I hope this question is good to be here.

Let $J$ be the Jacobian of a hyperelliptic curve $H$ of genus $2$ given by $y^2 = x^5 + h$.

I was calculating an explicit formula in Mumford coordinates of $[\sqrt{5}]\in \text{End}(J)$. When I say Mumford coordinates I mean divisors $D:=(x_1+x_2, x_1x_2, \tfrac{y_1-y_2}{x_1-x_2},\tfrac{x_2y_1-x_1y_2}{x_1-x_2})=:(A,B,C,D)\in J$ where $\text{Supp}(D)=\{(x_1,y_1), (x_2,y_2)\}$ and $(x_i,y_i)\in H$.

I used these coordinates in order to work with MAGMA.

The thing is, I got the formulae, the problem is that I do not know how to handle an special situation.

What I did is to calculate $[\zeta_5+\zeta_5^4] \in \text{Aut}(J)$ where the action of $\zeta_5^i$ on the curve $H$ is given by $\zeta_5^i(x,y)=(\zeta_5^i x,y)$ and then extend to the whole Jacobian.

After this, I calculated the formula for $2[\zeta_5+\zeta_5^4]+1=[\sqrt{5}]$ using Mumford coordinates and then if $D=(A,B,C,D)$ then I got $\sigma_i\in k(J)$ where $[\sqrt{5}](D)=(\sigma_1(A,B,C),\sigma_2(A,B,C), \sigma_3(A,B,C), \sigma_4(A,B,C))\in J$

Also I have formulae when the input divisor is of the form $[(x,y)-\infty]\in \Theta\subset J$

My problem is when the input is not in $\Theta$ (an ordinary, generic Divisor) but the output is in $\Theta$.

My question is, if $D:=(A,B,C,D)$ how can I compute if $[\sqrt{5}](D)\in \Theta$, and more... how can I compute the resulting divisor?

Detecting it I think is easy, I just need to calculate a determinant of a $4\times 4$ matrix in terms of $A,B,C$. But I do not know how to compute the divisor using my coefficients $(A,B,C,D)$. Is there a way to get explicit formulae for this "exceptional case" when is already known that the image will be in $\Theta$?



Here is a trick that might help: Compute $\sqrt{5}D$ as $\sqrt{5}(D+R)-\sqrt{5}R$ where $R$ is in the large set you know how to compute.

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  • $\begingroup$ Thanks for this suggestion @Watson, but the problem if I understood correctly your suggestion, is the same, since the resulting divisors in the substraction will be fine, but calculating the difference, even if I fix $R$ wont give me information about $\sqrt{5}D$ since its support may be different from the support of $R$ $\endgroup$ – Eduardo R. Duarte Aug 28 '17 at 9:36
  • $\begingroup$ So there is an explicit algorithm for doing computations in the Jacobian of a hyperelliptic curve which does normalize the divisors. $\endgroup$ – Watson Ladd Aug 28 '17 at 16:03

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