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The inertia group of a finite Galois extension of local fields is the kernel of the action of its Galois group on the residue field. In lower numbering it is the zeroth ramification group $G_0$.

For an odd integer $x=2a+1$, $x^2=1+4a(a+1)\equiv 1(\mathrm{mod}8)$. The same conclusion holds for every 2-adic unit by reduction modulo eight. Thus an odd square must be congruent to one modulo eight.

Conversely let $u\in {\mathbb{Z}}_2^{\times }$ with $u\equiv 1(\mathrm{mod}8)$. For digit lifting for square roots of 2-adic units, start with $x_3=1$. If $x_n$ is odd and $x_n^2\equiv u(\mathrm{mod}{2}^n)$, considerFor $n\ge 3$,since the remaining term is divisible by $2^{2n-2}$ and $2n-2\ge n+1$. The odd multiplier $x_n$ lets the choice of ${\epsilon }_n$ correct exactly the next binary digit. Thus $x_{n+1}^2\equiv u(\mathrm{mod}{2}^{n+1})$. The corrections have valuations tending to infinity, so completeness gives $x=\lim x_n\in {\mathbb{Z}}_2^{\times }$ with $x^2=u$.

The unit $3$ is congruent to one modulo two but is not a square, and $5$ is congruent to one modulo four but is not a square, by the necessary modulo-eight condition. Therefore

Every nonzero P-adic number in ${\mathbb{Q}}_2$ is uniquely $2^{m}u$ with $m\in \mathbb{Z}$ and $u$ a 2-adic unit. A square has even valuation, and the unit square criterion shows that the unit class is determined by its odd residue modulo eight. The four residues form $C_2\times C_2$, generated by $-1$ and $5$; the valuation parity supplies the third factor, generated by $2$. Hence the square-class group of the 2-adic numbers isIts eight classes are represented by $1,-1,5,-5,2,-2,10,-10$. They are distinct: the first four have even valuation and distinct unit residues modulo eight, while multiplication by $2$ gives odd valuation and the same four distinct unit residues.

In characteristic different from two, completing the square in a quadratic minimal polynomial makes every quadratic extension $K(\sqrt{d})$ for a nonsquare $d\in K$. If $K(\sqrt{d})=K(\sqrt{e})$, write $\sqrt{e}=a+b\sqrt{d}$ and square. The coefficient $2ab$ must vanish. Since $e$ is nonsquare, $b\ne 0$, so $a=0$ and $e/d=b^2$. Conversely a square ratio plainly gives the same field. Thus the seven quadratic extensions of the 2-adic numbers are exactly

The unique unramified extension in this list is ${\mathbb{Q}}_2(\sqrt{5})$. To verify this, $\alpha =(1+\sqrt{5})/2$ is integral and satisfies $X^2-X-1=0$. Its reduction $X^2+X+1$ is irreducible over ${\mathbb{F}}_2$, so its residue field contains ${\mathbb{F}}_4$. The standard degree formula for finite extensions of complete discretely valued fields with perfect residue field is $[L:K]=e(L/K)f(L/K)$. Here the field degree is two, so the residue degree is two and the ramification index is one.

Conversely an unramified quadratic extension has residue field ${\mathbb{F}}_4$. A root there of $X^2+X+1$ lifts by Hensel lemma to a root of $X^2-X-1$, whose derivative is a unit. For this lift $\beta$, $(2\beta -1{)}^2=5$, so the extension contains ${\mathbb{Q}}_2(\sqrt{5})$ and must equal it. Every other field in the list is therefore ramified.

For the requested ramification groups, choose $L={\mathbb{Q}}_2(\sqrt{2})$ and $\pi =\sqrt{2}$. We use uniqueness of the extension of the valuation to a finite extension, together with the degree formula just stated. Since $v_2(\pi )=1/2$, the extension has ramification index two and residue degree one. Normalize $v_L(\pi )=1$, giving $v_L(2)=2$. Every element is $a+b\pi$ with $a,b\in {\mathbb{Q}}_2$, and the two summands have valuations of opposite parity:Thus $a+b\pi$ is integral exactly when $a,b\in {\mathbb{Z}}_2$, proving ${\mathcal{O}}_L={\mathbb{Z}}_2\oplus {\mathbb{Z}}_2\pi$ directly.

