Description of the decomposition and the inertia group in terms of the product $\mathbb{Z}^*_{p^k}\times \mathbb{Z}^*_n $

Question

Let $\omega^{\frac{2\pi}{m}}$, we fix a prime p and write $m=p^kn$ with $p\not| \, n$.

We know that the Galois group of $\mathbb{Q}[\omega]$ over $\mathbb{Q}$ is isomorphic to $\mathbb{Z}^*_m$ that is isomorphic to $\mathbb{Z}^*_{p^k}\times \mathbb{Z}^*_n$.

How can I describe $D$ and $E$ (corresponding to $p$) in terms of this product?

Here $D:=D(Q|P)$ is the decomposition group and $E=E(Q|P)$ is the inertia group.

Answer 1

In general, suppose that $L/K$ is a Galois extension and $\mathfrak P$ is a prime of $L$ lying above a prime $\mathfrak p$ of $K$. Let $G=\mathrm{Gal}(L/K)$, $D$ be the decomposition group of $\mathfrak {P/p}$ and $I$ the inertia group. Then there is a tower of field extensions $$K=L^G\subset L^D\subset L^I\subset L.$$ Let $\mathfrak P_D,\mathfrak P_I$ be the primes of $L^D, L^I$ lying below $\mathfrak P$. Then:

• $\mathfrak p$ splits completely in $L^D$
• $\mathfrak P_D$ is inert in $L^I$
• $\mathfrak P_I$ is totally ramified in $L$.

Moreover, $L^I$ is the maximal subextension of $L/K$ in which $\mathfrak p$ is unramified, and $L^D$ is the maximal subextension in which $\mathfrak p$ is unramified and splits completely.

---

In your case, $K = \mathbb Q$ and $L =\mathbb Q(\zeta_m)$, where $\zeta_m$ is a primitive $m$-th root of $1$.

The inertia group is relatively easy: we have $\mathbb Q\subset\mathbb Q(\zeta_{n})\subset\mathbb Q(\zeta_m)$. Since $p\nmid n$, $p$ is unramified in $\mathbb Q(\zeta_n)$. On the other hand, any prime above $p$ in $\mathbb Q(\zeta_n)$ is totally ramified in $\mathbb Q(\zeta_m)$. It follows that $$I = \mathrm{Gal}(\mathbb Q(\zeta_m)/\mathbb Q(\zeta_n))\subset G.$$ Via your isomorphism, we have $I = (\mathbb Z/p^k\mathbb Z)^\times$.

The decomposition group is a bit harder: you need to determine how $p$ splits in $\mathbb Q(\zeta_n)$. You can do this by writing $\mathbb Q(\zeta_n)$ as the compositum of the fields $\mathbb Q(\zeta_{p_i^{a_i}})$ where $n = \prod_{i}p_i^{a_i}$ and using this question.

The decomposition group will be $(\mathbb Z/p^k\mathbb Z)^\times\times D'$ where $D'$ is the decomposition group of $p$ in $\mathbb Q(\zeta_n)$.

Answer 2

Let me call $L = \mathbb{Q}[\omega]$, $\alpha = \omega ^{n}$ and $\beta=\omega^{p^k}$. Moreover, let me call $M = \mathbb{Q}[\alpha]$ and $N=\mathbb{Q}[\beta]$. Then $M$ is the $p^k$-th cyclotomic fiel, and $N$ is the $n$-cyclotomic field. Recall that we are working with abelian extension, then $D$ and $E$ do not depend on the prime $Q$ in $L$ lying over $p$. We know that $ref = \phi(m)$, where $e = \phi(p^k)$ and $f$ is the smallest integer such that $p^f \equiv 1 \bmod $.

Claim: $D(Q|p) \cong D( (1-\alpha)|p)_{\alpha} \times D(P|p)_{\beta}$, where the subscripts point out the obvious belonging.

Indeed, $(1-\alpha)$ is the unique prime lying over $p$ (I'm following the proof of Theorem 26 in Marcus's book). In particular $p \mathbb{Z}[\alpha] = (1-\alpha)^e$. then $|D_{\alpha}| = e$. Furthemore, $p$ does not divide $n$, thus in $N$, $p$ is unramified, i.e. $p \mathbb{Z}[\beta] = P_1 \cdots P_r$. Following again the proof of Theorem 26, we know that $|D_{\beta}|= f$. Then $|D_{\alpha} \times D_{\beta}|= ef = |D|$. Therefore, there is a little hope that my claim is true.

By assumption, $D \cong ( D \cap G(L/M)) \times (D \cap G(L/N)) = D(Q|(1-\alpha)) \times D(Q|P)$ (it comes from $\mathbb{Z}_m^{*} \cong \mathbb{Z}_n^{*} \times \mathbb{Z}_{p^k}^{*}$ i.e. $G \cong G(L/M) \times G(L/N)$). Now, using exercise 10 chapter 4 in Marcus, you find out $D(Q|(1-\alpha)) \cong D_{\beta}$ and $D(Q|P) \cong D_{\alpha}$.

Arguing as before, you are able to prove $E \cong E_{\alpha} \times E_{\beta}$.

Moreover, it is worth noting $E \cong D_{\alpha} \cong E_{\alpha} \cong \mathbb{Z}_{p^k}^{*}$ and hence I suggest you to take a look at this question