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An elliptic curve is defined over the field of real numbers:

$y^2=x^3 + ax + b$

A point P and scalar n can be multiplied using a combination of point doubling and adding.

What about point division? Given point P on the curve, what procedures exist for computing point P/n?

This question usually comes up in the context of finite fields, but I'm curious about the field of real numbers. Is division even possible in this case, and if so, under what conditions?

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    $\begingroup$ Over the reals there are elliptic logarithms and elliptic exponentials that convert the problem to division in real numbers. The only issue is that you can only divide by 2 (or any even number) if you are in the connected component of the identity. This is not an MO question, though. $\endgroup$
    – Felipe Voloch
    Commented Nov 20, 2014 at 17:02
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    $\begingroup$ If you are looking for "procedures" to compute it for general fields, then look at sage's function P.division_points(n). It starts by finding roots of the division polynomial in the field. $\endgroup$
    – Chris Wuthrich
    Commented Nov 20, 2014 at 17:21
  • $\begingroup$ @FelipeVoloch, thanks for the insight. What are the options for dividing by an odd number? $\endgroup$
    – Rich Apodaca
    Commented Nov 20, 2014 at 18:18
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    $\begingroup$ @ChrisWuthrich, sage's division_points function was helpful (although I'm not a sage user). I'm not so much interested in finding a procedure as understanding how feasible the process is. For example, how does division_points perform as n and the coordinates of P increase in value? $\endgroup$
    – Rich Apodaca
    Commented Nov 20, 2014 at 18:30
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    $\begingroup$ @RichApodaca The process is very feasible, in the following sense. I'll describe one case. Often $E(\mathbb{R})$ is isomorphic as a group to $\mathbb{R}^*/q^{\mathbb Z}$ for a real $q$ that's easy to compute to lots of decimal places. Further, the isomorphism is given by an explicit, rapidly converging series. So we have an isomorphism $f:\mathbb{R}^*/q^{\mathbb Z}\to E(\mathbb{R})$. You're taking a point $P\in E(\mathbb{R})$ and asking to solve $f(nt)=P$. So it's really a problem of numerical analysis: how easily can one solve such equations for an explicit power series $f(t)$? $\endgroup$
    – Joe Silverman
    Commented Nov 20, 2014 at 19:14

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The multiplication by $n$ map is surjective on the set of complex points, so you can always divide there, and a point has $n^2$ complex inverse images. Over the reals, $E(\mathbb{R})$ is isomorphic, as a real Lie group, to either the circle group $S^1$ or to two copies $S^1\times\mathbb{Z}/2\mathbb{Z}$. In the former case, every real point has a exactly $n$ real $n$'th roots, in the latter it depends on whether $n$ is odd or even and on whether the point is on the identity component or not. One can easily compute these roots using the standard formulas that parametrize $E(\mathbb{C})$, or even better, by an isomorphism $E(\mathbb{C})\cong\mathbb{C}^*/q^{\mathbb{Z}}$ with $q$ real. See for example Chapter V Section 2 of my book "Advanced Topics in the Arithmetic of Elliptic Curves," but of course the theory has been well known for a very long time. (Felipe is probably right, this question would be better suited for MathStackexchange.)

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