Non-isogenous elliptic curves and hyperelliptic jacobians

TL;DR

This paper investigates conditions under which hyperelliptic Jacobians are isogenous, revealing that certain cases imply CM type with specific multiplication, while others prove non-isogeny based on Galois group properties.

Contribution

It establishes new criteria for isogeny and non-isogeny of hyperelliptic Jacobians based on polynomial reducibility and Galois group structures.

Findings

If one polynomial is irreducible and the other reducible, isogenous Jacobians have CM by the n-th roots of unity.

When both polynomials are irreducible with disjoint splitting fields, certain Galois group conditions imply Jacobians are not isogenous.

The paper characterizes the CM type of Jacobians in specific algebraic settings.

Abstract

Let $K$ be a field of characteristic different from $2$, $\bar{K}$ its algebraic closure. Let $n \ge 3$ be an odd prime such that $2$ is a primitive root modulo $n$. Let $f(x)$ and $h(x)$ be degree $n$ polynomials with coefficients in $K$ and without repeated roots. Let us consider genus $(n-1)/2$ hyperelliptic curves $C_f: y^2=f(x)$ and $C_h: y^2=h(x)$, and their jacobians $J(C_f)$ and $J(C_h)$, which are $(n-1)/2$-dimensional abelian varieties defined over $K$. Suppose that one of the polynomials is irreducible and the other reducible. We prove that if $J(C_f)$ and $J(C_h)$ are isogenous over $\bar{K}$ then both jacobians are abelian varieties of CM type with multiplication by the field of $n$th roots of $1$. We also discuss the case when both polynomials are irreducible while their splitting fields are linearly disjoint. In particular, we prove that if $char(K)=0$, the Galois group of one of the polynomials is doubly transitive and the Galois group of the other is a cyclic group of order $n$, then $J(C_f)$ and $J(C_h)$ are not isogenous over $\bar{K}$.