The nonidentity automorphism $\sigma$ negates $\pi$. For any integral $a+b\pi$,of valuation at least three, with equality when $b$ is a unit. By the definition $G_i=\{\sigma :v_L(\sigma x-x)\ge i+1\text{ for every }x\in {\mathcal{O}}_L\}$, the ramification in the quadratic extension generated by the square root of two is thereforeThe lower ramification break is two; the valuation used here is the integral normalization on $L$.

Fix a non-Archimedean local field $K$, write $Q=|k_K|$, and use the arithmetic Frobenius normalization throughout: a uniformizer maps to the automorphism inducing $a\mapsto a^Q$ on the maximal unramified extension. The Local Artin map, or local reciprocity map, is the canonical continuous group homomorphismwhere $K^{\mathrm{ab}}$ is the maximal abelian extension inside a fixed algebraic closure with its separable part understood. Its finite projection to every finite abelian extension $L/K$ is onto and has kernel exactly the norm subgroup of a local field extension:In particular, the quotient has order $[L:K]$. This is how local class field theory turns abelian extension problems into computations inside a multiplicative group.

One precise definition of the finite reciprocity map uses the cup-product construction of local reciprocity. For a finite Galois extension with group $G$, the local fundamental class $u_{L/K}\in H^2(G,L^{\times })$ is specified by local Brauer invariant $1/[L:K]$. Cup product gives an isomorphismThe finite Local Artin map is its inverse, composed with the quotient map from $K^{\times }$. For abelian $G$ the target is $G$ itself. Taking the compatible inverse limit over finite abelian extensions defines ${\mathrm{Art}}_K$.

The essential inputs to this cohomological construction have concrete arithmetic content. The local Brauer invariant identifies $\mathrm{Br}(K)$ with $\mathbb{Q}/\mathbb{Z}$; restriction to $L$ multiplies the invariant by $[L:K]$, so its kernel is cyclic of that order and has the distinguished generator $u_{L/K}$. For every subgroup $H\subseteq G$, Hilbert theorem 90 gives $H^1(H,L^{\times })=0$, while the same invariant calculation over the fixed field gives $H^2(H,L^{\times })$ cyclic of order $|H|$. These are the hypotheses of the algebraic cup-product isomorphism in Tate cohomology of a finite group. It is obtained by dimension shifting in group cohomology with a projective resolution of the trivial module and the extension representing the fundamental class. Thus the norm-kernel statement comes from the degree-zero definition of Tate cohomology of a finite group, rather than from an arbitrary assignment of an automorphism to each element. In an unramified extension of degree $d$, the fundamental class is the cyclic algebra with arithmetic Frobenius generator and parameter a uniformizer; the isomorphism sends that Frobenius to the class of the uniformizer. This explains the chosen sign of reciprocity.

Several properties follow immediately from this construction. If $L\supseteq E\supseteq K$ are finite abelian extensions, restricting ${\mathrm{Art}}_{L/K}(a)$ to $E$ gives ${\mathrm{Art}}_{E/K}(a)$. The Local Artin map is therefore compatible with the finite projections defining its infinite target. For any finite separable extension $E/K$ and $a\in E^{\times }$, its norm compatibility isHere $K^{\mathrm{ab}}E/E$ is abelian, so the displayed restriction makes sense. An outline of the compatibility proof is to use the corestriction map in group cohomology: on multiplicative coefficients its degree-zero map is the norm. The cup-product projection formula, together with restriction and corestriction of the local fundamental class, gives exactly the commuting reciprocity diagram. The compatibility is an arithmetic instance of the functoriality of the cohomological definition.

