Let $X,Y$ be smooth projective varieties over a field $k$. Assume that there is a finite Galois extension $L/k$ such that there is an isomorphism $f : X \times_k L \cong Y \times_k L$.
Is it true that $X$ and $Y$ are birationally equivalent over $k$?
Notice that $f$ does not need to be the base change of a morphism $X \to Y$. Also, it does not need to be Galois-equivariant. I don't know if one could see $X$ as a Zariski-dense subscheme of $X \times_k L$, because I only know that fiber products provide projection maps $X \times_k L \to X$ (so it does not go in the right direction…).
Best Answer
No. Consider $k=\Bbb R$, $L=\Bbb C$, with $X=\Bbb P^1_{\Bbb R}$ and $Y=\operatorname{Proj} \Bbb R[x,y,z]/(x^2+y^2+z^2)$. After base change to $\Bbb C$, both $X$ and $Y$ are isomorphic to $\Bbb P^1_{\Bbb C}$. On the other hand, the $\Bbb R$-rational points of $X$ are dense while $Y$ has no $\Bbb R$-rational points. Any $\Bbb R$-morphism preserves $\Bbb R$-rational points, so no open subset of $X$ can be isomorphic to any open subset of $Y$.