I have some questions on following exercise sheet from a former course on Local Fields and Class Field Theory. this is the original source.
Questions:
I try to prove Exercise 1(a): Let $K= \mathbb{Q}_p(\zeta_{p^n}) $ for $n \ge 1$. Ex 1(a) is to prove that $K/\mathbb{Q}_p$ is Galois and totally ramified. (geven hint: to prove that $\zeta_{p^n}-1$ is a uniformizer.)
two questions on this exercise:
1): why $K/\mathbb{Q}_p(\zeta_{p^n})$ is totally ramified?
2): below I showed that $K$ is Galois without using that $\zeta_{p^n}-1$ is a uniformizer. nevertheless I would like to know what is the way the show that $K$ is Galois using (!) that $\zeta_{p^n}-1$ is uniformizer explicitly.
what I tried: on 1): by definition an extension $L/F$ of non-archimedian local field is totally ramified iff $L$ and $F$ have the same residue fields; for example see this pdf. $\mathbb{Q}_p$ has residue field $\mathbb{F}_p$ and $K$ has $\mathbb{F}_p[\zeta_{p^n}]$. but since $p^n $ not divides $p-1 = \vert \mathbb{F}_p^* \vert$, we conclude $\zeta_{p^n} \not \in \mathbb{F}_p$ and therefore the both residue fields of $K$ and $\mathbb{Q}_p$ not coinside and thus thus the extension $K/\mathbb{Q}_p(\zeta_{p^n})$ can't be totally ramified. I'm a bit confused about that. what I'm doing wrong at this point?
on 2): the minimal polynomial of $\zeta_{p^n}$ over $\mathbb{Q}_p$ divides $X^{p^n}-1$ and is therefore separable, since $char(\mathbb{Q}_p)=0$. additionally, $X^{p^n}-1$ splits over $K$, since $\zeta_{p^n} \in K$ by definition and all roots of $X^{p^n}-1$ are of the form $\zeta_{p^n}^k$ for all $1 \le k \le p^n$. therefore, the minimal polonomial of $\zeta_{p^n}$ over $\mathbb{Q}_p$ splits also in $K$. therefore $K$ is Galois.
now I have keen interest to find out the alternative way to prove, which seemingly was proposed, using the information that $\zeta_{p^n}-1$ is uniformizer explicitly.
Best Answer
For your question 1, you can use the fact that $\pi= \zeta_{p^n}-1$ is a root of the polynomial : $F(X)=(1+X)^{(p-1)p^{n-1}}+(1+X)^{(p-1)p^{n-2}} + \dots + 1$ which is an Eisenstein polynomial over $\mathbb{Q}_p$, and hence irreducible. But $\mathbb{Q}_p(\pi) = \mathbb{Q}_p(\zeta_{p^n}) = K$, so $[K : \mathbb{Q}_p] = \phi(n)$.
On the other side, to show that $K/\mathbb{Q}_p$ is totally ramified, you can notice that $F(0) = p = \prod_{\sigma \in Gal(K/\mathbb{Q}_p)} \sigma(\pi)$. But we have also : $|\pi|_K = |\sigma(\pi)|_K$, hence $|p|_K = |\pi^{\phi(n)}|_K$. So, we have : $p \in \pi^{\phi(n)}O_K^{\times}$. But as $\phi(n)$ is also the degree of $K/\mathbb{Q}_p$, you can conclude directly that $\pi$ is a uniformizer and $K/ \mathbb{Q}_p$ is totally ramified. Moreover, we can deduce from everything above that the conjuguate of $\zeta_{p^n}$ is the $(\zeta_{p^n}^{i})_{gcd(i,p) = 1, i \in \{ 0, \dots, p^n-1 \}}$, and then the Galois group of $K/\mathbb{Q}_p$ is clearly isomorphic to $(\mathbb{Z}/p^n\mathbb{Z})^{\times}$.
Finally, to find $O_K$, you can prove that $O_K \cap \mathbb{Q}_p[\pi] = \mathbb{Z}_p[\pi]$.