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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.

Choose $\beta$ with ${\beta }^{p(p-1)}=-p$ and put $N=p(p-1)$. The Eisenstein polynomial $T^N+p$ shows that $L={\mathbb{Q}}_p(\beta )$ has degree $N$, is totally ramified, and has uniformizer $\beta$. Its subfield ${\mathbb{Q}}_p({\beta }^p)$ is the field $K$ from the previous part. Because $p$ is odd,Thus $L$ contains $a$ and ${\zeta }_p$, and contains all roots $a{\zeta }_p^j$ of $T^p-p$. Conversely $[{\mathbb{Q}}_p(a):{\mathbb{Q}}_p]=p$, while $[K:{\mathbb{Q}}_p]=p-1$. Their coprime degrees force their compositum to have degree $N$. It is contained in $L$ and hence equals it. Therefore $L$ is exactly the splitting field, not merely an extension containing it.

The Galois group has a normal subgroup $H=\mathrm{Gal}(L/K)$ of order $p$, acting by $\beta \mapsto {\zeta }_p^j\beta$. The $(p-1)$st roots of unity already lie in ${\mathbb{Q}}_p$ by the Hensel lemma. The maps $\beta \mapsto \omega \beta$, with ${\omega }^{p-1}=1$, supply a complement of order $p-1$. Its conjugation acts faithfully on $H$, so $G\cong C_p⋊{\mathbb{F}}_p^{\times }$.

Normalize $v_L(\beta )=1$. Total ramification and the uniformizer criterion for lower ramification groups reduce the calculation to $i_G(\sigma )=v_L(\sigma \beta -\beta )$. For nonidentity $\sigma \in H$,since $e(L/K)=p$ and $v_K({\zeta }_p^j-1)=1$. For $\sigma \notin H$, write $\sigma \beta =\omega {\zeta }_p^j\beta$ with $\omega \ne 1$. Its multiplier has residue $\overline{\omega }\ne 1$, so $i_G(\sigma )=1$. The lower ramification numbering is thereforeAll later groups are trivial. The wild lower break is $p$, not one. In the upper ramification numbering, the Herbrand function sends this break to $p/(p-1)$: $G^0=G$, $G^u=H$ for $0<u\le p/(p-1)$, and $G^u=1$ above it.

As an independent consistency check, the different exponent from ramification groups is $(N-1)+p(p-1)=2N-1$. The derivative of the Eisenstein polynomial gives the same answer, $v_L(N{\beta }^{N-1})=N+(N-1)$, since $v_L(p)=N$.

The filtration and its meaning. Let $L/K$ be a finite Galois extension of non-Archimedean local fields, with group $G$, normalized valuation $v_L$, and residue fields $k_L/k_K$ of characteristic $\ell$. The lower ramification numbering isThe higher ramification groups are normal, decrease with $j$, and eventually become trivial: for each nonidentity automorphism some integral element is moved by a nonzero amount. The first group $G_0$ is the inertia group, withIts fixed field is the maximal unramified extension inside $L$. The wild inertia group is $G_1$. For a uniformizer ${\pi }_L$, there are injective homomorphismsThe kernels are the indicated next groups; the additive homomorphism assertion follows by expanding a product of automorphisms modulo ${\pi }_L^{j+2}$. These maps show that $G_0/G_1$ is cyclic of order prime to $\ell$, while $G_1$ is an $\ell$-group. Thus tame ramification is exactly the case $G_1=1$.

To calculate the groups, the uniformizer criterion for lower ramification groups saysIndeed, ${\mathcal{O}}_L$ is generated by ${\pi }_L$ over the integers of the maximal unramified subextension, which $G_0$ fixes. Differences of powers of ${\pi }_L$ are divisible by $\sigma {\pi }_L-{\pi }_L$, so the one test implies all the defining inequalities.

Eisenstein facts used in the calculations. For a complete discrete valuation ring $A$ with uniformizer $\varpi$, a monic polynomialis an Eisenstein polynomial if $\varpi \mid a_j$ for every $j<d$ and ${\varpi }^2\nmid a_0$. The Eisenstein criterion makes it irreducible. If $\theta$ is a root, its valuation relative to the base normalization is $1/d$, the extension has degree and ramification index $d$, its residue field is unchanged, andThe root valuation follows by comparing the terms of its equation. The ramification index must then be at least $d$, hence exactly $d$. In the basis $1,\theta ,\dots ,{\theta }^{d-1}$, the valuations of $b_j{\theta }^j$ are distinct modulo $d$, so an integral linear combination has every $b_j\in A$. This proves the integer-ring assertion. Conversely, a uniformizer in a totally ramified extension generates the field, and its minimal polynomial is Eisenstein; this follows from its value $1/d$, equal valuations of its conjugates, and the valuation one of its norm.

