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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 !

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  • $\begingroup$ The inertia group usually depends on the prime. Anyway, it gives information only about the prime w.r.t. it is defined. $\endgroup$ – Jyrki Lahtonen Oct 25 '13 at 9:31
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    $\begingroup$ Or more precisely. If you consider the set of prime ideals of $L$ lying above a given prime ideal of $K$, their respective inertia groups are all conjugate in the Galois group, so if one is unramified, then all are. But if you consider the primes of $L$ lying above another prime of $K$, then there is no reason to expect those inertia groups to be related to the former ones. $\endgroup$ – Jyrki Lahtonen Oct 25 '13 at 9:37
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    $\begingroup$ An easy way to know that the Galois group can't tell you global information is to work with examples. For example, a Galois extension of $\mathbb{Q}$ is ramified (since every extension of $\mathbb{Q}$ is ramified!), but there are only finitely many such primes that are so, and all of those other primes will have trivial inertia group. As Jyrki pointed out, the Galois action really only works fiber wise (above a particular prime), and so is unable to tell you global information (at least in this case). $\endgroup$ – Alex Youcis Oct 25 '13 at 9:39
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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 $\mathfrak{p}$.

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