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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.

The polynomial is Eisenstein at every prime number $p$, so a polynomial root $\alpha$ generates a totally ramified cubic field $F/{\mathbb{Q}}_p$. The splitting-field Galois group is a transitive subgroup of $S_3$, hence $C_3$ or $S_3$. Its polynomial discriminant isThe product of the three polynomial root differences is a square root of $\Delta$, transformed by the sign of the polynomial root permutation. Thus the group is $C_3$ exactly when $\Delta$ is a square, and otherwise it is $S_3$. In the latter case the splitting field isIndeed adjoining one polynomial root leaves a quadratic factor, whose polynomial discriminant is $\Delta /f^{\prime }(\alpha {)}^2$, so the square root of $\Delta$ supplies its remaining polynomial roots.

Suppose first that $p>3$. The unit $-(4p+27)$ reduces to $-27$, and Hensel's lemma makes it a square exactly when $-3$ is a square modulo $p$. Quadratic reciprocity givesIf $p\equiv 1(\mathrm{mod}3)$, the field generated by $\alpha$ is already the cyclic splitting field, totally ramified of degree three. If $p\equiv 2(\mathrm{mod}3)$, the nonsquare unit gives an unramified quadratic field $E={\mathbb{Q}}_p(\sqrt{\Delta })$. Since $F$ and $E$ have coprime degrees, $[FE:{\mathbb{Q}}_p]=6$. Its residue degree is at least two and its ramification index at least three; their product is six, so they are exactly two and three. Thus its inertia group is the normal $C_3$ inside $S_3$.

In both cases the ramification index is three and is not divisible by $p$. The wild inertia group is trivial, so all positive lower groups are trivial. As a different check, $f^{\prime }(\alpha )=3{\alpha }^2+p$ has valuation two in $F$ when $p>3$, agreeing with tame different exponent $3-1$.

For $p=3$ the polynomial discriminant is $-351=-3^3\cdot 13$, which is nonsquare. Hence $G=S_3$. The quadratic subfield can be written $E={\mathbb{Q}}_3(\beta )$ with ${\beta }^2=-39$. It is ramified of degree two, by the Eisenstein polynomial $T^2+39$. Together with the totally ramified cubic field this forces $e(L/{\mathbb{Q}}_3)$ to be divisible by both two and three. Since $[L:{\mathbb{Q}}_3]=6$, the splitting field is totally ramified. In its normalized valuation,

All differences between distinct polynomial roots have the same valuation, since $S_3$ permutes the polynomial root pairs and preserves the absolute value on a field. Alsohas valuation six, so each polynomial root difference has valuation three. A nonidentity three-cycle fixes $\beta$, and thereforeThe transposition fixing $\alpha$ negates $\beta$, so sends $\pi$ to $-\pi$ and has displacement valuation $v_L(2\pi )=1$. The other transpositions have the same value by conjugacy. The uniformizer criterion now determines every group:For the wild case the different exponent is $5+2=7$. Independently, the cubic subfield has different exponent $v_F(3({\alpha }^2+1))=3$, and the quadratic extension $L/F$ is tame with exponent one. The tower formula gives $1+2\cdot 3=7$, confirming the filtration. Its positive upper break is $1/2$, because $[G_0:G_1]=2$. This completes ramification of the splitting field of X3 plus pX plus p for every odd prime number.

Write $G=\mathrm{Gal}(F/K)$ andEvery member of this decomposition group preserves ${\mathcal{O}}_F$ and $\mathfrak{q}$, so reduction defines a homomorphismIts kernel is the inertia group. We prove surjectivity of decomposition group reduction without presupposing a lift of Frobenius.

If the two residue fields coincide, surjectivity is immediate. Otherwise choose a nonzero $\overline{a}\in k_{\mathfrak{q}}$ generating it over $k_{\mathfrak{p}}$; a generator exists by the primitive element theorem for finite fields. Use the Chinese remainder theorem in ${\mathcal{O}}_F$ to choose $a$ whose residue is $\overline{a}$ at $\mathfrak{q}$ and zero at every other prime above $\mathfrak{p}$. ConsiderIts coefficients are integral and fixed by $G$, hence lie in ${\mathcal{O}}_K$: an element of $K$ integral over ${\mathcal{O}}_K$ belongs to this integrally closed ring of integers.

