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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?

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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

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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.

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