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 $H^{1}(\mathbf {F} _{q},G)=H_{\text{ét}}^{1}(\operatorname {Spec} \mathbf {F} _{q},G)$   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.

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

Proof

Define

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

Then we have: (identifying the tangent space at a with the tangent space at the identity element)

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

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 $f_{1}$.

See also

• Lang–Steinberg theorem

References

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