The Local Artin map has dense image, because every finite abelian quotient receives a surjective map. Its kernel is trivial, and it extends to an isomorphismwhere this profinite completion uses the open finite-index subgroups of the topological multiplicative group. The original map is generally not onto: on the unramified quotient it sends the integer valuation into $\hat{\mathbb{Z}}$, rather than producing every profinite integer. Its units map onto the abelian inertia group. For a finite abelian extension, the corresponding unit image is its inertia group, while the valuation accounts for its residue extension. Indeed $v_K(Nz)=f(L/K)v_L(z)$, so the valuation quotient has order $f$; the full norm quotient has order $ef$, leaving unit image of order $e$. Since unit images act trivially on the residue field, they give precisely the inertia group. More finely, the image of the higher principal-unit group $1+{\mathfrak{m}}_K^r$ in a finite abelian extension is its rth upper ramification group, for each integer $r\ge 1$. The use of upper ramification numbering is essential for compatibility with quotients.

The existence theorem of local class field theory states that every open finite-index subgroup $H\subseteq K^{\times }$ is $N_{L/K}(L^{\times })$ for a unique finite abelian extension $L/K$. Inclusions are reversed: larger extensions have smaller norm subgroups. In the established reciprocity isomorphism, the proof is simply to take the fixed field of the open subgroup corresponding to $H$. Conversely a finite extension gives an open kernel of a finite continuous projection. Uniqueness follows because that kernel specifies its fixed field. The substantive existence problem is therefore to prove that the reciprocity isomorphism realizes all open finite-index subgroups; it is not just this final fixed-field argument.

The Lubin–Tate description of local reciprocity provides an explicit route to that existence problem. Choose a uniformizer $\pi$ and a Lubin–Tate series $f(T)=\pi T+T^Q$, with its associated Lubin–Tate formal group $F$. Solving $f(F(X,Y))=F(f(X),f(Y))$ successively by total degree constructs the law with linear part $X+Y$; the analogous equations construct the endomorphisms $[a]$ with linear coefficient $a$. The congruence $f(T)\equiv T^Q(\mathrm{mod}\pi )$ supplies the divisibility needed at each step. These endomorphisms satisfy $[a][b]=[ab]$ and make the torsion an ${\mathcal{O}}_K$-module.

If $f_r$ denotes the rfold iterate, the primitive level-r Lubin–Tate torsion points are roots ofReduction modulo $\pi$ is $T^{Q^{r-1}(Q-1)}$, and the constant term is $\pi$, so this is an Eisenstein polynomial. Its roots therefore generate a totally ramified extension $K_{\pi ,r}$ of degree $(Q-1)Q^{r-1}$. A primitive point generates the entire torsion module, and all points are its images under the integral convergent endomorphisms $[a]$. The torsion field is therefore a Galois extension of $K$. The action map into $({\mathcal{O}}_K/{\pi }^r{)}^{\times }$ is injective, and the computed degree equals the order of this unit group, proving it is an isomorphism. Passing to the inverse limit gives $\mathrm{Gal}(K_{\pi }/K)\simeq {\mathcal{O}}_K^{\times }$ for the Lubin–Tate tower $K_{\pi }$.

The unramified tower and $K_{\pi }$ are linearly disjoint field extensions: a finite common subextension would be both unramified and totally ramified, hence trivial. The Lubin–Tate description of local reciprocity on their compositum isThe inverse on the unit action goes with the arithmetic Frobenius convention; changing to geometric Frobenius inverts the whole reciprocity map. The reciprocity theorem identifies this explicit action with the finite cup-product maps, independently of the chosen uniformizer. Its proof compares the fundamental class with the norm relations between consecutive torsion fields. The construction makes the possible finite kernels concrete: on the compositum of the degree-d unramified field with $K_{\pi ,r}$ the kernel isThese subgroups are cofinal among open finite-index subgroups. Indeed any such $H$ contains $1+{\pi }^r{\mathcal{O}}_K$ for sufficiently large $r$, and contains ${\pi }^d$ for some positive $d$, since the class of $\pi$ in $K^{\times }/H$ has finite order. Taking the appropriate fixed subgroup in this explicit finite compositum realizes $H$. This proves the open-subgroup part of existence once the norm compatibility of the explicit and cohomological constructions is established. Together with the finite reciprocity maps it also shows that $K^{\mathrm{ab}}=K^{\mathrm{nr}}{K}_{\pi }$.

