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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).

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    $\begingroup$ The compositum of infinitely many subgroups is well-defined, it is just the smallest subgroup containing all of them. I think it should do the job (maybe add "smallest closed" for the Galois correspondence). $\endgroup$ – user208649 Apr 19 '20 at 17:14
  • $\begingroup$ Yes, you're right... And is this isomorphic to the direct sum of all of them? If so, this could also help for (3), since $\prod_{v\not\in \Sigma}\text{Hom}(I_v,T)\cong \text{Hom}(\oplus_{v\not\in \Sigma} I_v,T)$, and so maybe $\oplus_{v\not\in \Sigma} I_v=\text{Gal}(K_{\Sigma}/K)$... But I think that, to be true, they need to have trivial intersections. $\endgroup$ – LStefanello Apr 19 '20 at 19:06
  • $\begingroup$ I don't think the elements in different inertia subgroups commute, so I don't think you can get a direct sum. $\endgroup$ – user208649 Apr 19 '20 at 19:24
  • $\begingroup$ Ye, the inertia are normal but not of trivial intersection $\endgroup$ – LStefanello Apr 19 '20 at 19:25
  • $\begingroup$ One last thought is that the inertia subgroups should generate the absolute galois group of $K_\Sigma$ and so homomorphisms from the absolute galois group to $T$ are determined by their actions on each $I_v$. I don't feel comfortable enough with homology to address (3) but maybe that works for the twisted homomorphisms in $H^1$ and that would at least be a start for passing from the second to the third lines. $\endgroup$ – user208649 Apr 20 '20 at 5:51
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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.

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