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

asked 2016-04-26 19:06:01 +0200

s3binator gravatar image

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.


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answered 2016-04-27 21:00:21 +0200

nbruin gravatar image

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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Asked: 2016-04-26 19:06:01 +0200

Seen: 278 times

Last updated: Apr 27 '16