As a direct check, on a finite unramified extension the construction reduces toThe earlier elementary unit-norm proof shows its kernel is exactly ${\pi }^{d\mathbb{Z}}{\mathcal{O}}_K^{\times }$. At $K={\mathbb{Q}}_p$, the Lubin–Tate cyclotomic example takes $[p](T)=(1+T{)}^p-1$ and $F(X,Y)=X+Y+XY$. Its torsion is $\zeta -1$, so a unit $u$ acts by $\zeta \mapsto {\zeta }^{u^{-1}}$. These examples exhibit both the valuation and unit parts of local reciprocity with consistent signs.

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:

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.

Let $G=\mathrm{Gal}(L/K)$. The lower ramification groups are $G_{-1}=G$ and, for $i\ge 0$,

If $L/K$ is totally ramified and ${\pi }_L$ is a uniformizer, then ${\mathcal{O}}_L={\mathcal{O}}_K[{\pi }_L]$. Factoring $F(\sigma {\pi }_L)-F({\pi }_L)$ by $\sigma {\pi }_L-{\pi }_L$ for $F\in {\mathcal{O}}_K[X]$ proves the uniformizer criterion for lower ramification groups

For $\sigma \in G_0$, defineThe inertia group $G_0$ acts trivially on $k_L$, so $\theta (\sigma \tau )=\theta (\sigma )\theta (\tau )$. Its kernel consists exactly of those $\sigma$ for which $\sigma ({\pi }_L)/{\pi }_L\equiv 1$ modulo the maximal ideal, namely $G_1$. The first isomorphism theorem therefore gives an injection

For $G=\mathrm{Gal}(L/K)$ defineThese are the lower ramification groups. If $\sigma$ lies in every $G_i$, then $\sigma ({\pi }_L)={\pi }_L$; the equivalent definition using all $a\in {\mathcal{O}}_L$ then gives $\sigma (a)=a$, so $\sigma =1$.

For $\sigma \in G_0$, setBecause inertia acts trivially on $k_L$, $\theta$ is a homomorphism. Its kernel is exactly $G_1$, so it induces an injection $G_0/G_1\hookrightarrow k_L^{\ast }$.

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 $7$.

For an odd prime $p$, this Eisenstein polynomial generates a totally ramified cubic root field. Its discriminant is $-p^2(4p+27)$. For $p>3$ the unit part is a square exactly when $p\equiv 1(\mathrm{mod}3)$. The splitting field then has cyclic Galois group of order three; otherwise it has symmetric group $S_3$, with an unramified extension of degree two and inertia group $C_3$. Both cases are tamely ramified, so all positive ramification groups are trivial.

At $p=3$, adjoining $\beta =\sqrt{-39}$ gives a ramified quadratic subfield, and the splitting field has total ramification index six. A root $\alpha$ has normalized valuation two, while $\beta$ has valuation three, so $\pi =\beta /\alpha$ is a uniformizer. The derivative $3{\alpha }^2+3$ has valuation six, and all distinct root differences have valuation three. A three-cycle fixes $\beta$, giving $v_L(\sigma \pi -\pi )=2$. A transposition fixing $\alpha$ negates $\beta$, giving valuation one; conjugation handles the other transpositions. Hence $G_0=S_3$, $G_1=C_3$, and $G_i=1$ for $i\ge 2$. The different exponent is seven.

For a degree-$d$ polynomial, reflecting the usual coefficient-index axis replaces $(j,v(a_j))$ by $(d-j,v(a_j))$. The reflected lower polygon has slopes equal to root valuations, rather than their negatives. Horizontal lengths, and therefore multiplicities, are unchanged. This convention makes positive lower ramification groups breaks appear as positive slopes of a reduced ramification polygon.

The wild inertia group is the first ramification group $G_1$. It is the unique Sylow subgroup of the inertia group for the residue characteristic.