Let $V$ be an algebraic variety over a field $K$. Is there a constant $d = d(V) \in \mathbb{N}$ such that for any variety $W$ defined over $K$ and isomorphic to $V$ over the algebraic closure of $K$, there is a field extension $F/K$ with $[F:K] \leq d$ such that $V$ and $W$ are isomorphic over $F$? 



Minimal fields of isomorphism for varieties
Yes if $K=\mathbb{R}$ for example, but no in general.


Namely this fails for curves of genus $1$, over $\mathbb{Q}$, say. Given an elliptic curve $E$ over $\mathbb{Q}$ and a positive integer $d$, a general $d$-covering of $E$ will contain a divisor of degree $d^2$, but no divisors of smaller degree. In particular, it will only become isomorphic to $E$ over a field extension of degree $d^2$.


