It is clear from the isomorphism between elliptic curves over $\mathbb{C}$ and complex tori that the sum of the $m$-torsion points is the identity in the group law of the elliptic curve. How generally does this hold, and how can one see it (not using the Lefschetz principle, please)?
$\begingroup$
$\endgroup$
1
-
1$\begingroup$ Possibly helpful: If you have an abelian group of odd order, then the product of all the elements is the identity. $\endgroup$ – Dylan Moreland Jun 10 '12 at 18:46
Add a comment
|
$\begingroup$
$\endgroup$
It's not true if $m$ is not prime to the characteristic of the field (e.g. take an ordinary elliptic curve in characteristic 2; it will have exactly one non-trivial 2-torsion point).
We also need the field to be algebraically closed, although you may have been assuming that anyway (e.g. take an elliptic curve over $\mathbb{R}$ whose real points have only one connected component - then there's a uniqut non-trivial two-torsion point).
Once we make these two assumptions on the ground field, the torsion is isomorphic to $(\mathbb{Z}/m)^2$ as an abelian group, and this is a property of that group.
$\begingroup$
$\endgroup$
2
This more-or-less follows from abstract group theory:
Let $P$ = sum of all points of order $m$. Then $P$ is itself a point of order $m$. Now, let $k$ be any integer co-prime to $m$ and note that multiplication by $k$ is a permutation of $E[m]$. So $[k]P$ = sum of all points in $E[m] = P$. So $[k-1]P = 0$. Since $[m]P = 0$ too, then if $gcd( k-1, m ) = 1$ then it follows that $P = 0$.
The case $m = 2$ escapes this proof since $k$ is odd and $gcd( k-1, m ) = 2$.
-
$\begingroup$ It really helps to format your answers using MathJax (see FAQ). I updated your answer as an example. Regards $\endgroup$ – Amzoti Mar 25 '13 at 4:06
-
$\begingroup$ I think you mean to exclude the case $k\equiv 1\pmod m$, otherwise your argument fails. And, in the case $m=2$, your argument can never work, as you have pointed out. $\endgroup$ – awllower Mar 25 '13 at 4:07
