ag.algebraic geometry - 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.