I'm searching for a recommendable reference dealing with theory of 
non-Archimedean local fields where I can find proofs of the following claims about
finite extensions $L/K$ of non-Archimedean local fields with finite 
residue fields $l / k$. I'm pretty sure that this request might not have a research level but up to now I haven't obtained a satisfying answer asking the same at MSE.


Firstly, we use notations: We denote by $q \in O\_K$ a uniformizer of $O\_K$
and $\pi \in O\_L$ a uniformizer of $O\_L$. Then $n= [L:K]=e \cdot f$ with
$f= [l:k]$ and $(q)O\_L= (\pi)^eO\_L$.


Now I'm looking for rigorous proofs of the following statements:



> 
> (i) Case $L/K$ is Galois & unramified ($\Rightarrow$ $e=1$ & $[L:K]= [l:k]$)
>  and $p$ the characteristic of finite field $k$:
> 
> 
> 


The claim is $G= \operatorname{Gal}(L/K)= \operatorname{Gal}(l/k)=g$. I'm looking for a proof making an
explicit construction of the "lift" $\operatorname{Gal}(l/k) \to \operatorname{Gal}(L/K)$, i.e. if we have
a $k$-automorphism of $l= k(a)= k[X]/ (X^f-1)$, how can it be uniquely lifted
to a $K$-automorphism of $L$?



> 
> (ii) If $L/K$ is unramified and $K= \mathbb{Q}\_p$, then $L$ is cyclotomic:
>  $L= \mathbb{Q}\_p(\zeta\_n)$, for $\zeta\_n$ an appropriate root of unity.
> 
> 
> (iii)(1) Case $L/K$ totally ramified with $n =e$ coprime to $p=\operatorname{char}(k)$: there exists $b \in K$ with $L= K(\sqrt[e]{b})$.
> 
> 
> 


How can this $b$ be 
constructed? Can the $b$ be chosen as a uniformizer: i.e. $b= uq$ with $u \in O\_K^\times$?



> 
> (2) If moreover $p \mid \lvert k \rvert -1$, why is $L/K$ cyclic & Galois extension?
> 
> 
> 



Theory of extensions of non-archimedian local fields
See [Fesenko and Vostokov - Local fields and their extensions](http://www.ams.org/books/mmono/121). (i) is Proposition 3.3(2). (ii) is Proposition 3.2(1). (iii)(1) is Proposition 3.5(1) (and, yes, $b$ may be chosen as a uniformiser). For (iii)(2), I think you meant $e \mid \lvert k\rvert - 1$, not $p \mid \lvert k\rvert - 1$ (which is impossible). Then ($k$, hence) $K$ contains an $e$th root of unity, so $L = K(\sqrt[e]q)$ is (Galois and) cyclic over $K$.