Reduce at $\mathfrak{q}$. When $\sigma \notin D_{\mathfrak{q}}$, the chosen residue of $a$ at ${\sigma }^{-1}\mathfrak{q}$ is zero, so $\sigma a$ reduces to zero at $\mathfrak{q}$. When $\sigma \in D_{\mathfrak{q}}$, its residue is $\overline{\sigma }(\overline{a})$, which is nonzero. ThereforeAlthough this product is written in $k_{\mathfrak{q}}[T]$, its coefficients belong to $k_{\mathfrak{p}}$. Put $q=|k_{\mathfrak{p}}|$. The finite-field Frobenius automorphism $b\mapsto b^q$ therefore permutes its nonzero roots. In particular, ${\overline{a}}^q$ is $\overline{\sigma }(\overline{a})$ for some $\sigma \in D_{\mathfrak{q}}$. Since $\overline{a}$ generates $k_{\mathfrak{q}}$, that $\overline{\sigma }$ agrees with Frobenius on the whole field. The Galois group of a finite field extension is cyclic and generated by Frobenius, so its generator lies in the image. Thus

The three quadratic subfields are $\mathbb{Q}(\sqrt{-2})$, $\mathbb{Q}(\sqrt{-3})$ and $\mathbb{Q}(\sqrt{6})$, with respective fundamental discriminants $-8,-3,24$. Their associated quadratic Dirichlet characters are the Kronecker symbolsThe regular representation of $C_2\times C_2$ is the direct sum of the trivial character and these three nontrivial characters. At an unramified prime the character is the sign by which Frobenius automorphism acts on the square root defining its quadratic subfield; at a ramified prime the linear character has no inertia group invariants. The local factor of an Artin L-function consequently givesThe zero values of a quadratic Dirichlet character at conductor primes mean that its local factor is one, not that the whole Artin L-function vanishes there.

Only the rational primes $2,3,5,7$ can contribute to coefficients with index at most ten. Their character values areFor the odd unramified primes these entries follow by checking squares: modulo five, both $-8$ and $-3$ are the nonsquare $2$, while $24$ is the square $4$; modulo seven they are respectively $6,4,3$. At $3$, $-8\equiv 1$ is a square. At $2$, the $-3$ character is $-1$ because the integral quadratic generator has polynomial $X^2+X+1$, irreducible modulo two. The other entries at $2$ and $3$ are zero exactly where the corresponding fundamental discriminant is divisible by that prime. This verifies every local factor, including the ramified ones.

The Euler product makes $a_n$ a multiplicative arithmetic function. Expanding the first two factors givesTherefore $a_2=0$, $a_4=1$, $a_8=0$, $a_3=2$ and $a_9=3$. Both factors for $5$ and $7$ have zero coefficient of $T$, so $a_5=a_7=0$. Multiplicativity gives $a_6=a_2{a}_3=0$ and $a_{10}=a_2{a}_5=0$, while $a_1=1$. The answer isAs an arithmetic check, the regular representation Artin L-function is the Dedekind zeta function of $F$. The local factors say there is one prime of norm four above two and two primes of norm three above three. Their powers give respectively one ideal of norm four and three ideals of norm nine, matching the nonzero coefficients above.

Write $\Gamma =\mathrm{Gal}(K_{\infty }/K)\cong {\mathbb{Z}}_p$. Its only finite subgroup is trivial. In particular, a real place cannot acquire complex inertia in this Zp-extension: the possible nontrivial inertia at a real place has order two. Thus every infinite place splits in the tower.

Suppose that no finite prime ramified. Every finite layer would then be an abelian everywhere unramified extension of number fields, including splitting at real places. Every layer would lie in the ordinary Hilbert class field of $K$, a finite extension. Their degrees are unbounded, a contradiction. Therefore some finite prime must ramify.

