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Let $m,n$ be coprime square-free integers, $m\equiv 3(\mathrm{mod}4)$ and $n\equiv 2(\mathrm{mod}4)$. For the biquadratic extension $L=\mathbb{Q}(\alpha ,\beta )$, with ${\alpha }^2=m$ and ${\beta }^2=n$, an integral basis is $1,\alpha ,\beta ,(\beta +\alpha \beta )/2$. To prove spanning, write an algebraic integer as $(a+b\alpha +c\beta +d\alpha \beta )/2$. The field traces to the three quadratic subfields and the ring of integers of a quadratic field formula force $a,b,c,d\in \mathbb{Z}$. The field norm to $\mathbb{Q}(\alpha )$ has coefficients $(a^2+m{b}^2-n{c}^2-mn{d}^2)/4$ and $(ab-ncd)/2$. Their integrality gives $ab\equiv 0(\mathrm{mod}2)$ and $a^2-b^2-2(c^2+d^2)\equiv 0(\mathrm{mod}4)$. Thus $a,b$ are even and $c\equiv d(\mathrm{mod}2)$, exactly the asserted integer span. Conversely, the half-sum $w$ has $w^2=n(1+m)/4+(n/2)\alpha \in \mathbb{Z}[\alpha ]$, so integral dependence is transitive proves it is an algebraic integer. Rational linear independence is immediate from its nonzero $\alpha \beta$ coefficient.

Let $m,n$ be coprime square-free integers, $m\equiv 2(\mathrm{mod}4)$ and $n\equiv 1(\mathrm{mod}4)$, with $n\ne 1$. Put ${\alpha }^2=m$, ${\beta }^2=n$, and $\omega =(1+\beta )/2$. Then $1,\alpha ,\omega ,\alpha \omega$ is an integral basis of $\mathbb{Q}(\alpha ,\beta )$. All four entries are algebraic integers. For spanning, field traces to the three quadratic subfields show that an arbitrary algebraic integer has form $(a+b\alpha +c\beta +d\alpha \beta )/2$ with $a,b,c,d\in \mathbb{Z}$. Its field norm to $\mathbb{Q}(\beta )$ is $A/4+(ac-mbd)\beta /2$, where $A=a^2+n{c}^2-m{b}^2-mn{d}^2$. The ring of integers of a quadratic field formula forces $A$ even and $A\equiv 2ac-2mbd(\mathrm{mod}4)$. These give $a\equiv c(\mathrm{mod}2)$ and then $b\equiv d(\mathrm{mod}2)$. The element is consequently $(a-c)/2+((b-d)/2)\alpha +c\omega +d\alpha \omega$. The coefficient determinant $1/4$ proves rational linear independence.

The power basis $1,\theta ,{\theta }^2$ is an integral basis. The defining polynomial is irreducible by Eisenstein's criterion at two and its discriminant is $-140$. The discriminant-index formula for an integral lattice restricts its index in the ring of integers to one or two. An index of two would supply an algebraic integer $\eta =(a+b\theta +c{\theta }^2)/2$ with $a,b,c\in \{0,1\}$ not all zero. Its field trace $(3a-4c)/2$ forces $a=0$. If $b=0$ this gives ${\theta }^2/2$ directly; if $b=1$, the identity $\theta \eta ={\theta }^2/2-c\theta -c$ gives the same integrality. But ${\theta }^2/2=-1-{\theta }^{-1}$, while $N({\theta }^{-1})=-1/2$ excludes integrality by the field norm. Thus the index is one, proving the claim.

In $L={\mathbb{Q}}_p(\theta )$ with ${\theta }^2=p$, the Eisenstein polynomial $X^2-p$ gives $v_L(\theta )=1$ and $v_L(p)=2$. The element $x=\theta /p$ has field trace zero, but valuation $-1$, so is not in the valuation ring. Cancellation in a field trace contrasts with the multiplicative test in unit norm detects units in a local extension.

Write $x=r+s\sqrt{d}$ with $r,s\in \mathbb{Q}$. The field trace and field norm areThe element is an algebraic integer if and only if $T,N\in \mathbb{Z}$: necessity follows from its two conjugates; sufficiency follows from the monic polynomial $X^2-TX+N$.

Suppose $x$ is an algebraic integer. Set $a=2r\in \mathbb{Z}$. ThenWrite $2s=h/k$ in lowest terms, with $k>0$. Then $k^2\mid d$, and the hypothesis that $d$ is a square-free integer forces $k=1$. Thus $b=2s\in \mathbb{Z}$, and the remaining condition isThe only squares modulo four are zero and one. For $d\equiv 2(\mathrm{mod}4)$, an odd $b$ is impossible, after which $a$ is even. For $d\equiv 3(\mathrm{mod}4)$, an odd $b$ would require $a^2\equiv 3(\mathrm{mod}4)$, again impossible. For $d\equiv 1(\mathrm{mod}4)$, the condition is precisely $a\equiv b(\mathrm{mod}2)$.

