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Let $E/Q$ be an elliptic curve. $E(\bar{Q})$ is a complicated abelian group, which equals to all closed points of $\bar{E}$, and also, a $G_Q$-Galois module. Its torsion part, $E(\bar{Q})_{tor}$ is a subgroup ( $G_Q$-submodule ) of $E(\bar{Q})$, as an abstract group is isomorphic to $(Q/Z)^2$.

Consider $E(\bar{Q})/E(\bar{Q})_{tor}$. It is naturally a $Q$-vector space, as it is uniquely divisible, also, a $G_Q$-module. Then we can talk about the invariant subspace under $G_Q$ action, $E(\bar{Q})/E(\bar{Q})_{tor}^{G_Q}$. Obviously, $E(Q)\otimes Q$ is a subspace of $E(\bar{Q})/E(\bar{Q})_{tor}^{G_Q}$.

Are they equal?

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    $\begingroup$ Isn't your space just $E(\bar{\mathbb{Q}})\otimes_{\mathbb{Z}} \mathbb{Q}$ as the map from $E(\bar{\mathbb{Q}})$ to it is surjective? Then it is clear that the Galois-invariant part is $E(\mathbb{Q})\otimes \mathbb{Q}$, isn't it? $\endgroup$
    – Chris Wuthrich
    Commented May 7, 2021 at 8:37
  • $\begingroup$ @Chris Wuthrich Yes, you are right... In this case if an element $P$ in $E(\bar{Q})\otimes Q$ is stable under $G_Q$ action, then there is an integer $n$ such that $nP$ in $E(\bar{Q})$ is stable under $G_Q$ action. So they are indeed equal. But generally, are tensor functor and taking invariant functor always commute with each other? $\endgroup$
    – Yuan Yang
    Commented May 7, 2021 at 11:50

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