Return to Revisions

One possible answer to this could be [Lang's theorem][1]: it says that if $G/\mathbb{F}_q$ is a smooth connected algebraic group, then $H^1(\mathbb{F}_q,G)$ is trivial, or otherwise put every $G$-torsor has an $\mathbb{F}_q$-rational point. This generalizes your example: if $X/\mathbb{F}_q$ is a smooth projective variety such that $X_{\bar{\mathbb{F}}_q}$ is isomorphic to an abelian variety, then $X$ is a torsor under its Albanese variety $A = Alb(X)$. If $X$ has dimension $1$ then requiring the genus of $X$ to be $1$ is enough. If $X$ has dimension $2$ then by the classification of surfaces it is enough to assume that e.g. the canonical bundle of $X$ is trivial and $X$ is not simply connected. [1]: https://en.wikipedia.org/wiki/Lang%27s_theorem