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There is a statement about the number of division points, which I've read in a few papers, but it never seems to have any references where it comes from or why it is true. The statement is the following:

Let $[n]$ be the multiplication-by-n map and let $Q\in E(\mathbb{\overline K})$ be a point on an elliptic curve, then there are $n^2$ points $P_i$, where $i=1,\dots,n^2$, for which it holds that $$ [n]P_i = Q. $$

I am currently working with cyclic groups on elliptic curves over finite fields, and this statement doesn't really seem to hold there, or maybe I'm just understanding it wrong. But let me make an example, and maybe you'll figure out what I'm doing wrong.

So let $$\langle A \rangle=\{A,[2]A,[3]A,[4]A,[5]A,[6]A,[7]A,O\}$$ be a cyclic group of order 8 on $E$ generated by $A$, where $O$ is the neutral element (point at infinity). If, for example, I choose $Q=[5]A$ and $n=3$, I can only find a single $P_i$ that satisfies the upper equation, namely $$ [3]P_i = [3]([7]A) = [5]A = Q. $$

My question is, where are the rest of the $n^2=9$? Can there even be 9 when my group only has eight elements? Or can these elements be in different groups? Intuitively, if they are points of the same order, they should be in the same cyclic subgroup, right? What if $\langle A\rangle$ is the largest order subgroup on $E$?

Or do you know the source/proof of this statement by any chance? The papers I have found this statement in are the following

Thank you for any help!

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    $\begingroup$ I guess you need to consider the algebraic closure of ${\Bbb K}$. $\endgroup$ – Wuestenfux May 10 at 11:50
  • $\begingroup$ For the notation : given an elliptic curve $E/\Bbb{F}_p : y^2=x^3+ax+b$ then $E[n] = \{ P \in E(\overline{\Bbb{F}_p}), nP = O\}$ and $E(\Bbb{F}_{p^k})[n] =\{ P \in E(\Bbb{F}_{p^k}), nP = O\}$ the latter has less than $n^2$ points, the former has $n^2$ points whenever $p \nmid n$, the proof being that by induction $\deg(n) = n^2$ and $n$ is a separable endormophism because it is not of the form $f \circ \phi$ with $\phi(x,y) = (x^p,y^p)$ since the latter has degree $p$ $\endgroup$ – reuns May 10 at 12:23
  • $\begingroup$ I changed it to be defined over the algebraic closure, but I don't see how this affects my question. Or am I missing something? $\endgroup$ – Gemeis May 11 at 16:52
  • $\begingroup$ anything still unclear ? without knowing the curve you are working with we can't help more.. $\endgroup$ – reuns May 14 at 1:02
  • $\begingroup$ @reuns The question is independent of the underlying curve, but rather a property of the multiplication-by-m map, which is surjective. And I can see why it works for torsion points, as you pointed out, but I don't see how it can be taken as a general assumption or how it would work in the example I provided. $\endgroup$ – Gemeis May 14 at 19:21
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Try with $E:y^2=x^3+x+1$ over $\Bbb{F}_5$ in http://magma.maths.usyd.edu.au/calc/

       p := 5; Fp := GF(p);
    E := EllipticCurve([Fp | 1, 1]); E;
    O := Points(E)[1]; DivisionPoints(O,3);

    #DivisionPoints(O,3);

3

    q := 5^2; Fq<a> := GF(q); E_Fq := BaseChange(E,Fq); 
    O_Fq := Points(E_Fq)[1]; DivisionPoints(O_Fq,3);

    #DivisionPoints(O_Fq,3);

9

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