TITLE: When $k = \mathbb{F}_q$ finite field, $X$ always has $k$-rational point, and so $A \simeq X$?
QUESTION [11 upvotes]: Let $k$ be an arbitrary field. Let $(A, e)$ be an abelian variety over $k$, and let $X$ be a torsor for $A$, i.e. $X$ is a proper smooth $k$-variety, and there is an $A$-action acting $:A \times X \to X$ such that for any $k$-scheme $L$ and a point $x \in X(L)$, the induced "orbit" map $A_L \to X_L$ given by $a \mapsto a + x$ is an isomorphism. When $k = \mathbb{F}_q$ is a finite field, how do I see that $X$ always has a $k$-rational point, and thus $A \simeq X$?

REPLY [14 votes]: A bit of overkill, but it follows from the Weil conjectures. The structure of cohomology ($H^i = \wedge^i H^1$) is computed over the algebraic closure and it follows that the number of points is $\prod(\alpha_i-1)$ where the $\alpha_i$ are the eigenvalues of Frobenius on $H^1$ so $|\alpha_i| = q^{1/2}$ and the product is therefore not zero.