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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.

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2 Answers

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 :

  1. Artin's Reciprocity Laws.
  2. A fantastic description of Artin's-L-Functions by J.W.Cogdell which is here.
  3. Interesting proofs found at papers of Artin "Uber eine neue Art von L-Reihen " and " Zur theorie der L-Reihen mit allgemeinen Gruppencharackteren".

Thats all I know,

Thank you.

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Thanks, that's sort of confirms what I expected. What I'm interested in is really seeing how one can find such a Hecke character. How to make it explicit. There's supposedly some nice isomorphisms of the relevant groups through CFT. I suspect this might be a bit tricky. –  pki Jan 16 '12 at 17:27
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@pki : No problem, but it took me 45 minutes to write up the entire thing, I think I need to learn some tricks in LaTeX to do the things much easily, and I think many great people have sent me a much more big-answer when I had some doubts, so I must help this community as far as I can by sharing the things I know , until I live. Take care. –  Iyengar Jan 16 '12 at 17:57
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Let $L/K$ be an abelian extension of global fields. If $C_K$ denotes idèle class group of $K$, then global class field theory gives an isomorphism between $C_K/N_{L/K}C_L$ on the one hand, and the Galois group $G_{L/K}$ on the other hand, called Artin's reciprocity law. Using this isomorphism, you can regard any character of $G_{L/K}$ as a character of $C_K$, which gives you the interpretation of Artin $L$-functions as Hecke $L$-functions.

This correspondence is as explicit, as the reciprocity law, i.e. fairly explicit. To define a homomorphism $C_K\rightarrow G_{L/K}$, it is enough to say what it does at each local place, and then extend multiplicatively. Artin's reciprocity map sends the uniformiser $\pi$ at an unramified place $\mathfrak{p}$ to the Frobenius element at $\mathfrak{p}$, i.e. the unique element of $G_{L/K}$ that acts as the Frobenius on the extension of finite fields that you get by reducing $K$ modulo $\mathfrak{p}$ and $L$ modulo a prime above it. There are tricks for computing the Frobenius element at a given prime, but if you want to do it by hand for a few primes in a low-tech way, you just reduce and check what each element of $G_{L/K}$ reduces to.

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Sir, I am not that great to keep my answer along with your one, if you think my answer has no-worth, please let me know, so that I delete it. –  Iyengar Jan 17 '12 at 14:16
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