Why is the local polynomial of an Artin representation independent of the choice of the Frobenius element? I will take the notation from this paper on which I am currently working.
Let $K/\mathbb{Q}_p$ be finite (i.e. K is a local field) and $F/K$ be a Galois extension. We shall denote by $\mathbb{F}_K$ and $\mathbb{F}_F$ the residue field of $K$ and $F$, respectively.
Let $\operatorname{Frob}_{F/K}$ be a Frobenius element, i.e. any element $g \in \operatorname{Gal}(F/K)$ that induces the Frobenius automorphism $x \mapsto x^q$ under the restriction homomorphism $\operatorname{Gal}(F/K) \to \operatorname{Gal}(\mathbb{F}_F/\mathbb{F}_K)$ where $q = |\mathbb{F}_K|$.
For a chosen Frobenius element, we denote by $\Phi_{F/K} = \operatorname{Frob}_{F/K}^{-1}$ the geometric Frobenius.
Let $\rho : \operatorname{Gal}(\bar{K}/K) \to \operatorname{GL}(V)$ denote an Artin representation. Then we can define its local polynomial:
$$ P(\rho,T) = \det(1-\Phi_{F/K}T | \rho^{I_{F/K}})$$
where $\rho$ factors through $F/K$ (i.e. $\operatorname{Gal}(\bar{K}/F) \subset \ker{\rho})$ and 
$$ \rho^{I_{F/K}} = V^{I_{F/K}} = \{ x \in V | \rho_\sigma(x) = x \: \forall \sigma \in I_{F/K} \} $$
is the subspace of $\rho$ of $I_{F/K}$-invariant vectors.
Question: Why is the local polynomial independent of the choice of $\operatorname{Frob}_{F/K}$ resp. $\Phi_{F/K}$?
I have a hard time to figure this out: If we choose any basis $B$ of $\rho^{I_{F/K}}$, how does it affect the representation matrix $[\Phi_{F/K}]^B_B$, so that $P(\rho,T) = \det(I - T \cdot [\Phi_{F/K}]^B_B)$ does not change if we take another Frobenius element?
Could you please explain this to me? Thank you.
 A: Suppose that $V$ is a vector space with an action of a group $G$ (say a local Galois group), suppose that $I$ is a normal subgroup of $G$ (say the inertia group), and say that $\phi \in G/I$ is any element (say the Frobenius element). Let $V^{I}$ be the subspace of $V$ fixed by $I$. Let $g$ and $g'$ be any two lifts of $\phi$ to $G$. Since they both map to $\phi$ in $G/I$, it follows that $g^{-1} g'$ maps to the identity in $G/I$, and so $g^{-1} g' = x \in I$, and so $g' = g x$ for some $x \in I$. But now compare the action of $g'$ and $g$ on an vector $v \in V^{I}$. Since $v \in V^{I}$ is fixed by $I$, and so $x(v) = v$. It follows that
$$g'(v) = g x(v) = g(x (v)) = g(v).$$
Thus $g$ and $g'$ have the same action on $V^{I}$.
The next part is how does the construction depend on a choice of basis for $V^{I}$. Here the point is that for a vector space $W$ and an invertible linear operator $M$ on $W$, one has
$$\begin{aligned}
\det(I - M T | W) =  & \ \det(M^{-1})(\det(M - T | W)) \\
 = & \ \det(M^{-1}) \cdot \mathrm{char.poly}(M^{-1},T).
\end{aligned}$$
and both the determinant and characteristic polynomial of  a linear operator $M^{-1}$ do not depend on the choice of basis.
