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We consider an abelian variety $A$ defined over the rational numbers $\mathbb{Q}$. For a torsion point $P\in A(\bar{\mathbb{Q}})$, consider the field $\mathbb{Q}(P)$ obtained by adjoining to $\mathbb{Q}$ the coordinates of the point $P$.

For a prime $p$, let $\mu_p$ be a primitive $p$-th root of unity and $P$ a $p$-torsion point of $A$.

Is it true that if $\mu_p\notin \mathbb{Q}(P)$, then $\mu_p\notin \mathbb{Q}(nP)$, for $n=1,\dots,p-1$?

If not, how can we characterize this fact?

Answers and comments are welcome. Thanks in advance.

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    $\begingroup$ If $A$ is defined over $\mathbb{Q}$, meaning that the multiplication map and the coordinates of the group identity are defined over $\mathbb{Q}$, then for every pair of closed points $P$ and $R$ of $A$, the compositum $\mathbb{Q}(P,R)$ of $\mathbb{Q}(P)$ and $\mathbb{Q}(R)$ admits a natural field extension $\mathbb{Q}(P+R)\to \mathbb{Q}(P,R)$. In particular, $\mathbb{Q}(nP)$ is a subfield of $\mathbb{Q}(P)$. So if some element $\zeta_p$ is not contained in $\mathbb{Q}(P)$, then it is also not contained in $\mathbb{Q}(nP)$. $\endgroup$ – Jason Starr Nov 27 '18 at 13:10
  • $\begingroup$ Whence $\mathbb{Q}( P)=\mathbb{Q}( nP)$. $\endgroup$ – reuns Nov 28 '18 at 10:06

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