$\DeclareMathOperator\Gal{Gal}$Let $E$ be an elliptic curve defined over a number field $K$. Put $G:=\Gal(\bar{K}/K)$, and for each valuation $v$ of $K$, put $G_v:=\Gal(\bar{K_v}/K_v)$. Consider the following exact sequence: $$ 1 \to E_\text{tor} \to E \to E/E_\text{tor} \to 1.$$ Then we have a diagram: $$\require{AMScd}\begin{CD} 1 @>>> E(K)_\text{tor} @>>> E(K) @>>> (E(\bar{K})/E(\bar K)_\text{tor})^{G} @>>> H^1(K,E(\bar{K})_\text{tor}) @>>> H^1(K,E) @>>> \cdots \\ @. @VVV @VVV @VVV @VVV @VVV \\ 1 @>>> \prod_v E(K_v)_\text{tor} @>>> \prod_v E(K_v) @>>> \prod_v (E(\bar{K_v})/E(\bar{K_v})_\text{tor})^{G_v} @>>> \prod_v H^1(K_v,E(\bar{K_v})_\text{tor}) @>>> \prod_v H^1(K_v,E) @>>> \cdots . \end{CD}$$ I am wondering is there any descriptions about the above cohomology groups $(E/\text{tor})^{G_{-}}$ and $H^1(-,-)$ (maybe when $E$ has good & multiplicative reductions at $v$)?
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$(E(\bar{K})/tor)^{G}$ is simply $E(K) \otimes \mathbb Q$ since every element of $E(\bar{K})$ has finite orbit and if every two elements of the orbit differ by a torsion point then there exists some $n$ whose order is a multiple of the order of the torsion points and then multiplying by $n$ gives something in $E(K)$, and vice versa, every element of $E(K)$ can be divided by $n$ in $(E(\bar{K})/tor)^{G}$.
In particular, if $E(K)$ has rank $r$, then $(E(\bar{K})/tor)^{G}$ is isomorphic to $\mathbb Q^r$.
An analogous statement is true for the local version - it is isomorphic to $E(\bar{K})^{G_v} \otimes \mathbb Q$, which is a somewhat weird group (it is not, for example, the same as $E(K_v)\otimes \mathbb Q$, as one might assume).
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$\begingroup$ I am struggling to understand what you are saying in the local case. When you write $E(\bar{K})^{G_v}\otimes \mathbb{Q}$, did you intentionally write $\bar{K}$ instead of $\bar{K}_v$? If so, can you give more details on how we end up with $\bar{K}$ here instead of $\bar{K}_v$. $\endgroup$– SnaccCommented Oct 21 at 23:46
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$\begingroup$ @Snacc The question was edited by LSpice to change the formulation at the same time as I wrote my answer. Before that, the local version in the question also had $E(\overline{K})$, and I was just copying it. $\endgroup$ Commented Oct 22 at 0:48
AMScd; I hope that that was all right. You also had $\prod_v (E(\bar K)/\text{tor})$ in the bottom row, which I'm pretty sure was meant to be $\prod_v (E(\bar{K_v})/\text{tor})$; I edited accordingly. $\endgroup$