For a monogenic separable integer ring $A[\theta ]$, the different ideal is generated by $f^{\prime }(\theta )$. For a Galois extension, the different exponent from ramification groups isIn the totally ramified case these agree directly: $f^{\prime }({\pi }_L)={\prod }_{\sigma \ne 1}({\pi }_L-\sigma {\pi }_L)$, and an automorphism with displacement valuation $h$ contributes once to each of $G_0,\dots ,G_{h-1}$.

Upper numbering. Extend the lower indexing to real $s\ge 0$ by $G_s=G_{\lceil s\rceil }$, and setOn $[-1,0]$ take $\phi (s)=s$. This Herbrand function slows the indexing as the groups shrink. Lower numbering is compatible with subgroups, while upper ramification numbering is compatible with quotients: for normal $H$, $(G/H{)}^u=G^{u}H/H$. This is the fact that upper ramification groups commute with quotients. Thus upper breaks are particularly useful when comparing intermediate Galois extensions.

The eighth-root cyclotomic extension. Put $\zeta ={\zeta }_8$ and $\pi =\zeta -1$. The shifted cyclotomic polynomialis Eisenstein at two. Hence $L={\mathbb{Q}}_2(\zeta )$ is totally ramified of degree four, $v_L(2)=4$, and $\pi$ is a uniformizer. Its Galois group consists of ${\sigma }_a(\zeta )={\zeta }^a$ for $a=1,3,5,7$, and is $C_2\times C_2$.

Since $\zeta$ is a unit,For $a=3,7$, this is the valuation of $i-1$ or $-i-1$, namely two, because their squares are units times two. For $a=5$, it is $v_L(-2)=4$. The ramification groups of the eighth-root cyclotomic extension of the 2-adic field are thereforeThe lower breaks are one and three, andThus the upper breaks are one and two:The different exponent from ramification groups is $3+3+1+1=8$, agreeing with $v_L(4{\zeta }^3)=8$ from the derivative of $T^4+1$.

The cubic splitting field at three. Let ${\alpha }^3=2$, $\zeta ={\zeta }_3$, and putThe polynomialsare Eisenstein at three. They give totally ramified subextensions of degrees three and two. Their intersection is the base field, so $L={\mathbb{Q}}_3(\alpha ,\zeta )$ has degree six. Its ramification index is divisible by both three and two, hence equals six; therefore $L/{\mathbb{Q}}_3$ is totally ramified. It is the splitting field of $T^3-2$, with Galois group $S_3$.

Normalize $v_L$ so that $v_L(3)=6$. Then $v_L(\beta )=2$ and $v_L(\lambda )=3$, makinga uniformizer. Let $\sigma (\alpha )=\zeta \alpha ,\sigma (\zeta )=\zeta$ and $\tau (\alpha )=\alpha ,\tau (\zeta )={\zeta }^2$. They generate $S_3$. Since $\tau (\lambda )=(\zeta +1)\lambda$,For the order-three automorphism,Here $\alpha$ is a unit and $v_L(\zeta \alpha +1)=v_L(\sigma \beta )=2$. The same calculation gives valuation two for ${\sigma }^2$. The three transpositions are conjugate and the defining filtration is normal, so they all have displacement valuation one. The ramification groups of the splitting field of T3 minus 2 over Q3 areThe lower breaks are zero and one. Since $[G_0:G_s]=2$ for $0<s\le 1$,and the upper groups areThe different exponent from ramification groups is $5+2=7$. As an independent check, the cubic subfield has different exponent $v_{{\mathbb{Q}}_3(\alpha )}(3{\alpha }^2)=3$, and the quadratic extension above it is tame with different exponent one. Transitivity of the different ideal gives $1+2\cdot 3=7$, as required.

Let $a=\sqrt[3]{3}$ and let $\zeta$ be a primitive cube root of unity. The splitting field is $L={\mathbb{Q}}_3(a,\zeta )$. The polynomial $X^3-3$ is an Eisenstein polynomial, so $[{\mathbb{Q}}_3(a):{\mathbb{Q}}_3]=3$. Meanwhile $(1+2\zeta {)}^2=-3$ shows that ${\mathbb{Q}}_3(\zeta )$ is quadratic: $-3$ is not a square because its valuation is odd. A quadratic field cannot lie in the degree-three field. Consequently $[L:{\mathbb{Q}}_3]=6$, and the action on the three roots identifies its Galois group with the symmetric group $S_3$.

A useful uniformizer isSince $(\zeta -1{)}^3=-3\zeta (\zeta -1)$, we obtainThus $\pi$ is a root of the Eisenstein polynomial $X^6+3$. Its degree is six, so $L={\mathbb{Q}}_3(\pi )$, the extension is totally ramified, and $\pi$ is a uniformizer. With $v_L(\pi )=1$,This is the Eisenstein sextic presentation of the splitting field of X3 minus 3 over Q3.

