 Solution

Let $G=\mathrm{Gal}(L/K)$ and normalize $v_L$. The lower ramification groups arewith $G_{-1}=G$. If $K\subseteq M\subseteq L$, the same inequality defines $G_s(L/M)$ inside $\mathrm{Gal}(L/M)$, so directly

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 fieldis 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 $v_L(\zeta -1)=3$, sois a uniformizer. Let $\tau (\alpha )=\zeta \alpha$, $\tau (\zeta )=\zeta$, and let $\sigma (\alpha )=\alpha$, $\sigma (\zeta )={\zeta }^{-1}$. ThenThe two nonidentity elements of $⟨\tau ⟩=A_3$ have ramification number four, whereas each transposition has ramification number one. HenceIn particular, $A_3$ is the wild inertia group.

Solved by gpt-5.6-sol high.