Can the relative degree and ramification index can be read off the characteristic polynomial? Let $L/K$ be a Galois extension of global fields with Galois group $G$. Assume that for a prime $p$ of $K$ we are given the datum $(L_{p}(s,\rho))_{\rho}$, where $\rho$ varies over the irreducible representations of $G$ and $L_{p}(s,\rho)$ is the Euler factor in the sense of Artin; i.e. it is the reciprocal of the characteristic polynomial of the Frobenius $\sigma_p$ after we restrict $\rho$ to the decomposition group $D$ and then we replace $\rho$ with the representation of $D/I$ on $V^I$, where $I$ is the inertia group. So in formulas $$L_{p}(s,\rho) = \det ({\rm id}_{V^I}- X\cdot \rho(\sigma_p)|_{V^I} )^{-1},$$ where $X=N(p)^{-s}$. (Here $V=V_{\rho}$ is the underlying space of the representation.)
My question: 
Does this datum encode the relative degree and ramification index of the prime $p$? 
 A: Yes. The regular representation of $G$ decomposes as the sum of $\rho$'s with multiplicities $\dim\rho$, hence the product of $L(s,\rho)^{\dim\rho}$ over the various $\rho$'s equals the Dedekind zeta function $\zeta_L(s)$. That is, if $e$ (resp. $f$) is the ramification (resp. inertia) degree in $L/K$ of a given prime $p$ of $K$, then the Euler factors at $p$ reveal that
$$ \prod_\rho L_p(s,\rho)^{\dim\rho}=(1-N(p)^{-fs})^{n/(ef)},$$
where $n=(L:K)$. So your datum determines $f$ and $n/(ef)$, hence also $e$, because it clearly determines $n$.
A: I will try to answer my question, at least on how to recover $e$, following a discussion with Chantal David on this. 
First put $n=[L:K]$, $e_p$ the ramification index of $p$, $f_p$ the relative degree (aka, inertia degree, aka residue degree), and $g_p$ the number of primes of $L$ lying above $p$. Then 
$$
\frac{\zeta_L(s)}{\zeta_K(s)} = \prod_{p} (1-N(p)^{-f_p s})^{-g_p}(1-N(p)^{-s}).
$$
(Here $N(p) = |O_K/p|$.)
On the other hand, Artin proved that 
$$
\frac{\zeta_L(s)}{\zeta_K(s)} = \prod_{\rho\neq \rho_0}\prod_p  L_p(s,\rho)^{\dim \rho},
$$
where $\rho_0$ is the trivial irreducible representation. 
So comparing these locally (at our $p$) and introducing a new variable $u=N(p)^{-s}$, we get that 
$$
(1-u^{f_p})^{-g_p}(1-u) = \prod_{\rho\neq \rho_0} L_p(s,\rho)^{\dim \rho}.
$$
Comparing degrees in $u$ gives that 
$$
1-g_pf_p = \sum_\rho \dim\rho \deg_u L_p(s,\rho)
$$
so that we recovered $g_pf_p$ and thus also $e_p=\frac{n}{g_pf_p}$.
