In algebraic geometry, Lang's theorem, introduced by Serge Lang, states: if G is a connected smooth algebraic group over a finite field
F
q
{\displaystyle \mathbf {F} _{q}}
, then, writing
σ
:
G
→
G
,
x
↦
x
q
{\displaystyle \sigma :G\to G,\,x\mapsto x^{q}}
for the Frobenius, the morphism of varieties
G
→
G
,
x
↦
x
−
1
σ
(
x
)
{\displaystyle G\to G,\,x\mapsto x^{-1}\sigma (x)}
is surjective. Note that the kernel of this map (i.e.,
G
=
G
(
F
q
¯
)
→
G
(
F
q
¯
)
{\displaystyle G=G({\overline {\mathbf {F} _{q}}})\to G({\overline {\mathbf {F} _{q}}})}
) is precisely
G
(
F
q
)
{\displaystyle G(\mathbf {F} _{q})}
.
The theorem implies that
H
1
(
F
q
,
G
)
=
H
e
´
t
1
(
Spec
F
q
,
G
)
{\displaystyle 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
Spec
F
q
{\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 (e.g., elliptic curves.) In fact, this application was Lang's initial motivation. If G is affine, the Frobenius
σ
{\displaystyle \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
σ
{\displaystyle \sigma }
that induces a nilpotent operator on the Lie algebra of G.
Contents
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
→
G
,
f
a
(
x
)
=
x
−
1
a
σ
(
x
)
.
{\displaystyle f_{a}:G\to G,\quad f_{a}(x)=x^{-1}a\sigma (x).}
Then, by identifying the tangent space at a with the tangent space at the identity element, we have:
(
d
f
a
)
e
=
d
(