Let $E$ be an elliptic curve over a number field $F$. Assume that $E$ has a $F$-rational non-torsion point $Z$. For each prime $p$, let $\frac{1}{p}Z$ be the set of $X\in E(\bar{F})$ such that $pX=Z$. Is it possible to have


$F(\frac{1}{p}Z)=F(E[p])$


for infinitely many prime $p$?



$p$-th root of non-torsion points on elliptic curves
No. Lemma 3.7 on page 11 from [my paper here](http://users.wfu.edu/rouseja/cv/draft33.pdf) implies that if $\mathcal{T}\_{1} = {\rm Gal}(K(E[p])/K)$ and there is a normal subgroup $H \unlhd \mathcal{T}\_{1}$ with order coprime to $p$ for which $E[p]^{H} = 0$, then $E(K) \cap pE(K(E[p])) = pE(K)$. (Full disclosure - I learned the proof of Lemma 3.7 from a very conscientious referee.) This implies that if there is such a subgroup $H$, and the point $Z \ne pW$ for some $W \in E(K)$, then $F(\frac{1}{p} Z) > F(E[p])$.


Now, for any elliptic curve $E$, the subgroup $H = \{ kI : k \in \mathbb{F}\_{p}^{\times} \}$ of scalar multiples of the identity will be contained in the image of $\rho\_{p} : {\rm Gal}(K(E[p])/K) \to GL\_{2}(\mathbb{F}\_{p})$ if $p$ is large enough. (If $E$ is non-CM, then this follows from Serre's theorem that $\rho\_{p}$ will be surjective if $p$ is large enough. If $E$ has CM, then for large $p$ the image of $\rho\_{p}$ will be a Cartan subgroup or its normalizer, depending on whether $F$ contains ${\rm End}(E) \otimes \mathbb{Q}$ or not.)


