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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.

Herbrand quotient and local norm indices. Let $G=⟨\sigma ⟩$ have order $n$. For an additive $G$-module $M$, put $D=\sigma -1$ and $N=1+\sigma +\cdots +{\sigma }^{n-1}$. The relevant Tate cohomology of a cyclic group groups areIf both groups are finite, their size ratio is the Herbrand quotientMultiplicative modules use products for $N$ and $D(m)=\sigma (m)/m$. A finite module has quotient one: $\#\ker N=\#M/\#NM$, while $\#DM=\#M/\#M^G$, and the two cohomology orders are equal. The six-term periodic cohomology sequence shows $h(B)=h(A)h(C)$ for a short exact sequence $0\to A\to B\to C\to 0$ when these groups are finite. Thus the quotient is unchanged by finite-index changes of lattices. For the trivial $G$-module $\mathbb{Z}$, the zeroth group is $\mathbb{Z}/n\mathbb{Z}$ and the negative first group is zero, so $h_G(\mathbb{Z})=n$.

Now let $L/K$ be a cyclic extension of p-adic fields of degree $n$. For a sufficiently deep principal unit subgroup $U_L^r=1+{\mathfrak{m}}_L^r$, the p-adic logarithm is a $G$-equivariant isomorphism with the additive p-adic lattice ${\mathfrak{m}}_L^r$; it suffices to take $r>v_L(p)/(p-1)$. The normal basis theorem makes $L$ a regular $K[G]$-module. Consequently ${\mathfrak{m}}_L^r$, as a ${\mathbb{Z}}_p[G]$-lattice, is commensurable with a direct sum of $[K:{\mathbb{Q}}_p]$ copies of ${\mathbb{Z}}_p[G]$. These regular lattices have zero Tate groups: its invariants are the multiples of the sum of the basis elements, every such element is a norm, and vectors with coefficient sum zero are images of $D$. Commensurability and the finite-module calculation give $h_G(U_L^r)=1$. Since $U_L/U_L^r$ is finite, $h_G(U_L)=1$. These comparisons also establish finiteness of the Tate groups concerned.

The valuation exact sequence $1\to U_L\to L^{\times }\to \mathbb{Z}\to 0$ now gives $h_G(L^{\times })=n$. Hilbert theorem 90 makes ${\hat{H}}^{-1}(G,L^{\times })$ trivial. One can prove the cyclic statement directly: for $a$ of norm one set $A_0=1$, $A_i=a\sigma (a)\cdots {\sigma }^{i-1}(a)$ and choose $c$ for which $b={\sum }_i{A}_i{\sigma }^i(c)\ne 0$. Such a $c$ exists by linear independence of distinct field automorphisms. Then $\sigma (b)=b/a$, so $a=b/\sigma (b)$ is a coboundary. Therefore the local cyclic norm index isThis is the central use of the Herbrand quotient: it calculates a norm index without first constructing the local reciprocity map.

Let $e,f$ be the ramification index and residue degree. With normalized integer valuations, $v_K(Nx)=f{v}_L(x)$. A norm is a unit exactly when its preimage is a unit, and the norm valuations fill $f\mathbb{Z}$. Hence there is an exact sequenceThe unit norm index is $\boxed{[{\mathcal{O}}_K^{\times }:N{\mathcal{O}}_L^{\times }]=e}$. In an unramified extension all units are norms and the obstruction is the valuation modulo $n$; in a totally ramified cyclic extension the entire index comes from units. The norm subgroup is open: on deep principal units, logarithm carries the norm to the field trace, and the trace of a full p-adic lattice contains a sufficiently deep p-adic lattice in $K$.

