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 ${\displaystyle \mathbf {F} _{q}}$, then, writing ${\displaystyle \sigma :G\to G,\,x\mapsto x^{q}}$ for the Frobenius, the morphism of varieties

${\displaystyle G\to G,\,x\mapsto x^{-1}\sigma (x)}$ 

is surjective. The theorem implies that ${\displaystyle H^{1}(\mathbf {F} _{q},G)=H_{\text{ét}}^{1}(\operatorname {Spec} \mathbf {F} _{q},G)}$   vanishes, and, consequently, any G-bundle on ${\displaystyle \operatorname {Spec} \mathbf {F} _{q}}$ 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. In fact, this application was Lang's initial motivation.

Proof

Define

${\displaystyle f_{a}:G\to G,\quad f_{a}(x)=x^{-1}a\sigma (x)}$ .

Then we have:

${\displaystyle (df_{a})_{e}=d(h\circ (x\mapsto (x^{-1},a,\sigma (x))))_{e}=dh_{(e,a,e)}\circ (-1,0,d\sigma _{e})=-1+d\sigma _{e}}$  

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

See also

• Lang–Steinberg theorem

References

• T.A. Springer, "Linear algebraic groups", 2nd ed. 1998.