Let $X$ be the Fermat quartic $x^4+y^4+z^4+w^4=0$ in $\mathbb P^3$. It is known that $X$ contains infinitely many $(-2)$-curves, that is, smooth rational curves. (One way to obtain in infinitely many is to use the various elliptic fibrations on $X$, and use translations in the fibers.) Note however that these curves are typically not defined over $\mathbb Q$, even though $X$ is. My question is the following: 


*Is there a finite extension $K$ of $\mathbb Q$ such that all of the smooth rational curves are defined over $K$ and have a $K$-rational point?*


I suspect that the answer is no, but I don't see how to prove it.



Rational curves on the Fermat quartic surface
I am posting my comments as an answer. I am concerned that I misunderstand the OP, so let me state first the result. There exists a finite field extension $K/\mathbb{Q}$ such that for every closed immersion $\mathbb{P}^1\_\mathbb{C} \hookrightarrow X\otimes\_{\mathbb{Q}}\mathbb{C}$, there exists a closed immersion $\mathbb{P}^1\_K \hookrightarrow X \otimes\_{\mathbb{Q}} K$ with the same image after base change to $X\otimes\_{\mathbb{Q}}\mathbb{C}$. 


To prove this, first observe that the geometric Picard group $\text{Pic}(X\otimes\_{\mathbb{Q}}\mathbb{C})$ is finitely generated. Thus, there exists a finite field extension after which a finite generating set of the geometric Picard group is defined over that field. 


Second, the geometric automorphism group of the complex scheme, $\text{Aut}\_{\mathbb{C}}(X\otimes\_{\mathbb{Q}}\mathbb{C})$, is a discrete group that is finitely generated by Corollary 2.4 of Chapter 15, p. 315 of the following; result due to H. Sterk -- Finiteness results for algebraic K3 surfaces. Math. Z., 189(4):507–513, 1985. 


Daniel Huybrechts   

*Lectures on K3 Surfaces*   

Part of Cambridge Studies in Advanced Mathematics   

September 2016   

ISBN: 9781107153042   

<http://www.math.uni-bonn.de/people/huybrech/K3Global.pdf>


Thus, after a further finite field extension, there exists a finite subset of $\text{Aut}\_K(X\otimes\_{\mathbb{Q}} K)$ whose images generate the geometric automorphism group. Thus, $\text{Aut}\_K(X\otimes\_{\mathbb{Q}} K)$ is already the full geometric automorphism group.


Third, by Corollary 4.6 of Chapter 8, p. 161 -- result again proved by Sterk in the article cited above -- the action of the geometric automorphism group on the set of smooth rational curves has only finitely many orbits. For each of those finitely many orbits, after a degree $2$ field extension, every curve in that orbit acquires a rational point. Thus, after a further finite field extension, every smooth, genus $0$ curve on $X$ acquires a rational point.


Please note, all of this is valid for every K3 surface over every characteristic $0$ field, not only for the Fermat quartic surface over $\mathbb{Q}$. Using extensions of Sterk's results by Lieblich and Maulik, this should also be true in positive characteristic.


