= Solution
Let $G=\operatorname{Gal}(L/K)$ and normalize $v_L$. The lower <ramification groups> are
$$
G_s(L/K)=\{\sigma\in G:v_L(\sigma(x)-x)\geq s+1\text{ for every }x\in\mathcal O_L\},
\qquad s\geq0,
$$
with $G_{-1}=G$. If $K\subseteq M\subseteq L$, the same inequality defines $G_s(L/M)$ inside $\operatorname{Gal}(L/M)$, so directly
$$
G_s(L/M)=G_s(L/K)\cap\operatorname{Gal}(L/M).
$$
Now suppose $M/K$ is <Finite Galois extension>. The <inertia group> is the kernel of the action on the residue field. Restriction sends $G_0(L/K)$ into $G_0(M/K)$. Conversely, the maximal unramified subextension of $M/K$ is the intersection of $M$ with the maximal unramified subextension of $L/K$. The Galois correspondence therefore shows that the restriction image is all of $G_0(M/K)$.
For the explicit extension, take $\alpha^3=3$ and a primitive cube root of unity $\zeta$. The polynomial $X^3-3$ is Eisenstein over $\mathbb Q_3$, while $\mathbb Q_3(\zeta)/\mathbb Q_3$ is a ramified quadratic extension. Its splitting field
$$
L=\mathbb Q_3(\alpha,\zeta)
$$
is therefore a totally ramified extension of degree six with Galois group $S_3$. With $v_L(3)=6$, one has $v_L(\alpha)=2$ and $v_L(\zeta-1)=3$, so
$$
\pi=\frac{\zeta-1}{\alpha}
$$
is a uniformizer. Let $\tau(\alpha)=\zeta\alpha$, $\tau(\zeta)=\zeta$, and let $\sigma(\alpha)=\alpha$, $\sigma(\zeta)=\zeta^{-1}$. Then
$$
v_L(\tau(\pi)-\pi)=v_L((\zeta^{-1}-1)\pi)=4,
\qquad
v_L(\sigma(\pi)-\pi)=1.
$$
The two nonidentity elements of $\langle\tau\rangle=A_3$ have ramification number four, whereas each transposition has ramification number one. Hence
$$
G_{-1}=G_0=S_3,
\qquad
G_1=G_2=G_3=A_3,
\qquad
G_s=1\quad(s\geq4).
$$
In particular, $A_3$ is the <wild inertia group>.
Solved by gpt-5.6-sol high.
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