought about first studying $H^1(k,\mathrm{Pic}(X_{\bar{k}}))$ using a wild idea as follows:

We apply the (étale) cohomology functor $H^i(X_{\bar{k}},-)$ to the Kummer sequence $$0 \rightarrow \mu_n \rightarrow \mathbb{G_m} \rightarrow \mathbb{G}_m \rightarrow 0$$ to obtain the long cohomology sequence $$0 \rightarrow \mu_n(\bar{k}) \rightarrow \bar{k}^* \rightarrow \bar{k}^* \rightarrow H^1(X_\bar{k},\mu_n) \rightarrow \mathrm{Pic}(X_{\bar{k}}) \rightarrow \mathrm{Pic}(X_{\bar{k}}) \rightarrow H^2(X_{\bar{k}},\mu_n) \rightarrow H^2(X_{\bar{k}},\mathbb{G}_m)=0.$$

Then we apply $H^1(k,-)$ to $$H^1(X_\bar{k},\mu_n) \rightarrow \mathrm{Pic}(X_{\bar{k}}) \rightarrow \mathrm{Pic}(X_{\bar{k}}) \rightarrow H^2(X_{\bar{k}},\mu_n)$$ and study the result. But I'm not sure if there are any spectral sequences we can use, or if this approach is even feasible.

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    $\begingroup$ To answer the general question in the title: If $M$ is a $G_k$-module that is free as a $\mathbb{Z}$-module, then $H^1(k,M)$ is the quotient of $(M\otimes \mathbb{Q}/\mathbb{Z})^{G_k}$ by $M^{G_k}\otimes\mathbb{Q}/\mathbb{Z}$. That is what computer algebra systems like magma do behind the scene, $\endgroup$
    – Chris Wuthrich
    Commented Jul 11, 2021 at 11:14
  • $\begingroup$ @ChrisWuthrich I'm just curious here... Could you explain how does $\mathbb{Q}/\mathbb{Z}$ come about in this instance? This abelian additive group appears quite a lot in what I'm studying, for example, it contains the Brauer group of any local field, or that this is the colimit of all the $n$-th roots of unity. Your result looks neat, are there any suitable references? $\endgroup$
    – oleout
    Commented Jul 11, 2021 at 16:09
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    $\begingroup$ It just follows from the short exact sequence $0\to M \to M\otimes\mathbb{Q} \to M\otimes\mathbb{Q}/\mathbb{Z}\to 0$. $\endgroup$
    – Chris Wuthrich
    Commented Jul 11, 2021 at 16:13
  • $\begingroup$ So I apply the functor $H^i(k,-)$ to your exact sequence and obtained $$0 \rightarrow M^{G_k} \rightarrow (M \otimes \mathbb{Q})^{G_k} \rightarrow (M \otimes \mathbb{Q}/\mathbb{Z})^{G_k} \rightarrow H^1(k,M) \rightarrow H^1(k, M \otimes \mathbb{Q}).$$ Is the second last map surjective, i.e., is the last term zero? Otherwise I don't see how $H^1(k,M)$ can be realised as a quotient of $(M \otimes \mathbb{Q}/\mathbb{Z})^{G_k}$. $\endgroup$
    – oleout
    Commented Jul 12, 2021 at 3:10
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    $\begingroup$ This cohomology group is both torsion and divisible, hence trivial. $\endgroup$
    – Daniel Loughran
    Commented Jul 12, 2021 at 8:18

1 Answer 1

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I will focus attention on smooth projective varieties $X$ over $k$ with $\mathrm{Pic}(X_{\bar{k}})$ a free finitely generated abelian group, as they illustrate all the essential behaviour relevant to your question.

Here $\mathrm{Pic}^0(X_{\bar{k}})$ is trivial so it is certainly not true in general that the map $H^1(k,\mathrm{Pic}^0(X_{\bar{k}})) \rightarrow H^1(k,\mathrm{Pic}(X_{\bar{k}}))$ is surjective. Your argument has problems as the Neron-Severi group is just equal to the Picard group here, and it is not isomorphic to $\mathbb{Z}$ with trivial Galois action in general.

As for your second question on how to calculate $H^1(k,\mathrm{Pic}(X_{\bar{k}}))$. It depends a lot on the geometry of your specific situation. There are no cheap tricks which can help you using e.g. Kummer theory. If you are able to write down explicit generators for $\mathrm{Pic}(X_{\bar{k}})$ and you know how Galois acts on them, then you can actually compute this cohomology group using standard commands for group cohomology in Magma. This is a tried and tested approach which appears in many papers; particularly common examples which appear in the literature include del Pezzo surfaces and K3 surfaces. You could have a look at the PhD Thesis of Martin Bright for example to see a detailed treatment of this method.

In general I should add that the Hochschild-Serre spectral sequence identifies $H^1(k,\mathrm{Pic}(X_{\bar{k}}))$ with the so-called algebraic part of the Brauer group of $X$. It is for this reason people are usually interested in computing $H^1(k,\mathrm{Pic}(X_{\bar{k}}))$, and any paper which calculates the Brauer group of a specific variety over a number field will almost certainly start by calculating $H^1(k,\mathrm{Pic}(X_{\bar{k}}))$.

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  • $\begingroup$ I'll have a look at the thesis as you mentioned, thank you. The neat part of the "projective" assumption is that the varieties would satisfy $\bar{k}[X]^* = \bar{k}^*$, and so the Hochschild-Serre spectral sequence reduces to $$\mathrm{Br}(k) \rightarrow \mathrm{Br}_1(X) \rightarrow H^1(k,\mathrm{Pic}(X_{\bar{k}})) \rightarrow H^3(k,\bar{k}^*) = 0.$$ Thus one has the identification as you mentioned. However, removing the properness condition would instead need us to understand the groups $H^i(k,\bar{k}[X]^*)$ for $i =2,3$, at least. This is related to some other questions I asked on this site. $\endgroup$
    – oleout
    Commented Jul 11, 2021 at 8:31
  • $\begingroup$ The spectral sequence still tells you something in the non-projective case. See Lemma 6.3 of Sansuc - Groupe de Brauer et arithmetiques des groupes algebriques lineaires sur un corps de nombres $\endgroup$
    – Daniel Loughran
    Commented Jul 11, 2021 at 8:45

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