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For a finite point $p$, use the valuation $v_p$ with local parameter $u=t-p$; at infinity use $u=1/t$. Its extensions $w$ to $E$ are permuted transitively by $G$. The completed residue field is $F$, because $F$ is algebraically closed, so the local decomposition group equals its inertia group.

Write $e=e(w/v_p)$. Characteristic zero makes all local ramification tame. One can choose a uniformizer $\pi$ in the completion for whichIndeed, initially $u$ is ${\pi }^e$ times a unit; that unit has an $e$th root in $F[[\pi ]]$, by its algebraically closed residue field and the invertibility of $e$. Absorb it into $\pi$. The local automorphisms are then $\pi \mapsto \zeta \pi$, $\zeta \in {\mu }_e(F)$, and form a cyclic group of order $e$. Define $e_p=e$.

To specify an element rather than a subgroup, fix a compatible choice of primitive roots of unity ${\zeta }_e\in F$. Define ${\tau }_w$ by its uniformizer characterand let $C_p$ be its conjugacy class in $G$. Over $\mathbb{C}$, take ${\zeta }_e=e^{2\pi i/e}$, corresponding to positive local continuation around $p$. At an unramified point, $e_p=1$ and $C_p=\{1\}$.

Changing the uniformizer does not affect this character. If ${\pi }^{\prime }=a\pi$ with $a$ a unit, inertia fixes its residue, soThe character is injective: a nonidentity finite-order automorphism with derivative one would have $\tau (\pi )=\pi +b{\pi }^r+\cdots$ for $r\ge 2$, and its $N$th iterate would have coefficient $Nb$ at that first nonzero displacement. Characteristic zero contradicts finite order. Thus the chosen primitive character identifies a unique generator.

For another place $w^{\prime }=\sigma w$, conjugation gives $D_{w^{\prime }}=\sigma D_w{\sigma }^{-1}$, preserves its order, and carries ${\tau }_w$ to ${\tau }_{w^{\prime }}$, since $\sigma$ fixes the constants and hence ${\zeta }_e$. This proves that $e_p$ and the branch-cycle conjugacy class are independent of the place and uniformizer. Without the stated roots-of-unity convention, only the cyclic inertia-subgroup conjugacy class is intrinsic: replacing ${\zeta }_e$ by another primitive root replaces the generator by a coprime power, which need not lie in the same conjugacy class.

Put $u=t-p$ and write $a(t)=u^{r}b(u)$, where $b(0)\ne 0$. The unit $b$ has a square root in $\mathbb{C}[[u]]$. Dividing the radical by that square root and by $u^{\lfloor r/2\rfloor }$ reduces the completed extension toIf $r$ is even, it is unramified and locally splits; if $r$ is odd, adjoining $u^{1/2}$ gives ramification index two. At points where $a(p)\ne 0$, $r=0$ and there is no branch. ThereforeFor infinity set $u=1/t$. If $D=\mathrm{deg}a$,The same quadratic branch parity criterion applies to the negative valuation $-D$, soThe generic irreducibility excludes $a$ being a square. Local splitting at the even-order places is compatible with that global irreducibility. The number of odd finite orders plus the possible odd infinity order is even, as also follows from the sum of orders of a rational function being zero.

The supplied integral equation has elliptic-curve discriminant $\Delta =-11$. It therefore has good reduction at every $p\ne 11$. At eleven its discriminant has valuation one. Under a change of Weierstrass equation of an elliptic curve, discriminant valuations differ by a multiple of twelve. A good integral model would have valuation zero, which is impossible. This is the discriminant valuation obstruction to good reduction; it also shows the displayed model is minimal at eleven. Hence

For a local field $F$ containing the Nth roots of unity, fix the Local Artin map with arithmetic Frobenius convention. The Hilbert norm residue symbol can be defined byIt is independent of the chosen root, continuous and bimultiplicative, factors through $F^{\times }/(F^{\times }{)}^N$ in each variable, and gives a nondegenerate pairing on that finite group. It satisfies $(a,b{)}_{F,N}=(b,a{)}_{F,N}^{-1}$ and $(a,1-a{)}_{F,N}=1$ for $a\ne 1$. Its norm interpretation isFor $N=2$ the quadratic Hilbert symbol is symmetric and sign-valued. It is also characterized by the existence of a nonzero solution of $z^2=a{x}^2+b{y}^2$. Moreover $(a,-a{)}_F=1$, and therefore $(a,a{)}_F=(a,-1{)}_F$. We now compute both the square-class groups and the pairing explicitly.

For odd $p$, write $a=p^{\alpha }u$ with $u\in {\mathbb{Z}}_p^{\times }$. An element is a square exactly when $\alpha$ is even and the residue of $u$ is a square in ${\mathbb{F}}_p^{\times }$. Necessity follows by reduction; sufficiency follows from the simple-root form of Hensel lemma applied to $X^2-u$, since $2x$ is then a unit. If $\eta$ is any unit with nonsquare residue, this givesThe nonsquare unit defines an unramified quadratic extension. Its field norms include every unit and have even valuation, as follows from the local norm calculation below. Thus $(\eta ,u{)}_p=1$ for every unit, but $(\eta ,p{)}_p=-1$. Finally $(p,p{)}_p=(p,-1{)}_p=(-1/p)$, by symmetry and the preceding unit calculation. Bimultiplicativity now proves explicit quadratic Hilbert symbols over the p-adic numbers:Here the factors are Legendre symbols of the nonzero residues. The formula works for arbitrary integer valuations $\alpha ,\beta$, since only their parity matters.

