It is known that the group of $K$-rational points of an elliptic curve $E$ is finitely generated if $K$ is a number field of finite degree over $\mathbb{Q}$.
Much less is known if $K$ is infinite-dimensional over $\mathbb{Q}$.


The rank of $E$ over $\mathbb{Q}(\sqrt{-1},\sqrt{2},\sqrt{3},\sqrt{5},...)$ is infinite. It would however suffice for my purposes if I knew that for every prime $p$ there is an infinite family $Q$ of primes, $p\notin Q$, so that:


(1) The rank of $E(\mathbb{Q}(\sqrt{-q}))$ equals that of $E(\mathbb{Q})$ for all $q\in Q$.


(2) $p$ has the same prescribed splitting behavior (i.e. inert, split or ramified) in $\mathbb{Q}(\sqrt{-q})/\mathbb{Q}$ for all $q\in Q$.


Is it known that such a family exists?



Is there an infinite family of primes $q_{1},q_{2},...$ so that the rank of $E(\mathbb{Q}(\sqrt{-q_{i}}))$ equals that of $E(\mathbb{Q})$?
Pasten has answered the question: Murty (MS1106677, Corollary to Theorem 2) has shown that the quadratic twist of $E$ by a prime $q$ has rank zero for infinitely many primes $q$, if GRH holds.


