I must be misunderstanding something about Artin reciprocity. Let $K/k$ be an abelian extension of number fields with Galois group $G$, $I_k$ the ideles of $k$, and $P$ a prime of $k$ (with $v$ a place corresponding to $P$) which is unramified in $K$. If $\mathfrak P$ is a prime of $K$ which lies over $P$, then $\mathcal O_K/ \mathfrak P$ is an extension of $\mathcal O_k/P$ having degree $f = f(\mathfrak P/P)$. The Galois group of any finite extension of finite fields is cyclic.

Shouldn't the decomposition group $G_P = \{ \sigma \in G : \sigma \mathfrak P = \mathfrak P\}$ also be cyclic then? I thought $G_P$ modulo the inertia group (which is trivial, since $P$ is unramified) is isomorphic to the Galois group of residue fields.

And if that's true, shouldn't the restricted Artin map $$k_v^{\ast} \rightarrow I_k \rightarrow G_P$$ which sends an $x$ to $Frob(P,K/k)^{\nu_P(x)}$, be automatically surjective? The Frobenius element is the thing that generates $G_P$ when $P$ is unramified, and any $x$ with value $1$ maps to it. Serge Lang goes to some length to prove this (Theorem 3, chapter 11), but I don't see why it's necessary.

  • $\begingroup$ I haven't used Lang, but it seems to me that you are looking at the proof for the global Artin map. The local Artin map comes much earlier, and the corresponding proof is much simpler. $\endgroup$ Nov 29 '14 at 4:39
  • $\begingroup$ This is definitely the local Artin map. I think we had already proved the surjectiity of the global Artin map. My question is why the proof of Theorem 3 is so long $\endgroup$
    – D_S
    Nov 29 '14 at 20:57

The answer is that Theorem 3 is, as I asserted, completely trivial when $v$ is an unramified place. However, the result is also true when $v$ is ramified, and in this case we wouldn't have an explicit formula for the effect of the Artin map on $v$ unless we knew more about the extension of fields $K/k$.


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