Definition of inertia group in infinite extensions In algebraic number theory we learn the definition of inertia group of a finite Galois extension. What is the definition in the case of an infinite extension? (say in the algebraic closure of Q).
 A: If you define inertia groups via valuation theory, then the exact same definition that works for finite extensions works also for infinite extensions.  Namely, If $K/k$ is a (possibly infinite) Galois extension of global fields, and $v$ is place of $K$, then the inertia group, say $I(K/k)$, of $K/k$ consists of those $\sigma\in \operatorname{Gal}(K/k)$ such that 
$$
v(\sigma(\alpha)-\alpha)>0
$$
for all $\alpha\in K$ satisfying $v(\alpha)\geq 0$.  Incidentally, it is easy to check that this agrees with the other natural definition, that it is the inverse limit of the inertia subgroups $I(L/k)$ as you range over all finite galois subextensions $k/L/K$.
A: as the global case was only parenthetically asked for, I will answer the (1-dimensional) local case:
local case: say $K$ is a discrete valuation field. then one can attempt the same definition as in the above answer but $K$ must be assumed to be henselian (e.g. in the case of a so-called 'local field', $K$ is then assumed to be complete (hence henselian) [and with perfect (or simply finite) residue field]). Then, assuming $K$ henselian, if $K^s$ denotes a fixed separable closure of $K$ and $G_K$ the Galois group $Gal(K^s/K)$, one can define the inertia subgroup $I_K\leq G_K$ as:
$$I_K = Gal(K^s/K^{ur}),$$ where $K^{ur}$ is 'the' maximal unramified extension of $K$ in $K^s$. I'll leave following as a fun and instructive exercise:
Suppose $K$ is a discrete valuation field and consider the henselization $K^h$ and completion $\hat{K}$ of $K$ with respect to the valuation on $K$. Is it true that $I_{K^h} = I_{\hat{K}}$?  
