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

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    $\begingroup$ When lifiting $\overline{\beta}$ a primitive root of $x^{p^f-1}-1$ you obtain $\beta$ whose order divides $p^f-1$ and is $\ge$ than the order of $\overline{\beta}$, so $\beta$ is a primitive root of $x^{p^f-1}-1$. And the author meant $q = p^k$ with $\mathbb{F}_q$ being (isomorphic to) the residue field of $K$ $\endgroup$ – reuns Oct 23 '18 at 19:41
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    $\begingroup$ There's no mistake in saying the residue field of $K$ is $\Bbb F_q$: $q$ need not equal $p$. As $K$ is a finite extension of $\Bbb Q_p$, its residue field is a finite extension of that of $\Bbb Q_p$, that is a finite extension of $\Bbb F_p$. This is $\Bbb F_q$ where $q$ is a power of $p$ (possibly $q=p$ but not necessarily so). $\endgroup$ – Lord Shark the Unknown Oct 23 '18 at 19:50
  • $\begingroup$ Damn it, I didn't realize that haha. So my proof is correct? I see that I need only to change my $p$ to $q$, where of course, $q$ is a power of $p$ since the residue field of $K$ is a finite extension of the residue field of $\mathbb{Q}_{p}$, which is $\mathbb{F}_{p}$. $\endgroup$ – Camacho Camachito Oct 23 '18 at 21:03

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