Conversely, $\sqrt{d}$ satisfies $X^2-d$, and when $d\equiv 1(\mathrm{mod}4)$, the element $\omega =(1+\sqrt{d})/2$ satisfiesTheir integer spans therefore consist of algebraic integers. In the last case,This proves the ring of integers of a quadratic field formula. The required integral basis isTheir rational linear independence follows from $\sqrt{d}\notin \mathbb{Q}$.

Put $\alpha =\sqrt{m_1}$, $\beta =\sqrt{m_2}$, andThe proposed integral basis for coprime biquadratic radicands congruent to two and one isEach entry is an algebraic integer, by the ring of integers of a quadratic field formula and closure under products. Their coefficient matrix in $1,\alpha ,\beta ,\alpha \beta$ has determinant $1/4$, so they are rationally linearly independent. It remains to show that they span every algebraic integer.

Write $\gamma =(a+b\alpha +c\beta +d\alpha \beta )/2$ initially with rational $a,b,c,d$. The field traces to $\mathbb{Q}(\alpha )$ and $\mathbb{Q}(\alpha \beta )$ are $a+b\alpha$ and $a+d\alpha \beta$. Since $m_1\equiv m_1{m}_2\equiv 2(\mathrm{mod}4)$, the corresponding integral basis in each subfield gives $a,b,d\in \mathbb{Z}$. The field trace to $\mathbb{Q}(\beta )$ is $a+c\beta$. Membership in $\mathbb{Z}[\omega ]$ says that $2a,2c$ are integers of equal parity. As $a$ is already an integer, this forces $c\in \mathbb{Z}$ too.

Now the field norm to $\mathbb{Q}(\beta )$ isBy the ring of integers of a quadratic field formula, $A/2$ is an integer and has the same parity as $ac-m_{1}bd$. Thus $A$ is even, giving $a\equiv c(\mathrm{mod}2)$, andUsing $m_1\equiv 2$, $m_2\equiv 1$ and $a\equiv c(\mathrm{mod}2)$, this becomesHence $b\equiv d(\mathrm{mod}2)$. Finally,an expression with integer coefficients. This proves the asserted integral basis without assuming a general theorem about products of elements of an integral basis.

Again put $\alpha =\sqrt{m_1}$ and $\beta =\sqrt{m_2}$. The ring of integers of each of the three quadratic subfields has integral basis $1,\alpha$, $1,\beta$, or $1,\alpha \beta$, respectively: their square-free radicands are respectively congruent to $3,2,2$ modulo four. The field traces of the algebraic integer $\gamma$ to those subfields areThey are algebraic integers, so comparison with each integral basis givesThis use of all three subfields is needed to control $d$ as well as the other coefficients.

The field norm to $\mathbb{Q}(\alpha )$ isIts two coefficients must be integers. The second condition says $ab$ is even, since $m_2$ is even. Reducing the numerator of the first coefficient modulo four givesReduction modulo two first gives $a\equiv b(\mathrm{mod}2)$. Since $ab$ is even, both $a$ and $b$ are even. The same congruence now gives $c^2+d^2\equiv 0(\mathrm{mod}2)$, or $c\equiv d(\mathrm{mod}2)$. Therefore every algebraic integer has the formwith four integer coefficients.

For the converse, $1,\alpha ,\beta$ are algebraic integers, andHere $4\mid 1+m_1$ and $2\mid m_2$, so both coefficients are integers. The monic polynomial $X^2-w^2$ and integral dependence is transitive show that $w$ is an algebraic integer. The four displayed generators are rationally linearly independent, since their coefficient matrix in $1,\alpha ,\beta ,\alpha \beta$ has determinant $1/2$. Hence the integral basis for coprime biquadratic radicands congruent to three and two isIn this formula $\sqrt{m_1{m}_2}$ denotes the product $\alpha \beta$ of the chosen square roots; this convention also handles negative radicands.

Put $\alpha =\sqrt{m_1}$, $\beta =\sqrt{m_2}$ and $F=\mathbb{Q}(\alpha )$. The hypotheses make $L/\mathbb{Q}$ a biquadratic extension. To check the degree of a field extension is four, if $\beta =x+y\alpha$ with rational $x,y$, squaring gives $xy=0$. The case $y=0$ would make $m_2$ a rational square; the case $x=0$ would make $m_2/m_1$ a rational square. Neither can occur for the specified distinct coprime square-free integers. Hence $[L:F]=2$.

