I'm studying algebraic number theory for quals (reading Neukrich, first two chapters), and I wonder if there's a nice nontrivial example that gives a good intuition for decomposition, inertia, and ramification groups (and their fixed fields). More precisely, I'm seeking for a (finite) extension of a (discrete) valuation field $(L, w)/(K, v)$ where all the inclusions
$$
\{1\} \subset R_{w} \subset I_{w} \subset G_{w} \subset G
$$
are proper.
For example, when $K$ is a complete DVR with respect to $v$ (or if $K$ is an Henselian field), then $G_{w} = G$ since valuation extends uniquely.
So I want $K$ to be non-Henselian so that the valuation $v$ does not extends uniquely to $L$. So $K$ should be a global field (right?).
For $I_{w} \neq G_{w}$, we should have $l\neq k$ where $l, k$ are residue fields of $L, K$ respectively.
For $I_{w}\neq \{1\}$, the extension $L/K$ should be ramified over $v$, and the two proper inclusions $\{1\} \subset R_{w}$ and $R_{w} \subset I_{w}$ suggest that the ramification degree $e(w|v) = [w(L^{\times}):v(K^{\times})]$ should be divisible by $p$ (characteristic of the residue field) but also contains some non $p$ factor.
If we choose $p = 3$ and we denote $r, e, f$ to be the standard quantities related to $w|v$, then I want to find an extension with $r = 2, f = 2, e = 3e' = 6$, so the extension degree $[L:K]$ should be at least $24$.
I tried a cyclotomic extension and it seems work for $L = \mathbb{Q}(\zeta_{72})$. However, I can't find the above groups explicitly.
First, from the isomorphism $$\mathrm{Gal}(L/\mathbb{Q}) \simeq (\mathbb{Z}/72\mathbb{Z})^{\times}\simeq (\mathbb{Z}/9\mathbb{Z})^{\times} \times (\mathbb{Z}/8\mathbb{Z})^{\times}\simeq (\mathbb{Z}/6\mathbb{Z})\times(\mathbb{Z}/2\mathbb{Z})\times(\mathbb{Z}/2\mathbb{Z}),$$
I found the generators of the Galois group (by hand) and they are
$$
\sigma_{-7} = \sigma_{65},\quad \sigma_{19},\quad \sigma_{37}
$$
where $\sigma_{a}(\zeta) = \zeta^{a}$ for $(a, 72) = 1$. Also, their orders are 6, 2, 2 respectively.
In our case, we have $e = 6, f = 2, r = 2$ for the prime $3\in \mathbb{Q}$ and the prime 3 factors in $L$ as
$$
(3)\mathbb{Z}[\zeta_{72}] = (3, \zeta_{72}^{2} + \zeta_{72} – 1)^{6}(3, \zeta_{72}^{2} – \zeta_{72} – 1)^{6} = \mathfrak{p}_{1}^{6}\mathfrak{p}_{2}^{6}
$$
since $\Phi_{72}(x) = x^{24} – x^{12} + 1 = (x^{2} +x -1)^{6}(x^{2} – x -1)^{6}$ in $\mathbb{F}_{3}$.
Since $L/K$ is an abelian extension, $G_{w} = G_{\mathfrak{p}_{1}} = G_{\mathfrak{p}_{2}} = G_{3}$ is an index 2 subgroup of $G = \mathrm{Gal}(L/K)$. But I can't figure out which of the above generators are in $G_{3}$ and which of them aren't.
For example, how can we figure out $\sigma_{-7} \in G_{3}$ or not, which is equivalent to
$$
\sigma_{-7}(\zeta_{72}^{2} + \zeta_{72} – 1) = \zeta_{72}^{-14} + \zeta_{72}^{-7} – 1 \in \mathfrak{p}_{1} = (3, \zeta_{72}^{2} + \zeta_{72} – 1)
$$
Best Answer
$\def\Gal{\mathrm{Gal}}$ $\def\Q{\mathbf{Q}}$ $\def\Z{\mathbf{Z}}$
Let's start off with $L = \Q(\zeta_8)$, so $\Gal(L/\Q) = (\Z/8 \Z)^{\times}$.
It's certainly the case that $L$ is unramified at $3$. It follows that the decomposition group at $3$ is generated by Frobenius at $3$. But, in cyclotomic fields, the Frobenius element at $3$ is just $[3]: \zeta \mapsto \zeta^3$. So in this case, $I_3(L/\Q)$ is trivial, $G_3(L/\Q) = \langle [3] \rangle$, and $G(L/\Q) = (\Z/8 \Z)^{\times}$. You can also check that the fixed field of $\langle [3] \rangle$ is $K = \mathbf{Q}(\zeta + \zeta^3) = \mathbf{Q}(\sqrt{-2})$.
Now let $M = \Q(\zeta_{72})$. Then $\Gal(M/\Q) = (\Z/72 \Z)^{\times}$. Now check:
$M/L$ is totally ramified at $3$, and $L/\Q$ is unramified, so $I_3(M/\Q)$ is the order six group $\Gal(M/L)$ consisting of elements in $(\Z/72 \Z)^{\times}$ which are $1 \bmod 8$. It is generated by $-7 \bmod 72$, and the wild inertia group is generated by $(-7)^2 = 49 \bmod 72$.
$G_3(M/\Q)$ is given by $\Gal(L/K)$, so it is generated by $I_3$ together with (any lift) of $[3] \in (\Z/8 \Z)^{\times}$, e.g. $-7 \bmod 72$ and $11 \bmod 72$.