What's the relationship between between Artin $L$-functions and Dirichlet or Hecke $L$-functions if $L/K$ is an abelian extension? I've been told that one can interpret the Artin $L$-functions as these other $L$-functions through class field theory. Can anyone explain how this is done?
EDIT: I'll add this to make the harder part of the question more explicit. Given an Artin $L$-function $L(L/K,\rho,s)$ with $L/K$ abelian, how can one find a Hecke character $\chi$ s.t. $L(L/K,\rho,s)=L(s,\chi)$? I'm interested in seeing how this can be done explicitly through global CFT. I'm guessing the reciprocity law should make this somewhat "natural" given that it connects ray class groups and Galois groups.
Your question is a very good one, To add something I know it appears as follows :
" Its a well known result that Artin , through his famous " Reciprocity law " has proved that in the case of abelian extensions of number fields the Artin L-functions are Hecke-L-functions.
To say further Let $E⁄K$ be an abelian Galois extension with Galois group $G$. Then for any character $\sigma: G \to C^×$ (i.e. one-dimensional complex representation of the group $G$), there exists a Hecke character $\chi$ of $K$ such that
$$L^{\large\rm{Artin}}_{E/K}(\sigma,s)=L^{\large\rm{Hecke}}_{K}(\chi,s)$$
where the left hand side is the Artin $L$-function associated to the extension with character $\sigma$ and the right hand side is the Hecke $L$-function associated with $\chi$, The formulation of the Artin reciprocity law as an equality of $L$-functions allows formulation of a generalisation to n-dimensional representation, though a direct correspondence still lacking.
And to mention the whole proof is completely difficult and strenuous too, I try to mention the gist and big-picture.
A Hecke L-function as everyone know is an Euler product $L(s, χ)$ attached to a character $χ$ of $F^× \backslash I_F $ where $I_F$ is the group of ideles of $F$ . If $v$ is a place of $F$ then $F^{×}_{v}$ imbeds in $I_F$ and $χ$ defines a character $χ_v$ of $F^{×}_{v}$ . To form the function $L(s, χ)$ we take a product over all places of $F$:
$$L(s, χ) = \prod _v L(s, χ_v).$$
An Artin $L$-function is associated to a finite-dimensional representation $\rho$ of a Galois group $\rm{Gal}(K/F)$, $K$ being an extension of finite degree. It is defined arithmetically and its analytic properties are extremely difficult to establish. Once again $$L(s,\rho ) = L(s, \rho_v),$$ $\rho_{v}$ being the restriction of to the decomposition group. For our purposes it is enough to define the local factor when $v$ is defined by a prime $p$ and $p$ is unramified in $K$. Then the Frobenius conjugacy class $Φ_p$ in $\rm{Gal}(K/F)$ is defined, and $$L(s, \rho_{v}) = \frac{1}{\large det(I − \rho(Φ_p)/Np^s)} = \prod^{d}_{i=1}\frac{1}{\large 1 − β_i(p)/Np^s} $$, if $β_1(p), . . . , β_d(p)$ are the eigenvalues of $\rho(Φ_p)$.
Although the function $L(s,\rho )$ attached to $\rho$ is known to be meromorphic in the whole plane, Artin’s conjecture that it is entire when $\rho$ is irreducible and nontrivial is still outstanding. Artin himself showed this for one dimensional , and it can now be proved that the conjecture is valid for tetrahedral $\rho$ , as well as a few octahedral $\rho$ ( see the references I have searched in google and provided in below section ) . Artin’s method is to show that in spite of the differences in the definitions the function $L(s,\rho )$ attached to a one-dimensional $\rho$ is equal to a Hecke $L$-function $L(s, χ)$ where $χ = χ(\rho)$ is a character of $F^× \backslash I_F $. He employed all the available resources of class field theory, and went beyond them, for the equality of $L(s,\rho )$ and $L(s, χ(\rho))$ for all $\rho$ is pretty much tantamount to the Artin reciprocity law, which asserts the existence of a homomorphism for $I_F$ onto the Galois group $\rm{Gal}(K/F)$ of an abelian extension which is trivial on $F^×$ and takes $\omega_{p}$ to $Φ_p$ for almost all $p$.
The equality of $L(s,\rho )$ and $L(s, χ)$ implies that of $χ(\omega_p)$ and $\rho(Φ_p)$ for almost all $p$.
So you need to go through these things first :
Thats all I know,
Thank you.