= Solution
For the finite <Galois extension>, normalize the <discrete valuation> $v_L$ to have value group $\mathbb Z$. The <inertia group> is
$$
I=G_0=\ker\bigl(G\to\operatorname{Gal}(k_L/k_K)\bigr).
$$
The <lower ramification numbering> is
$$
G_i=\{\sigma\in G:v_L(\sigma x-x)\ge i+1\text{ for every }x\in\mathcal O_L\},\qquad i\ge0.
$$
The <wild inertia group> is $P=G_1$, and the <tame inertia quotient> is $I/P$. We will prove that $P$ is a $p$-group, where $p=\operatorname{char}k_K$, and that $I/P$ is cyclic of order prime to $p$. Tame inertia here is a quotient, not a specified canonical subgroup complementary to $P$.
First reduce the inertia calculation to a <totally ramified extension>. Choose $\bar b$ generating the finite residue extension $k_L/k_K$. Its minimal polynomial is separable. Lift that monic polynomial to $f\in\mathcal O_K[X]$, and use <Hensel lemma> in $\mathcal O_L$ to obtain a root $b$ with residue $\bar b$. Irreducibility of the reduction implies irreducibility of $f$ over $K$, by the monic form of the <Gauss lemma for polynomials> over a <discrete valuation ring>. Thus $K_0=K(b)$ has degree $[k_L:k_K]$ and residue field $k_L$; the degree formula makes $K_0/K$ unramified. An automorphism fixes $b$ precisely when it fixes $\bar b$: the nontrivial direction is uniqueness in <Hensel lemma>. Hence $I=\operatorname{Gal}(L/K_0)$, and $L/K_0$ is totally ramified.
Use the permitted integral-generation fact $\mathcal O_L=\mathcal O_{K_0}[\pi_L]$. Every element of $I$ fixes the coefficients in this expression. The identity $\sigma(\pi_L^j)-\pi_L^j=(\sigma\pi_L-\pi_L)\sum_{h=0}^{j-1}(\sigma\pi_L)^h\pi_L^{j-1-h}$ consequently proves the <uniformizer criterion for lower ramification groups>:
$$
G_i=\{\sigma\in I:v_L(\sigma\pi_L-\pi_L)\ge i+1\},\qquad i\ge0.
$$
Necessity follows by testing $x=\pi_L$, and the displayed power identity proves sufficiency for every integral polynomial in $\pi_L$.
Define the <uniformizer character of inertia>
$$
\theta_0:I\to k_L^\times,\qquad\theta_0(\sigma)=\overline{\sigma\pi_L/\pi_L}.
$$
It is multiplicative: writing $u_\tau=\tau\pi_L/\pi_L$, one has $\sigma\tau\pi_L/\pi_L=\sigma(u_\tau)(\sigma\pi_L/\pi_L)$, and $\sigma$ acts trivially on residue units. Its kernel is exactly $v_L(\sigma\pi_L-\pi_L)\ge2$, namely $G_1$. Changing the uniformizer to $u\pi_L$ leaves the residue character unchanged, since $\overline{\sigma(u)/u}=1$. Therefore
$$
I/P\hookrightarrow k_L^\times.
$$
Every finite subgroup of a field's multiplicative group is cyclic: if its exponent is $m$, every element is a root of $X^m-1$, so its order is at most $m$; a finite abelian group has an element of order its exponent, constructed by multiplying elements of maximal prime-power orders, so its order is at least $m$. Equality makes it cyclic. Here $k_L^\times$ has order $|k_L|-1$, prime to $p$. We have proved
$$
\boxed{\text{The tame inertia group }G_0/G_1\text{ is cyclic of order prime to }p.}
$$
To verify the assertion about wild inertia as well, for $i\ge1$ define
$$
\theta_i:G_i\to(k_L,+),\qquad\theta_i(\sigma)=\overline{\frac{\sigma\pi_L-\pi_L}{\pi_L^{i+1}}}.
$$
Write $\sigma\pi_L=\pi_L+a\pi_L^{i+1}$ and $\tau\pi_L=\pi_L+b\pi_L^{i+1}$. Composition gives $\sigma\tau\pi_L\equiv\pi_L+(a+b)\pi_L^{i+1}\pmod{\pi_L^{i+2}}$, because inertia fixes residues and $i\ge1$. Thus $\theta_i$ is additive with kernel $G_{i+1}$. Thus <positive ramification quotients are elementary abelian>: each $G_i/G_{i+1}$ is a subgroup of the additive finite field, hence an elementary abelian $p$-group.
Finally the filtration terminates: a nonidentity element of $I$ cannot fix $\pi_L$, since it already fixes $K_0$ and $\mathcal O_L=\mathcal O_{K_0}[\pi_L]$. Its displacement has finite valuation, and there are only finitely many automorphisms. Iterating the finite $p$-group quotients gives that $P$ is a $p$-group. It is normal in $I$ and has prime-to-$p$ quotient, so it is its unique Sylow $p$-subgroup, justifying the wild inertia terminology.
Back to article page
