How to construct random divisor once the set of rational points over the Jacobian of a hyperelliptic curve is created?

s3binator
33 ●5

I know how to deliberately create a divisor with knowledge of the mumford coordinates, but is there a way to generate a divisor randomly inside of X = J(FF) and extract its mumford coordinates as polynomials over FF?

Where divisors D = X([a,b]) are elements of X = J(FF), a and b are the mumford polynomials, and J is the Jacobian of a hyperelliptic curve.

Thanks!

answered 9 years ago

nbruin
4153 ●3 ●45 ●91

I assume FF is a finite field. One approach is to randomly choose the polynomial $a$ and then try and solve for a polynomial $b$ . The condition you have to meet is that for every irreducible factor $x_i$ of $a$ , a root $r_i$ of $a_i$ is an $x$ -coordinate of a point on the hyperelliptic curve defined over $FF(r_i)$ , say with $y$ -coordinate $y_i$ . So the probability you choose a polynomial $a$ that works is $(1/2)^m$ , where $m$ is the number of irreducible factors of $m$ .

Recovering $b$ is now a matter of interpolating the points $(x_i,y_i)$ and their conjugates. Make sure to properly randomize the sign choice in the square roots that you end up taking.

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How to construct random divisor once the set of rational points over the Jacobian of a hyperelliptic curve is created?