English

Non-isogenous elliptic curves and hyperelliptic jacobians

Number Theory 2022-03-04 v4 Algebraic Geometry

Abstract

Let KK be a field of characteristic different from 22, Kˉ\bar{K} its algebraic closure. Let n3n \ge 3 be an odd prime such that 22 is a primitive root modulo nn. Let f(x)f(x) and h(x)h(x) be degree nn polynomials with coefficients in KK and without repeated roots. Let us consider genus (n1)/2(n-1)/2 hyperelliptic curves Cf:y2=f(x)C_f: y^2=f(x) and Ch:y2=h(x)C_h: y^2=h(x), and their jacobians J(Cf)J(C_f) and J(Ch)J(C_h), which are (n1)/2(n-1)/2-dimensional abelian varieties defined over KK. Suppose that one of the polynomials is irreducible and the other reducible. We prove that if J(Cf)J(C_f) and J(Ch)J(C_h) are isogenous over Kˉ\bar{K} then both jacobians are abelian varieties of CM type with multiplication by the field of nnth roots of 11. 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)=0char(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 nn, then J(Cf)J(C_f) and J(Ch)J(C_h) are not isogenous over Kˉ\bar{K}.

Keywords

Cite

@article{arxiv.2105.03783,
  title  = {Non-isogenous elliptic curves and hyperelliptic jacobians},
  author = {Yuri G. Zarhin},
  journal= {arXiv preprint arXiv:2105.03783},
  year   = {2022}
}

Comments

24 pages. We include new results and examples. 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 an algebraic closure of the field $K$

R2 v1 2026-06-24T01:54:31.725Z