Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How does the strong approximation theorem for global function fields looks like?

For the number field $\mathbb{Q}$ it can be expressed as the surjection

$$ \mathbb{Q}^\times \times \mathbb{R}^\times \times \prod\limits_{p} \mathbb{Z}_p \twoheadrightarrow \mathbb{A}^\times.$$

I want to understand the image of the adelic norm.

share|improve this question
1  
Could you please provide some background, or at least a link where one could learn more? I, for one, would appreciate that. Thanks! –  William Apr 19 '12 at 3:01
    
    
@William: Is it better now? –  plusepsilon.de Apr 19 '12 at 8:08
    
Yes, great. Thank you. –  William Apr 19 '12 at 17:28
add comment

1 Answer

up vote 1 down vote accepted

Let $k$ be a global function field, i.e., a finite extension $\mathbb{F}_{q}(T)$, then $\left\| \cdotp \right\|_{\mathbb{A}} \twoheadrightarrow q^{\mathbb{Z}} \subset (0, \infty)$!

share|improve this answer
    
this is false! a good example to keep in mind (for a number field $k$; $S$ is the collection of all archimedean places) is that $\mathbb{A}_k^\times / (k^\times \prod_{v \nmid \infty} o_v^\times \prod_{v \mid \infty} k_v^\times)$ is the class group of $k$, by class field theory. (also note that strong approximation for algebraic groups, as in Prasad's article, only applies to simply connected semisimple groups, and $\mathbb G_m$ isn't semisimple.) –  fherzig Oct 19 '13 at 15:07
    
in fact, CFT also shows for totally complex $k$ that $\mathbb{A}_k^\times / \overline{k^\times \prod_{v \mid \infty} k_v^\times}$ is the galois group of the maximal abelian extension of $k$. so density fails very badly. –  fherzig Oct 19 '13 at 17:12
    
thm X.2.4 in artin-tate shows that $k^\times \prod_{v \in S} k_v^\times$ is never dense in $\mathbb A_k^\times$ (the closure has infinite image). –  fherzig Oct 19 '13 at 19:24
    
Thank you, fherzig. –  plusepsilon.de Oct 21 '13 at 9:04
    
you are welcome! –  fherzig Oct 21 '13 at 18:46
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.