What is the Galois group of $\mathbb{Q}_p[\zeta] / \mathbb{Q}_p$, where $\zeta$ is a $p^r$th root of unity? Let $\zeta$ be a $p^r$-th root of unity and $\mathbb{Q}_p$ the p-adic numbers. I would like to know the Galois automorphisms of $\mathbb{Q}_p[\zeta] / \mathbb{Q}_p$.
I already know,


*

*the degree of that extension is $\varphi(p^r) = p^{r-1} (p-1)$ 

*the extension is totally ramified


Furthermore I think the automorphisms are the same as in the global extension $\mathbb{Q}[\zeta] / \mathbb{Q}$ (namely, $\sigma_k: \zeta \mapsto \zeta^k$), but I do not know why. Furthermore this should be wrong if you switch to the $n$-th root of unity, where $n$ is not a prime power, but I also dont see any reason for this. I hope anyone can help.
Thanks
Laura
 A: Suppose $\zeta$ is a primitive $n$th root of unity. The difference between $\mathbb{Q_p}[\zeta]/\mathbb{Q}_p$ and $\mathbb{Q}[\zeta]/\mathbb{Q}$ is that $\mathbb{Q}_p$ in general may contain some extra relations between the $n$th roots of unity, in which case adjoining $\zeta$ will not enlarge it as much as it does $\mathbb{Q}$. 
For a general field $K$, an automorphism $\phi$ of $K[\zeta]/K$ is determined completely by where it sends $\zeta$. Since $\zeta$ is a root of unity, $\phi(\zeta)$ has to be one as well, so $\phi(\zeta)=\zeta^k$ for some integer $k$. Which integers $k$ are allowed depends on the minimal polynomial of $\zeta$, which depends on $K$. 
Over $\mathbb{Q}$ the minimal polynomial of $\zeta$ is the $n$th cyclotomic polynomial
$$ \Phi_n(X) = \prod_{ \substack{1 \leq k \leq n \\ gcd(k,n)=1}} (X - \zeta^k).$$
In other words the elements of ${\rm Gal}(\mathbb{Q}[\zeta]/\mathbb{Q})$ are $\phi_k(\zeta) = \zeta^k$ where $k$ belongs to $(\mathbb{Z} /n \mathbb{Z})^\times$. Then $k\mapsto \phi_k$ is an isomorphism of $(\mathbb{Z}/n\mathbb{Z})^\times$ onto ${\rm Gal}(\mathbb{Q}[\zeta]/\mathbb{Q})$ and so the latter has order $\varphi(n)$. For instance when  $n=p^r$, we have $[{\rm Gal}(\mathbb{Q}[\zeta]/\mathbb{Q})]=\varphi(p^r)=p^{r-1}(p-1)$.
Now over any other field $K$ of characteristic zero, the minimal polynomial of $\zeta$ has to divide $\Phi_n(X)$. If $n=p^r$ and $K=\mathbb{Q}_p$, the same argument used to prove the irreducibility of $\Phi_{p^r}$ over $\mathbb{Q}$, in other words Eisenstein's criterion, works over $\mathbb{Q}_p$. Here it is for $r=1$.
We have
$$ \Phi_{p}(X) = \frac{X^p - 1}{X-1}.$$
Setting $X = Y+1$, we can write
$$ \Phi_p(X)=\Phi_p(Y+1) = \frac{(Y+1)^p - 1}{Y} = Y^{p-1} + {p \choose p-1} Y^{p-2} + \cdots + {p \choose 2}Y + p.$$
All the coefficients after the leading one are divisible by $p$. If this polynomial were reducible, so that $\Phi_p(Y+1)=Q(Y)Q'(Y)$, then modulo $p$ we would get $\overline{Q}(Y)\overline{Q}'(Y) = Y^{p-1}$ over $\mathbb{F}_p$. That would mean that $Q(Y) = Y^r$ and $Q'(Y)=Y^s$ with $s+r = p-1$. It follows that $Q(Y)$ and $Q'(Y)$ must both have constant terms divisible by $p$. But then $Q(Y)Q'(Y)$ would have a constant term divisible by $p^2$, which $\Phi_p(Y+1)$ does not. Therefore $\Phi_p(X)$ is irreducible. This is (the proof of) Eisenstein's irreducibility criterion and in this case works the same over $\mathbb{Q}_p$ as over $\mathbb{Q}$. It shows that ${\rm Gal}(\mathbb{Q}_p[\zeta]/\mathbb{Q}_p) \cong {\rm Gal}(\mathbb{Q}[\zeta]/\mathbb{Q})$.
For $n=p^r$ with $r>1$ you can use the fact that $\Phi_{p^r}(X)=\Phi_p(X^{p^{r-1}})$ and use a similar argument.
If $n\neq p^r$, $\Phi_n(X)$ may not be irreducible over $\mathbb{Q}_p$, in which case $\mathbb{Q}_p[X]/(\Phi_n(X))$ splits into a direct product of isomorphic fields, one for each distinct prime of $\mathbb{Q}[\zeta]/\mathbb{Q}$ lying over $p$. Therefore the irreducibility of $\Phi_n(X)$ over $\mathbb{Q}_p$ is tied to the splitting behaviour of the prime $p$ in $\mathbb{Q}[\zeta]$. 
When $p\nmid n$, $p$ is unramified in $\mathbb{Q}_p[\zeta]/\mathbb{Q}_p$ and so ${\rm Gal}(\mathbb{Q}_p[\zeta]/\mathbb{Q}_p)$ is isomorphic to ${\rm Gal}(\mathbb{F}_p[\zeta]/\mathbb{F}_p)$, which is cyclic of order $n/gcd(n,p-1)$.
