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Let $k$ be an algebraically closed field of nonzero characteristic $p$, and let $E$ be an elliptic curve over $k$. Let $m \in \mathbb N$ such that $p$ doesn't divide $m$.

I want to show that, looking at $E$ as an abelian group, we have $$ mE(k) = E(k).$$ In other words, for every $P \in E(k)$ we can find a $Q \in E(k)$ such that $$ m Q = P.$$

I'm still a novice in elliptic curves, but one idea I had is to look at division polynomials which give coordinates of $mQ$ as rational functions of $x$-coordinate of $Q$. In that way, we can solve $mQ = P$ by solving some polynomial equations in $k$, and we can always do that because $k$ is an algebraically closed field.

Then the only problem is to show that everything about division polynomials holds up for characteristic $p$ and $m$ that isn't divided by $p$.

I was wondering is there a more ''fancy'' (read less computational) way to see this?

1 Answers 1

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I wish to help you. I will use algebraic geometry:

Let define a map $[m]:E(\overline{k})\to E(\overline{k})$ by form $[m]p=mp$ and we will proof the map $[m]$ is surjective homomorphism and $Ker[m]=E(\overline{k})[m]$.

$[m]$ is group homomorphism: $[m](p+p')=m(p+p')=mp+mp'=[m]p+[m]p'$

$[m]$ is surjectine homomorphism:

we have $E(\overline{k})=\{(x,y)\in \overline{k}^2; y^2=x^3+ax+b \}\cup\{\infty\}=V $ then $V$ is an algebraic set and let be $[m]:V\to W$.

Let be $q=(u,v)\in W$, then by Hilbert Nullstellensatz we have bijection between $(u,v)$ and maximal ideal $Q=(x-u,y-v) \in K[W]$, by theorem (1), we have a map $[m]^*:k[W] \to k[V] $ defined by form $[m]^*Q=[m]^*(x-u,y-v)=(x-[m]^*u,y-[m]^*v)=(x-\alpha,y-\beta)=P \in k[V]$ and from Hilbert Nullstellensatz we have bijection between maximal ideal $Q=(x-\alpha,y-\beta) \in K[V]$ and $p=(\alpha,\beta)\in V$ , then we have $\forall q\in W \exists p\in V; mp=q$ and then $mE(\overline{k})=E(\overline{k})$.

$Ker[m]=\{p\in V; [m]p=\infty \}=\{ p\in V; mp=\infty\}=E(\overline {k})[m] $

Theorem(1) Let be $V,W$ algebraic sets and $k[V], k[W]$ be coordinate rings ,if exsits $\phi:V\to W$,then exists unquie ring homomrphism $\phi^*:k[W] \to k[V]$

Note: If $k$ isnot algebraic closed, then we cannot use Hilbert Nullstellensatz theorem. Therefore $mE(k)\ne E(K)$