Unramified Extensions
The unramified extensions of a completion of a number field at a nonarchimedean prime are easily described and have a number of very special properties. We give just a few properties here.
Let $K$ be an algebraic number field, $R$ the ring of algebraic integers, $\mathfrak{p}$ a nonzero prime ideal of $R$, and $p$ the positive prime integer in $\mathfrak{p}$. Let $R = R/\mathfrak{p} \cong \text{GF}(q)$.
Theorem 3.9 Let $\hat{L}$ be a finite-dimensional extension of the completion $K_{\mathfrak{p}}$, such that $\hat{\mathfrak{p}}$ is unramified in $\hat{L}$. Let $f = [\hat{L}: K_{\mathfrak{p}}]$. Then
$$ \hat{L} = K_{\mathfrak{p}}(\beta) $$
where $\beta$ is a primitive $q^f - 1$ root of unity. Conversely, for any positive integer $f$ and primitive $q^f - 1$ root of unity $\beta$, the extension $K_{\mathfrak{p}}(\beta)$ is unramified and has degree $f$.
Proof Let $\hat{R}$ be the valuation ring in $K_{\mathfrak{p}}$ and $\hat{S}$ its integral closure in $\hat{L}$. By the unramified assumption, $\hat{\mathfrak{p}} \hat{S}$ is the maximal ideal of $\hat{S}$ and
$$ \frac{\hat{S}}{\hat{\mathfrak{p}} \hat{S}} \cong \text{GF}(q^f), \quad \frac{\hat{R}}{\hat{\mathfrak{p}} \hat{R}} \cong \text{GF}(q). $$
So $\hat{S}/\hat{\mathfrak{p}} \hat{S}$ contains a primitive $q^f - 1$ root of unity; by Hensel's lemma, there is a primitive $q^f - 1$ root of unity, $\beta$, in $\hat{S}$. The ring $\hat{R}(\beta)$ contains a full set of representatives of the cosets of $\hat{\mathfrak{p}} \hat{S}$ in $\hat{S}$. By Nakayama's Lemma (Chapter 1, following $\textbf{Proposition 1.4}$), we have $\hat{R}(\beta) = \hat{S}$, and hence $\hat{L} = K_{\mathfrak{p}}(\beta)$ follows.
Conversely, given a primitive $q^f - 1$ root of unity $\beta$, we know $K_{\mathfrak{p}}(\beta)$ is unramified because the discriminant of the polynomial $X^n - 1$, with $n = q^f - 1$, is not divisible by $p$. The change in residue class degree must be $f$ by consideration of the roots of unity in the finite fields. It follows that $[K_{\mathfrak{p}}(\beta): K_{\mathfrak{p}}] = f$.
Corollary 3.10 For any positive integer $f$, there is one and only one unramified extension of $K_{\mathfrak{p}}$ having dimension $f$ over $K_{\mathfrak{p}}$.
I was going through the proof of the above theorem and am unable to understand the proof of the converse:
Conversely given a primitive $q^f - 1$ root of unity $\beta$, we know $K_{\mathfrak{p}}(\beta)$ is unramified because the discriminant of the polynomial $X^n -1$, with $n = q^f-1$, is not divisible by $p$. The change in residue class degree must be $f$ by consideration of the roots of unity in the finite fields. It follows that $[K_{\mathfrak{p}}(\beta): K_{\mathfrak{p}}] = f$.
I am stuck in proving that the extension is unramified. I do not see the relevance in finding the discriminant of $X^n-1$ (with $n=q^f -1$), because that is not the minimal polynomial of $\beta$.
When it comes to proving that $[K_{\mathfrak{p}}(\beta): K_{\mathfrak{p}}] = f$, I was able to prove that $[K_{\mathfrak{p}}(\beta): K_{\mathfrak{p}}] \geq f$, because in $\hat S/ \hat {\mathfrak{p}} \hat S$, the polynomial $X^{q^f-1}-1$ splits, because it splits in $\hat S$, as $\beta\in \hat S$ is a primitive $(q^f-1)$th root of unity. But I was unable to prove the other direction of this inequality.
I would be grateful for any help to finish the proof.
