Ramification of a Galois extension I understand that an extension of number field $L/K$ is unramified if every non-zero prime ideal of $\mathcal{O}_K$ is unramified in $L$ (where a prime ideal $\mathfrak{p}$ of $\mathcal{O}_K$ is ramified if it has $e_i > 1$ for some $i$ when you write it as the decomposition of primes of $\mathcal{O}_L$).
However I am not sure of the situation when $L/K$ is Galois. For example, if the inertia group is trivial does it imply that the whole extension is unramified or just one of its primes ?  
Thanks for your help !

The simplification that occurs when $L/K$ is Galois is that for any fixed prime $\mathfrak{p}$ of $\mathcal{O}_K$, the Galois group acts transitively on the primes $\mathfrak{P}_i$ of $\mathcal{O}_F$ above $\mathfrak{p}$. So the decomposition groups $D_{\mathfrak{P}_i/\mathfrak{p}}$ and the inertia groups $I_{\mathfrak{P}_i/\mathfrak{p}}$ for different $i$ are all conjugate, and so the ramification indices $e_{\mathfrak{P}_i/\mathfrak{p}}$ are all the same, as are the residue field degrees.
This was for a fixed prime downstairs. The story for another fixed prime $\mathfrak{p}'$ of $\mathcal{O}_K$ is (more or less) independent of the story for $\mathfrak{p}$.
