Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is it always possible to find an isogeny from a hyperelliptic curve of genus 4, to a 'normal' elliptic curve (genus 1), or a product of elliptic curves?

Are such isogenies easy to compute?

share|cite|improve this question
What is an isogeny from a hyperelliptic curve? – Alex B. Sep 13 '11 at 7:54
@Alex B.: it's an isomorphism from its jacobian. – ted.k Sep 13 '11 at 7:56
up vote 4 down vote accepted

The short answer is "no". The Jacobian of a genus 4 curve is an abelian variety of dimension 4, so it cannot be isogenous to an abelian variety of dimension 1. However, it may happen, that it is isogenous to a product of 4 elliptic curves. This should be regarded as the exception, rather than the generic case.

For example, let us consider the simpler case of a genus 2 hyperelliptic curve $C$. As the answers in this related MO question explain, there is a degree $N^2$ isogeny from the Jacobian of such a curve to the product of two elliptic curves if and only if there is a degree $N$ map from $C$ to an elliptic curve. This is a pretty severe restriction. For example for $N=2$, this is equivalent to $C$ having a model of the form $y^2 = f(x^2)$ where $f$ is a cubic. On the other hand, an arbitrary hyperelliptic curve of genus 2 is given by $y^2=g(x)$, where $g$ is any degree 6 polynomial (in both cases no repeated roots and non-zero at 0). Most hyperelliptic curves won't admit any non-trivial morphism to an elliptic curve at all.

In this paper, examples are given for Jacobians of higher genus curves that are isogenous to products of elliptic curves.

I don't know much about the computational aspect of finding such an isogeny when it exists, so maybe somebody else will address that.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.