Artin mapArtinMap2002-04-14 20:29:432005-03-15 06:40:45DefinitionArtin symbolFrobenius automorphism% this is the default PlanetMath preamble. as your knowledge
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\newtheorem{definition}[theorem]{Definition}Let $L/K$ be a Galois extension of number fields, with rings of integers $\O_L$ and $\O_K$. For any finite prime $\P \subset L$ lying over a prime $\p \in K$, let $D(\P)$ denote the decomposition group of $\P$, let $T(\P)$ denote the inertia group of $\P$, and let $l := \O_L/\P$ and $k := \O_K/\p$ be the residue fields. The exact sequence
$$
1 \lra T(\P) \lra D(\P) \lra \Gal(l/k) \lra 1
$$
yields an isomorphism $D(\P)/T(\P) \cong \Gal(l/k)$. In particular, there is a unique element in $D(\P)/T(\P)$, denoted $[L/K,\P]$, which maps to the $q^{\rm th}$ power Frobenius map $\Frob_q \in \Gal(l/k)$ under this isomorphism (where $q$ is the number of elements in $k$). The notation $[L/K,\P]$ is referred to as the {\em Artin symbol} of the extension $L/K$ at $\P$.
If we add the additional assumption that $\p$ is unramified, then $T(\P)$ is the trivial group, and $[L/K,\P]$ in this situation is an element of $D(\P) \subset \Gal(L/K)$, called the {\em Frobenius automorphism} of $\P$.
If, furthermore, $L/K$ is an abelian extension (that is, $\Gal(L/K)$ is an abelian group), then $[L/K,\P] = [L/K,\P']$ for any other prime $\P' \subset L$ lying over $\p$. In this case, the Frobenius automorphism $[L/K,\P]$ is denoted $(L/K,\p)$; the change in notation from $\P$ to $\p$ reflects the fact that the automorphism is determined by $\p \in K$ independent of which prime $\P$ of $L$ above it is chosen for use in the above construction.
\begin{definition}
Let $S$ be a finite set of primes of $K$, containing all the primes that ramify in $L$. Let $I_K^S$ denote the subgroup of the group $I_K$ of fractional ideals of $K$ which is generated by all the primes in $K$ that are not in $S$. The {\em Artin map}
$$
\phi_{L/K}: I_K^S \lra \Gal(L/K)
$$
is the map given by $\phi_{L/K}(\p) := (L/K,\p)$ for all primes $\p \notin S$, extended linearly to $I_K^S$.
\end{definition}