# Unramified extensions, inertia groups and decomposition fields.

I am at the moment reading a paper where they use some Algebraic Number Theory, which I know little of and I am curious about the following:

Let $L/K$ be a finite Galois extension and let v be a place of K that does not ramify L. If v is a non-archimedean place, we have a notion of an inertia group and a decomposition group. What can we, in general, say about the relation between the inertia group here and the absolute galois group of L? The Galois Group of L over K? I know this is a vague question, but I am looking for what we can say here, for example, can we say that the inertia group is a normal subgroup of the galois group?

I assume $L$ and $K$ are supposed to be number fields (or global fields). There are in general several decomposition (inertia) groups in $\mathrm{Gal}(L/K)$ attached to a finite prime $v$ of $K$, one for each finite prime of $L$ lying over $v$. If $w$ is such a prime of $L$, then the decomposition group $G_w$ is the subgroup of $\sigma\in\mathrm{Gal}(L/K)$ such that $\sigma w=w$ (if you think of $w$ as a prime ideal in the ring of integers in $L$, this means that $\sigma$ fixes the prime ideal, though not necessarily pointwise). Every element of $G_w$ gives rise to a well-defined automorphism of the residue field $k(w)$ of $w$, which can be thought of as $\mathscr{O}_L$ modulo the corresponding prime ideal. This gives a homomorphism from $G_w$ to $\mathrm{Gal}(k(w)/k(v))$. Its kernel is, by definition, the inertia group $I_w$ of $w$ in $\mathrm{Gal}(L/K)$. So the inertia group is a normal subgroup of the decomposition group. If $w^\prime$ is another prime of $L$ lying over $v$, then there is an element $\sigma\in\mathrm{Gal}(L/K)$ with $\sigma w=w^\prime$, and then the corresponding decomposition groups are conjugate (via $\sigma$). So, unless, e.g., there is only one prime of $L$ lying over $v$, the decomposition group $G_w$ will not be normal in $\mathrm{Gal}(L/K)$. The same goes for inertia groups.
To say that $v$ is unramified in $L/K$ in this context is to say that $I_w$ is trivial for every $w\mid v$ (or equivalently, for any $w\mid v$). So in this case the inertia group of $w$ is (trivially) a normal subgroup of $\mathrm{Gal}(L/K)$. When $G$ is abelian, conjugation by any element is the identity, so there really is a well-defined decomposition (resp. inertia) group attached to any prime $v$ of $K$, and they are of course normal in $\mathrm{Gal}(L/K)$.
I'm not really sure what you're asking about the absolute Galois group of $L$. This is a different object. Namely it is $\mathrm{Gal}(L_s/L)$ where $L_s$ is a choice of separable closure of $L$. The Galois group $\mathrm{Gal}(L/K)$ is can be regarded as a quotient of $\mathrm{Gal}(K_s/K)$ once an embedding of $L$ into $K_s$ is chosen. One can also define decomposition groups and inertia groups for infinite Galois extensions like $K_s/K$, and they have similar properties to the ones defined for finite Galois extensions. Also, if $\bar{w}$ is a prime of $K_s$ lying over the prime $w$ of $L$ which itself lies over $v$, then the image of the decomposition group $D(\bar{w})\subseteq\mathrm{Gal}(K_s/K)$ in $\mathrm{Gal}(L/K)$ is the decomposition group $D(w)$. The same goes for inertia groups.
• This happens when the extension of completions $L_w/K_v$ is totally ramified, meaning that the ramification index of $v$ in $L/K$ is equal to $[L_w:K_v]$, or equivalently, the residue extension $k(w)/k(v)$ is trivial, that is, $k(w)=k(v)$. – Keenan Kidwell Apr 2 '13 at 16:19