Constructing Jacobians of rank 1

Abstract Let 𝐾 be a number field, let g ≥ 1 g\geq 1 be an integer and let f ⁢ ( x ) = ( x − a 1 ) ⁢ ⋯ ⁢ ( x − a 2 ⁢ g + 1 ) ∈ O K ⁢ [ x ] see text f(x)=(x-a_{1})\cdots(x-a_{2g+1})\in O_{K}[x] be a polynomial that splits into 2 ⁢ g + 1 2g+1 distinct linear factors. Write 𝐶 for the hyperelliptic curve given by C : y 2 = f ⁢ ( x ) C:y^{2}=f(x) and write J = Jac ⁡ ( C ) J=\operatorname{Jac}(C) for its Jacobian. Under mild technical assumptions on 𝑓 that are satisfied almost always, we prove that there exists some d ∈ K × d\in K^{\times} such that the quadratic twist J d J^{d} has rank exactly equal to 1. As a consequence, we deduce that, for any positive integer 𝑔, there exists an absolutely simple abelian variety over 𝐾 with dimension equal to 𝑔 and rank equal to 1.