Let $k$ be an algebraically closed field of characteristic $0$, let $C\_{/k}$ be a nice (smooth, projective, geometrically integral curve), let $K = k(C)$, and let $\overline{K}$ be an algebraic closure of $K$. Let $E\_{/K}$ be an elliptic curve with $j(E) \notin k$. Let $P \in E(\overline{K}) \setminus E(K)$ be a point of infinite order. Let $K(P)$ be the field of definition of $P$ (equivalently, the field obtained by adjoining to $K$ the coordinates of $P$ in a Weierstrass equation for $E$). Then 


$\langle P, E(K) \rangle \subset E(K(P))$. 


Must we have equality? 


Comments:


1) For my application, I may assume that $E\_{/K}$ is semistable, so please feel free to address the question under that additional hypothesis if it helps. (But I don't see how it does...)


2) I am not able to assume anything about $C$.


3) For my application, I have already dealt with the case in which $P$ has finite order, and in that case I could assume that $C = X\_m(n)$ -- the elliptic modular curve parameterizing $\mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$ torsion structures -- and $E$ is the universal elliptic curve over $C$ (the four pairs $(m,n) \in \{(1,1), (1,2), (1,3), (2,2)\}$ in which there is no universal elliptic curve are excluded). In this case the result follows from work of Shioda and Cox-Parry. 


4) If this is true, it seems to be closely related to Shioda's landmark 1972 paper on elliptic modular surfaces. I confess that I am asking this question before I have fully absorbed this important paper: I have several collaborators who would be happy if my plate were cleaner. 


**Added**: If I may, I'll try one variant of the question: what if $[n]P \in E(K)$ for some $n \in \mathbb{Z}^+$? 



Mordell-Weil of an elliptic surface after adjoining a nontorsion section: as small as possible?
I do not believe you have equality in general. I sketch a counterexample below, which is a geometric version of the fact that if $E/K$ is an elliptic curves such that the quadratic twist $E^{(d)}/K$ has Mordell-Weil rank at least 2 then the Mordell-Weil of $E/K(\sqrt{d})$ is at least two bigger than the Mordell-Weil rank of $E/K$.


Take an elliptic K3 surface $X\_0$ with two $I\_{\nu}^\*$-fibers, Mordell-Weil rank at least two and such that all further singular fibers are semistable. (Such surfaces exist by the surjectivity of the period map for K3 surfaces). Take now the quadratic twist $X$ of $X\_0$ by the two \*-fibers. Then $X$ is a rational elliptic surface and all singular fibers are semistable. 


The base curve $C$ of each of the elliptic fibrations is $\mathbb{P}^1$. 
Let now $C'=\mathbb{P}^1$ and let $C'\to C$ be the double cover ramified over the two points where the $I\_{\nu}^\*$ fibers are.


Then the relatively minimal smooth models of $X\times\_{C}C'$ and $X\_0\times\_{C} C'$ coincide and call this model $Y$. The Mordell-Weil rank of $Y$ equals the sum of the Mordell-Weil ranks of $X$ and $X\_0$ and hence the elliptic fibration $X\to C$ is a counterexample. 


[If you assume that $X$ has at least 6 singular fibers then $X\_0$ is also a counterexample, but $X\_0$ is not semistable.]


