2
$\begingroup$

Let $E$ be an elliptic curve, given by a Weierstrass equation $y^2 + a_1 x y + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6$.

If $P = (x, y)$ is a point on $E$ and $n$ is a (positive) integer, then the point $nP$ has coordinates $(x_n, y_n)$, which can be expressed as: $$x_n = \frac{F_n(x)}{G_n(x)}, y_n = \frac{U_n(x)}{V_n(x)}y,$$ where $F_n, G_n, U_n, V_n$ are polynomials in $x$.

It SEEMS that:

  1. $F_n$ has degree $n^2$, leading coeffcient $1$, and all other coefficients are polynomials of $a_1, \cdots, a_6$;
  2. $G_n$ has degree $n^2 - 1$, leading coefficient $n^2$, and all other coefficients are polynomials of $a_1, \cdots, a_6$;
  3. $U_n$ and $V_n$ have same degree $m$, given by: $$ m = \begin{cases} 3n^2/2, & 2 \mid n; \\ 3(n^2 - 1)/2, & 2 \nmid n.\end{cases}$$

I checked these results for some random curves and small values of $n$.

But how can I prove them?


I tried to use the formal group attached to the elliptic curve. It gives some information, but I'm still far from a proof ...

$\endgroup$
  • 6
    $\begingroup$ Did you try to use division polynomials? See in particular exercise 3.7 in Silverman's The arithmetic of elliptic curves. $\endgroup$ – François Brunault Aug 23 '17 at 15:25
  • $\begingroup$ I believe a nice starting point are (very) old formulas for $\wp(nz)$. $\endgroup$ – Alex Gavrilov Aug 26 '17 at 11:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.