2
$\begingroup$

The Artin $L$-series for Abelian extensions are known to coincide with Hecke $L$-series, which in particular implies that if $E/K$ is a Abelian extensions, and $\chi$ is a non-trivial simple character of $\textrm{Gal}(E/K)$, then $L(E/K,\chi,s)$ admits an analytic continuation holomorphic on $\mathbb{C}$. This proves the Artin conjecture for Abelian extensions.

It is said that this settles the Artin conjecture for all degree 1 representations. But how?

Say we have $\textrm{Gal}(E/K) \cong S_3$. How does the above fact imply that $L(E/K,\chi,s)$ is entire for any non-trivial simple degree 1 character $\chi$?

All help or input would be highly appreciated.

$\endgroup$
1
  • $\begingroup$ If $\rho $ is a representation of $Gal(E/K)$ let $H=\ker(\rho)=\{ g,\rho(g)=I\}$, it is normal in $Gal(E/K)$ and $\tilde{\rho}(gH) = \rho(g)$ is a representation of $Gal(E/K)/H=Gal(E^H/K)$ and $L(E/K,\rho,s)= L(E^H/K,\tilde{\rho},s)$. If $\rho(x)\rho(y)=\rho(y)\rho(x)$ then $Gal(E/K)/H$ is abelian and $L(E/K,\rho,s)$ is a product of Hecke L-functions of $K$. $\endgroup$
    – reuns
    Jan 3, 2020 at 18:47

1 Answer 1

Reset to default
4
$\begingroup$

Let $G=\textrm{Gal}(E/K)$

By the universal property of the abelianization, $\chi:G \to \Bbb C^\times$ factors over $G^{ab}$, so we obtain $\overline{\chi}:\textrm{Gal}(L/K) \to \Bbb C^\times$ where $L=E^{[G,G]}$. Now $L/K$ is abelian and $\chi$ is the inflation of $\overline{\chi}$, so by functoriality of Artin L-functions with respect to inflations we have $$L(E/K,\chi,s)=L(L/K,\overline{\chi},s)$$

$\endgroup$
2
  • $\begingroup$ By "$\chi: G \to \mathbb{C}^{\times}$ factors over $G^{ab}$", do you mean that every simple degree 1 character of $G$ can be expressed as $\chi \circ \pi$, where $\chi$ is a simple character of $G_{\textrm{Ab}}$ and $\pi: G \to G_{\textrm{Ab}}$ is the projection? $\endgroup$
    – Heinrich Wagner
    Jan 3, 2020 at 11:59
  • 1
    $\begingroup$ @HeinrichWagner yes, this is precisely what I mean $\endgroup$
    – Lukas Heger
    Jan 3, 2020 at 19:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .