I have been looking at several different proofs that Artin $L$-series of Abelian extensions coincide with Hecke $L$-series.
In Serge Lang's $\textit{Algebraic Number Fields}$ (XII §2) and Jürgen Neukirch's $\textit{Algebraische Zahlentheorie}$ (VII §10) they use a property which I paraphrase in the following manner:
Let $E/K$ be an Abelian extension, $G:=\textrm{Gal}(E/K)$, and let $\chi$ be a simple character of $G$. We then have $G/\textrm{Ker}(\chi) \cong \textrm{Gal}(E_{\chi}/K)$, where $E_{\chi}$ is the fixed subfield corresponding to $\textrm{Ker}(\chi) \vartriangleleft G$. By inflation, we may view $\chi$ is a character of $\textrm{Gal}(E_{\chi}/K)$, and we have:
$$
L(E_{\chi}/K,\chi,s) = L(E/K,\chi,s)
$$
I have some scruples with this argument.
I understand the property of inflation to mean the following:
Let $E/K$ be a Galois extension, $G:=\textrm{Gal}(E/K)$, and let $E'/K$ be a bigger Galois extension ($E \subset E'$), $G':=\textrm{Gal}(E'/K)$, and let $\chi$ be a simple character of $G$. We then have:
$$
L(E'/K,\chi',s) = L(E/K,\chi,s)
$$
$\chi' = \chi \circ \pi$, where $\pi: G' \to G$ is the canonical projection.
That is, inflation allows us to pass $\textit{from a smaller Galois extension to a bigger one}$.
But Lang and Neukirch (et al.) seem to be going the other way: They take a character of a bigger Galois extension and pass to a smaller one.
But this is manifestly impossible. Take $\textit{e.g.}$ $\mathbb{Q} / \mathbb{Q}$ for the smaller extension and $\mathbb{Q}(i) / \mathbb{Q}$ for the bigger one. The above would imply that:
$$
L(\mathbb{Q}(i) / \mathbb{Q}, \chi, s) = L( \mathbb{Q} / \mathbb{Q} , \chi , s) = \zeta(s)
$$
On the left, we could let $\chi$ be the non-trivial sign character. On the right, we must necessarily have the trivial character.
How is this to be understood?
Thank you for your attention.
Best Answer
The converse to the inflation is that, if $\rho$ is a representation of $G=Gal(E/K)$ then $H=\ker(\rho)$ is normal and $\tilde{\rho}(gH)=\rho(g)$ is a (faithful) representation of $G/H=Gal(E/K)/Gal(E/E^H)=Gal(E^H/K)$ and $$L(E/K,\rho,s)=L(E^H/K,\tilde{\rho},s)$$