Let K be a number field, let g 1 be an integer and let f (x) = (x - a₁) (x - a₂₆ + ₁) OKx be a polynomial that splits into 2g + 1 distinct linear factors. Write C for the hyperelliptic curve given by C: y² = f (x) and write J = Jac (C) for its Jacobian. Under mild technical assumptions on f that are satisfied almost always, we prove that there exists some d K^ such that the quadratic twist Jᵈ 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.
Koymans et al. (Mon,) studied this question.