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Is it true that the group $$H^1(Gal(K^{ab}/K)/\mu_{\nu}(Gal(K_{\nu}^{ab}/K_{\nu})),E_{p^n})$$ is always p-divisible? Or are there any conditions which, when satisfied, guarantee its p-divisibility? Here, $()^{ab}$ denotes the maximal Abelian extension of a field, $K_{\nu}$ is the localization of a field number field $K$ at a finite prime $\nu$, $\mu_{\nu}:Gal(K_{\nu}^{ab}/K_{\nu})\rightarrow Gal(K^{ab}/K)$ is induced by the injection $K\rightarrow K_{\nu}$ and $E_{p^n}$ are the $p^n$-torsion points of a (non-singular) elliptic curve $E$ defined over $K$.

I am specifically interested in the cases

(A.) $E$ has bad reduction at $\nu$.

(B.) $\nu$ divides $p$.

(C.) None of the above, and $E(K_{\nu})_{p^{\infty}}=E(\overline{K_{\nu}})_{p^{n}}$

Can somebody give me a good reference?

I am not an expert on the subject, so please have mercy on my soul.

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  • $\begingroup$ Why is the image of $\mu_\nu$ a normal subgroup of ${\rm Gal}(K^{\rm ab}|K)$ ? Sorry if I missed something obvious. $\endgroup$ May 16, 2015 at 14:26
  • $\begingroup$ Because $Gal (K^{ab}/K) $ is Abelian. $\endgroup$ May 16, 2015 at 14:40
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    $\begingroup$ As the group is killed by $p^n$, it can only be $p$-divisible if it is trivial. $\endgroup$
    – anon
    May 16, 2015 at 15:37
  • $\begingroup$ The group in question only makes sense if all poitns in $E_{p^n}$ are defined over $K^{ab}$. So you have to be in a rather particular situation unless they are already defined over $K$. $\endgroup$ May 16, 2015 at 16:47
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    $\begingroup$ No. IF the $m$-torsion points are defined over $K$, then all $Q\in E(\bar{K})$ with $mQ\in E(K)$ are defined over an abelian extension. -- Instead the extension adjoining all $m$-torsion points is a subgroup of $GL_2(\mathbb{Z}/m\mathbb{Z})$. It is very often equal to the full (very non-abelian) group. $\endgroup$ May 17, 2015 at 14:28

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