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 $p$? 



Can the relative degree and ramification index can be read off the characteristic polynomial?
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 $n$.


