Do the local polynomials of Weil representations coincide if they are Artin representations (factor through a finite quotient)? Let 


*

*$K$ be a local field, 

*$G_K$ its absolute Galois group, 

*$I_K$ the inertia subgroup of $G_K$,

*$\operatorname{Frob}_K \in G_K$ be a Frobenius element, i.e. any element of $G_K$ acting as $x \mapsto x^{|k|}$ on $\bar{k}$, the algebraic closure of the residue fiel $k$ of $K$,

*$W_K$ be the Weil group of $K$ and

*$\rho$ be a Weil representation, i.e. it is a representation $\rho: W_K \to \operatorname{GL}_n(\mathbb{C})$ with $\rho(I_K)$ being finite.


The local polynomial $P(\rho,T)$ is the inverse characteristic polynomial of $\operatorname{Frob}_K^{-1}$ on the inertia invariants of $\rho$, i.e. $$ P(\rho,T) = \det(1-\operatorname{Frob}_K^{-1} T \, | \, \rho^{I_K}). $$
We say that $\rho$ factors through a finite quotient if there is a finite Galois extension $F/K$ such that $\operatorname{Gal}(\bar{K}/F) \subset \ker{\rho}$ ($\rho$ is also called an Artin representation then). This means that $\rho$ comes from a representation $\bar{\rho}: \operatorname{Gal}(F/K) \to \operatorname{GL}_n(\mathbb{C})$.
We can define a local polynomial for $\bar{\rho}$ the same way:
$$ P(\bar{\rho},T) = \det(1-\operatorname{Frob}_{F/K}^{-1} T \, | \, \bar{\rho}^{I_{F/K}}) $$
where $I_{F/K}$ is the inertia subgroup of $\operatorname{Gal}(F/K)$ and $\operatorname{Frob}_{F/K} \in \operatorname{Gal}(F/K)$ is any Frobenius element.
Question Do we have $
P(\rho,T) = P(\bar{\rho},T)$?
For me, it is especially difficult to understand the first definitions without $F/K$ because I am not able to compute them explicitly.
Could you please help me with this question? Thank you in advance!
 A: $\DeclareMathOperator{\Gal}{Gal}$$\DeclareMathOperator{\Frob}{Frob}$
This comes down to checking a few things.  Let $\kappa(K), \kappa(F)$ be the residue fields of $K$ and $F$.
Let $\rho: \operatorname{Gal}(\overline{K}/K) \rightarrow \operatorname{GL}(V)$ be a continuous, finite dimensional representation of the Weil group.  Suppose that the kernel of $\rho$ contains $\operatorname{Gal}(\overline{K}/F)$, so we have a well defined homomorphism $\overline{\rho}: \Gal(F/K) \rightarrow \operatorname{GL}(V)$.  The inertia group $I_K$ is the kernel of the surjective homomorphism
$$\Gal(\overline{K}/K) \rightarrow \Gal(\kappa(K)^{\operatorname{sep}}/\kappa(K))$$
and the inertia group $I_{F/K}$ is the kernel of the surjective homomorphism
$$\Gal(F/K) \rightarrow \Gal(\kappa(F)/\kappa(K))$$
First, a given $\sigma \in \Gal(\overline{K}/K)$ induces the Frobenius on $\Gal(\kappa(K)^{\operatorname{sep}}/\kappa(K))$ if and only if its image in $\Gal(F/K)$ induces the Frobenius on $\Gal(\kappa(F)/\kappa(K))$.
Second, the image of $I_K$ in $\Gal(F/K)$ is equal to $I_{F/K}$.  Thus
$$\{v \in V : \rho(\sigma)v = v \textrm{ for all } \sigma \in I_K\} = \{v \in V : \overline{\rho}(\sigma)v = v \textrm{ for all } \sigma \in I_{F/K}\}$$
or $\rho^{I_K} = \rho^{I_{F/K}}$.
