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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.

Both forms have discriminant of a binary quadratic form $-60$. The reduction algorithm for a positive definite binary quadratic form gives $|b|\le a\le c$ and $a\le \sqrt{60/3}<5$ for a reduced positive definite representative. Enumerating $a=1,2,3,4$ givesThe second and fourth forms are imprimitive and represent only even integers. Every odd prime represented by a form of discriminant of a binary quadratic form $-60$ must therefore be represented by $f$ or $g$.

For a prime $p\nmid 30$, quadratic reciprocity givesEvery prime $p\equiv 1(\mathrm{mod}15)$ has this symbol equal to one and hence is represented by one of the two primitive forms, by part (b). But an odd prime represented by $g$ and different from three is $2(\mathrm{mod}3)$, so these primes must be represented by $f$. Similarly, primes $p\equiv 2(\mathrm{mod}15)$ have both symbols equal to $-1$, and are represented by $g$, since a prime represented by $f$ is $1(\mathrm{mod}3)$. Both residue classes are coprime to 15, so Dirichlet theorem on primes in arithmetic progressions proves each form represents infinitely many primes.

The same modulo-three distinction excludes every prime other than three from being represented by both. The form $f$ does not represent three; $g$ represents three and five but neither is represented by $f$, and two is represented by neither. Therefore there are no primes represented by both forms.

For an odd prime $p$, the Legendre symbol isGauss lemma says that if $p\nmid a$ and $N$ of the least positive residues ofexceed $p/2$, then $(a/p)=(-1{)}^N$.

For $a=2$, none of $2,4,\dots ,p-1$ needs reduction modulo $p$, and the terms exceeding $p/2$ are those with $j>p/4$. HenceChecking $p\equiv 1,3,5,7(\mathrm{mod}8)$ shows that $N$ has the same parity as $(p^2-1)/8$. Therefore

By quadratic reciprocity, and because $149\equiv 1(\mathrm{mod}4)$,The three factors are $-1,1,1$, respectively, so

Write an integral binary quadratic form as $[a,b,c]=a{x}^2+bxy+c{y}^2$, with discriminant of a binary quadratic form $D=b^2-4ac$. It is positive definite when $a>0$ and $D<0$. Two forms are equivalent if one is obtained from the other by an integral linear change of variables whose matrix lies in ${\mathrm{SL}}_2(\mathbb{Z})$. A positive definite form is reduced whenwith $b\ge 0$ when $|b|=a$ or $a=c$.

To reduce a form, apply a unimodular shear $x\mapsto x+ny$ with $n$ chosen so that the new middle coefficient has $|b^{\prime }|\le a$. If the new final coefficient $c^{\prime }$ is smaller than $a$, exchange the variables, using a determinant-one rotation, so that the positive leading coefficient decreases from $a$ to $c^{\prime }$. Repeating must stop because a positive integer cannot decrease indefinitely. At termination $|b|\le a\le c$, and elementary sign changes impose the boundary convention. This is the reduction algorithm for a positive definite binary quadratic form, and proves that every positive definite form is equivalent to a reduced one.

The displayed forms have discriminant $1-36=-35$, so the intended discriminant is $-35$. For any reduced form of this discriminant, enumeration of reduced binary quadratic forms givesThe boundary convention and direct checking of $a=1,2,3$ leave only

Suppose first that a prime $p>7$ is represented by $f_1$. The identityshows modulo both $5$ and $7$ that $4p$ is a nonzero square. HenceSimilarly,Modulo $5$, this says $2p$ is a square, while modulo $7$ it says $5p$ is a square. Since $(2/5)=(5/7)=-1$, the multiplicativity of the Legendre symbol gives

Conversely, suppose the two displayed Legendre symbols have the same sign. By quadratic reciprocity,Choose an odd integer $B$ satisfying $B^2\equiv -35(\mathrm{mod}p)$. Since $B$ is odd, $B^2+35$ is also divisible by $4$, sois a positive definite integral form of discriminant $-35$ that represents $p$. Reducing $F$ yields either $f_1$ or $f_2$, and equivalence preserves represented integers. The necessary symbol calculation above determines which one: the positive pair selects $f_1$, and the negative pair selects $f_2$. Therefore

If the positive odd integer $n$ has prime factorizationits Jacobi symbol iswhere each factor is a Legendre symbol; by convention $(a/1)=1$. Thus the symbol is zero exactly when $\gcd (a,n)>1$.

The quadratic reciprocity law for the Jacobi symbol says that for coprime positive odd integers $m,n$,To prove it, write $m={\prod }_i{p}_i^{e_i}$ and $n={\prod }_j{q}_j^{f_j}$. The quadratic reciprocity law for distinct odd primes statesMultiplying this identity over $i,j$ with multiplicities $e_i{f}_j$ gives the required product of Jacobi symbols. Its sign exponent modulo two isbecause multiplication of odd integers adds the quantities $(r-1)/2$ modulo two. This proves the law.

For the requested value, two applications of reciprocity givesince $2019\equiv 7(\mathrm{mod}503)$, $503\equiv 6(\mathrm{mod}7)$, and $6$ is a quadratic nonresidue modulo $7$.