Let $K/\mathbb{Q}$ be an algebraic extension, and let $E\_1,E\_2/\mathbb{Q}$ be elliptic curves. Is it possible that the Mordell-Weil rank of $E\_1(K)$ is finite while that of $E\_2(K)$ is infinite?



Ranks of elliptic curves depend only on the field?
The answer is yes, at least if you assume that Tate-Shafarevich conjecture.


Let $E$ be an elliptic curve over a number field $k$. Under some mild hypothesis (see Corollary 1.10 of <http://arxiv.org/abs/0904.3709>) there exist quadratic twists of $E$ which have trivial Mordell-Weil rank. In other words, there exists quadratic extensions in which the rank of $E$ does not go up. Assuming in addition the Tate-Shafarevich conjecture there are also quadratic twists of $E$ whose Mordel-Weil rank is $1$ (see Corollary 1.12 loc. cit.), and hence quadratic extensions in which the rank of $E$ goes up. Generalizing the methods of loc. cit. along the line of <http://arxiv.org/abs/1504.02343> one can show that (assuming the Tate-Shafaretich conjecture) if $E\_1$ and $E\_2$ are elliptic curves over $k$ satisfying certain independence conditions then one can find a quadratic extension in which the rank of $E\_1$ stays the same and the rank of $E\_2$ goes up. Applying this construction successively one can construct an infinite extension $K/k$ such that the rank of $E\_1(K)$ is finite and the rank of $E\_2(K)$ is infinite.


