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).
[Math] Definition of inertia group in infinite extensions
algebraic-number-theory
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1) Let $G=$ Gal$(L/K)$, $p \in K$. Suppose $p$ is unramified and nonsplit. (Since only finitely many primes are ramified, it suffices to show that this cannot occur.) Since $p$ is unramified and nonsplit and $efg=|G|$, we see that $f=|G|$ and the decomposition group $D_p$ is isomorphic to $G$. But we also have that $D_p$ is isomorphic to the Galois group of the residue field of $L/K$ at $p$, which is cyclic of order $f$. This contradicts our hypothesis on $G$.
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.
Best Answer
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 $k/L/K$.