I hope this is a good question.

Recently I worked with genus two curves $H$ that have multiplication by $[\zeta_5]\in \text{Aut}(H)$, that is, multiplication by $e^{2\pi i/5}$. This automorphism is naturally extended to the Jacobian of $H$. Since $\zeta_5 + \zeta_5^{-1}=\tfrac{-1+\sqrt{5}}{2}$ and the Endomorphism ring of the Jacobian of $H$ is isomorphic to $\mathbb{Z}[\zeta_5]$ there is a $\sqrt{5}$ isogeny, which I computed explicitly to work over Finite fields using the curve $y^2 = x^5 + c$.

The question is, I was reading Paul Gaudry's paper on how to do scalar multiplication on the Kummer Surface. In his paper he uses Rosenhain form of hyperelliptic curves (and my curve is not in that form obviously).


I am a little lost but, can someone point me out if I can do this scalar multiplication by $\sqrt{5}$ using the Kummer surface ? More precisely, if $D_1,D_2\in\mathcal{J}$ are divisors in the Jacobian of a hyperelliptic curve of genus 2. Is there a way to obtain analog formulas like the biquadratic forms for $D_1+D_2$ or $2D_1$ in the Kummer surface but for $\sqrt{5}D_1$ in the Kummer?


  • $\begingroup$ It is not clear what you are asking for. What means "doing this scalar multiplication"? $\endgroup$ – js21 Nov 14 '17 at 10:44
  • $\begingroup$ What I mean, if it is possible to get formula for $[\sqrt{5}]$D in the Kummer Surface in an analog way multiplication by $2$ or $n$ are obtained by Gaudry using biquadratic forms. I edited the last part of my question. Thanks for the remark $\endgroup$ – Eduardo R. Duarte Nov 14 '17 at 10:57
  • $\begingroup$ There definitely is a formula that gives multiplication by $\sqrt{5}$ on the Kummer surface; it will be given by homogeneous polynomials of degree 5. If I have some time, I'll try to figure them out for $y^2 = x^5 + c$ (not today, though). $\endgroup$ – Michael Stoll Nov 14 '17 at 15:55
  • $\begingroup$ @MichaelStoll In fact, I was thinking to use the usual bilinear forms $\Phi_{ij}$ found in Cassels' Prolegomena. If $D$ is the generic point of the Jacobian, I could get the divisors $A:=2\zeta_5 +D$ and $B:=2\zeta_5^{-1}$ , we have that $A+B=\sqrt{5}D$. But I think this could be done directly as you say with degree 5 polynomials. $\endgroup$ – Eduardo R. Duarte Nov 14 '17 at 17:27

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