is surjective. Note that the kernel of this map (i.e., ) 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 (e.g., elliptic curves.) 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 (see below for the precise statement.)
The proof (given below) actually goes through for any that induces a nilpotent operator on the Lie algebra of G.
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.
Then (identifying the tangent space at a with the tangent space at the identity element) 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 .