Constructing Jacobians of rank 1

Let K be a number field, let g ≥ 1 be an integer and let f(x) = (x - a₁) ⋯ (x - a2g + 1) ∈ OK[x] 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 $1$.