The nonidentity field automorphism of $L/F$ fixes $\alpha$ and sends $\beta$ to $-\beta$. Thus, for $\gamma =(r+s\alpha )+(t+u\alpha )\beta$, the field norm and field trace areIf $\gamma$ is an algebraic integer, so is its conjugate, and their sum and product are algebraic integers. Conversely, put $T={\mathrm{Tr}}_{L/F}(\gamma )$ and $N=N_{L/F}(\gamma )$. The identityshows that $\gamma$ is integral over $\mathbb{Z}[T,N]$. If $T,N$ are algebraic integers, the ring generated by algebraic integers $\mathbb{Z}[T,N]$ is integral over $\mathbb{Z}$. Since integral dependence is transitive, $\gamma$ is an algebraic integer. This proves the trace and norm integrality criterion for a quadratic extension, including elements already in $F$.

Let $\alpha =\sqrt[3]{3}$. The polynomial $X^3-3$ is Eisenstein at three, so the number field has degree of a field extension three. Use the permitted ring of integers $R=\mathbb{Z}[\alpha ]$, whose integral basis is $1,\alpha ,{\alpha }^2$. The field traces $\mathrm{Tr}(1)=3$, $\mathrm{Tr}(\alpha )=\mathrm{Tr}({\alpha }^2)=0$, $\mathrm{Tr}({\alpha }^3)=9$, and $\mathrm{Tr}({\alpha }^4)=0$ giveThere is one real field embedding and one conjugate pair of nonreal field embeddings. The Minkowski bound for ideal classes is thereforeEvery ideal class has an integral ideal representative of ideal norm at most four, so it is enough to show that every prime ideal of ideal norm at most four is a principal ideal.

Modulo two,where the quadratic factor is an irreducible polynomial over the finite field ${\mathbb{F}}_2$. The Dedekind factorization theorem givesNowThus $(\alpha -1)=P_2$, by containment and equality of ideal norms, and cancellation in the second identity gives $({\alpha }^2+\alpha +1)=Q_2$. This explicitly handles the prime ideal of ideal norm four, as well as the one of ideal norm two.

Modulo three the polynomial is $X^3$, soThese exhaust the prime ideals of ideal norm at most four. Every bounded representative factors into the listed principal ideals, and hence is a principal ideal. Therefore the Class group of Q of cube root of three is

Set $L=K_{\mathfrak{P}}$ and $F=k_{\mathfrak{p}}$, with valuation rings $S,R$. Let $e$ be the ramification index, $f$ the residue-field degree, and $p$ the residue characteristic. A base uniformizer $\pi$ generates ${\mathfrak{P}}^e$ in $S$, up to a unit. The free $R$-module $S$ has rank $ef$. For an integral $a\in S$, its field trace is the matrix trace of multiplication by $a$ on this module. Reduction modulo $\pi$ preserves the matrix trace.

The quotient $S/\pi S$ has a filtrationEach of the $e$ successive quotients is a one-dimensional vector space over ${\kappa }_L=S/\mathfrak{P}$. On each, multiplication by $a$ acts as multiplication by its residue $\bar{a}$, whose trace over ${\kappa }_F=R/\pi R$ is ${\mathrm{Tr}}_{{\kappa }_L/{\kappa }_F}(\bar{a})$. A basis adapted to the filtration makes the multiplication matrix block triangular, so its trace is the sum of these quotient traces. This proves trace reduction in a ramified local extension:Since the hypothesis says $p\mid e$, the right side is zero in ${\kappa }_F$. ThusNo normality hypothesis is needed. The original PDF only requires divisibility of $e$, not divisibility exactly once; the converted TeX has corrupted that condition and several local subscripts.

Use $R={\mathcal{O}}_k$, $S={\mathcal{O}}_L$, $T={\mathcal{O}}_K$, and denote their relative codifferents by $C_{L/k}$, $C_{K/L}$ and $C_{K/k}$. Transitivity of the field trace givesFor the forward direction, testing an element $t\in T$ against every $s\in S$ is legitimate because $st\in T$. For the reverse direction, take $s=1$ and use trace transitivity.

Localize at a nonzero base prime ideal. The intermediate ring is semilocal Dedekind domain, so its invertible fractional ideal $C_{L/k}$ is principal there, say $C_{L/k}=d^{-1}S$ locally. This principality follows by choosing, with the Chinese remainder theorem for ideals, a generator at each of its finitely many maximal ideals. The displayed criterion now becomesInvert these fractional ideals using Solution 1. Equality at every base localization gives the global different in a tower formulaThe second factor is extended to the upper ring of integers; that extension is implicit in the printed formula. At a local prime the exponents consequently satisfy $d(K/k)=d(K/L)+e(K/L)d(L/k)$.

