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For $a,b\in {\mathbb{Q}}_p^{\times }$, the quadratic Hilbert symbol $(a,b{)}_p$ is one if $b$ is a field norm from ${\mathbb{Q}}_p(\sqrt{a})$, and minus one otherwise; a square $a$ gives the trivial extension and symbol one. It depends only on the two square classes. Equivalently, the conic $z^2=a{x}^2+b{y}^2$ has a nonzero rational point precisely when the symbol is one. Interchanging $a$ and $b$ in the conic gives $(a,b{)}_p=(b,a{)}_p$. The Hilbert norm residue symbol is a group homomorphism in each argument; for nonsquare $a$, the cyclic local norm index makes its group kernel a subgroup of index two. The resulting pairing on square classes is nondegenerate. Alsosince $N(\sqrt{a})=-a$. In the language of local class field theory, $(a,b{)}_p={\mathrm{Art}}_p(b)(\sqrt{a})/\sqrt{a}$; in the Brauer group, it records whether the quaternion cyclic algebra $(a,b)$ splits. These are compatible norm-obstruction descriptions.

An unramified extension of local fields of degree two has norm group consisting precisely of elements with even valuation. Indeed,and the norms on units are surjective. For the latter assertion, the norm on the finite residue fields is surjective. On successive higher principal-unit groups, the congruenceallows one to correct a proposed unit norm one digit at a time: the residue field trace is surjective, and completeness makes the corrections converge. An unramified quadratic field is unique up to isomorphism. Therefore, if $d$ is its defining nonsquare class,

For odd $p$, choose the nonsquare unit $u_0$ from part (i). The polynomial $T^2-u_0$ has irreducible separable reduction, so it defines the unramified quadratic extension. ThusThe trivial class is included in ${\Delta }_2$, since a degree-one extension is unramified. The two units pair trivially, while $(u_0,p{)}_p=-1$. Any class outside ${\Delta }_1$ has odd valuation and consequently pairs nontrivially with $u_0$. This proves directly

For $p=2$, the unit square classes from part (i) give ${\Delta }_1=\{1,-1,5,-5\}$. The element $\omega =(1+\sqrt{5})/2$ satisfies $T^2-T-1$; its reduction $T^2+T+1$ is irreducible and separable over ${\mathbb{F}}_2$. This shows that ${\mathbb{Q}}_2(\sqrt{5})$ is unramified of degree two, givingFor a fully explicit pairing check, use the generators $(-1,5,2)$. We have $(-1,5{)}_2=1$ and $(-1,2{)}_2=1$, since $5=N(1+2i)$ and $2=N(1+i)$ from ${\mathbb{Q}}_2(i)$. But $(-1,-1{)}_2=-1$: a norm equal to $-1$ would be a sum of two squares. If either summand had negative valuation, the sum would have negative valuation (equal negative valuations give valuation $1-2r<0$); if both were integral, their squares could not sum to $-1\equiv 3(\mathrm{mod}4)$. Finally, the unramified norm criterion gives $(5,2{)}_2=-1$. The identity $(a,b{)}_2=(b,a{)}_2$ and $(a,a{)}_2=(a,-1{)}_2$ determine the remaining entries. Writing $(a,b{)}_2=(-1{)}^{B(a,b)}$, the exponent matrix isFor $b=(-1{)}^s{5}^t{2}^r$, annihilating both generators $-1,5$ of ${\Delta }_1$ requires $s=r=0$, so $b\in \{1,5\}$. Annihilating $5$ alone requires $r=0$, so $b$ is a unit class. HenceThese are the Hilbert-symbol annihilators of units and unramified square classes. Annihilators here are taken with respect to this pairing, or equivalently in the character group of a finite abelian group after identifying square classes with their Hilbert-symbol characters.

We give the detailed norm-index argument for a cyclic extension $L/K$ of p-adic fields; thus $K$ is a finite extension of ${\mathbb{Q}}_p$. Write $G=⟨\sigma ⟩$, $|G|=n$. For an additive $G$-module $M$, put $D=\sigma -1$ and $N=1+\sigma +\cdots +{\sigma }^{n-1}$. The two periodic Tate cohomology of a cyclic group groups and the Herbrand quotient arewhen these groups are finite. For multiplicative modules, $N$ is the product of conjugates and $D(z)=\sigma (z)/z$.

The periodic complex alternating $D$ and $N$ gives a six-term exact sequence for every short exact sequence of modules. Taking the alternating product of the finite group orders in that exact sequence provesFor a finite module, the counting fibers gives $|NM|=|M|/|\ker N|$ and $|DM|=|M|/|\ker D|$, so the two Tate groups have equal order. This proves the Herbrand quotient of a finite module is one. It follows that replacing a module by a finite-index submodule does not change its quotient, whenever the cohomology groups are finite.

Two basic calculations drive the norm computation. For the trivial module $\mathbb{Z}$, $D=0$ and $N$ is multiplication by $n$, so $h_G(\mathbb{Z})=n$. For a regular lattice $R[G]$, with $R={\mathbb{Z}}_p$, invariants are constant coefficient vectors and every such vector is a norm: put its constant coefficient in just one coordinate before applying $N$. The group kernel of $N$ consists of the coefficient vectors of sum zero. Successively taking partial sums solves the cyclic difference equation $Dx=y$ for every such vector $y$. Thus both Tate groups vanish and $h_G(R[G])=1$. This is cyclic cohomology of a regular lattice.

Let $U_L={\mathcal{O}}_L^{\times }$ and $U_L^r=1+{\mathfrak{P}}_L^r$. For sufficiently large $r$, the p-adic logarithm and p-adic exponential are inverse $G$-equivariant isomorphismsTo make the range explicit, $r>v_L(p)/(p-1)$ suffices: the higher terms in the logarithm and exponential have greater valuation than their leading terms, their series converge there, and the identities $\log (\exp x)=x$ and $\exp (\log (1+x))=1+x$ remain valid. Thus a deep multiplicative unit group becomes an additive ${\mathbb{Z}}_p$-lattice in $L$.

