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This is from J.S.Milne's Elliptic Curves book.

We have $H^1(\mathbb{Q}, E_2) \cong (\mathbb{Q}^× / \mathbb{Q}^{×2})^2$ because $Gal(\mathbb{Q}^{al} / \mathbb{Q})$ acts trivially on $E_2(\mathbb{Q}^{al})$, so

$$ H^1(\mathbb{Q}, E_2) \cong H^1(\mathbb{Q}, \mu_2)^2 \cong (\mathbb{Q}^× / \mathbb{Q}^{×2})^2$$

Here, $ \mu_n(L) = \{x \in L | x^n =1\}, E_n(L)=$ the set of points of order $n$ in the Elliptic Curve $E(L)$ and $H^1(G,M)$ is the first cohomological group

My question is, how can I prove a similar isomorphism: $H^1(K, E_2) \cong (K^×/ K^{×2})^2$? ($K$ is a number field) Especially since I don't know how $E_2(K)$ looks like, which I know for $K = \mathbb{Q}$

Thanks!

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    $\begingroup$ I don't have Milne's book handy, but the isomorphism you quote is only valid if all of the 2-torsion is defined over $\mathbb Q$. In that case, $E_2(\mathbb Q)\cong(\mathbb Z/2\mathbb Z)^2$ as Galois modules. This works exactly the same over any number field as long as $E_2\subset E(K)$. And if $E_2\not\subset E(K)$, it is generally not true that $H^1(K,E_2)$ is isomorphic to $(K^\times/{K^\times}^2)^2$. OTOH, it is always true that $H^1(K,\mu_n(\bar K))\cong (K^\times/{K^\times}^n)$. $\endgroup$
    – Joe Silverman
    Jul 22, 2019 at 18:56
  • $\begingroup$ @Joe Silverman Thanks, that was helpful and I failed to see how obvious it was before (the being isomorphic part) $\endgroup$
    – Shreya
    Jul 23, 2019 at 8:44

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