Inertia and Decomposition Field in $\mathbb{Q}(\omega=exp(2\pi i/m)$, where $m=p^{k}n,(p,n)=1$.

Question

let $\omega=exp(2\pi i/m),m=p^{k}n,(p,n)=1$. We know that the Galois group $\mathbb{Q}(\omega)$ over $\mathbb{Q}$ is isomorphic to $(\mathbb{Z}_{m})^\times$, which is naturally isomorphic to $\mathbb{Z}_{p^k}^{\times}\times(\mathbb{Z}_{n})^{\times}$. My question is: how does one describe the decomposition group, $D$ and inertia group $E$ with respect to $p$ in terms of $(\mathbb{Z}_{p^k})^\times$ and $(\mathbb{Z}_{n})^\times$?

Here is my attempt: Let $Q$ be any prime lying over $p$ in $\mathbb{Q}(\omega)$. $p$ splits completely in $\mathbb{Q}(\omega^{p^k})$. Hence, $\mathbb{Q}(\omega^{p^k})$ lies in the fixed field of the decomposition group $L_{D}$, which is a subfield of $L_{E}$ (the fixed field of the inertia group). Therefore, we have $[L_{E}:\mathbb{Q}]\geq [\mathbb{Q}(\omega^{p^{k}}):\mathbb{Q}]=\phi(n)$. But we know that $[L:L_{E}]=\phi(p^{k})$, which implies $L_{E}=\mathbb{Q}(\omega^{p^{k}})$. Hence, $E$ is isomorphic to $\mathbb{Z}_{p^k}$ (need to use Galois theory).

The part that i am having trouble with is how does one describe $D$? I know that $D/E$ will be a subgroup of $\mathbb{Z}_{n}^\times$, and also cyclic, but i am not sure how to describe it. Thanks.

Answer 1

With $\Phi_m(x)$ the minimal polynomial of $\omega_m$ you'll have $x^{p^k}-1 = (x-1)^{p^k} \bmod p$ so $\omega_{p^k} = 1 \bmod p,\omega_m = \omega_n \bmod p$ and $\Phi_m(x) =\Phi_n(x)^{\phi(p^k)}\bmod p$.

Then $p \Bbb{Z}[\omega_n] = \prod_{j=1}^{\phi(n)/a} (p,\omega_n-\zeta_j)$ where $a$ is the order of $p$ modulo $n$ and the $\zeta_j= \omega_n^{d_j},gcd(n,d_j)=1$ are the roots of $\Phi_n(x)$ in $ \Bbb{Z}[\omega_n]/(p)$ whose $p$-th powers are distinct, ie. $\Phi_n(x) = \prod_{j=1}^{\phi(n)/a}\prod_{l=1}^a (x-\zeta_j^{p^l}) \in \Bbb{Z}[\omega_n]/(p)[x]$ and $$p \Bbb{Z}[\omega_m] = \prod_{j=1}^{\phi(n)/a} (p,\omega_m-\zeta_j)^{\phi(p^k)} = \prod_{ r \in \Bbb{Z}/(n)^\times / \langle p \rangle} (p,\omega_m-\omega_n^r)^{\phi(p^k)}$$

Since we are in abelian extension the decomposition/inertia group are the same for each of those prime ideal.

The inertia group are the $\sigma\in Gal(\Bbb{Z}[\omega_m]/\Bbb{Z}), \sigma(\omega_m) = \omega_m^r$ such that $\omega_m^{p^kr}=\sigma(\omega_n) = \omega_n=\omega_m^{p^k}$ ie. $r \equiv 1 \bmod n$, and the decomposition group are the $\sigma\in Gal(\Bbb{Z}[\omega_m]/\Bbb{Z})$ such that $\sigma(\omega_n) = \omega_n^{p^l}$ ie. $r \equiv p^l \bmod n$, that is $D_p$ is the preimage in $\Bbb{Z}/(m)^\times $ of $ \langle p \rangle \subset \Bbb{Z}/(n)^\times$.

Then the game is to show $$\zeta_{\Bbb{Z}[\omega_m]}(s) = \prod_\mathfrak{q} \frac{1}{1-N(\mathfrak{q})^{-s}} = \prod_{\chi \bmod m} L(s,\tilde{\chi})$$ where $L(s,\tilde{\chi})$ is the Dirichlet L-function of the primitive character underlying each character $\chi$ modulo $m$.