4
$\begingroup$

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 $\ker(\pi)$?

$\endgroup$
1
  • 1
    $\begingroup$ In homogeneous coordinates, i posted an answer. However i think the points of $\ker(\pi)$ in affine coordinate. What we can say about valuation of $x$ and $y$ when $(x,y)\in \ker(\pi)$? Silverman answer is $v(x),v(y)<0$ but i can't figure out why $\endgroup$
    – user253407
    Aug 10, 2015 at 16:10

2 Answers 2

Reset to default
2
$\begingroup$

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 $v(y)=0$.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
5
  • 1
    $\begingroup$ by the way we do not need the ordinarity hypothesis; this is also true for supersingular curves. Also, the above write-up uses a weierstrass model which is only valid for $p \neq 2, 3$. $\endgroup$
    – hunter
    Aug 10, 2015 at 17:11
  • $\begingroup$ This is a great answer! thanks! I have a question on $2v(y)=3v(x)$ how recover it? From the equation for me its clear $2v(y)=v(x^3+\ldots)$. Finally if $y$ is not an integer then $v(y)<0$ in particular $m<0$. For have an appropriate triple you should multiply for $\pi^{-m}$,isn't it? $\endgroup$
    – user253407
    Aug 10, 2015 at 17:39
  • 1
    $\begingroup$ @user253407 whoops, you are right about $m$ versus $-m$, i'll edit that into the answer! How to get $2v(y) = 3v(x)$? Remember that we are assuming in this paragraph that both $v(y)$ and $v(x)$ are less than zero (otherwise this doesn't work). The trick is that $v(x^3 + ax + b)$ has to be $v(x^3)$, because $ax+b$ has to have larger valuation than $x^3$ (its valuation is at least $v(x)$), and so there can be no cancellation. $\endgroup$
    – hunter
    Aug 10, 2015 at 18:12
  • 1
    $\begingroup$ @user253407 to spell the last thing I said out: if we add two terms whose valuations are different, the valuation of the sum is always the minimum of the valuations of the term. Since we assume$v(x) < 0$, we have $v(x^3) < v(x)$. Since $a$ and $b$ are integers, we also have that $v(x) \leq v(ax+b)$, and therefore we must have $v(x^3 + (ax + b)) = v(x^3)$. $\endgroup$
    – hunter
    Aug 10, 2015 at 18:16
  • 1
    $\begingroup$ @user253407 sorry for all these comments! my sign error also gave a wrong inequality in my answer... making another edit. $\endgroup$
    – hunter
    Aug 10, 2015 at 18:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .