Mordell-Weil rank of an elliptic curve over $\mathbb{Q}(\sqrt{-1},\sqrt{2},\sqrt{3},\sqrt{5},...)$? 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 $\mathbb{Q}$.
The picture is less clear if $K$ is infinite-dimensional over $\mathbb{Q}$.
I believe the best result is that of Kobayashi who proved (modulo the usual conjectures on Hasse-Weil $L$-functions and BSD) that $\operatorname{rank}(E(\mathbb{Q}^{\operatorname{ab}}))=\infty$.
What is known about the rank of $E$ over $\mathbb{Q}(\sqrt{-1},\sqrt{2},\sqrt{3},\sqrt{5},...)$?
EDIT: I have a follow-up question that is more precise:
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 $E(\mathbb{Q})$?
 A: The rank is probably unbounded. If you can arrange for the quadratic twist $E_d$ of $E$ to have positive rank (e.g. by making the sign of the functional equation be $-1$ and assuming BSD) then $E$ has a point of infinite order in $\mathbb{Q}(\sqrt{d})$ and, by doing that for infinitely many $d$, you get infinite rank over your field.
A: I think the answers and comments in Argument for unboundedness of integral points of elliptic curves over number fields show the rank is unbounded in finite extensions of the rationals and in infinite extensions it is infinite.
Basically the question takes $f(x)=y^2$ and takes points $x,\sqrt{f(x)}$, giving example.
JSE's answer shows linear independence of the example, which scales infinitely as far as I can tell.
JSE's comment about unboundedness of rank in number fields:

It should do. Just take your r points defined over disjoint quadratic fields, whose compositum is K; the MW rank over THAT field is finite, so you can certainly choose an integer x such that (sqrt(f(x))) generates a further quadratic extension of K, then you have rank r+1 and you just keep going. – JSE Apr 28 '14 at 17:33 

