Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be a connected commutative algebraic group over $\mathbb{F}_q$. If $\text{Fr}_q : G \to G$ denotes the $q$-Frobenius morphism, we define the Lang isogeny $L_q$ to be the endomorphism of $G$ given by $g \mapsto \text{Fr}_q(g)g^{-1}$.

I have two questions about this important map.

  1. It is not too hard to see that $L_q$ is etale by computing its differential, but why is $L_q$ finite?
  2. Granted that $L_q$ is a finite Galois covering with group $G(\mathbb{F}_q)$, we get a surjection $\pi_1(G,\overline{1}) \to G(\mathbb{F}_q)$. What can we say about the kernel of this homomorphism?

Is there a good modern reference for basic results about $L_q$? If not, would someone kindly explain these two points to me?

share|improve this question
2  
So $G$ is assumed to be smooth? You should step away from the specificity of the Lang isogeny and commutativity and prove to yourself that any homomorphism $f:G \rightarrow H$ between finite type group schemes over a field (with no smoothness or commutativity hypotheses) is finite if its kernel has finitely many geometric points. When $G$ and $H$ are moreover smooth then you should show to yourself that such an $f$ is flat. One needs such facts to have a robust theory in positive characteristic, and they are good exercises in understanding quotients. –  user30379 Jan 28 '13 at 0:03
add comment

2 Answers

up vote 5 down vote accepted
  1. Every etale morphism is finite over some nonempty open set. (For instance, locally somewhere write it as a standard etale morphism $(A[t]/f(t))_{g(t)}$, then consider the open set where the norm of $g$, that is, the resultant of $f$ and $g$, is nonzero.) If a group homomorphism $H \to G$ is finite over some nonempty open set, it is finite over all translates of that set, so it is finite everywhere.

  2. It is isomorphic to $\pi_1(G,\overline{1})$. In general, if $X \to Y$ is finite etale and Galois, the kernel of the map $\pi_1(Y) \to Gal(X/Y)$ is $\pi_1(X)$.

share|improve this answer
    
Oops, that was easier (or at least more general) than I would have expected. –  Justin Campbell Jan 27 '13 at 22:12
add comment

Your questions are expressed in modern language, but keep in mind that Lang's original paper in Amer. J. Math. (1956) used the language of Rosenlicht and Weil. So it's a challenge to translate the ideas. A slightly more contemporary proof of Lang's theorem, incorporating Steinberg's useful generalization, is given by Springer in his textbook Linear Algebraic Groups (4.4.17). In reality not very much is involved in the proof, which relies as you note on computing the differential of the Lang map. What you see is that the map is "finite" in the older sense, via the differential criterion. I'm not sure yet how to read your second question, but this too should be easy to answer in that older language.

I should add that your question about commutative algebraic groups is more special than the setting of Lang's original theorem, where he considered arbitrary connected algebraic groups. Apart from the language issue, it's worth looking at the original paper (available online from JSTOR), where a number of interrelated results on algebraic groups over finite fields are first worked out.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.