Let $v$ have residue characteristic $\ell \ne p$. Apply local class field theory to the corresponding decomposition group, a closed subgroup of $\Gamma$. The image of the unit group ${\mathcal{O}}_{K_v}^{\times }$ is the inertia group. Its first principal-unit subgroup is a pro-l group, so its continuous image in the pro-p group $\Gamma$ is trivial. The residue-unit quotient is the finite group ${\mathbb{F}}_{q_v}^{\times }$. Its image is finite, and is therefore also trivial in the torsion-free group ${\mathbb{Z}}_p$. The entire inertia group is trivial. HenceEquivalently, the tame ramification relation with a Frobenius element would force a tame inertia generator to satisfy ${\tau }^{q_v-1}=1$, which is impossible nontrivially in ${\mathbb{Z}}_p$.

For the final assertion, the cyclotomic Zp-extension of $K$ is $K\mathbb{Q}[p^{\infty }]$. Fix $v\mid p$ and put $E=K_v$, a finite extension of ${\mathbb{Q}}_p$. Let $C_n$ be the completion of $\mathbb{Q}[p^n]$ at its unique prime over $p$. This is a totally ramified extension of ${\mathbb{Q}}_p$ of degree $p^n$. In a compatible local algebraic closure,Here the inequality follows from the tower through $C_n$. The right side is unbounded. These composita occur among the completions of the cyclotomic tower over $K$, so $v$ has unbounded ramification index, in particular nontrivial inertia group. ThusThis argument allows an arbitrary finite intersection between $K$ and the rational cyclotomic tower; it does not assume disjointness.

Let $F_n$ be the degree-$p^n$ layer, ${\Gamma }_n={\Gamma }^{p^n}$, andConjugation makes $Y$ a compact $\Lambda$-module. The goal is to show that it is finitely generated and has Iwasawa-module rank zero. This proof applies to every Zp-extension, not only a cyclotomic one, and makes no Leopoldt conjecture assumption.

By the ramification argument above, only primes over $p$ ramify in $F_{\infty }/F$, and at least one does. Every nonzero closed inertia group in $\Gamma \cong {\mathbb{Z}}_p$ is open. After replacing $F$ by a finite layer, every prime which ramifies is totally ramified in the remaining tower. The number $s\ge 1$ of such primes then stays constant. This replacement does not affect whether $Y$ is a torsion module: ${\mathbb{Z}}_p[[\Gamma ]]$ is finite free over the Iwasawa algebra of an open subgroup, and the two module ranks vanish together.

Let ${\mathcal{G}}_n=\mathrm{Gal}(L_{\infty }/F_n)$, and let $D_n$ be its maximal abelian quotient. The abelianization over a Zp-extension formula gives an exact sequenceChoose a lift of a generator of ${\Gamma }_n$ if a splitting is desired. We will bound the ${\mathbb{Z}}_p$-rank of the middle term independently of $n$.

At each of the $s$ ramified primes, inertia in $L_{\infty }/F_n$ maps isomorphically onto ${\Gamma }_n$: its kernel is inertia in the unramified extension $L_{\infty }/F_{\infty }$, hence trivial, and the map onto the totally ramified base inertia is surjective. Thus each image in $D_n$ is procyclic and has ${\mathbb{Z}}_p$-rank at most one. No other finite prime contributes inertia. Quotienting $D_n$ by the closed subgroup generated by these $s$ images gives an abelian extension of $F_n$ unramified at all finite primes. Its Galois group is finite by class field theory, using the ordinary or narrow ideal class group according to the treatment of real places. Any infinite-place inertia is finite and does not affect the rank bound. Therefore $D_n$ is finitely generated over ${\mathbb{Z}}_p$ andThe preceding exact sequence now yields the uniform boundIt does not assert that these coinvariant modules are always finite when several primes ramify.

At $n=0$, $Y/TY$ is finitely generated over ${\mathbb{Z}}_p$, so $Y/(p,T)Y$ is finite. Lift a finite basis of this quotient to $Y$. The Compact Nakayama lemma shows that these lifts generate $Y$ over $\Lambda$. Briefly, the quotient by their compact generated image satisfies $Q=(p,T)Q$. Every finite continuous $\Lambda$-quotient of $Q$ has nilpotent action by $(p,T)$: $p$ is nilpotent, and modulo $p$ a finite pro-$p$ action makes $\gamma -1$ nilpotent. Such a quotient must be zero. Finite quotients separate points of a compact pro-p group, so $Q=0$.

