Let C be a geometrically integral curve over a number field K and let K' be a number field containing K. Does there exist a number field L containing K such that


* $L \cap K' = K$, and
* $C(L) \neq \emptyset$?


Note that the hypotheses on C are necessary -- the curve x^2 + y^2 = 0, with the origin removed, is not geometrically integral, but gives a counterexample for K = Q and K' = Q(i).


Also, I can prove that this is true when C has prime gonality. It would be odd, though, for this to be a necessary hypothesis.



Existence of points on varieties which avoid a given number field.
Yes, this follows from a Theorem of Moret-Bailly, see for example Corollary 1.5


<http://math.stanford.edu/~conrad/vigregroup/vigre05/mb.pdf>


Roughly speaking, given a finite set $S$ of primes with $C(K\_v)$ is non-empty,
this produces a field $L$ with $C(L) \neq \emptyset$ and $L\_v = K\_v$ for all $v \in S$.


To guarantee that $L \cap K' = K$, one may as well assume that $K'/K$ is Galois
with Galois group $G$. Then for every conjugacy class $g \in G = \mathrm{Gal}(K'/K)$,
let $v$ be a prime such that $\langle \mathrm{Frob}\_v \rangle = \langle g \rangle \in G$
and $C(K\_v) \ne \emptyset$. (The existence of such $v$ follows from Cebotarev, the Weil conjectures, and Hensel's Lemma.) If $S$ is the resulting set, then one may find $L$
with $C(L)$ non-empty and $L\_v = K\_v$ for all $v \in S$, and so (by Cebotarev) that
$L \cap K' = K$.


This theorem gets used all the time in "potential modularity" theorems.