Hilbert norm residue symbol and the local-to-global principle. More generally, for a local field $F$ containing ${\mu }_m$, fix the local reciprocity map ${\mathrm{rec}}_F$ with uniformizers acting as arithmetic Frobenius on unramified extensions. The Hilbert norm residue symbol isThis is independent of the chosen root, is bilinear, and has value one exactly when $b$ is a norm from $F(a^{1/m})$. These are standard consequences of Local Artin reciprocity. The case $m=2$, for which the values are signs, is the one directly governing quadratic forms. The quadratic Hilbert symbol at a place $v$ is defined for $a,b\in K_v^{\times }$ bywith value one for all $b$ when $a$ is a square. Equivalently it is one precisely when $z^2=a{x}^2+b{y}^2$ has a nonzero solution over $K_v$. For nonsquare $a$, a solution has $y\ne 0$ and gives $b=(z/y{)}^2-a(x/y{)}^2$; the converse follows from the same norm identity. When $a$ is square the conic is already isotropic. Symmetry follows from this conic criterion. The local cyclic norm index gives a norm subgroup of index two; its sign character is multiplicative in $b$, and symmetry gives multiplicativity in $a$. Thus the symbol is a nondegenerate bilinear pairing on the square-class group of a field, since every nonsquare $a$ gives a nontrivial norm character. Also $(a,-a{)}_v=1$, because $N(\sqrt{a})=-a$, and $(a,a{)}_v=(a,-1{)}_v$.

For an odd-residue-characteristic p-adic field with residue size $q$, write $a={\pi }^{r}u$, $b={\pi }^{s}w$, and let $\chi$ be the quadratic character of the residue units. ThenThe unramified quadratic extension has every unit as a norm and only even norm valuations; in a ramified quadratic extension the norm of a unit has square residue. The preceding unit norm index is two, so the square-residue condition in the ramified case is also sufficient. These facts, together with $N(\sqrt{\pi })=-\pi$, determine the formula on the generators $u,\pi$ of the square-class group. At a real place the symbol is negative exactly when both arguments are negative, and at a complex place it is always one. For ${\mathbb{Q}}_2$, with $u,w$ odd units, the dyadic formula isOnly residue classes modulo eight and the parities of $r,s$ enter; other dyadic fields retain the norm definition.

The Hilbert reciprocity law states ${\prod }_v(a,b{)}_v=1$ for $a,b\in K^{\times }$. Only finitely many factors can be nontrivial: outside the places above two, the Archimedean places, and the finite places where $a$ or $b$ is not a unit, the odd-residue formula gives one. For $K=\mathbb{Q}$ this product formula is a formulation of quadratic reciprocity, including its supplementary laws. It forces local norm obstructions to occur with compatible parity. It is a necessary compatibility law, not by itself a substitute for the following local-to-global theorem.

The Hasse-Minkowski theorem states that a nondegenerate quadratic form over a number field has a nonzero isotropic vector if and only if it does over every completion. Equivalently, two nondegenerate quadratic forms are isometric globally if and only if they are isometric at every place. The forward directions are immediate; the reverse directions are the substantive global theorem. Over non-Archimedean completions, local isometry classes are determined by dimension, determinant square class and the Hasse invariant of a quadratic formAt real places one uses signature, and at complex places dimension suffices. Reciprocity gives ${\prod }_v{ϵ}_v(q)=1$ for a globally diagonalized form. These invariants make the theorem practically usable: local square classes and norm characters replace an unrestricted search for rational solutions. In particular every quadratic form of dimension at least five over a p-adic field is isotropic; for such a form over a number field the only isotropy obstructions are definite signatures at real places.

As a concrete application, take nonsquare $a\in K^{\times }$. Then $b$ is a global norm from $K(\sqrt{a})$ if and only if $(a,b{)}_v=1$ at every place. The local conditions make the ternary form $⟨1,-a,-b⟩$ isotropic everywhere. Hasse-Minkowski theorem gives a global solution of $z^2-a{x}^2=b{y}^2$; since $a$ is nonsquare, $y$ cannot be zero, and division by $y$ gives the global norm. For example $-1$ fails to be a norm from $\mathbb{Q}(i)$ already at the real place. Thus the quadratic Hilbert symbol detects local norm solvability, while the Hasse-Minkowski theorem turns solvability at all places into a global quadratic solution. The reciprocity and local-to-global theorems in this essay are stated as standard results; the norm interpretation, bilinearity and application are derived above.