Image of Frobenius So here is the setting: I have a Galois extension $L/K$, and a prime ideal $\mathfrak{P}$ of $\mathcal{O}_L$ lying over a prime ideal of $\mathcal{O}_K$. There is a map
$
D_{\mathfrak{P}} \rightarrow \operatorname{Gal}((\mathcal{O}_L/\mathfrak{P})/(\mathcal{O}_K/ \mathfrak{p})) = G
$
and we find that the galois group is cyclic generated by the Frobenius automorphism. Now say I have a representation
$$
\rho: G \rightarrow \operatorname{GL}_2(A)
$$
where $A$ is basically $\mathbb{C}$ or $\overline{\mathbb{F}}_p$ or $\overline{\mathbb{Q}}_p$  (I do not think it matters for my question). Then I do not understand what $\rho(\operatorname{Frob}_p)$ represents (for $p$ unramified). Many articles/books talk of $\operatorname{Tr}\rho(\operatorname{Frob}_p)$ so it would be very useful to know.
Also, if anyone knows of any "introductory" textbook on this subject, it would be great.
Thank you !
 A: From the context I assume your $K$ is meant to be a number field.
Let $\mathfrak{P}$ be a prime of $L$. Then we have a decomposition group $D_\mathfrak{P}$ and an inertia group $I_{\mathfrak{P}}$, with $I_\mathfrak{P} \subset D_\mathfrak{P} \subset \operatorname{Gal}(L / K)$, and the quotient $D_\mathfrak{P} / I_\mathfrak{P}$ is generated by the Frobenius $\operatorname{Frob}_{\mathfrak{P}}$ (or topologically generated, if your $L/K$ isn't assumed to be a finite extension)
In particular, if I have a representation $\rho$ of $\operatorname{Gal}(L / K)$, and I know that $\rho$ is trivial on $I_\mathfrak{P}$, it follows that $\rho(\operatorname{Frob}_{\mathfrak{P}})$ is well-defined (EDIT: because $\rho$ factors through $G / I_{\mathfrak{P}} \supset D_\mathfrak{P}/I_{\mathfrak{P}} \ni \operatorname{Frob}_{\mathfrak{P}}$). 
Now, here is the more delicate part. If $\mathfrak{P}$ and $\mathfrak{P}'$ are two primes of $L$ above the same prime $\mathfrak{p}$ of $K$, then there is an element of $\operatorname{Gal}(L / K)$ which sends $\mathfrak{P}$ to $\mathfrak{P}'$, and conjugating by this element sends $D_\mathfrak{P}$, $I_\mathfrak{P}$ and the Frobenius $\operatorname{Frob}_{\mathfrak{P}}$ to their analogues for $\mathfrak{P}'$. So $\rho$ is unramified at $\mathfrak{P}$ iff it's unramified at $\mathfrak{P}'$, and we can make sense of $\operatorname{Tr} \rho(\operatorname{Frob}_{\mathfrak{p}})$: it's a shorthand notation for ``$\operatorname{Tr} \rho(\operatorname{Frob}_{\mathfrak{P}})$ for any choice of prime $\mathfrak{P} \mid \mathfrak{p}$'' -- different choices of $\mathfrak{P}$ will give different elements in the image of $\rho$, but they'll all be conjugate, and thus they'll all have the same trace.
Does that clear up your confusion?