Let $\tau (a)=\zeta a$, $\tau (\zeta )=\zeta$, and let $s(a)=a$, $s(\zeta )={\zeta }^{-1}$. Then $\tau$ has order three, $s$ has order two, and $s\tau s={\tau }^{-1}$. Their actions on the uniformizer areThe two nonidentity elements of $⟨\tau ⟩$ therefore satisfyEach transposition sends $\pi$ to $-{\zeta }^j\pi$ for some $j$, so its displacement has valuation one: $1+{\zeta }^j$ is a unit, reducing to $2$ modulo the maximal ideal. The uniformizer criterion for lower ramification groups now givesFor real indices, this is $S_3$ on $[-1,0]$, $C_3$ on $(0,3]$, and $1$ for $t>3$. In particular the lower breaks are $0$ and $3$.

On $(0,3]$ the index $[G_0:G_t]$ is two; beyond three it is six. Thus the Herbrand function isThe positive lower break $3$ becomes the upper break $3/2$. ThereforeThe fractional upper break is allowed because the full extension is nonabelian. As a check, the different exponent from ramification groups is $5+3\cdot 2=11$. The derivative of the minimal polynomial $X^6+3$ gives the same result: $v_L(6{\pi }^5)=6+5=11$. This is the $p=3$ case of ramification groups of the splitting field of Xp minus p over the p-adic numbers.

Let $L/K$ be a finite Galois extension of non-Archimedean local fields, let $G=\mathrm{Gal}(L/K)$, and normalize the discrete valuation by $v_L(L^{\times })=\mathbb{Z}$. The lower ramification numbering isHere $G_0$ is the inertia group and $G_1$ is the wild inertia group. For real $s\ge -1$, put $G_s=G_{\lceil s\rceil }$; thus $G_s=G_0$ for $-1<s\le 0$. Define the Herbrand function and its inverse byextending $\phi (s)=s$ on $[-1,0]$. The upper ramification numbering is $G^u=G_{\psi (u)}$ for $u\ge -1$.

Let ${\alpha }^p-\alpha =t^{1-p}$ over $K={\mathbb{F}}_p((t))$, with $t$ its uniformizer. The Artin–Schreier polynomial has derivative $-1$ in characteristic $p$, so all its roots are distinct. If $\alpha$ is one root and $\beta$ another, then $(\beta -\alpha {)}^p=\beta -\alpha$, so $\beta -\alpha \in {\mathbb{F}}_p$. Conversely every $\alpha +a$, $a\in {\mathbb{F}}_p$, is a root. Thus $L=K(\alpha )$ contains all roots and is a splitting field of a separable polynomial; it is Galois. Every automorphism has the form ${\sigma }_a(\alpha )=\alpha +a$, and its Galois group embeds in the additive group of ${\mathbb{F}}_p$.

There is no root in $K$. If $v_K(z)\ge 0$, then $z^p-z$ cannot have negative valuation. If $v_K(z)<0$, the ultrametric inequality gives $v_K(z^p-z)=p{v}_K(z)$, divisible by $p$, whereas $v_K(t^{1-p})=1-p$ is not. The Galois group is consequently nontrivial. Its order divides the prime $p$, so

Let $e=e(L/K)$ be the ramification index. The equation forces $v_L(\alpha )<0$, and henceAs $\gcd (p,p-1)=1$, $p\mid e$. Since $e\le [L:K]=p$, we get $e=p$, residue-field degree one, and $v_L(\alpha )=1-p$. Thus this is a totally ramified extension, andThis is a uniformizer. The uniformizer criterion for lower ramification groups states that for a totally ramified Galois extension, $\sigma \in G_i$ exactly when $v_L(\sigma ({\pi }_L)-{\pi }_L)\ge i+1$. Here every nonidentity automorphism satisfiesTherefore the ramification break of an Artin–Schreier pole occurs at $p-1$, and the lower groups areThis includes $p=2$, where the break is one. There is no off-by-one shift: the condition is $v_L(\sigma {\pi }_L-{\pi }_L)\ge i+1$.

For the upper groups, compute the Herbrand function explicitly:Hence the upper break is also $p-1$ and, with the stated real-index convention,As a consistency check, ${\pi }_L$ satisfies the Eisenstein polynomial $X^p-t^{p-1}X-t$. Its derivative in characteristic $p$ is $-t^{p-1}$, giving different exponent $p(p-1)$, equal to ${\sum }_{i\ge 0}(|G_i|-1)=p(p-1)$ from the different exponent from ramification groups.