The original PDF has cube roots throughout this question; the ninth roots in the converted TeX are transcription errors. Put $\alpha =\sqrt[3]{m_1{m}_2^2}$ and $\beta ={\alpha }^2/m_2$. The given condition excludes the trivial cube case. Both generators are integral, becauseMoreoverso $A=\mathbb{Z}+\mathbb{Z}\alpha +\mathbb{Z}\beta$ is an order in the cubic field. The field traces of $\alpha ,\beta ,{\alpha }^2,{\beta }^2$ are zero, while $\mathrm{Tr}(\alpha \beta )=3m_1{m}_2$. Thus the trace pairing matrix of this basis isWe must prove that this is the maximal order, not just compute the discriminant of a suborder. The discriminant-index formula for an integral lattice restricts any index prime to those dividing $3m_1{m}_2$. If $p\mid m_1$, the cubic for $\alpha$ is Eisenstein at $p$, while $m_2$ is a unit there and $A\otimes {\mathbb{Z}}_p={\mathbb{Z}}_p[\alpha ]$. If $p\mid m_2$, use instead the Eisenstein polynomial for $\beta$, and $\alpha ={\beta }^2/m_1$. In either case an Eisenstein generator is a uniformizer of the totally ramified local field and generates its full valuation ring. Indeed expansion in powers of that uniformizer using base residue representatives, and grouping powers modulo three, expresses every integral local element in its power basis.

These arguments also cover three when $3\mid m_1{m}_2$. If $3\nmid m_1{m}_2$, every unit modulo nine has cube congruent to $1$ or $-1$. Therefore the condition $m_1\not\equiv \pm m_2(\mathrm{mod}9)$ implies $m_1{m}_2^2\not\equiv \pm 1(\mathrm{mod}9)$: multiply a contrary congruence by the inverse of $m_2^2$ and use $m_2^3\equiv \pm 1$. Choose $r\in \{1,-1\}$ with $r^3\equiv m_1{m}_2^2(\mathrm{mod}3)$. The polynomialis Eisenstein at three, since its constant term is divisible by three but not nine. Its root $\alpha -r$ again generates the full local valuation ring. There is no index prime left. We have proved the integral basis of a cubefree pure cubic field and the requested discriminant:For $K=\mathbb{Q}(\sqrt[3]{6})$ write ${\alpha }^3=6$. Here ${\mathcal{O}}_K=\mathbb{Z}[\alpha ]$ and $d_K=-972$. The norm of $x+y\alpha +z{\alpha }^2$ is the determinant of its multiplication matrix, givingIn particularAn integral element of norm one has an integral inverse, by its monic characteristic polynomial. Here direct multiplication gives $u^{-1}=109+60\alpha +33{\alpha }^2$. Nontriviality also follows from linear independence of $1,\alpha ,{\alpha }^2$.

The signature is $(1,1)$, so the Minkowski bound for ideal classes isEvery class therefore has an integral representative of norm at most eight. Its prime factors have norms at most eight, and lie above $2,3,5,7$. Since the index of $\mathbb{Z}[\alpha ]$ is one, reduction of $X^3-6$ gives their factorizations by the Kummer-Dedekind theorem. At two and three the polynomial is $X^3$, so there is one prime of norm $2$ or $3$, respectively. They are principal, becauseAt five the factorization is $(X-1)(X^2+X+1)$ with the quadratic irreducible. The norm-five prime is $(\alpha -1)$ because $N(\alpha -1)=5$; the norm-twenty-five prime is outside the bound.

At seven the roots are $3,5,6$, giving three primes ${\mathfrak{q}}_r=(7,\alpha -r)$ of norm seven. The elements $\alpha +1$ and ${\alpha }^2+\alpha -5$ have norm seven and vanish at roots six and three, respectively; hence they generate ${\mathfrak{q}}_6$ and ${\mathfrak{q}}_3$. Since ${\mathfrak{q}}_3{\mathfrak{q}}_5{\mathfrak{q}}_6=(7)$, the remaining prime is principal too. Every prime which can divide a Minkowski representative is now principal, soThis proves class number one for Q of cube root six with all primes within the bound accounted for.

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.

