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Let $E$ be an elliptic curve over $\mathbb{Q}$. Does the image of $\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ under the mod $p$ Galois representation tell us whether or not $E$ has rational $p$-torsion or not?

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Let $E$ be an elliptic curve defined over $\mathbb{Q}$, and let $p$ be a prime. The mod $p$ representation attached to $E$ is a homomorphism $$\rho_{E,p} : \text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \to \text{Aut}(E[p])$$ given by $\sigma \mapsto (\phi_\sigma : E[p] \to E[p])$, such that $\phi_\sigma(R)=\sigma(R)$ for every $R\in E[p]$. If we fix a $\mathbb{Z}/p\mathbb{Z}$-basis $\{P,Q\}$ of $E[p]$, then $\text{Aut}(E[p])\cong \text{GL}(2,\mathbb{Z}/p\mathbb{Z})$, and so we can regard $\rho_{E,p}$ as $$\rho_{E,p} : \text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \to \text{GL}(2,\mathbb{Z}/p\mathbb{Z}).$$ Now, $R$ is in $E(\mathbb{Q})[p]$ if and only if $\sigma(R)=R$ for all $\sigma \in \text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, if and only if $R$ is fixed by every $\phi_\sigma \in \text{Aut}(E[p])$, if and only if $R=\lambda P+\mu Q$ and the vector $(\lambda, \mu)$ is a common eigenvector of every matrix in $\text{GL}(2,\mathbb{Z}/p\mathbb{Z})$ in the image of $\rho_{E,p}$, with eigenvalue $1$. In particular, $E(\mathbb{Q})[p]$ is non-trivial if and only if there is a $\mathbb{Z}/p\mathbb{Z}$-basis of $E[p]$ such that the image of $\rho_{E,p}$ in $ \text{GL}(2,\mathbb{Z}/p\mathbb{Z})$ is a subgroup of $$\left\{ \left(\begin{array}[cc] $1 & b \\ 0 & c \end{array}\right): b\in \mathbb{Z}/p\mathbb{Z},\ c\in (\mathbb{Z}/p\mathbb{Z})^\times\right\}.$$

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  • $\begingroup$ do you know some reference for your answer?? $\endgroup$ – danihelovij May 8 '20 at 10:29
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    $\begingroup$ There is really no reference specifically for this kind of basic material on Galois representations attached to elliptic curves, but of course, this is something one indirectly acquires while reading Silverman's "The Arithmetic of Elliptic Curves". $\endgroup$ – Álvaro Lozano-Robledo May 8 '20 at 16:51
  • $\begingroup$ thank you. I read "rational points on elliptic curve by J.Silverman$ but he use $Gal(\mathbb{Q(E[n])/\mathbb{Q})$ and not $Gal(\overline{\mathbb{Q}})$. Do you know why? $\endgroup$ – danihelovij May 8 '20 at 17:28
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    $\begingroup$ The subgroup Gal($\overline{\mathbb{Q}}/\mathbb{Q}(E[p])$) of Gal($\overline{\mathbb{Q}}/\mathbb{Q}$) is the kernel of $\rho_{E,p}$ that I discuss above. So $\rho_{E,p}$ induces a representation Gal$(\overline{\mathbb{Q}}/\mathbb{Q}(E[p]))\to \text{GL}(2,\mathbb{Z}/p\mathbb{Z})$. $\endgroup$ – Álvaro Lozano-Robledo May 9 '20 at 14:18
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Yes, it does, and in a rather straightforward way. The $p$-torsion in $E(\mathbb{Q})$ is precisely the fixed vectors under the Galois action. In particular, $E$ has full rational $p$-torsion if and only if the mod $p$ representation is trivial.

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