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For a Finite Galois extension of $K(T)$ with algebraically closed characteristic-zero constants, the local decomposition and inertia groups coincide and are cyclic. A fixed compatible choice of primitive roots of unity selects a generator through its uniformizer character of inertia. Changing the prime above a base point conjugates this generator, and changing the uniformizer does not change its character. Without the roots-of-unity convention, only the inertia-subgroup conjugacy class is intrinsic. Over $\mathbb{C}$, positive local orientation gives the standard convention.

For the finite Galois extension, normalize the discrete valuation $v_L$ to have value group $\mathbb{Z}$. The inertia group isThe lower ramification numbering isThe 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 $p=\mathrm{char}{k}_K$, 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 $I=\mathrm{Gal}(L/K_0)$, 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: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 inertiaIt 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 $v_L(\sigma {\pi }_L-{\pi }_L)\ge 2$, namely $G_1$. Changing the uniformizer to $u{\pi }_L$ leaves the residue character unchanged, since $\overline{\sigma (u)/u}=1$. ThereforeEvery 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

To verify the assertion about wild inertia as well, for $i\ge 1$ defineWrite $\sigma {\pi }_L={\pi }_L+a{\pi }_L^{i+1}$ and $\tau {\pi }_L={\pi }_L+b{\pi }_L^{i+1}$. Composition gives $\sigma \tau {\pi }_L\equiv {\pi }_L+(a+b){\pi }_L^{i+1}(\mathrm{mod}{\pi }_L^{i+2})$, because inertia fixes residues and $i\ge 1$. 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.