I'm stucked with a theorem withouth proof saw in a Book about G-Functions by Dwork, I will appreciate any hint, also I provide a ''proof'' of that theorem, but a feel that is too ''bla bla'' and I think is wrong. Here is the theorem:
Let $K$ be a finite extension of $\mathbb{Q}_{p}$ with residue field $\overline{K}=\mathbb{F}_{q}$ and let $f$ be a positive integer. Then the unique unramified extension of $K$ is $K(\xi)$ where $\xi$ is a primitive $(q^{f}-1)$ root of unity.
First, I think that the autor made a mistake saying that the residue field is $\mathbb{F}_{q}$, since we know that the residue field of the $p-$adic numbers are $\mathbb{F}_{p}$, so when the autor write $q\,\, $ I will understand is $p$, so my proof uses $p$ instead of $q$. Here is my attempt:
Let $F$ be the unique unramified extension of $K$ of degree $f$ (This fact was learned on the proof of the existence of unramified extension, the degree of the extension is equal to the degree of the separable extension of the residue fields (of course, in this case any extension of $\mathbb{F}_{p}$ is separable (Galois)) ).
We look at the extension $\overline{F}=\mathbb{F}_{p^{f}}$ of $\mathbb{F}_{p}$. We know that the multiplicative group of units is cyclic, lets say that $\overline{\alpha}$ is a generator, and we know that is a primitive $(p^{f}-1)$ root of unity, so satisfy $\overline{f}(x)=x^{p^{f}-1}-1\in\mathbb{F}_{p}[x]$ (observe that this polynomial is separable). Let $\overline{m}(x)$ be the mynimal polynomial of $\overline{\alpha}$ over $\mathbb{F}_{p}$ which has degree $f$. I know that the lifting of $\overline{m}$ is also irreducible. Now I pick $\alpha\in\mathcal{O}_{F}$ (the ring of integers of F) such that $(\alpha\mod \mathfrak{p}_{F})=\overline{\alpha}$. Here $\mathfrak{p}_{F}$ denotes the unique maximal ideal of the ring of integers.
Note that $\overline{m(\alpha)}=\overline{m}(\overline{\alpha})=0$ and $\overline{m^{\prime}(\alpha)}=\overline{m}^{\prime}(\overline{\alpha})\neq 0$ (this last fact is due to the separability of $\overline{m}$), so by hensel Lemma we can find a root of $m$, say $\beta$ whose reduction modulo $\mathfrak{p}_{T}$ is $\overline{\alpha}$. Then the extension $K(\beta)/K$ is unramified (Here I'm repeating the arguments of the proof of the existence of unramified extensions), and the degree of the extension is $f$. Also, notice nath $\beta$ is actually a primitive $(p^{f}-1)-$rooth of unity, and by the uniqueness of the extension $F$ we must have $F=K(\beta)$.
My only questionf of my approach is how do I know that $\beta$ is actually a primitive rooth of unity. For me is clear that is a rooth of unity, but not a primitive rooth.
I had a second approach: Clearly $E=K(\xi)$ is the spliting field of $f(x)=x^{p^{f}-1}-1$. To see that this extension is unramified I only need to check that $[\overline{K(\xi)}:\overline{K}]=[K(\xi):K]$. What I can say is that $[\overline{K(\xi)}:\overline{K}]=f$ but I can't show tha last part.
I will appreciate any hint or correction. Thanks