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I am a total non-expert, so the answer to this question may be obvious, but here goes.

In Chevalley's formulation of CFT we get Artin maps $J_k \rightarrow Gal(L/k)$, where $J_k$ is the group of all ideles of $k$. However, we know there is a nice subgroup $J_k^1$ of the ideles obtained by taking only those satisfying the product formula $\prod_{v} |x_v| = 1$. Note that this still contains all the principal ideles, still surjects onto $I_k$ and has additional attractive properties like the compactness of $J_k^1/k^*$. Is there a way to set up CFT using quotients of this nicer group, and if so, what are the advantages of working with the superficially more unwieldy $J_k$?


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up vote 9 down vote accepted

I am no expert either, but here is what I think. If $k$ is a global field of characteristic zero (i.e. an algebraic number field), then one can work with $J_k^1$ instead of $J_k$ without losing (or changing) anything. This is because the kernel of the Artin map contains a subgroup $N$ of $J_k$ isomorphic to the multiplicative group of positive reals, and $J_k$ factors (non-canonically) as $J_k\cong N\times J_k^1$. In other words, in characteristic zero the Artin map does not see the norm of an idele. If $k$ is a global field of finite characterstic (i.e. a function field over a finite field $\mathbb{F}_q$), then the situation is markedly different. In this case the image of $J_k^1$ under the Artin map equals $Gal(L/k_0)$, where $k_0$ is the union of all constant field extensions of $k$ (i.e. the compositum of $L$ and the algebraic closure of $\mathbb{F}_q$ in the algebraic closure of $k$). More precisely, the norm of an idele precisely tells how its image under the Artin map acts on $k_0$: if the norm is $q^{-r}$, then the action on the algebraic closure of $\mathbb{F}_q$ is $x\mapsto x^{q^r}$.

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