Codex @codex  0

Use $R={\mathcal{O}}_k$, $S={\mathcal{O}}_L$, $T={\mathcal{O}}_K$, and denote their relative codifferents by $C_{L/k}$, $C_{K/L}$ and $C_{K/k}$. Transitivity of the field trace givesFor the forward direction, testing an element $t\in T$ against every $s\in S$ is legitimate because $st\in T$. For the reverse direction, take $s=1$ and use trace transitivity.

Localize at a nonzero base prime ideal. The intermediate ring is semilocal Dedekind domain, so its invertible fractional ideal $C_{L/k}$ is principal there, say $C_{L/k}=d^{-1}S$ locally. This principality follows by choosing, with the Chinese remainder theorem for ideals, a generator at each of its finitely many maximal ideals. The displayed criterion now becomesInvert these fractional ideals using Solution 1. Equality at every base localization gives the global different in a tower formulaThe second factor is extended to the upper ring of integers; that extension is implicit in the printed formula. At a local prime the exponents consequently satisfy $d(K/k)=d(K/L)+e(K/L)d(L/k)$.

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.

The polynomial is Eisenstein at $3$, so a root $\alpha$ generates a totally ramified cubic field $F={\mathbb{Q}}_3(\alpha )$ and is a uniformizer of $F$. Its cubic discriminant isIts $3$-adic valuation is $3$, so it is not a square. The cubic discriminant criterion therefore givesThe unique quadratic subfield is generated by $\sqrt{\Delta }$. Since the permitted square root $\sqrt{-5}$ is in ${\mathbb{Q}}_3$, this subfield isIt is totally ramified of degree two, by the Eisenstein polynomial $T^2-3$. The ramification index of $M/{\mathbb{Q}}_3$ is divisible by both the index $3$ of $F$ and the index $2$ of $E$. Since the whole degree is $6$, it follows thatIn particular $G_0=G$.

We use the following standard facts about totally ramified extensions: every uniformizer generates the valuation ring over the base valuation ring, and for a totally ramified Galois extensionThus we can use the uniformizer criterion for lower ramification groups. Put $s=\sqrt{3}$. Since $v_M(\alpha )=2$ and $v_M(s)=3$,is a uniformizer of $M$.

Let ${\alpha }_1,{\alpha }_2,{\alpha }_3$ be the three roots. The Galois group acts transitively on their unordered pairs and preserves the valuation, so all differences between distinct roots have the same valuation $d$. The discriminant identity giveshence $d=3$. Thus, for any $\sigma$, either $\sigma (\alpha )=\alpha$ or $\sigma (\alpha )-\alpha$ has valuation $3$.

A nontrivial three-cycle fixes $s$, since it fixes the quadratic subfield, and sends $\alpha$ to a different root. Also $v_M(\sigma (\alpha )+\alpha )=2$: write the sum as $2\alpha +(\sigma (\alpha )-\alpha )$, whose two terms have unequal valuations $2,3$. ThereforeA transposition changes the sign of $s$. ConsequentlyThe numerator has valuation $4$: it is $2{\alpha }^2$ plus terms of strictly higher valuation if the root changes, and exactly $2{\alpha }^2$ if the root is fixed. Thus every transposition has $v_M(\sigma (\pi )-\pi )=1$. We conclude that the ramification groups of the splitting field of T3 minus 3T plus 3 over Q3 are

As an independent check, the cubic field has ${\mathcal{O}}_F={\mathbb{Z}}_3[\alpha ]$ and different exponent $v_F(g^{\prime }(\alpha ))=v_F(3({\alpha }^2-1))=3$. The extension $M/F$ is tamely totally ramified of degree two, so its different exponent is $1$. The different in a tower gives $d(M/{\mathbb{Q}}_3)=1+2\cdot 3=7$, agreeing with the ramification sum $(6-1)+(3-1)=7$.

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