Normalize the discrete valuation by $v_L({\pi }_L)=1$, and put $v_L(0)=+\infty$. For $\sigma \ne 1$, define its ramification numberand set $i_G(1)=+\infty$. The minimum exists: all finite values are nonnegative integers, and a nonidentity automorphism moves some integral element. The lower ramification numbering isThus $G_s=G_{\lceil s\rceil }$, $G_0$ is the inertia group, and $G_1$ is the wild inertia group. This states the convention at integer endpoints as well as between them. Automorphisms preserve $v_L$, so these are normal subgroups of $G$.

Define the Herbrand function and its inverse byThe integrand is positive and piecewise constant. Since $G_t=1$ for sufficiently large $t$, the function is continuous, strictly increasing and unbounded, and its inverse is defined on all nonnegative reals. The upper ramification numbering isIn particular, the denominator in the Herbrand integral involves $G_0$, even when $L/K$ is not totally ramified. These conventions are also given in Milne's Class Field Theory, Chapter I, §4.

Now use the monogenic hypothesis ${\mathcal{O}}_L={\mathcal{O}}_K[\alpha ]$. Since $L/K$ is Galois, its minimal polynomial has the distinct conjugates $\sigma (\alpha )$ as roots, soTaking valuations proves the first equality.

For every polynomial $P\in {\mathcal{O}}_K[T]$, the difference $P(\sigma (\alpha ))-P(\alpha )$ is divisible in ${\mathcal{O}}_L$ by $\sigma (\alpha )-\alpha$. Every element of ${\mathcal{O}}_L$ is such a polynomial value, and taking the element $\alpha$ attains the resulting lower bound. HenceFor an integer $i_G(\sigma )=m\ge 0$, the element $\sigma$ lies in exactly the groups $G_0,\dots ,G_{m-1}$; if $m=0$, it lies in none. Summing this count over the nonidentity elements and interchanging the two finite sums givesThis proves the ramification-group sum for a monogenic integer ring. Equivalently it is the different exponent from ramification groups, since the different ideal is generated by $f^{\prime }(\alpha )$ under the same monogenic hypothesis.

Let $\alpha$ be a root ofThe polynomial is Eisenstein at $3$, so $E={\mathbb{Q}}_3(\alpha )$ is totally ramified of degree three and ${\mathcal{O}}_E={\mathbb{Z}}_3[\alpha ]$. Its discriminant isIts odd valuation makes $\Delta$ nonsquare in ${\mathbb{Q}}_3$, so the Galois group of an irreducible cubic shows that the splitting field $L$ has Galois group $S_3$. The quadratic extension obtained by adjoining $\sqrt{\Delta }$ is ramified, so $L/{\mathbb{Q}}_3$ is totally ramified. Thereforebecause the wild inertia group is the unique Sylow $3$-subgroup of $S_3$.

It remains to find the wild break. Sinceand ${\alpha }^2+1$ is a unit, the different ideal of $E/{\mathbb{Q}}_3$ has exponent $v_E(f^{\prime }(\alpha ))=3$. The extension $L/E$ is a tamely ramified quadratic extension and has different exponent one. The different in a tower therefore gives different exponentOn the other hand, the different exponent from ramification groups iswhere $b$ is the last index for which $G_b=A_3$. Thus $b=1$, and

For ${\mathbb{Q}}_2({\zeta }_8)/{\mathbb{Q}}_2$, the shifted polynomial $(1+T{)}^4+1$ is Eisenstein and ${\zeta }_8-1$ is a uniformizer. The three nonidentity automorphisms have displacement valuations two, four and two. Thus the lower breaks are one and three, the upper breaks one and two, and the different exponent from ramification groups is eight.

For $L={\mathbb{Q}}_3({\zeta }_3,\alpha )$ with ${\alpha }^3=2$, the shifted cubic for $\alpha +1$ and quadratic for ${\zeta }_3-1$ are Eisenstein. Their compositum is totally ramified of degree six. The element $\pi =({\zeta }_3-1)/(\alpha +1)$ has valuation one. Order-three automorphisms move it with valuation two and transpositions with valuation one. The lower breaks are zero and one, the upper breaks zero and one-half, and the different exponent from ramification groups is seven.

For the splitting field $M/{\mathbb{Q}}_3$ of $T^3-3T+3$, the cubic is Eisenstein and its discriminant is $-135$. The quadratic subfield is ${\mathbb{Q}}_3(\sqrt{3})$, and the degree-six extension is totally ramified. If $\alpha$ is a root and $s=\sqrt{3}$, then $\pi ={\alpha }^2/s$ is a uniformizer. Transpositions have $v_M(\sigma \pi -\pi )=1$, while three-cycles have value $2$, giving the displayed ramification groups. Its different exponent from ramification groups is $7$.