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Let $E$ be an elliptic curve over $k=\mathbb{Q}$. Consider $H^1(k,E)$. In this answer Daniel Loughran writes: "I'm pretty sure that this cohomology group has elements of arbitrarily large order". I would be happy to have an explanation of this fact and/or a reference.

(I usually work with linear algebraic groups. For an abelian linear algebraic group $A$ there exists a natural number $d=d(A_{\bar k})$ such that the order of any element of $H^1(k,A)$ is $\le d$.)

EDIT: Cassels, § 10 and § 27, writes that this result was first proved by Shafarevich in 1957.

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    $\begingroup$ For a $g$-dimensional abelian variety $A$ over a number field $k$, $A[N](\overline{k})$ has size $N^{2g}$ rather than just $N^g$ as for tori. This causes even $Ш^1_S(k,A)[p^{\infty}]$ (!) to be infinite when $g[k:\mathbf{Q}]>{\rm{rank}}(A(k))$ and $S$ is the set of places of $k$ over $p$; for details, see Example 7.5.1 of math.stanford.edu/~conrad/papers/cosetfinite.pdf $\endgroup$
    – nfdc23
    Commented May 21, 2016 at 16:59
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    $\begingroup$ Cassels, Diophantine Equations with special reference to elliptic curves, JLMS 1966 § 27. $\endgroup$
    – Felipe Voloch
    Commented May 22, 2016 at 2:05
  • $\begingroup$ @FelipeVoloch: Thank you for the reference to the paper of Cassels. It is an excellent survey! However it seems that his proof of Shafarevich's theorem in § 27 is not complete. Indeed, it is not clear why his group Б in the exact sequence (27.1) cannot have infinitely many elements of order $m$. One needs some finiteness result for Б. $\endgroup$
    – Mikhail Borovoi
    Commented May 24, 2016 at 18:09

2 Answers 2

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Here is the kind of method I had in mind.

We have the elliptic curve Kummer sequence $$0 \to E[n] \to E \to E \to 0,$$ Here I denote by $E[n]$ the $n$-torsion group scheme of $E$. Applying Galois cohomology we obtain $$0 \to E(\mathbb{Q})/nE(\mathbb{Q}) \to H^1(\mathbb{Q}, E[n]) \to H^1(\mathbb{Q}, E)[n] \to 0.$$

By the Mordell-Weil theorem, the group $E(\mathbb{Q})/nE(\mathbb{Q})$ is finite (its cardinality grows roughly like $n^{\mathrm{rank}(E)}$).

Thus it suffices to show that $H^1(\mathbb{Q}, E[n])$ is infinite. I think that this should be some general property of Galois cohomology for non-trivial finite abelian group schemes over number fields, which probably you already know about.

Anyway, the argument should go as follows: Choose a splitting field $k/\mathbb{Q}$ for $E[n]$. We then apply inflation-restriction to obtain $$0 \to H^1(\mathrm{Gal}(k/\mathbb{Q}), E[n]) \to H^1(\mathbb{Q}, E[n]) \to H^1(k, (\mathbb{Z}/n\mathbb{Z})^2)^{\mathrm{Gal}(k/\mathbb{Q})} \to H^2(\mathrm{Gal}(k/\mathbb{Q}), E[n]).$$ The first and the latter group are finite. The group $H^1(k, (\mathbb{Z}/n\mathbb{Z})^2)$ is clearly infinite, and I think that it is still infinite after taking Galois invariants. Though this last step is the part I did not fully check. Is it clear to you?

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  • $\begingroup$ If one assumes $\mu_n \subseteq k$, one is left with proving that $(k^\times/n)^{\mathrm{Gal}(k/\mathbf{Q})}$ is infinite. $\endgroup$
    – user19475
    Commented May 21, 2016 at 17:01
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    $\begingroup$ I think that one can prove that $(\mathbf{Q}^\times/n)^G = \mathbf{Q}^\times/n$ shares an infinite quotient with $(k^\times/n)^G$. $\endgroup$
    – user19475
    Commented May 21, 2016 at 17:14
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    $\begingroup$ Let $k$ be a number field, $M$ a nonzero finite discrete ${\rm{Gal}}(\overline{k}/k)$-module, and $S$ a finite set of places $v$ of $k$ containing all $v|\infty\#M$ and $v$ where $M$ is ramified, so $M$ splits over the maximal extension $k_S/k$ unramified outside $S$. Hence, $H^1(k,M)$ is the direct limit along injective inflation of the $H^1(k_S/k,M)$'s, so it suffices to show $h^1(k_S/k,M)$ is unbounded as $S$ varies. By the global Euler char. formula and 9-term exact sequence, it suffices to show $\prod_{v\in S} h^2(k_v,M)$ is unbounded as $S$ varies. By local duality and Chebotarev, QED. $\endgroup$
    – nfdc23
    Commented May 22, 2016 at 2:56
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    $\begingroup$ @nfdc23: Thank you! I have posted a separate question. I would appreciate if you write an answer rather than just a comment.... $\endgroup$
    – Mikhail Borovoi
    Commented May 22, 2016 at 5:13
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    $\begingroup$ @DanielLoughran: Now it is clear for me, see my answer! $\endgroup$
    – Mikhail Borovoi
    Commented May 24, 2016 at 18:13
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Theorem 6 of this paper gives a generalization of Shafarevich's Theorem:

Theorem: Let $K$ be a Hilbertian field. Let $A_{/K}$ be a nontrivial abelian variety. If $n > 1$ is indivisible by the characteristic of $K$ and $A(K)/nA(K)$ is finite, then $H^1(K,A)$ has infinitely many elements of order $n$.

As explained in the paper, the hypothesis that $A(K)/nA(K)$ be finite cannot be omitted, but the hypothesis that $n$ is indivisible by the characteristic of $K$ can be weakened to: $A(K^{\operatorname{sep}})$ contains a point of order $n$. Via the Kummer sequence, the proof quickly reduces to showing that $H^1(K,A[n])$ contains infinitely many elements of order $n$, which is related to your other question and (yet more closely) to Lior Bary-Soroker's answer to it.

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