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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 $E(\mathbb{Q})$?

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    $\begingroup$ If I recall correctly, there was something a bit like this in Drew Sutherland's talk at the recent Cremona birthday conference, notes available here: math.mit.edu/~drew/Cremona60.pdf $\endgroup$
    – Martin Bright
    Commented Jul 13, 2016 at 9:16

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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

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    $\begingroup$ The first sentence is a bit misleading. There are plenty of infinite extensions where the curve has still finite rank. Even among the composites of quadratic extensions. $\endgroup$
    – Chris Wuthrich
    Commented Jul 13, 2016 at 11:39
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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.

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  • $\begingroup$ "and, by doing that for infinitely many $d$, you get infinite rank over your field." Can you elaborate a bit on this, please? $\endgroup$
    – user19475
    Commented Jul 13, 2016 at 10:41
  • $\begingroup$ 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. $\endgroup$
    – Chris Wuthrich
    Commented Jul 13, 2016 at 11:41
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    $\begingroup$ @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. $\endgroup$
    – Wojowu
    Commented Jul 13, 2016 at 15:18

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