Points of degree $n$ on an elliptic curve

arithmetic-geometryelliptic-curves

Let $E$ be an elliptic curve over a number field $k$. It is known that $E(k)$ is a finitely generated abelian group and has a structure of the form $\mathbb{Z}^r \oplus T$, where $T$ is torsion. The integer $r \geq 0$ is called the rank of $E$. For a closed point $P \in E(\bar{k})$, we define the degree of $P$ to be the smallest degree of the finite extension $K/k$ such that $P$ is $K$-rational. Take note that we are NOT talking about the order of $P$ here.

Suppose $E$ has rank $r \geq 1$, i.e., it has infinitely many $k$-points. Does it follow that $E$ has infinitely many points of degree $n$, for $n \geq 2$?

I'm thinking about this question because suppose we fix a point $P$ of degree, say, $2$. Consider the automorphisms of $E$ $\varphi_i: P \mapsto Q_i$ for some $Q_i$ of degree $2$ different from $P$ (this is because automorphisms send points to points of similar degree). If $r \geq 1$, are there infinitely many distinct $Q_i$?

Best Answer

This is true for $n=2$ for every elliptic curve over $k$ (regardless of rank). Say $E$ is given by the equation $y^2=x^3+ax+b$. It is a theorem of Cassels that we can find a prime $p$ so that $k$ embeds into $\mathbb{Q}_p$ and $|a|_p = |b|_p =1$, provided $a$ and $b$ are nonzero. If either one happens to be zero the argument will go through all the same. Now let $c=1+ap^2+bp^3$. Since $v_p(c/p)=-1$, $c/p$ is not a square in $\mathbb{Q}_p$ and hence not in $k$ either. Let $L=\mathbb{Q}_p(\sqrt{c/p})$, which is totally ramified over $\mathbb{Q}_p$ with index 2. Thus the valuation $w$ on $L$ satisfies $w(z)=2v_p(z)$ for all $z\in\mathbb{Q}_p$. The point $P=(\frac{1}{p}, \frac{\sqrt{c/p}}{p})$ satisfies our defining equation and is defined over $k(\sqrt{c/p})$ (resp. $L$) but not over k (resp. $\mathbb{Q}_p$). Now, every point $P'=(x', y')$ on $E$ defined over $L$ can be written in the form $(x',y')=(\zeta\pi^{-2n},\eta\pi^{-3n})$ where $\zeta$ and $\eta$ are integral in $L$, $\pi=\sqrt{p/c}$ is a uniformizer for $L$, and $n=\text{max}\{0, -\frac{1}{2}w(x')\}=\text{max}\{0, -\frac{1}{3}w(y')\}$. In this way we obtain a number $n(P')$ for every $P'\in E(L)$. Note that if $P'$ is defined over $\mathbb{Q}_p$, then $n(P')$ is even. Finally, it is a theorem of Lutz that this function $n$ has the following property: if $n(P')\geq 1$, then $n(mP')=n(P')+2v_p(m)$ for every $m\neq 0$. Our point $P=(\frac{1}{p}, \frac{\sqrt{c/p}}{p})$ satisfies $n(P)=1$ and therefore $n(mP)=1+2v_p(m)$ is odd and therefore not defined over $\mathbb{Q}_p$ for $m\neq 0$ and a fortiori it is not defined over $k$ either. In this way we have produced infinitely many points of $E$ which are defined over a quadratic extension of $k$, but not over $k$ itself.

Much of this argument comes from Frey and Jarden's paper "Approximation theory and the rank of abelian varieties over large algebraic fields." I imagine a closer inspection of their paper would yield a positive result to your question for every $n\geq 2$, but I couldn't say for certain at the moment.