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

Question

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 $A \simeq X$?

Answer 1

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 $x_1\in X(k)$.

Answer 2

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.

Answer 3

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