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 $A \simeq X$?



When $k = \mathbb{F}_q$ finite field, $X$ always has $k$-rational point, and so $A \simeq X$?
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.




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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 $x\_1\in X(k)$.


