Abelian extension and Dirichlet characters $\newcommand\Q {\Bbb Q}
\newcommand\Z[1]{\Bbb Z/#1 \Bbb Z}
\newcommand\Gal{\mathrm{Gal}}
\newcommand\K[1]{\Bbb Q(\zeta_{#1})}$
Let $K/\Q$ be an abelian extension with Galois group $G$ and let $\rho : G \to \Bbb C^*$ be a character.
Let $\sigma_p \in \Gal(K/\Q)$ such that $\sigma_p(x)-x^p \in P$ for any $x \in \mathcal O_K$, where $P$ is any prime of $K$ above $p$.

Is it true that there is an integer $k \geq 1$ and a Dirichlet character $\chi : (\Z k)^{\times} \to \Bbb C^*$ such that
  $$\rho(\sigma_p)\Big\vert_{\Bbb C^{\,I_P}} = \chi(p)$$
  for any rational prime $p$ ?

where
$\Bbb C^{\,I_P}$ is the $0$ or $1$-dimensional subspace of $\Bbb C$ invariant under the inertia group $I(P/p) \leq \Gal(K/\Bbb Q)$.
Actually I want to show that there is an integer $k \geq 1$ and a Dirichlet character $\chi : (\Z k)^{\times} \to \Bbb C^*$ such that
$L(\rho,s)=L(\chi,s)$ for any $s>1$, where the LHS is the Artin L function and the RHS is the Dirichlet L function. 

Ideas:
— By Kronecker–Weber, there is an integer $k \geq 1$ such that $K \subset \K k$.
Let $\pi : \Gal(\K k / \Q) \to \Gal(K/\Q)$ be the projection and define  $\chi = \psi \circ \pi \circ \rho$, where $\psi$ is an isomorphism $(\Z k)^{\times} \to \Gal(\K k / \Q)$.
If $p$ is unramified, then $I(P/p) = 0$ for any $P$ above $p$, and we can take $\sigma_p$ to be the restriction of $\zeta_k \mapsto \zeta_k^p$ (where $(p,k)=1$). Then $\sigma_p$ is mapped to $p \pmod k$ via $\psi$, and $\chi(p) = \rho(\sigma_p)$ as desired.
But what happens for ramified primes? If $p$ is ramified, then $\chi(p) = 0$, so we would need $\Bbb C^{I_P}$ to be $0$. It doesn't seem to be true.
 A: $\newcommand\Q {\Bbb Q}\newcommand\Gal{\operatorname{Gal}}$Thanks to Mathmo123's comments, I can provide an answer.
I will show that $L(\rho,s)=L(\chi,s)$, by showing that
$$\det(1-\rho(\sigma_P)p^{-s}, \mathbb{C}^{I_P}) =  1-\chi(p)p^{-s}$$ for any prime $p$.
Without loss of generality, we can assume that $\rho$ is faithful. Let $H = \mathrm{ker}(\rho)$, and $\pi_H : G \to G/H$ the projection, where $G=\Gal(K/\Q)$. Then, from theory about Artin L functions, we have
$$\mathcal L(K/\Q, \rho,s)=\mathcal L(K^H/\Q, \phi,s)$$
where $\rho = \phi \circ \pi_H$ (and $K^H/\Q$ is abelian).
So we can replace $\rho$ and $K$ by $\phi$ and $K^H$ if necessary.
— If $p$ is not ramified in $K$, then $\sigma_p$ corresponds to $p$ as I said above. We get
$$\det(1-\rho(\sigma_P)p^{-s}, \mathbb{C}^{I_P}) =  1-\chi(p)p^{-s}.$$
— If $p$ is ramified, then $I(P/p) \neq \{id\}$ and since $\rho$ is injective, we have $\rho(I(P/p)) \neq \{1\}$. Therefore $\mathbb{C}^{I(P/p)} = 0$ and $\det(1-\rho(\sigma_P)p^{-s}, \mathbb{C}^{I_P})=1$.
But $p$ is also ramified in $\mathbb{Q}(\zeta_k) \supset K$, so $p$ divides $k$, so that $\chi(p)=0$. We also have $1-\chi(p)p^{-s}=1$ i.e. we also get
$$\det(1-\rho(\sigma_P)p^{-s}, \mathbb{C}^{I_P}) =  1=1-\chi(p)p^{-s}$$
which concludes the proof.
