I am doing a small project using elliptic curve in cryptography. My doubt is, can I directly convert a projective to a Jacobian coordinate system without using the affine conversion in elliptic curve cryptography? Actually, I am stuck at this point…


1 Answer 1


If the question is how to take a point in projective coordinates (a triplet $(A,B,C)$, where the corresponding affine coordinate is $(AC^{-1}, BC^{-1})$), and you want to convert it into the same point in Jacobian coordinates $(D, E, F)$ (where the affine coordinate is $(DF^{-2}, EF^{-3})$), I believe this works:

$$D = AC$$

$$E = BC^2$$

$$F = C$$

We can easily see that $DF^{-2} = AC^{-1}$ and $EF^{-3} = BC^{-1}$, hence $(D,E,F)$ refers to the same point as the original, and this conversion takes only three multiplies.

  • $\begingroup$ Thank you @poncho. To reverse the operation will the below equations work? $$ A = DF $$ $$ B = E $$ $$ C = F^{3} $$ $\endgroup$
    – vijita
    Jun 30, 2014 at 8:39
  • $\begingroup$ @vijita: that reversed operation also preserves $DF^{-2}=AC^{-1}$ and $EF^{-3} = BC^{-1}$, and so, yes, those equations also "works". $\endgroup$
    – poncho
    Jun 30, 2014 at 12:42

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