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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.

Because $K$ has characteristic $p$, the prime field ${\mathbb{F}}_p$ lies in $K$, andfor every $c\in {\mathbb{F}}_p$. These are all $p$ roots. Moreover $f^{\prime }(t)=-1$, so $f$ is a separable polynomial.

Let $m(t)$ be the minimal polynomial of an algebraic element of $\alpha$ over $K$. Translation by $c\in {\mathbb{F}}_p$ sends an irreducible factor $g(t)$ of $f(t)$ to the irreducible factor $g(t-c)$. The additive group of ${\mathbb{F}}_p$ therefore acts on the irreducible factors, and every orbit has size $1$ or $p$. An orbit of size $p$ would already contribute $p\mathrm{deg}m$ to the degree-$p$ polynomial $f$, forcing $\mathrm{deg}m=1$. Then $\alpha \in K$, all roots $\alpha +c$ lie in $K$, and $L=K$, contrary to the hypothesis.

Thus $m(t-c)=m(t)$ for every $c\in {\mathbb{F}}_p$. If $1\le d=\mathrm{deg}m<p$, comparison of the coefficient of $t^{d-1}$ in $m(t-c)-m(t)$ for $c\ne 0$ gives $-dc$ times the nonzero leading coefficient, a contradiction. Hence $\mathrm{deg}m=p$, and since $m$ is a monic divisor of the monic degree-$p$ polynomial $f$, we have $m=f$. Thus $f$ is irreducible over $K$.

All roots $\alpha +c$ already belong to $K(\alpha )$, so the splitting field isAs the splitting field of a separable polynomial, $L/K$ is a Galois extension. For each $c\in {\mathbb{F}}_p$, irreducibility gives a distinct $K$-automorphismThese exhaust the $p=[L:K]$ automorphisms. This is an Artin–Schreier extension, and

Let $K$ be an equal-characteristic-p local field and let $c\in K$ have $v_K(c)=-m<0$, with $p\nmid m$. If ${\alpha }^p-\alpha =c$, the Artin–Schreier extension $L=K(\alpha )$ has degree $p$ and is totally ramified: a root in $K$ would force the pole order to be divisible by $p$, and $p{v}_L(\alpha )=-me(L/K)$ forces $e=p$. Choose $1\le s<p$ and an integer $r$ with $rp-sm=1$. For a base uniformizer $t$, the element ${\pi }_L=t^r{\alpha }^s$ has valuation one. Under ${\sigma }_a(\alpha )=\alpha +a$, $a\in {\mathbb{F}}_p^{\times }$, the leading term in ${\sigma }_a({\pi }_L)-{\pi }_L$ has valuation $rp-(s-1)m=m+1$. The uniformizer criterion for lower ramification groups therefore gives the single lower break $m$. The Herbrand function is the identity up to that break, so the upper break is also $m$.