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Jul 13, 2016 at 15:18 comment added Wojowu @TimoKeller If the set of points generated by these points of infinite order (each one in $\Bbb Q(\sqrt{d})\setminus\Bbb Q$ for distinct primes (say) $d$) were finitely generated, then all the generators would be elements of a single finite extension of $\Bbb Q$, which would imply that a finite extension contains infinitely many square roots of primes, which isn't the case.
Jul 13, 2016 at 11:41 comment added Chris Wuthrich Actually, I would think that this is proven. Anayltic methods should give a large density of negative $d$ for which the twist vanishes to order $1$. Then Heegner point constructions should yield a new point of infinite order.
Jul 13, 2016 at 10:50 vote accept The Thin Whistler
Jul 13, 2016 at 10:50
Jul 13, 2016 at 10:41 comment added user19475 "and, by doing that for infinitely many $d$, you get infinite rank over your field." Can you elaborate a bit on this, please?
Jul 13, 2016 at 9:15 history answered Felipe Voloch CC BY-SA 3.0