Question about inertia groups and unramified extensions

Question

Let $K$ be a number field, and $v$ a finite place. If $\bar{K}$ is a separable closure of $K$, then in $G_K=\text{Gal}(\bar{K}/K)$ we can find the decomposition group of (a place over) $v$, which is isomorphic to the Galois group of $\bar{K_v}/K_v$, with $K_v$ the completion at $v$.

It is well know that the fixed field of the inertia $I_v$ in $\bar{K_v}$ is the maximal unramified extension of $K_v$. Is it also true that the fixed field of the inertia in $\bar{K}$ is the maximal extension of $K$ unramified at $v$? I think this is true, since we can easily move to the finite case where is true, but also I could appreciate a check.

After this, if we consider the maximal extension of $K$ unramified at $v$ and $v'$, with $v\ne v'$, then it is the intersection of the maximal extension unramified at $v$ with the one unramified at $v'$ (it is true? It seems to me obvious), therefore, by Galois correspondence, the product of the inertia $I_vI_{v'}$ is the group corresponding to that field.

But what happens if we consider the maximal extension unramified outside a finite set of places, so unramified at an infinite set of places? The infinite intersection would correspond to an infinite product of subgroups, which of course make no sense. So have we to compute it with, maybe, inverse limit, or something like this?

(My final goal is to understand a proof in Rubin's book Euler systems: he proved that, given a Galois representation $T$ with coefficients in the valuation ring $O$ of a finite extension of $\mathbb{Q}_p$, and a finite set of primes $\Sigma$ containing all primes where $T$ ramifies, primes above $p$ and infinite places, then the Selmer group $S^{\Sigma}(K,T)$ is equal to $H^1(K_{\Sigma}/K,T)$, where $K_{\Sigma}$ is the maximal extension unramified outside $\Sigma$.

The proof is the following: $$\begin{split}S^{\Sigma}(K,T)&\overset{(1)}{=}\ker \left(H^1(K,T)\to \prod_{v\not\in\Sigma} H^1(K_v,T)/H_f^1(K_v,T)\right)= \\&\overset{(2)}{=}\ker\left(H^1(K,T)\to\prod_{v\not\in\Sigma}\text{Hom}(I_v,T)\right)=\\&\overset{(3)}{=}\ker\left(H^1(K,T)\to H^1(K_{\Sigma},T)\right){=}H^1(K_{\Sigma}/K,T). \end{split}$$ (1) is the definition. In (2) we need $\text{Hom}(I_v,T)^{Fr}$, the fixed points of Frobenius, but not a big deal, since we want the kernel,so we can enlarge the codomain. But my big problem is in (3), with which the question is related).

Answer 1

Pick a place $w$ of $\overline{K}$ above $v$. The absolute Galois group acts transitively on those places so we don't care which $w$ we chose.

By density of $\overline{K}$ in $\overline{K}_w\cong \overline{K_v}$ then $Gal(\overline{K}_w/K_v)$ is the $\sigma\in Gal(\overline{K}/K)$ that are continuous for $w$, ie. $D_w$.

$$I_w = Gal(\overline{K}_w/K_v^{nr}), \qquad K_v^{nr}=\bigcup_{n\ge 1} K_v(\zeta_{p^n-1})$$ For a finite Galois extension $L/K$, iff $L$ is fixed by $I_w$ then $L\subset K_v^{nr}$ to that $L/K$ is unramified at $w|_L$. And since $L/K$ is Galois we have unramified at $w|_L$ iff unramified at all the places above $v$.

$\overline{K}^{I_w}$ is the largest extension of $K$ unramified at $w$, not the same as unramified at all the places above $v$.

When considering the largest extension unramified at a set of places, consider the subfield fixed by the all corresponding intertia groups, ie. the subfield fixed by the subgroup of $Gal(\overline{K}/K)$ generated by those intertia groups.