Relationship between Galois extensions of local fields and Galois extensions of numbers fields Can someone give me a reference where I can find a proof of the following result :
Let $L'/K'$ be a Galois extension of local fields, then there exists a Galois extension $L/K$ of numbers field such that : $L\subset L'$ and $K\subset K'$  and places $\frak{p}$$\in P(K),$ $\frak P$$\in P(L)$ such that : $L'=L_{\frak P}$ and $K'=K_{\frak p}.$ We may even assume that  $Gal(L'/K')=Gal(L/K).$
thank you for your help.
 A: The first statement just follows from Krasner's lemma which says the following:

Theorem(Krasner): Let $K$ be a complete, nonarchimedean field. Let $\overline{K}$ be an algebraic closure of $K$, and give $\overline{K}$ the unique absolute value extending that on $K$. Then, if $\alpha\in K^{\text{sep}}$. Then, if $\beta\in\overline{K}$ satisfies $|\beta-\alpha|<|\beta-\alpha'|$ for all (non-identity) conjugates of $\alpha$, then $K(\alpha)\subseteq K(\beta)$.

The key is the following corollary of Krasner's lemma:

Corollary: Let $K$ be a complete nonarchimedean field, and let $f(T)\in K[T]$ be monic, irreducible, and separable of degree $n$. Then, there exists $\delta>0$ such that if $g\in K[t]$ is monic of degree $n$ and satisfies $|g-f|_{\text{Gauss}}<\delta$, then $g$ is irreducible and for any root $\alpha$ of $f$ there is a root $\beta$ of $g$ satisfying $K(\alpha)=K(\beta)$.

To see why this does it, we must first show that your field, $K$ say, is a finite extension of some $\mathbb{Q}_p$. You probably already know this, but regardless.
Let $p=\text{char }k$ (where $k$ is the residue field of $K$). Then, since $p\mathcal{O}_K/\mathfrak{p}=0$, we evidently have that $p\in\mathcal{O}_k$, and so by scaling the absolute value we may assume that $|p|=p^{-1}$. Then, the restriction of the absolute value on $K$ to $\mathbb{Q}$ must be the $p$-adic valuation $|\cdot|_p$. Since $K$ is complete, this implies that $\mathbb{Q}_p\hookrightarrow K$.
Note then that obviously $[k:\mathbb{F}_p]=f(K/\mathbb{Q}_p)$ and $[|K^\times|:p^\mathbb{Z}]=e(K/\mathbb{Q}_p)$ must be finite (the latter since $|K^\times|$ is cyclic). Thus, from basic number theory we know that $K/\mathbb{Q}_p$ is finite, and since $\mathbb{Q}_p$ is complete, the absolute value on $K$ must be the unique one.
Now, since $\mathbb{Q}_p$ is perfect, we can write $K=\mathbb{Q}_p(\alpha)$. Let $f(T)\in\mathbb{Q}_p[T]$ be the minimal polynomial of $\alpha$. By the corollary to Krasner's lemma, there exists a $\delta$ such that if $g(t)\in K[T]$ is such that $f(t)-g(t)$ has Gauss norm less than $\delta$, then all that good stuff happens. But, since $\mathbb{Q}$ is dense in $\mathbb{Q}_p$, we can always find a $g(t)\in \mathbb{Q}[t]$ which has close enough Gauss norm to $f(t)$, no matter how small $\delta$ is. Thus, there is some $g(t)\in \mathbb{Q}[t]$ which is irreducible, and for which there is a root $\beta$ of $g(t)$ such that $\mathbb{Q}_p(\beta)=\mathbb{Q}_p(\alpha)=K$.
Consider then $M=\mathbb{Q}(\beta)$. Clearly $M$ is a number field, and $M\subseteq K$. The restriction of the norm on $K$ to $M$ must be of the form $|\cdot|_\mathfrak{P}$ for some prime $\mathfrak{P}$ of $M$ by Ostrowski's theorem for number fields. But, note that since $K$ is complete, the completion of $M$ at $|\cdot|_\mathfrak{P}$ is just the closure of $M$ in $K$. But, note that evidently $M=\mathbb{Q}(\beta)$ is dense in $\mathbb{Q}_p(\beta)=K$, so that the closure of $M$ in $K$, i.e. the completion of $(M,|\cdot|_\mathfrak{P})$, is just $K$ as desired.
All of this material can be found pretty easily on the internet--just google Krasner's lemma. It's in Pete L Clark's notes for example.
Now, I think your question about Galois groups is incorrect, unless I am misreading it. Take any extension of number fields $L/K$ which is Galois, but with non-solvable Galois group. Take a prime $\mathfrak{p}$ of $K$, and a prime $\mathfrak{P}$ of $L$ lying above it. Then, consider $L_\mathfrak{P}/K_\mathfrak{p}$. Clearly then $\text{Gal}(L_\mathfrak{P}/K_\mathfrak{p})\not\cong \text{Gal}(L/K)$ since every Galois group of an extension of local fields is solvable.