Take $B$ to be linear in its first argument and conjugate-linear in its second, with $\bar{a}=a^q$. For a Hermitian form over a quadratic finite field, $B(v,v)\in {\mathbb{F}}_q$. We first establish the quadratic finite-field norm and trace surjectivity needed for orthonormalization of a finite-field Hermitian form. The field norm $N(a)=a^{q+1}$ maps ${\mathbb{F}}_{q^2}^{\ast }$ onto ${\mathbb{F}}_q^{\ast }$ because the multiplicative group of a finite field is cyclic of order $q^2-1$. The field trace $\mathrm{Tr}(a)=a+a^q$ is an ${\mathbb{F}}_q$-linear map into ${\mathbb{F}}_q$. It is nonzero: a nonzero polynomial of degree $q$ cannot vanish at all $q^2$ elements. It is therefore onto, including in characteristic two.

There must be $v$ with $B(v,v)\ne 0$. Otherwise, for any $v,w,a$, expanding the diagonal value givesIf $B(v,w)\ne 0$, varying $a$ contradicts the surjectivity of the field trace. Thus all pairings would vanish, contrary to nondegeneracy. Choose $a$ with $N(a)=B(v,v{)}^{-1}$; then $v_1=av$ has norm one. Every $w$ decomposes uniquely asThe restricted Hermitian form over a quadratic finite field on the orthogonal complement for a sesquilinear form is nondegenerate: a vector there orthogonal to that complement is also orthogonal to $v_1$, hence to all of $V$, and is zero. Induction produces an orthonormal basis $v_1,\dots ,v_n$.

The unitary group over a finite field is the group of invertible ${\mathbb{F}}_{q^2}$-linear maps preserving $B$. In the orthonormal basis just constructed, its concise description isThe notation here uses the size of the matrix field; it is also frequently written $U_n(q)$. There is no determinant-one condition in this definition.

Let $\pi \in K$ be a uniformizer. Because $L/K$ is unramified, the same $\pi$ is a uniformizer of $L$, and $v_L{|}_K=v_K$. The conjugates of $z\in L^{\times }$ all have the same valuation, soThis proves that every field norm has valuation divisible by $n$. Conversely $N_{L/K}({\pi }^r)={\pi }^{nr}$ for every integer $r$, so it remains to prove surjectivity of unit norms in unramified extensions.

Write $|k_K|=Q$, so $|k_L|=Q^n$. Reduction of the field norm on units is the finite-field normThe multiplicative group of a finite field is cyclic. The displayed power map on a group of order $Q^n-1$ has image of order $Q-1$, and is therefore onto $k_K^{\times }$. Given a unit $u\in {\mathcal{O}}_K^{\times }$, first choose $b_1\in {\mathcal{O}}_L^{\times }$ with $N(b_1)\equiv u(\mathrm{mod}\pi )$.

We can correct the norm at each subsequent power of $\pi$. For $r\ge 1$ and $a\in {\mathcal{O}}_L$, expanding the product over the cyclic Galois group givesAll products involving at least two correction terms are divisible by ${\pi }^{2r}$, hence by ${\pi }^{r+1}$. The residue field trace is onto even if the residue characteristic divides $n$. Indeed its polynomial $T(X)=X+X^Q+\cdots +X^{Q^{n-1}}$ is nonzero of degree $Q^{n-1}<Q^n$; it cannot vanish at every element of $k_L$. Its values lie in $k_K$, since raising $T(a)$ to $Q$ cyclically permutes the summands. Thus it is a nonzero $k_K$-linear map to the one-dimensional space $k_K$, and is surjective.

Suppose $N(b_r)\equiv u(\mathrm{mod}{\pi }^r)$. Choose ${\bar{a}}_r$ whose trace is the coefficient of ${\pi }^r$ in $u/N(b_r)-1$, and put $b_{r+1}=b_r(1+{\pi }^r{a}_r)$. Then $N(b_{r+1})\equiv u(\mathrm{mod}{\pi }^{r+1})$, and $b_{r+1}\equiv b_r(\mathrm{mod}{\pi }^r)$. By completeness the units $b_r$ converge to a unit $b$. The field norm is continuous, being a polynomial in coordinates of multiplication, so $N(b)=u$. Multiplying such a unit by a power of $\pi$ provesThe proof uses no local class field theory, and works in either characteristic.

False. This is an example of integral trace need not imply integrality. Take $K={\mathbb{Q}}_p(\theta )$ with ${\theta }^2=p$, the splitting field of $X^2-p$. Its two conjugates are $\theta ,-\theta$, and its normalized discrete valuation satisfies $v_K(\theta )=1$, $v_K(p)=2$, by the Eisenstein polynomial criterion. Forthe field trace is zero, but $v_K(x)=-1$. Therefore $x$ is not in the valuation ring, despite having integral field trace. Cancellation in a sum of conjugates explains why a field trace test fails.

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.