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Nov 10, 2018 at 12:34 comment added Chris Wuthrich @deanjana That is a completely different question. Wlog we may also fix the Galois group $G$ of the Galois closure of $K$ and even the representation $\rho$. Then you ask how often is the $\rho$-part of the Mordell-Weil group of dimension larger than $\dim\rho$. Root numbers sometimes force rank growth, but that will never force more than one copy of $\rho$ I believe. I would tend to believe that it is very hard to find growing fine Mordell-Weil groups. So maybe yes, the corank of your kernel stays bounded. I doubt one can prove much about this.
Nov 9, 2018 at 15:04 comment added debanjana Since you mentioned that this map is as injective as it can be, should one expect that for elliptic curves $E/\mathbb{Q}$ of rank 0 or 1, as one varies over those $K/\mathbb{Q}$ of fixed degree where the rank of $E$ is still $\leq 1$, the kernel stays bounded?
Nov 9, 2018 at 12:57 vote accept debanjana
Nov 9, 2018 at 12:12 comment added Chris Wuthrich .. and to answer the actual question. A condition on $K$ only will never do. Just take a rank $2$ curve over $\mathbb{Q}$ then the map will be non-injective for all $K$.
Nov 9, 2018 at 9:46 history answered Chris Wuthrich CC BY-SA 4.0