# Lang's theorem

In algebraic geometry, Lang's theorem, introduced by Serge Lang, states: if G is a connected smooth algebraic group over a finite field $\mathbf {F} _{q}$ , then, writing $\sigma :G\to G,\,x\mapsto x^{q}$ for the Frobenius, the morphism of varieties

$G\to G,\,x\mapsto x^{-1}\sigma (x)$ is surjective. Note that the kernel of this map (i.e., $G=G({\overline {\mathbf {F} _{q}}})\to G({\overline {\mathbf {F} _{q}}})$ ) is precisely $G(\mathbf {F} _{q})$ .

The theorem implies that $H^{1}(\mathbf {F} _{q},G)=H_{\mathrm {{\acute {e}}t} }^{1}(\operatorname {Spec} \mathbf {F} _{q},G)$ vanishes, and, consequently, any G-bundle on $\operatorname {Spec} \mathbf {F} _{q}$ is isomorphic to the trivial one. Also, the theorem plays a basic role in the theory of finite groups of Lie type.

It is not necessary that G is affine. Thus, the theorem also applies to abelian varieties (e.g., elliptic curves.) In fact, this application was Lang's initial motivation. If G is affine, the Frobenius $\sigma$ may be replaced by any surjective map with finitely many fixed points (see below for the precise statement.)

The proof (given below) actually goes through for any $\sigma$ that induces a nilpotent operator on the Lie algebra of G.

## The Lang–Steinberg theorem

Steinberg (1968) gave a useful improvement to the theorem.

Suppose that F is an endomorphism of an algebraic group G. The Lang map is the map from G to G taking g to g−1F(g).

The Lang–Steinberg theorem states that if F is surjective and has a finite number of fixed points, and G is a connected affine algebraic group over an algebraically closed field, then the Lang map is surjective.

## Proof of Lang's theorem

Define:

$f_{a}:G\to G,\quad f_{a}(x)=x^{-1}a\sigma (x).$ Then (identifying the tangent space at a with the tangent space at the identity element) we have:

$(df_{a})_{e}=d(h\circ (x\mapsto (x^{-1},a,\sigma (x))))_{e}=dh_{(e,a,e)}\circ (-1,0,d\sigma _{e})=-1+d\sigma _{e}$ where $h(x,y,z)=xyz$ . It follows $(df_{a})_{e}$ is bijective since the differential of the Frobenius $\sigma$ vanishes. Since $f_{a}(bx)=f_{f_{a}(b)}(x)$ , we also see that $(df_{a})_{b}$ is bijective for any b. Let X be the closure of the image of $f_{1}$ . The smooth points of X form an open dense subset; thus, there is some b in G such that $f_{1}(b)$ is a smooth point of X. Since the tangent space to X at $f_{1}(b)$ and the tangent space to G at b have the same dimension, it follows that X and G have the same dimension, since G is smooth. Since G is connected, the image of $f_{1}$ then contains an open dense subset U of G. Now, given an arbitrary element a in G, by the same reasoning, the image of $f_{a}$ contains an open dense subset V of G. The intersection $U\cap V$ is then nonempty but then this implies a is in the image of $f_{1}$ .