We shall now consider a number field ${k}$ and an abelian extension ${L}$. Let ${S}$ be a finite set of primes (nonarchimedean valuations) of ${k}$ containing the ramified primes, and consider the group ${I(S)}$ of fractional ideals prime to the elements of ${S}$. This is a free abelian group on the primes not in ${S}$. We shall define a map, called the Artin map from ${I(S) \rightarrow G(L/k)}$.

1. How does this work?

Specifically, let ${\mathfrak{p} \notin S}$ be a prime in ${k}$. There is a prime ${\mathfrak{P}}$ of ${L}$ lying above it. If ${A,B}$ are the rings of integers in ${k,L}$, respectively, then we have a field extension ${A/\mathfrak{p} \rightarrow B/\mathfrak{P}}$. As is well-known, there is a surjective homomoprhism of the decomposition group ${G_{\mathfrak{P}}}$ of ${\mathfrak{P}}$ onto ${G(B/\mathfrak{P} / A/\mathfrak{p})}$ whose kernel, called the inertia group, is of degree ${e(\mathfrak{P}|\mathfrak{p})}$.

But, we know that the extension ${B/\mathfrak{P} / A/\mathfrak{p}}$ is cyclic, because these are finite fields. The Galois group is generated by a canonically determined Frobenius element which sends ${a \rightarrow a^{|A/\mathfrak{p}|}}$. We can lift this to an element ${\sigma_{\mathfrak{p}}}$ of ${G_{\mathfrak{P}}}$, still called the Frobenius element. (more…)