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Although I understand the Mumford representation of points on the Jacobian (of a genus 2 hyperelliptic curve), I don't understand how Magma represents such points. I would guess the confusion arises because Magma represents hyperelliptics in a weighted projective space.

Here is an example from the Magma Handbook:

Example CrvHyp_point_creation_jacobian (H125E15)

Points on $y^2 = x^6 - 3x - 1$ and their images on the Jacobian.

> _<x> := PolynomialRing(Rationals());
> C := HyperellipticCurve(x^6-3*x-1);
> J := Jacobian(C);
> ptsC := Points(C : Bound := 100);
> ptsC;
{@ (1 : -1 : 0), (1 : 1 : 0), (-1 : -1 : 3), (-1 : 1 : 3) @}
> ptsJ := [ ptsC[i] - ptsC[1] : i in [2,3,4] ];
> ptsJ;
[ (1, x^3, 2), (x + 1/3, x^3, 2), (x + 1/3, x^3 + 2/27, 2) ]

For example, how does (x + 1/3, x^3 + 2/27, 2) encode (-1:1:3) - (1:-1:0)?

The relevant page in the Magma handbook is:

http://magma.maths.usyd.edu.au/magma/handbook/text/1432#15869

Thanks in advance!

wishcow
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davidbrent
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1 Answers1

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It is probably two years too late to matter, but here is an answer to your question:

Let us denote the four points that you found by $P_{-\infty}=(1 : -1 : 0), P_{+\infty}=(1 : 1 : 0), P_1=(-1 : -1 : 3)$ and $P_2=(-1 : 1 : 3)$. Note that the points as given are in weighted homogenous space, and are NOT the coordinates for $x$ and $y$.

What you want to prove is that the divisor $D_1=P_2-P_{-\infty}$ is equal or linearly equivalent to the divisor $D_2 = <x + 1/3, x^3 + 2/27, 2>$ given in Mumford representation.

We follow the recipe in the Magma handbook http://magma.maths.usyd.edu.au/magma/handbook/text/1464#16435 :

First we need to homogenize $a(x)=x + 1/3$ to degree $2$ (this is the third component of the Mumford representation). We get $A(x,z) = xz+1/3z^2$. We homogenize $b(x)=x^3 + 2/27$ to degree $genus+1=3$ to get $B(x,z) = x^3 + 2/27z^3$.

Solving $A(x,z)=0, y=B(x,z)$ we get that if $z=0$ then $y=x^3$ so that the point $P_{+\infty}$ is a solution. If $z\neq{0}$ then we scale to $z=1$ and get $x=-1/3$ and $y=1/27$ which is point $P_2$. So the divisor represented by the Mumford representation is $D_2 = P_2 + P_{+\infty} - P_{+\infty} - P_{-\infty}$. But this is clearly equal to $D_1$.

wishcow
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