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Let $E$ be an elliptic curve over $\mathbb{Q}_p$ and let $E(\mathbb{Q}_p) \supset E^0(\mathbb{Q}_p) \supset E^1(\mathbb{Q}_p) \supset \cdots$ be its $p$-adic filtration, where $E^n(\mathbb{Q}_p) = \{P \in E^0(\mathbb{Q}_p) : \dfrac{x(P)}{y(P)} \in p^n \mathbb{Z}_p\}$ for $n \geq 1$ and $E^0(\mathbb{Q}_p)$ is the set of all points that get mapped to a nonsingular point on the reduced curve $\overline{E}(\mathbb{F}_p)$.

I want to show that $E^1(\mathbb{Q}_p)$ is torsion free. Apparently we only need to consider points of order $p$ (cf Milne). Why is this?

Here's what I know: if $(m,p) = 1$ then $P \mapsto mP$ is a bijection from $E^1$ to $E^1$. This clearly implies that we only need to consider the case where $p \mid m$. But why can we skip right ahead and only consider $p=m$?

By the way I am following http://www.jmilne.org/math/Books/ectext5.pdf.

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    $\begingroup$ You reduce to order $p$ by letting $m$ be any torsion order for a point $P$, then let $p|m$ be prime and $P^{m/p}$ has order $p$. $\endgroup$ – Adam Hughes Jul 10 '14 at 19:45

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