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 $A \simeq X$?

This is a theorem of Lang's from 1956. Here's an online document giving a proof (in the form $H^1(A,k)=0$):

Lecture 14: Galois Cohomology of Abelian Varieties over Finite Fields, William Stein. http://wstein.org/edu/2010/582e/lectures/582e-2010-02-12/582e-2010-02-12.pdf

Stein notes that there is a "more modern proof" in the first few sections of
Chapter VI of Serre's
*Algebraic Groups and Class Fields*.

The original article is Serge Lang, "Abelian varieties over finite fields," *Proceedings of the National Academy of Sciences* **41.3** (1955): 174-176. It is available at http://www.pnas.org/content/41/3/174.short. It's very short, but phrased in the "old-style" Weil language of algebraic geometry.

Here's a quick sketch of a proof with $k=\mathbb F_q$. Choosing a point of $X(\overline{k})$, we can make $X$ into an abelian variety over $\overline{k}$. Then the $q$-power Frobenius map $\phi_q:X\to X$ is the composition of an isogeny and a translation, say $\phi_q(x)=f(x)+x_0$ with $f:X\to X$ an isogeny (defined over $\overline{k}$) and $x_0\in X(\overline{k})$. The fact that $\phi_q$ is inseparable implies that $f$ is inseparable, and hence $(1-f)^*$ acts as the identity map on differentials. Thus $1-f$ has finite kernel, so it is surjective, and thus there is a point $x_1\in X(\overline k)$ satisfying $(1-f)(x_1)=x_0$. This implies that $\phi_q(x_1)=x_1$, and hence that $x_1\in X(k)$.

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.

This is a theorem of Serge Lang, proved in a paper called "Abelian varieties over finite fields".