Galois group of unramified extension of $\mathbb{Q}_p$ is cyclic

Question

I have seen that there for every $f$ there exists a unique unramified extension to $\mathbb{Q}_p$ of degree $f$, and it is $\mathbb{Q}_p(\delta)$ where $\delta$ is a primitive $(p^f-1)$-root of unity. Looking at the polynomial \begin{equation*} p(x)=x^{p^f-1}-1 \in \mathbb{Q}_p[X] \end{equation*} we see that it is separable with the roots $\{1,\delta,\ldots,\delta^{p^f-2}\}$ and splitting field being $\mathbb{Q}_p(\delta)$ so $\mathbb{Q}_p(\delta)/\mathbb{Q}_p$ is a Galois extension of degree $f$ with basis $\{1,\delta,\ldots,\delta^{f-2}\}$. An automorphism $\sigma \in \mbox{Gal}(\mathbb{Q}_p(\delta)/\mathbb{Q}_p)$ is determined by where it sends $\delta$ and since $p(\delta)=0$ we see that $\sigma$ takes $\delta$ to a new $(p^f-1)$-root of unity. This means that $\sigma(\delta) = \delta^i$ where $i=1,2\ldots,p^f-2$. One can therefore look at the group homomorphism \begin{equation*} \phi:\mbox{Gal}(\mathbb{Q}_p(\delta)/\mathbb{Q}_p) \rightarrow (\mathbb{Z}/(p^f-1)\mathbb{Z})^\times \end{equation*} defined by $\phi(\sigma) = i$ where $\sigma(\delta)=\delta^i$. This homomorphism is injective so $\mbox{Gal}(\mathbb{Q}_p(\delta)/\mathbb{Q}_p)$ is isomorphic to a subgroup of $(\mathbb{Z}/(p^f-1)\mathbb{Z})^\times$. Unfortunately, $(\mathbb{Z}/(p^f-1)\mathbb{Z})^\times$ is not cyclic so my proof doesn't really work. In most notes I have read they use other results to prove this but I am interested in a more straightforward approach, if there is one.

Answer 1

Let $\zeta$ be a primitive $(p^k-1)$-th root of unity. Then, it is true that $\mathbb{Q}_p(\zeta)/\mathbb{Q}_p$ is unramified of degree $k$. That said, the roots of the minimal polynomial of $\zeta$ over $\mathbb{Q}_p$ are $\zeta,\zeta^p,\ldots,\zeta^{p^{k-1}}$. Note then that for any $\sigma\in\mathrm{Gal}(\mathbb{Q}_p(\zeta)/\mathbb{Q}_p)$ you have that $\sigma(\zeta)=\zeta^{p^d}$ for some unique $d\in \{0,\ldots,k-1\}$. We denote this Galois element by $\sigma_d$. Since this extension is Galois it's not hard to see that $d\mapsto \sigma_d$ is an isomorphism $\mathbb{Z}/k\mathbb{Z}\to\mathrm{Gal}(\mathbb{Q}_p(\zeta)/\mathbb{Q}_p)$.