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.
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 $K/k$.