Inertia fields and decomposition fields in cyclotomic extensions

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.

• Consider the obvious subfields of $n$-th and $p^k$-th roots of unity; determine the groups there and think about what this means. – franz lemmermeyer Mar 11 at 6:01
• 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. – JonHales Mar 13 at 2:49