By the normal basis theorem, $L\cong K[G]$ as a $K[G]$-module. As a ${\mathbb{Q}}_p[G]$-module it is therefore ${\mathbb{Q}}_p[G{]}^{[K:{\mathbb{Q}}_p]}$. Transport the regular lattice $M_0={\mathbb{Z}}_p[G{]}^{[K:{\mathbb{Q}}_p]}$ into $L$. Both $M_0$ and ${\mathfrak{P}}_L^r$ are full $G$-stable ${\mathbb{Z}}_p$-lattices, so their intersection has finite index in each. The exact sequences with finite quotients show that the Tate groups of ${\mathfrak{P}}_L^r$ are finite and thatSince $U_L/U_L^r$ is finite, the exact-sequence formula for the Herbrand quotient gives $h_G(U_L)=1$. Finally, the valuation sequencehas trivial $G$-action on $\mathbb{Z}$. Hence the Herbrand quotient of the local multiplicative group is

For completeness, the denominator vanishes by Hilbert theorem 90. If $N_{L/K}a=1$, put $c_0=1$ and $c_j=a\sigma (a)\cdots {\sigma }^{j-1}(a)$. Independence of distinct field automorphisms lets us choose $t\in L$ with $b={\sum }_{j=0}^{n-1}{c}_j{\sigma }^j(t)\ne 0$. Since $c_n=1$, shifting the sum gives $\sigma (b)=b/a$. Thus $a=b/\sigma (b)$, a multiplicative coboundary. Therefore ${\hat{H}}^{-1}(G,L^{\times })=1$. The invariant subgroup of $L^{\times }$ is $K^{\times }$, soThis proves the cyclic local norm index without assuming local reciprocity. If the extension has ramification index $e$ and residue degree $f$, the valuation formula $v_K(Nz)=f{v}_L(z)$ shows that the valuation contribution to the index is $f$. A norm is a unit only when $z$ is a unit, so the remaining contribution isIn particular, unit norms are surjective in an unramified cyclic extension. The norm subgroup is open: on deep units its logarithm is ${\mathrm{Tr}}_{L/K}({\mathfrak{P}}_L^r)$, a nonzero full lattice in $K$, and therefore contains a sufficiently deep principal-unit group. This connects the index calculation to the open finite-index norm subgroups classified by local class field theory.

To finish, the Hasse norm theorem for a cyclic extension of number fields states that a nonzero element is a global norm if and only if it is a norm at every finite and infinite completion. Here is its Brauer-theoretic proof outline. Associate to $a\in K^{\times }$ the cyclic algebra $A=(L/K,\sigma ,a)$, with relations $z\ell =\sigma (\ell )z$ and $z^n=a$. The splitting criterion for a cyclic algebra says that it splits exactly when $a=Nc$: if $a=Nc$, multiplication by $L$ and the operator $c\sigma$ realize it as ${\mathrm{End}}_K(L)$; conversely, in a split matrix representation, the underlying $n$-dimensional space is one-dimensional over $L$, so $z$ must act as $c\sigma$ and $z^n=Nc$.

Apply this criterion at every completion, using the restricted cyclic character and any one completion of $L$ above that place. If $a$ is a local norm everywhere, every localized Brauer class is zero. Injectivity in the Albert-Brauer-Hasse-Noether theorem forces $[A]=0$ in $\mathrm{Br}(K)$; the global splitting criterion gives $a=Nc$. Conversely, a global norm is a local norm after completion, since the norm on $L{\otimes }_K{K}_v$ is the product of the component norms, whose images coincide in the Galois case. The cyclic hypothesis is essential; the stated local-global principle does not hold for arbitrary extensions.

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.

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.

The main theorem of local class field theory consists of Local Artin reciprocity and its existence theorem. There is a canonical continuous reciprocity mapwith dense image, normalized so that a uniformizer maps to a chosen Frobenius. For every finite abelian extension $L/K$, it induces an isomorphismMoreover, the existence theorem of local class field theory says that $L\mapsto N_{L/K}(L^{\times })$ is an inclusion-reversing bijection between finite abelian extensions of $K$ in $K^{\mathrm{ab}}$ and finite-index open subgroups of $K^{\times }$.

The main theorem of local class field theory gives a continuous Local Artin mapwith dense image, normalized by sending a uniformizer to a chosen Frobenius. For every finite abelian extension $L/K$, it induces the Local Artin reciprocity isomorphismThe existence theorem of local class field theory says that the finite-index open subgroups of $K^{\times }$ are exactly the norm subgroups $N_{L/K}(L^{\times })$ for finite abelian extensions $L/K$, and that the extension is uniquely determined inside $K^{\mathrm{ab}}$.

Write $e=e(L/K)$ and $f=f(L/K)$. The valuation of a field norm satisfiesso the valuation image of the norm subgroup is $f\mathbb{Z}$. The exact sequence obtained from $v_K:K^{\times }\to \mathbb{Z}$ therefore givesThe left side is $[L:K]=ef$ by Local Artin reciprocity. Cancelling $f$ proves

At least one finite prime ramifies in a Zp-extension; otherwise the entire tower would lie in a finite Hilbert class field. Ramification can occur only over $p$: local class field theory makes inertia away from $p$ an image of a unit group with finite maximal pro-$p$ quotient, whereas ${\mathbb{Z}}_p$ has no nontrivial finite subgroup. Nonzero inertia is open in ${\mathbb{Z}}_p$, so it becomes total after passing to a sufficiently high finite layer.