Suppose now that $Y$ had positive $\Lambda$-rank $r$. The Iwasawa module structure theorem provides a pseudo-isomorphism towith $E$ a finitely generated torsion module. Passing to ${\Gamma }_n$-coinvariant modules leaves a finite cokernel. Butis free of rank $p^n$ over ${\mathbb{Z}}_p$. Thus the coinvariant modules would have rank at least $r{p}^n$, contradicting the uniform bound $s-1$. Equivalently their ranks have asymptotic form $r{p}^n+O(1)$, with the torsion elementary divisors contributing only a bounded rank.

It follows that $r=0$. Since $\Lambda$ is an integral domain, rank zero means every element is killed by a nonzero scalar. ThereforeThis is the unramified Iwasawa torsion theorem. The crucial arithmetic inputs are finite class fields and the bounded number of ramified primes; neither vanishing of a $\mu$-invariant nor a cyclotomic main conjecture is required.

Let $F/K$ be a finite Galois extension, choose a prime ideal $\mathfrak{q}$ above $\mathfrak{p}$, and write $k_{\mathfrak{q}}$ and $k_{\mathfrak{p}}$ for the residue fields. The decomposition group and inertia group areThe Frobenius automorphism on the finite field $k_{\mathfrak{q}}$ is $x\mapsto x^{N\mathfrak{p}}$. Its inverse image defines a distinguished coset in $D_{\mathfrak{q}}/I_{\mathfrak{q}}$; a Frobenius element is a lift of that coset. At an unramified prime, $I_{\mathfrak{q}}=1$ and the lift is unique. Changing the prime above $\mathfrak{p}$ conjugates it, so a Frobenius conjugacy class is intrinsically attached to $\mathfrak{p}$. These are arithmetic Frobenius conventions.

For a finite Galois extension of local fields, let $G=\mathrm{Gal}(L/K)$, $G_{-1}=G$, and, for integers $i\ge 0$, define the lower ramification groups byIn particular, $G_0$ is the inertia group. The identity satisfies every bound because $v_L(0)=\infty$.

If the extension is totally ramified, its residue fields agree. Choose representatives in ${\mathcal{O}}_K$ of their common residue field. Every $a\in {\mathcal{O}}_L$ has a convergent uniformiser expansion $a={\sum }_{j\ge 0}{c}_j{\pi }_L^j$ with these representatives $c_j$, all fixed by $G$. For $j\ge 1$,Each summand in the second factor has valuation $j-1$. Thus a bound $v_L(\sigma ({\pi }_L)-{\pi }_L)\ge i+1$ implies the same bound on $\sigma (a)-a$, by convergence and the valuation inequality. Necessity follows by testing $a={\pi }_L$. This proves the uniformizer criterion for lower ramification groups:

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.

Normalize the discrete valuation on $L$ by $v_L(L^{\times })=\mathbb{Z}$, set $v_L(0)=+\infty$, and write ${\mathcal{O}}_L$ for its valuation ring. For a finite Galois extension of local fields, define $G_{-1}=G=\mathrm{Gal}(L/K)$ and, for integers $i\ge 0$,Equivalently, $G_i$ is the kernel of the action on ${\mathcal{O}}_L/{\mathfrak{m}}_L^{i+1}$. These are the lower ramification numbering of the ramification groups. In particular, $G_0$ is the inertia group, the kernel of the action on the residue field, and $G_1$ is the wild inertia group. They form a decreasing sequence of normal subgroups, eventually trivial.

For a real index $t\ge -1$, extend by $G_t=G_{\lceil t\rceil }$. Define the Herbrand function byIt is continuous, strictly increasing and piecewise linear; denote its inverse by ${\psi }_{L/K}$. The upper ramification numbering isThe definition includes the placement of the groups at break endpoints: the group at a break is the group before the drop. Lower numbering is compatible with subgroups; upper numbering is the one compatible with quotients. In a totally ramified extension, the uniformizer criterion for lower ramification groups permits testing the defining valuation inequality on one uniformizer instead of on every integral element.

Galois cohomology records the obstruction to choosing Galois-invariant division points. Let $G_K=\mathrm{Gal}(\overline{K}/K)$ be the absolute Galois group and give its algebraic-point modules the discrete topology. A continuous $1$-cocycle with values in a Galois module $M$ is a function $c:G_K\to M$ satisfyingA group coboundary has the form $c_{\sigma }=\sigma a-a$. Quotienting cocycles by coboundaries defines $H^1(K,M)$. For the finite module $E[m]$ these cocycles have finite image and factor through finite data; continuity is essential because $G_K$ is a profinite group.

