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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.

The Hasse-Minkowski theorem says that a nondegenerate quadratic form over a number field has a nonzero isotropic vector over that field if and only if it has one over every completion. In particular, over $\mathbb{Q}$ one tests $\mathbb{R}$ and every ${\mathbb{Q}}_{\ell }$. For the present integral form, a rational isotropic vector can be multiplied by a common denominator to give an integer one. This is the Hasse-Minkowski principle.

Write $P=p$, $R=p+12$ and $S=p+24$. All three primes are odd and exceed three: $p=2$ or $3$ would make $p+12$ composite. The second supplementary law for quadratic reciprocity gives

Necessity at $P$. Scale any nonzero vector over ${\mathbb{Q}}_P$ so that its coordinates lie in ${\mathbb{Z}}_P$ and at least one is a unit. Reduction of its equation modulo $P$ gives $y^2=2z^2$, since twelve is invertible modulo $P$. If the Legendre symbol $(2/P)$ is $-1$, this forces $P\mid y,z$. The original equation then forces $P\mid x$, because its other two terms are divisible by $P^2$. This contradicts the normalization. Thus

Sufficiency at the coefficient primes. Suppose $(2/P)=1$. A nonzero solution modulo $P$ is $(x,y,z)=(0,\sqrt{2},1)$. Its $y$-derivative is nonzero modulo $P$, so fixing the other two coordinates and applying the Hensel lemma lifts it to ${\mathbb{Q}}_P$.

Since $S\equiv P(\mathrm{mod}8)$, the same Legendre symbol criterion gives $(2/S)=1$. Modulo $S$, the equation reduces to $y^2=2x^2$, and the vector $(1,\sqrt{2},0)$ is a simple zero in the $y$ coordinate. It lifts to ${\mathbb{Q}}_S$. Modulo $R$, the coefficients of $x^2$ and $z^2$ are respectively $-12$ and $12$, so $(1,0,1)$ is a zero with nonzero $x$-derivative. It lifts to ${\mathbb{Q}}_R$.

The remaining odd primes. If $\ell$ is odd and divides none of $PRS$, the reduced ternary quadratic form is nondegenerate. An elementary finite-field argument finds a zero with $z=1$: if its coefficients are $A,B,C$, the setsboth have $(\ell +1)/2$ elements, so intersect. The resulting vector has nonzero $z$-derivative $2C$, and the Hensel lemma lifts it. This isotropy of nondegenerate ternary quadratic forms over finite fields verifies every remaining odd completion without an extra congruence condition.

The dyadic and real places. Setting $y=1,z=2$ reduces the dyadic problem toThe right side is a 2-adic unit congruent to $-3-4=1$ modulo eight. The unit square classes of the p-adic integers computed above show that it is a square in ${\mathbb{Q}}_2$. This also explains the printed hint: its integer can be taken as $m=-(p+21)/2$. Over $\mathbb{R}$, the positive and negative coefficients make the form indefinite; for instance $x=0,z=1,y=\sqrt{S/R}$ gives a nonzero zero.

All completions now have a nonzero isotropic vector. The Hasse-Minkowski theorem, followed by clearing denominators, proves