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Let $L/K$ be a quadratic Galois extension of number fields. Let $E$ be an elliptic curve.

Consider the natural map $$ F: H^1(\operatorname{Gal}(L/K), E(L)) \to \bigoplus_{v \in M_K} H^1(\operatorname{Gal}(L_w/K_v), E(L_{v_L})), $$ where $\operatorname{Gal}(L_w/K_v)$ denotes the local Galois group corresponding to the place $v$.

For an abelian group $M$, let $M^*$ denote its Pontryagin dual.

In this context, the Tate-cohomology group is represented as $$ \hat{H}^0(\operatorname{Gal}(L/K),E(L)) = E(K)/(1+\sigma)E(L). $$(In other words, if we define a map $\operatorname{trace}:E(L)\to E(K), P\to P+P^\sigma$, it is $ \operatorname{coker}(\operatorname{trace})$.)

According to p214 of the cited paper (link), it's claimed that $$ (\operatorname{coker} F)^* \cong \hat{H}^0(\operatorname{Gal}(L/K),E(L)), $$ but this is asserted without further explanation.

Could you provide some insights or suggest strategies for proving this isomorphism?

N.B.

 The linked paper reads the dual of diagram $(3)$ on p 214 is the bottom diagram on p 214. lem 5 on p 214 follows immediately if we confirm the titled isomorphism.

 The notation used in this question is somewhat different from that of the paper. However, this question is still self-contained. I have provided a link to the paper for reference, but there is no mention of the isomorphism discussed in this question in the paper, even though it's described using slightly different notation.

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    $\begingroup$ Is $L/K$ a quadratic extension? If not, then your definitions of the trace map and the Tate cohomology is wrong! $\endgroup$ Commented Sep 27, 2023 at 14:21
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    $\begingroup$ If you indeed want to get an answer, please write your question with more details. What is $E$? $\endgroup$ Commented Sep 27, 2023 at 14:23
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    $\begingroup$ What is the relation between $L_w$ and $L_{v_L}$? $\endgroup$ Commented Sep 27, 2023 at 16:00
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    $\begingroup$ What is $F$? Is it related with $\mathcal F$ in the linked paper? $\endgroup$ Commented Sep 27, 2023 at 16:01
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    $\begingroup$ Lemma 4 on page 214 in the linked paper is indeed easy (though not obvious). If this is what you need, I can type a proof. $\endgroup$ Commented Sep 27, 2023 at 16:30

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$\newcommand{\coker}{{\rm coker}} $ I construct the required homomorphism. Let $A$ be an abelian variety over a number field $K$, and let $L/K$ be a finite Galois extension. We denote by $A'$ the dual abelian variety. Consider the following diagram with exact rows and columns:

$\require{AMScd}$ \begin{CD} 0 @. 0 \\ @VVV @VVV \\ H^1(L/K,A) @>F>> \bigoplus_{v\in M_K} H^1(L_{v_L}/K_v,A) \\ @VVV @VVV \\ H^1(K,A) @>>> \bigoplus_{v\in M_K} H^1(K_v,A)@>>> \Big(\widehat{A'(K)}\Big)^* @>>> 0 \\ @VVV @VVV @VVV \\ H^1(L,A) @>>> \bigoplus_{w\in M_L} H^1(L_w,A)@>>> \Big(\widehat{A'(L)}\Big)^* @>>> 0 \end{CD}

Here the second and third rows are the exact sequences of the global duality theorem; see, for instance, Milne, Arithmetic Duality Theorems, Remark I.6.14(b).

The first and second columns come from the restriction-inflation exact sequence; see, for instance, Proposition 4 in Section 5 of Atiyah and Wall, Cohomology of groups, in: Cassels and Fröhlich (eds.), Algebraic Number Theory, pages 94--115.

This diagram induces a homomorphism $$ \coker(F) \to \ker\Big( \widehat{A'(K)}^* \to \widehat{A'(L)}^*\Big).$$ Dualizing, we obtain a homomorphism $$\coker\Big( \widehat{A'(L)}\to \widehat{A'(K)}\Big)\,\longrightarrow\,(\coker\, F)^*.$$ Restricting, we obtain a homomorphism $$\coker\Big( {A'(L)}\to {A'(K)}\Big)\,\longrightarrow\,(\coker\, F)^*.$$ I leave to OP to check whether this homomorphism is an isomorphism.

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  • $\begingroup$ Let $f$ be the induced map $ker(\hat{A’(K)}^*\to \hat{A’(K)}^*$. I believe that if I can prove $cokerF\to kerf$ is surjective, its dual is injective, and its restriction is also injective. However, I am currently facing difficulty in proving that the induced map $ker(\hat{A’(K)}^*\to \hat{A’(K)}^*$ is surjective. Is it really surjective ? (If not, I have no tactics to prove injectivity of the last map) Any suggestions would be appreciated. $\endgroup$
    – Duality
    Commented Oct 2, 2023 at 7:09
  • $\begingroup$ I don't understand the formulas in your comment. Please re-type the comment. $\endgroup$ Commented Oct 2, 2023 at 10:34
  • $\begingroup$ You can click Edit under my answer, and then you can copy my LaTeX formulas. After that, don't forget to choose Cancel rather than Save edits. $\endgroup$ Commented Oct 2, 2023 at 10:34
  • $\begingroup$ $\newcommand{\coker}{{\rm coker}} $ Sorry for confusion of my notation. I want to prove $\coker\Big( {A'(L)}\to {A'(K)}\Big)\,\longrightarrow\,(\coker\, F)^*$ is injective. If I can prove $ \coker(F) \to \ker\Big( \widehat{A'(K)}^* \to \widehat{A'(L)}^*\Big)$ is surjective, then infectivity of $\coker\Big( {A'(L)}\to {A'(K)}\Big)\,\longrightarrow\,(\coker\, F)^*$ follows. But I I'm stuck at the part of proving$ \coker(F) \to \ker\Big( \widehat{A'(K)}^* \to \widehat{A'(L)}^*\Big)$ is surjective. $\endgroup$
    – Duality
    Commented Oct 2, 2023 at 10:37
  • $\begingroup$ Sorry, ×:infectivity $\to$ ○:injectivity. $\endgroup$
    – Duality
    Commented Oct 2, 2023 at 10:44

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