Reduction map on torsion points of an elliptic curve and their valuation

Question

Let $K$ be a field of characteristic zero, complete with respect a discrete valuation $v$. Assume that the residue field $k$ is of positive characteristic $p$.

Now take an elliptic curve $E$ defined over $K$ such that it has a good reduction at $v$ of height 1 (that is Hasse invariant not zero). There is the exact sequence (as abstract group) $$ 0 \longrightarrow \ker (\pi) \longrightarrow E[p](K) \overset{\pi}{\longrightarrow} \tilde E[p](k) \longrightarrow 0$$

where $\pi$ is the reduction map (described in homogeneus coordinate) $[x:y:z] \mapsto [\tilde x: \tilde y: \tilde z]$ ($\tilde t$ is the image of $t \in \mathcal{O}_K$ in k).

My question is: how describe $\ker(\pi)$ in terms of the valuation $v$? More specific, wich valuation have the coordinates of the points of $\ker(\pi)$?

Answer 1

The points $[x:y:z] \in \ker(\pi)$ are those mapped in $[0:1:0]$ so they must satisfy $v(x),v(z)>0$ and $v(y)=0$.

Answer 2

Answering the question as clarified in a comment.

Remember the process for passing between homogeneous and affine coordinates: a pair $(x, y)$ corresponds to $[x: y: 1]$. Remember also the process for reducing homogeneous coordinates: we scale by a power of the uniformizer (which I'll call $\pi$, which is mildly in conflict with your notation) until all of the coordinates are integers and at least one is a unit. We'll call a triple of coordinates "appropriate" if it satisfies this criterion. The reduction map is the map that reduces an appropriate triple mod $\pi$ (i.e. mod the maximal ideal).

Now, if $x$ and $y$ are integers, it is clear that $[x: y: 1]$ does not reduce to the origin (which is $[0: 1: 0]$) because the triple $(x, y, 1)$ is already "appropriate." So if we want a point $(x, y)$ to reduce to the origin, it will be necessary that at least one of $x$ and $y$ be non-integers. Since $$ y^2 = x^3 + ax + b $$ and since $a$ and $b$ are integers (good reduction) it follows that $x$ is an integer if and only if $y$ is an integer. So it is necessary that both $x$ and $y$ be non-integers.

Why is it sufficient? If $x$ and $y$ are both non-integers, the above equation gives $2v(y) = 3v(x)$ (there can be no cancellation otherwise -- see the comments for more details).

Let $m = -v(y)$. Then $[x: y: 1]$ can be written "appropriately" as $[\pi^mx: \pi^my:\pi^m]$. Now $\pi^m y$ is a unit, but $\pi^m$ isn't (because $y$ isn't an integer) and $\pi^m x$ isn't either (because $m = -v(y) = -\frac{3}{2}v(x) > -v(x)$). So this reduces to $[0:1:0]$, proving that Silverman's condition is also sufficient.