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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.

Normalize the discrete valuations so a uniformiser has value one, and write $k_K,k_L$ for the residue fields. The ramification index $e$ is specified by $v_L{|}_K=e{v}_K$, and the residue degree is $f=[k_L:k_K]$. For finite extensions of complete discretely valued fields, $[L:K]=ef$. One way to see the degree equality is that ${\mathcal{O}}_L$ is a finite free ${\mathcal{O}}_K$-module of rank $[L:K]$; reduction modulo a uniformiser of $K$ has $e$ successive quotients isomorphic to $k_L$, hence dimension $ef$ over $k_K$.

An unramified extension has $e=1$ and separable residue extension, equivalently $[L:K]=f$ with separable residue extension. A totally ramified extension has $f=1$, equivalently $[L:K]=e$. The separability condition in the unramified extension definition matters if the residue field is imperfect; it is automatic for finite residue fields.

Suppose $L/K$ is unramified. Choose a primitive element of a field extension $\bar{x}$ for the finite separable extension $k_L/k_K$ and lift it to $x\in {\mathcal{O}}_L$. Since the residue degree of $K(x)/K$ is at least $[k_K(\bar{x}):k_K]=[L:K]$, necessarily $K(x)=L$. Its monic minimal polynomial of an algebraic element $f_x$ has coefficients in ${\mathcal{O}}_K$. Its reduction has degree $[L:K]$ and annihilates $\bar{x}$, whose minimal polynomial over $k_K$ has that same degree. The two coincide, so ${\bar{f}}_x$ is separable.

Conversely, suppose $L=K(x)$, $x\in {\mathcal{O}}_L$, and ${\bar{f}}_x$ is separable. It must be irreducible: otherwise its coprime factors lift by the factorization form of Hensel's lemma, contradicting irreducibility of $f_x$. Hence $\bar{x}$ has degree $[L:K]$ over $k_K$. The degree equality forces $f=[L:K]$ and $e=1$, and the residue extension is separable. We have proved the unramified generator criterion with separable reduction:

Now let $k_L={\mathbb{F}}_q$, $q=p^f$. Every element of $k_L^{\times }$ is a simple root of $X^{q-1}-1$, whose derivative is a unit at each root. Hensel's lemma lifts each of these $q-1$ elements uniquely to a root in ${\mathcal{O}}_L^{\times }$. Thus $L$ contains all $q-1$ roots of $X^{q-1}-1$; these are the nonzero Teichmuller lifts.

More generally, a root of unity of order prime to $p$ reduces injectively into $k_L^{\times }$. Indeed, if such a root reduces to $1$, uniqueness in Hensel's lemma for $X^m-1$ makes it equal to $1$. Consequently every prime-to-$p$ order divides $q-1$.

A root whose order has a nontrivial $p$-part produces a primitive $p$th root of unity $\zeta$. It reduces to $1$ in characteristic $p$. Sinceall the factors have the same positive integral valuation. HenceThis is the ramification bound for a primitive pth root of unity. ThereforeFor $p=2$ the hypothesis $e<1$ cannot occur, so that instance is vacuous rather than a claim excluding the ever-present root $-1$.

For arbitrary positive integers $e,f$, the displayed extension has degree $ef$, ramification index $e$, and residue degree $f$. If $|t|=p^{-1}$, its unique extended absolute value on a field is $|\sum a_n{u}^n|=p^{-\min \{n:a_n\ne 0\}/e}$. A basis consists of ${\gamma }_i{u}^j$, where the ${\gamma }_i$ form an basis of ${\mathbb{F}}_{p^f}$ and $0\le j<e$. Regrouping a Laurent series by its exponent modulo $e$ proves the basis assertion. When $p$ divides $e$ the extension can be inseparable; the valuation calculation still applies.

For an integral generator $\alpha$ of a number field, let $A=\mathbb{Z}[\alpha ]$ and suppose $p$ is coprime to $[{\mathcal{O}}_K:A]$. Localizing at $p$ makes $A$ and the ring of integers equal, so reduction gives the displayed quotient ring. If $\overline{m}=\prod g_i^{e_i}$ with distinct monic irreducible polynomials, the Chinese remainder theorem gives local factors ${\mathbb{F}}_p[X]/(g_i^{e_i})$. Their residue fields have degrees $\mathrm{deg}{g}_i$ and their maximal ideals have nilpotence indices of an ideal $e_i$. Localizing the Dedekind domain ${\mathcal{O}}_K$ at the corresponding prime ideals gives discrete valuation rings, where this same nilpotence index of an ideal is the ramification index of a prime ideal. Thus $(p)=\prod (p,g_i(\alpha ){)}^{e_i}$ and each residue degree is $\mathrm{deg}{g}_i$.

Suppose ${\alpha }^m=a\in {\mathbb{Q}}^{\times }$ in a number field and $\mathfrak{p}$ lies above $p$. Its normalized integral valuation satisfies $m{v}_{\mathfrak{p}}(\alpha )=e_{\mathfrak{p}/p}{v}_p(a)$. If $m$ and $v_p(a)$ are coprime, then $m\mid e_{\mathfrak{p}/p}$. In a degree-$m$ field the fundamental identity for prime decomposition forces a unique prime, ramification index $m$, and residue degree one. This detects total ramification of a prime even when the displayed defining polynomial is not Eisenstein at $p$.

// Target: algebra.bigb

For a finite extension of local fields $L/K$, restriction multiplies the local Brauer invariant by $[L:K]$ and corestriction preserves it. For restriction, the residue-character value multiplies by the residue degree and the parameter valuation by the ramification index. For corestriction, lift the residue character using divisibility of $\mathbb{Q}/\mathbb{Z}$ and use the norm projection formula for cyclic Brauer pairings.

For an unramified extension of local fields $L/K$, an integral element whose residue generates $k_L/k_K$ generates $L/K$: its degree is at least the residue degree $f=[L:K]$. The reduced powers $1,\bar{\alpha },\dots ,{\bar{\alpha }}^{f-1}$ are a basis of the residue field extension. If $x=\sum c_j{\alpha }^j$, factor out a coefficient of least valuation. Reduction of the remaining sum is nonzero by that residue basis, so $v_L(x)={\min }_j{v}_K(c_j)$. Consequently $x$ is integral exactly when all $c_j$ are integral.