$L/K$ unramified at $v$ and $\widetilde{K}$ intermediate field $\Rightarrow\, \widetilde{K}/K$ unramified at $v$ Let $K$ be a number field, $L$ a field with $K\subset L\subset\mathbb{C}$ and $L/K$ Galois (not necessarily finite).
I'm trying to prove the following:

Let $L/K$ be unramified at a finite place $v$ and $\widetilde{K}$ an intermediate field $K\subset\widetilde{K}\subset L$ with $\widetilde{K}/K$ Galois, then $\widetilde{K}/K$ is unramified at $v$.

I think I know how to prove this for $L/K$ finite:
Let $p\subset\mathcal{O}_K$ be the prime associated to the finite place $v$. Since $L/K$ is unramified at $v$, then $p\mathcal{O}_L$ has the form $p\mathcal{O}_L=\mathfrak{P}_1...\mathfrak{P}_r$, where $\mathfrak{P}_1,...,\mathfrak{P}_r$ are primes above $p$. Letting $\mathfrak{Q}_i:=\mathfrak{P}_i\cap\mathcal{O}_{\widetilde{K}}$, we get:
$$p\mathcal{O}_{\widetilde{K}}=p\mathcal{O}_L\cap\mathcal{O}_{\widetilde{K}}=\mathfrak{Q}_1...\mathfrak{Q}_r$$
Therefore $p$ is unramified at $\widetilde{K}$, which is what we want.
Is this true for $L/K$ infinite? How do I prove it?
 A: Let $\tilde{\nu}$ be a prime in $\tilde{K}$ over $\nu$, and let $\beta$ be a prime over $\tilde{\nu}$, we have to prove that the restriction from $I_{\nu}^{\beta}$ to $I_{\nu}^{\tilde{\nu}}$ is surjective.
Consider the sets $A=\{I_{\nu}^{\mathfrak{a}_M}:\mathfrak{a}_{M}\subset \beta,\ \mathfrak{a}_M\ \text{is a prime over }\nu\ \text{of a finite Galois extension } M/K,\ M\subset L\}$ and 
$B=\{I_{\nu}^{\mathfrak{p}_N}:\ \mathfrak{p}_N\subset \tilde{\nu}, \ \mathfrak{p}_N\ \text{is a prime over }\nu \text{ of a finite Galois extension } N/K,\ N\subset \tilde{K}\}$.
This sets are ordered by $I_{\nu}^{\mathfrak{a}_{N_i}}\leq I_{\nu}^{\mathfrak{a}_{N_j}}$ if $N_i\subset N_j$, and 
if $I_{\nu}^{\mathfrak{a}_{N_i}}\leq I_{\nu}^{\mathfrak{a}_{N_j}}$ you have the restriction map, $Res:I_{\nu}^{\mathfrak{a}_{N_j}}\rightarrow I_{\nu}^{\mathfrak{a}_{N_i}}$, this restriction map is surjective because we are dealing with finite Galois extensions. You can prove these are projective systems and that $\lim\limits_{\leftarrow}A=I_{\nu}^{\beta}$ and $\lim\limits_{\leftarrow}B=I_{\nu}^{\tilde{\nu}}$. 
This means that $I_{\nu}^{\tilde{\nu}}\cong \{(\sigma_{N})_N\in\prod\limits_{K\subset N\subset \tilde{K}} I_N: \sigma_{N_i}=\sigma_{N_j}|_{N_i}\}$ and that $I_{\nu}^{\beta}\cong\{(\sigma_{M})_M\in\prod\limits_{K\subset M\subset L} I_M: \sigma_{M_i}=\sigma_{M_j}|_{M_i}\}$.
Since the restriction map between inertia groups of finite Galois extensions is surjective, the projections in the above product are surjective, and this means that the restriction $Res I_{\nu}^\beta\rightarrow I_{\nu}^{\mathfrak{p}_M}$ is surjective for any $M$ finite Galois subextension and $\mathfrak{p}_M$ a prime of $M$ with $\mathfrak{p}_M\subset \beta$.
With this you can see that the restriction from $I_{\nu}^{\beta}$ to $I_{\nu}^{\tilde{\nu}}$ is surjective.
