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 $E(\mathbb{Q})$?](https://mathoverflow.net/questions/244230/is-there-an-infinite-family-of-primes-q-1-q-2-so-that-the-rank-of-e)



Mordell-Weil rank of an elliptic curve over $\mathbb{Q}(\sqrt{-1},\sqrt{2},\sqrt{3},\sqrt{5},...)$?
I think the answers and comments in [Argument for unboundedness of integral points of elliptic curves over number fields](https://mathoverflow.net/questions/164629/argument-for-unboundedness-of-integral-points-of-elliptic-curves-over-number-fie) 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:



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> 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 
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