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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.

Let $L/K$ be a finite Galois extension of non-Archimedean local fields, with Galois group $G$, normalized valuation $v_L$, integer ring ${\mathcal{O}}_L$, and residue characteristic $p$. The lower ramification numbering isThese higher ramification groups measure successively finer agreement with the identity. The group $G_0$ is the inertia group, the kernel of the action on the residue field. Each $G_i$ is normal in $G$: conjugation permutes integral elements and preserves their valuations. The groups eventually become trivial, since each nonidentity automorphism moves some integral element and its displacement has finite valuation.

Let $K_0=L^{G_0}$, the maximal unramified subextension of a local field extension. Then $G/G_0$ is the Galois group of the residue extension, cyclic and generated by a Frobenius automorphism. For a uniformizer $\pi$ of $L$, the uniformizer criterion for lower ramification groups givesTo see why one uniformizer suffices, the extension $L/K_0$ is totally ramified and its integer ring is ${\mathcal{O}}_{K_0}[\pi ]$. Elements of $G_0$ fix its coefficients. Differences of powers of $\pi$ are divisible by $\sigma \pi -\pi$, so the valuation bound for $\pi$ implies it for every integral element. The converse follows by testing $\pi$ itself.

The tame inertia quotient is seen explicitly fromIt is a homomorphism because inertia acts trivially on the residue field, and its kernel is $G_1$. Thus $G_0/G_1$ is cyclic of order not divisible by $p$. For $i\ge 1$, the homomorphismhas kernel $G_{i+1}$. Indeed composing two automorphisms adds the leading displacements, since $\sigma \pi /\pi \equiv 1$ and inertia fixes residue coefficients. Hence each $G_i/G_{i+1}$ is an elementary abelian p-group. Eventual triviality makes $G_1$ a p-group; the prime-to-p quotient makes it the unique Sylow subgroup of $G_0$. This is the wild inertia group. Tame ramification means $G_1=1$.

The filtration also measures the different exponent. For the totally ramified extension over $K_0$, the minimal polynomial of an algebraic element $g$ of $\pi$ has polynomial roots $\sigma \pi$, $\sigma \in G_0$. The derivative formula givesFor a fixed $\sigma$, its displacement valuation is the number of indices $i\ge 0$ for which it belongs to $G_i$. Interchanging the finite sums therefore gives different exponent from ramification groups:The unramified part contributes zero. More generally different in a tower follows by taking products of bases of a module dual for the trace pairing, and givesThus a tame extension has different exponent $e-1$, while nontrivial wild groups contribute additional positive terms.

The lower numbering restricts to subgroups: if $H=\mathrm{Gal}(L/E)$, its groups are $H_i=H\cap G_i$, directly from the definition using the same top-field valuation. Quotients need a different indexing. For real $s\ge 0$ put $G_s=G_{\lceil s\rceil }$ and define the Herbrand functionThis upper ramification numbering stretches or compresses intervals according to the inertia indices. The quotient theorem says that for a normal subgroup $H$, upper ramification groups commute with quotients:In particular the same upper index has a meaning compatible with passing to a smaller Galois extension, which lower indices generally lack.

For a tame totally ramified extension the picture is simple: $G_0=G$, all $G_i$ with $i\ge 1$ are trivial, and the different exponent is $|G|-1$. The extension ${\mathbb{Q}}_p({\zeta }_p)$ for odd $p$ illustrates this, giving exponent $p-2$.

A mixed-characteristic wild family is the lower ramification filtration of a prime-power cyclotomic extension. Take $L={\mathbb{Q}}_p({\zeta }_{p^r})$ with odd $p$ and $\pi ={\zeta }_{p^r}-1$. If ${\sigma }_a(\zeta )={\zeta }^a$ and $a\ne 1$ modulo $p^r$, thenThe last equality follows because ${\zeta }^{a-1}$ has order $p^{r-v_p(a-1)}$ and its uniformizer valuation scales by the ramification index in that cyclotomic tower. The lower breaks are $0,p-1,p^2-1,\dots ,p^{r-1}-1$. Integrating the successive indices $(p-1),p(p-1),\dots$ gives upper breaks $0,1,\dots ,r-1$.

In equal characteristic there is an equally explicit wild example. Start with $K={\mathbb{F}}_q((t))$ and an Artin–Schreier extensionThere is no polynomial root in $K$: a negative-valuation element has $v(x^p-x)=pv(x)$, while the right side has valuation $-m$. Adjoining a polynomial root splits the polynomial, whose polynomial roots are $y+c$ for $c\in {\mathbb{F}}_p$, so the extension is cyclic of degree $p$. The valuation equation forces total ramification, $v_L(t)=p$ and $v_L(y)=-m$. Choose $1\le b\le p-1$ with $-bm\equiv 1(\mathrm{mod}p)$ and set $a=(1+bm)/p$. Then $\pi =t^a{y}^b$ is a uniformizer. For a nonidentity automorphism $y\mapsto y+c$,the first binomial coefficient $b$ is nonzero modulo $p$, and all subsequent terms have strictly higher valuation. ThereforeThis is the ramification break of an Artin–Schreier pole. It shows concretely that wild breaks, unlike tame inertia, can occur arbitrarily far out in the filtration.

For a finite Galois extension of non-Archimedean local fields, this is a homomorphism from the inertia group to $k_L^{\times }$ with kernel the wild inertia group. Inertia acts trivially on residue units, which proves multiplicativity and independence of the chosen uniformizer. Passing to the maximal unramified subextension reduces the kernel calculation to a totally ramified extension. Hence it embeds the tame inertia quotient into the cyclic multiplicative group of the residue field.