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, then, writing for the Frobenius, the morphism of varieties
is surjective. Note that the kernel of this map is precisely.
The theorem implies that
vanishes, and, consequently, any G-bundle on 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. If G is affine, the Frobenius may be replaced by any surjective map with finitely many fixed points
The proof actually goes through for any that induces a nilpotent operator on the Lie algebra of G.

The Lang–Steinberg theorem

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.
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:
Then we have:
where. It follows is bijective since the differential of the Frobenius vanishes. Since, we also see that is bijective for any b. Let X be the closure of the image of. The smooth points of X form an open dense subset; thus, there is some b in G such that is a smooth point of X. Since the tangent space to X at 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 then contains an open dense subset U of G. Now, given an arbitrary element a in G, by the same reasoning, the image of contains an open dense subset V of G. The intersection is then nonempty but then this implies a is in the image of.