TITLE: Fields of definition of a variety
QUESTION [23 upvotes]: Let $K$ and $L$ be two subfields of some field. If a variety is defined over both $K$ and $L$, does it follow that the variety can be defined over their intersection?

REPLY [19 votes]: No, not even for genus $0$ curves, if "$X$ is defined over $K$" means that the variety $X$, initially defined over an extension $F$ of $K$, is to be isomorphic to the base extension of some variety over $K$.  (Pete in his answer was implicitly assuming that he was allowed to base extend to an algebraic closure of $F$ before going down to $K$.)
For finite extensions $F \supseteq E$ of $\mathbb{Q}_p$, if $X$ is the genus $0$ curve over $F$ corresponding to the non-split quaternion algebra, then $X$ is a base extension of a genus $0$ curve over $E$ if and only if $[F:E]$ is odd (because $\operatorname{Br}(E)[2] \to \operatorname{Br}(F)[2]$ is multiplication by $[F:E]$ from $\frac{1}{2} \mathbb{Z}/\mathbb{Z}$ to itself).
So if $F$ is an $A_4$-extension of $k:=\mathbb{Q}_2$ (such an extension exists), and $K$ and $L$ are the subfields of $F$ fixed by two $3$-cycles in $A_4$, and $X$ is the genus $0$ curve over $F$ corresponding to the non-split quaternion algebra, then $X$ is a base extension of a genus $0$ curve over $K$, and similarly over $L$, but not over $K \cap L = \mathbb{Q}_2$.