Let "ur" be for "unramified maximal". I read as "well known" that if $L/K$ is an abelian totaly ramified extension of the local field $K$ then $Gal(\widehat{L_{ur}}/\widehat{K_{ur}})\simeq Gal(L/K)$ where $\widehat{ }$ is the completion of the corresponding fields.
I looked for this result in a tenth of books (Neukirsch, Lang, Cassels, Serre etc) I never saw anything about this. Can someone explain me how this result can be proved.
I think i got it but i would like to have a confirmation or infirmation.
In fact $Gal(\widehat{L_{ur}}/\widehat{K_{ur}})\simeq Gal(L/K)_{ram}=G_0$ and as $L/K$ is uramified $Gal(L/K)_{ram}=Gal(L/K)$
First Galois group of completions is the same : $Gal(\widehat{L_{ur}}/\widehat{K_{ur}})\simeq Gal(L_{ur}/K_{ur})$.
Next, one knows that every local field admits a unique unramified extension of degree $f$ (in a given separable closure). We note $K_f$.
It is also true (Lang prop 8 p. 49) that the compositum of an unramified extension by a finite one $L$ is unramified. Consequently $L.K_f$ is an unramified extension of $L$. As $[LK_f:L]=[K_f:K]=f$ we have $L_f=LK_f$.
Taking unions we deduce $L^{ur}=LK^{ur}$ (that is the maximal unramified extension).
We know too that a finite separable extension $L/K$ contains a unique maximal unramified subextension, $K_L$.
Now, $L\cap K^{ur}$ contains trivialy $K_L$ but is too an unramified subextension as it is a finite subextension of $K^{ur}$. From this $L\cap K^{ur}=K_L$.
$L^{ur}/L$ is abelian (of Galois group $\widehat{\Bbb Z}$), $L/K_L$ is abelian too (and even cyclic). Therefore $L^{ur}=L\,K^{ur}/K_L$ is abelian. $K^{ur}/K_L$ is abelian too so (Galois theory on the losange
$\begin{array}{rcccl} &&LK^{ur}=L^{ur}&&\\ &\swarrow&&\searrow\\ L& & & &K^{ur}\\ &\searrow&&\swarrow\\ &&L\cap K^{ur}=K_L \end{array} $
$L^{ur}=L\,K^{ur}/K^{ur}$ is Galois and $Gal(L^{ur}/K^{ur})\simeq Gal(L/K_L)$ that is the ramification or inertial group.