In characteristic zero, multiplication by $m\ge 2$ on the algebraic points of an elliptic curve is surjective, with kernel $E[m]\cong (\mathbb{Z}/m\mathbb{Z}{)}^2$. The short exact sequence of Galois modulesgives the Kummer exact sequence of an elliptic curveExplicitly, choose $Q$ with $mQ=P$ and set $\delta (P{)}_{\sigma }=\sigma Q-Q$. Changing $Q$ by an $m$-torsion point changes the cocycle by a coboundary. The class vanishes exactly when a suitable choice of $Q$ is Galois-fixed, that is, when $P\in mE(K)$. This proves the injection directly and identifies its arithmetic meaning.

The entire group $H^1(K,E[m])$ need not be finite. For example, Kummer theory gives $H^1(K,{\mu }_m)=K^{\ast }/K^{\ast m}$, which has classes supported on arbitrarily many different primes. The finite part needed for the Weak Mordell-Weil theorem comes from a ramification restriction on the image of $\delta$.

Choose a finite set $S$ of places containing the archimedean places, the primes dividing $m$ and all primes of bad reduction of an elliptic curve. At a finite place $v\notin S$, the curve has good reduction and $m$ is invertible in its valuation ring. The elliptic curve extends to a smooth proper group scheme, and $[m]$ on that model is finite etale. A point $P\in E(K_v)$ extends to an integral section by properness. Its division-point fibre is consequently finite etale over the valuation ring. Over the maximal unramified extension it has a point, so the Kummer cocycle restricts trivially to the inertia group. The module $E[m]$ itself is unramified there for the same reason. Thuswhere $H_S^1$ denotes classes unramified outside $S$.

Here is a proof that this restricted group is finite. Take a finite Galois extension $L/K$ containing all coordinates of $E[m]$ and the $m$th roots of unity, and enlarge $S$ by its ramified primes. Over $L$ the module is trivial and, after choosing a basis, is isomorphic to ${\mu }_m^2$. Hilbert theorem 90 and the multiplicative Kummer sequence therefore giveFor unramified Kummer classes with bounded prime support, a class unramified outside $S_L$ has valuations divisible by $m$ at every prime outside $S_L$: the valuation of an $m$th root in an unramified extension is integral. Consequently both coordinates lie in

To show this set finite, write the outside-$S_L$ divisor of $a$ as $mD$. The ideal class of $D$ belongs to the $m$-torsion of the ideal class group of the ring of $S_L$-integers. This gives the exact sequenceThe last map is onto: if $mD$ is principal in the $S_L$-ideal group, a generator represents a class with outside valuations divisible by $m$. The kernel consists exactly of S-units modulo $m$th powers. The Dirichlet unit theorem, with the finitely many inverted primes adjoined, makes the S-unit group finitely generated; the ideal class group is finite, and localization only quotients it. Both ends of the sequence are therefore finite.

Finally, the inflation-restriction exact sequence bounds the kernel of restriction from $H^1(K,E[m])$ to $H^1(L,E[m])$ by the finite group $H^1(\mathrm{Gal}(L/K),E[m])$. The unramified subgroup has finite image, contained in $L(S_L,m{)}^2$, and finite kernel, so it is finite. The Kummer injection now provesThis is the Weak Mordell-Weil theorem, proved without first assuming finite generation of $E(K)$.

For computation one refines the unramified group by local solvability. The Selmer group of an elliptic curve consists of classes whose restriction at every completion lies in the corresponding local Kummer image. It is finite and sits inwhere the Tate–Shafarevich group $\mathrm{Sha}$ measures classes in $H^1(K,E)$ that become trivial at every completion. Thus locally soluble descent equations can give an upper bound without every class coming from a rational point. No finiteness assumption on the whole Tate-Shafarevich group is needed for the weak theorem. Combining the finite quotient with the height descent lemma from the height essay yields the full Mordell-Weil theorem. Replacing multiplication by $m$ with a smaller isogeny gives the same cohomological framework for two-isogeny descent.

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 $u\ge -1$.