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Let $E/\mathbb{Q}_p$ be an elliptic curve and let $E^0(\mathbb{Q}_p)$ denote its nonsingular points. We accept that $E^0(\mathbb{Q}_p)$ is a subgroup of $E(\mathbb{Q}_p)$. Then let $\overline{E}^\text{ns}(\mathbb{F}_p)$ be reduced curve, also a group. We have a map $E^0(\mathbb{Q}_p) \to \overline{E}^\text{ns}(\mathbb{F}_p)$ given by $P \mapsto \overline{P}$.

The group identity in all groups considered is $(0:1:0)$.

Why is the kernel of this map all $(x:y:z) \in E^0(\mathbb{Q}_p)$ such that $x/y \in p \mathbb{Z}_p$? Woudln't this mean that $(p:1:1)$ would be in the kernel? (it's not)

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  • $\begingroup$ ... doesn't it depend on what point you take for the group identity ? $\endgroup$ – mercio Jun 26 '14 at 16:19
  • $\begingroup$ Right, thanks. I added that. We take $(0:1:0)$ to be the group identity in all groups considered. $\endgroup$ – nigel Jun 26 '14 at 16:20
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Seems to me that there’s something wrong with your formulation. I would have said that the condition for being in the kernel of reduction was that $z/y$ was in the maximal ideal.

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