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?



Fields of definition of a variety
Yes, if varieties are interpreted as subvarieties closed subschemes of base extensions of a fixed ambient variety scheme (e.g., affine space or projective space).


More precisely, suppose that $k \subseteq F$ are fields and the variety $X$ is an $F$-subvariety a closed subscheme of $\mathbf{P}^n\_F$. Say for a field $K$ with $k \subseteq K \subseteq F$ that "**$X$ is defined over $K$**" if $X$ is the base extension of some subvariety of $\mathbf{P}^n\_K$. Then $X$ has a minimal field of definition $E$ with $k \subseteq E \subseteq F$, characterized by the property that for any field $K$ with $k \subseteq K \subseteq F$, we have that $X$ is defined over $K$ if and only if $K$ contains $E$.


The same statement holds if $\mathbf{P}^n$ is replaced by any fixed $k$-variety $k$-scheme $Y$.


(Note: this answer does not contradict Pete's. This is just a different interpretation of the question.)


**EDIT:** As Brian points out, I was indeed assuming that my varieties were closed in the ambient space. The statement about minimal field of definition is not even true for open subschemes in characteristic $p$. For example, if $k=\mathbb{F}\_p$ and $F=k(t)$ and $Y=\operatorname{Spec} k[x]$ and $X=\operatorname{Spec} F[x,1/(x-t)]$, then $X$ is the base extension of $\operatorname{Spec} F^{p^n}[x,1/(x^{p^n}-t^{p^n})]$, and hence is definable over $F^{p^n}$ for all $n$, but not over the intersection of all these fields, which is just $k$. 


On the other hand, the intersection of any *finite* number of fields of definition is still a field of definition.


I have generalized to schemes as suggested by Brian.


