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In cryptography, it seems to be a common choice to use the so-called Jacobian coordinates to represent a point of an elliptic curve (see e.g. Elliptic Curves: Number Theory and Cryptography, L. C. Washington).

These coordinates differ from the standard projective coordinates in that we have $(x : y : z) ∼ (λ^2x : λ^3y : λz)$ instead of $(x : y : z) ∼ (λx : λy : λz)$.

As explained in this MIT course, using these coordinates, the multiplication-by-$n$ map has an expression in terms of polynomials: $nP = (φ_n : ω_n : ψ_n)$ where $φ_n, ω_n, ψ_n$ are polynomial expressions in the coordinates of a point $P$ which can be defined by induction.

Translating to the standard coordinates, that gives us the so-called division polynomials.

My questions are:

  • What is the motivation for using Jacobian coordinates to derive an expression for the multiplication-by-$n$ map? (compared, for instance, to using the projective coordinates)
  • Is this construction (Jacobian coordinates) part of a more general mechanism (e.g. in the study of Abelian varieties)?
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    $\begingroup$ The multiplication by $n$ is expressed with polynomials in standard projective space, too. In any embedding of the curve in a projective space, since it is an algebraic group. Now I don't understand your questions. $\endgroup$ Commented Oct 5 at 18:48
  • $\begingroup$ @ChrisWuthrich Indeed, we can always multiply by the denominators to have polynomial expressions in projective space. I reformulated my question to state clearly what I want to know: why use Jacobian coordinates specifically (and not standard projective coordinates)? $\endgroup$ Commented Oct 5 at 21:16
  • $\begingroup$ The answer is in the linked course notes. Although the formula for the sum in Jacobian coordinates is awful, there is equation (2) which expresses the $z$-coordinate of the sum in a simple way. That equation inspires the recursion formula for $\psi_n$. $\endgroup$ Commented Oct 6 at 9:39
  • $\begingroup$ @ChrisWuthrich I see, but I was wondering if there was a generic argument to chose the right weighted projective space. $\endgroup$ Commented Oct 6 at 10:19
  • $\begingroup$ I doubt it. For a smooth cubic in Weierstrass form the Jacobian coordinates are quite obvious. By Riemann-Roch, you have a function $x$ with a pole of order $2$ and another $y$ with a pole of order $3$ at $O$, which gives this choice. That is the underlying reason for (1:2:3). $\endgroup$ Commented Oct 6 at 10:50

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Given a projective variety $X$ and an ample line bundle $L$, the section ring $\bigoplus_{n=0}^\infty H^0(X, L^n)$ is finitely generated. We can always find a minimal set of homogeneous generators, say $f_1,\dots, f_r$, with $f_i \in H^0(X, L^{n_i})$.

Then $(f_1,\dots, f_r)$ defines an embedding of $X$ into weighted projective space with weights $n_1,\dots, n_r$. This gives a natural embedding of $X$ into weighted projective space.

For $X$ an elliptic curve, we can take $L$ any line bundle of degree $1$, which is necessarily of the form $\mathcal O(P)$ for a point $p$. The section ring is minimally generated by elements in degree $1,2,$ and $3$ (by the Riemann-Roch argument Chris Wuthrich mentions), justifying the Jacobian coordinates embedding.

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