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I've been trying to figure out the following problem. Given a prime $p$, and $\zeta_m = e^{2\pi i /m}$, and the factorization $m = p^kn$ with $(p,n) = 1$, then we know that the Galois group of $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is isomorphic to the group of units $(\mathbb{Z}/m\mathbb{Z})^{\times}$ which is in turn isomorphic to $(\mathbb{Z}/p^k\mathbb{Z})^{\times} \times (\mathbb{Z}/n\mathbb{Z})^{\times}$. I'm trying to see what the decomposition group $D$ and the inertia group $E$ both over $p$ look like when we consider the Galois group in terms of $(\mathbb{Z}/p^k\mathbb{Z})^{\times} \times (\mathbb{Z}/n\mathbb{Z})^{\times}$ instead of in terms of $(\mathbb{Z}/m\mathbb{Z})^{\times}$.

In the case where $k=0$, I think $E$ must be the trivial group since $p$ is unramified. I think the decomposition group $D$ would be the group of order $\phi(m)/f$ in $(\mathbb{Z}/m\mathbb{Z})^{\times}$ where $f$ is the order of $p \mod \phi(m)$. But I'm not sure where to go from here. Any tips or hints on where to proceed would be greatly appreciated.

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    $\begingroup$ Consider the obvious subfields of $n$-th and $p^k$-th roots of unity; determine the groups there and think about what this means. $\endgroup$ – franz lemmermeyer Mar 11 at 6:01
  • $\begingroup$ I can see what the decomposition and inertia fields look like in the subfield of $n-$th roots of unity since $p$ is unramified, but I'm not sure what the groups look like in the $p^{k}$-th roots of unity. My intuition is telling me that the inertia field in this second case should be trivial since $p$ should be ramified in any subfield of $\mathbb{Q}(\zeta_{p^{k}}$, and hence the decomposition field is also trivial. Is this right? Then how do I put those together? It feels like it should be the direct product, but this is not clear to me. $\endgroup$ – JonHales Mar 13 at 2:49

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