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Let $L/K$ be an abelian $p$-extension of number fields and $E$ be an elliptic curve over $\Bbb Q$. If $E[p](K)=0$, does it follow that $E[p](L)=0$ ? The converse is obviously true, but I don't have any reference for my problem.

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    $\begingroup$ Not sure but you may try : replace $L$ by $ K(E[p]) \cap L$, $G = Gal(L/K)$ is an abelian subgroup of $GL_2(\mathbb{F}_p)$. Does $g \in G$ of order $p$ mean $\langle g \rangle = u T u^{-1}$ where $T = \{\pmatrix{1 & b \\ 0 & 1 }\}$ ? Then only $DT = \{\pmatrix{ a & b \\ 0 & a }\}$ commute with $T$. So $G = u T u^{-1}$ which means $E(K)[p]$ contains a non-trivial element. $\endgroup$ – reuns Jan 21 at 15:03
  • $\begingroup$ The case of $char K=p$ is easier since $rank E[p](\bar{K})$ is $0$ or $1$. The group $Aut(\mathbb F_p)$ has no order $p$ elements. $\endgroup$ – eduard Jan 21 at 15:47

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