Geometric abelian class field theory

2011-02-09T15:24:35Z

There is a very nice geometric proof of Deligne for the Artin Reciprocity in the geometric setting, namely for a smooth, projective, geometrically irreducible curve $C$ over a finite field $\mathbb{F}_{q}$ , with function field $K=k(C)$, and idele group $\mathbb{I}_{K}:=\prod^{'}_{p\in|C|}K^{*}_{p}$ there is a one-to-one correspondence between the finite quotients of the double quotient space $k(C)^{*}\backslash\mathbb{I}_{K}/\prod_{p \in |C|}\widehat{\mathcal{O}_{p}^{*}}$ (which is isomorphic to $Pic_{C}(\mathbb{F}_{q})$ ) and the finite quotients of $\pi^{ab}_{1}(C)$ .

Now on the other hand the Artin Reciprocity Law for function fields states (e.g. in Artin-Tate: Class field theory) that the group $\mathbb{I}^{0}_{K}/K^{*}$ of norm 1 idele classes is isomorphic via the Reciprocity map to $Gal(\bar{K}^{ab}/K\bar{k})$ .

My questions would be:

These two statements seem to me first as different statements, don´t they?
If we put aside Deligne´s geometric proof for the geometric statement (not seriously and not so for long :-)) then how could one prove the geometric statement using the "number theoretic" Reciprocity Law for function fields?

Answer by Felipe Voloch for Geometric abelian class field theory

2011-02-09T21:19:04Z

They are different statements. What Deligne proves is the unramified case, i.e. the description of abelian extensions of $K$ unramified everywhere. If you could extend his argument to affine curves then you could possibly prove Artin reciprocity by his method. Going the other way should not be difficult. Have you looked at Serre's book "Groupes algebriques et corps de classes"?

Geometric abelian class field theory - MathOverflow

Geometric abelian class field theory

Answer by Felipe Voloch for Geometric abelian class field theory