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 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 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.