At $p=2$, the square of every odd integer is one modulo eight. Conversely, for an odd $u\equiv 1(\mathrm{mod}8)$, apply the strong form of Hensel lemma to $X^2-u$ at $1$: the error has valuation at least three, while the derivative has valuation one. Thus the unit square classes of the p-adic integers are precisely their four residues modulo eight. Combining this with parity of the valuation givesFor an odd unit putInspection of the residues $1,3,5,7$ gives $[u]=[-1{]}^{ϵ(u)}[5{]}^{\omega (u)}$. It remains to compute six basis pairings. Explicit field norms giveindeed $2=1^2+1^2$, $5=1^2+2^2$, $2=2^2-2\cdot 1^2$, and $5=5^2-5\cdot 2^2$.

In contrast $(-1,-1{)}_2=-1$. To justify the obstruction even for rational 2-adic coordinates, write $x=2^t{x}_0$, $y=2^t{y}_0$ with $x_0,y_0$ integral and at least one odd. Then $v_2(x_0^2+y_0^2)$ is zero if exactly one is odd, and one if both are odd. A sum of squares equal to the unit $-1$ consequently forces $t=0$ and exactly one coordinate odd. Its residue modulo four would then be one, whereas $-1$ is three.

Similarly $(2,5{)}_2=-1$. If $x^2-2y^2$ is a unit, the valuations of the two terms have opposite parity, so cannot cancel. It follows that $x$ is an odd integral element and $y$ is integral. Modulo eight its norm is then $1$ or $7$, never $5$. These signs, symmetry, and bimultiplicativity giveThe two boxed formulas determine the quadratic Hilbert symbol on all nonzero elements, including every rational prime $p$.

Write $\mathcal{O}$ for the valuation ring of $K$, $\pi$ for a uniformizer, $v(\pi )=1$, and $k={\mathbb{F}}_q$ for the residue field. A useful strong form of Hensel lemma is the following: if $f\in \mathcal{O}[X]$ and $a_0\in \mathcal{O}$ satisfythen there is a root $a\in \mathcal{O}$ with $v(a-a_0)>v(f^{\prime }(a_0))$, unique among roots in that ball. If $f(a_0)\ne 0$, its distance satisfies $v(a-a_0)=v(f(a_0))-v(f^{\prime }(a_0))$.

Here is a proof by successive correction. Put $s=v(f^{\prime }(a_0))$, $d_j=v(f(a_j))$, andStop if a root has already been reached. Otherwise the correction has valuation $d_j-s>s$. Integral Taylor expansions show that $v(f^{\prime }(a_{j+1}))=s$ andThus $d_j-2s$ grows at least geometrically, the corrections tend to zero, and completeness gives a limit $a$ with $f(a)=0$. The first correction has strictly smaller valuation than every later nonzero correction, proving the asserted distance. If $x,y$ are in the ball $v(x-a_0),v(y-a_0)>s$, thenThe second factor cannot vanish. Two roots in the ball must therefore coincide. This proves Hensel lemma; in particular a simple root modulo $\pi$ has a unique lift with that residue.

We use this to establish existence and uniqueness of unramified local extensions. For each integer $d\ge 1$, choose a monic irreducible polynomial $\bar{f}\in k[X]$ of degree $d$ and lift it to a monic $f\in \mathcal{O}[X]$. Reduction and Gauss lemma show that $f$ remains irreducible over $K$. LetThe ring $A$ is finite free and complete over $\mathcal{O}$, and $A/\pi A\cong {\mathbb{F}}_{q^d}$. An element outside $\pi A$ is invertible: lift an inverse from the residue field and invert the remaining $1+\pi z$ by a convergent geometric series. Every nonzero element of $A$ is ${\pi }^t$ times such a unit, since ${\bigcap }_t{\pi }^{t}A=0$ in a finite free module. Thus $A$ is a discrete valuation ring, its fraction field is $L$, andThis constructs an unramified extension of each degree.

Conversely, let $M/K$ be an unramified extension of degree $d$. Its residue field is the unique degree-$d$ finite field extension ${\mathbb{F}}_{q^d}/{\mathbb{F}}_q$. Choose a residue root of $\bar{f}$ in it. The root is simple, so Hensel lemma lifts it to a root of $f$ in $M$. This embeds $L$ into $M$, and equality of degrees makes it an isomorphism. All $d$ residue conjugates lift in $L$ itself; hence $L/K$ is Galois, and reduction identifies its Galois group with the cyclic Galois group of the residue field extension. Its distinguished Frobenius automorphism reduces to $x\mapsto x^q$.

There is exactly one unramified extension $K_d$ of each degree inside a fixed algebraic closure, andFor the middle equality, every nonzero element of ${\mathbb{F}}_{q^d}$ lifts uniquely to a $(q^d-1)$st root of unity, because that exponent is prime to $p$. A primitive such residue generates the finite field extension, so its lift generates $K_d$. Taking the union gives the maximal unramified extension, with Galois group $\hat{\mathbb{Z}}$.

We give the full local-field norm-index argument. Let $G=⟨\sigma ⟩$ be cyclic of order $n$. On an additive $G$-module set $D=\sigma -1$ and $N=1+\sigma +\cdots +{\sigma }^{n-1}$. The two cyclic Tate cohomology groups areWhen these are finite, the Herbrand quotient is $h_G(M)=|{\hat{H}}^0(G,M)|/|{\hat{H}}^{-1}(G,M)|$. The periodic complex alternating $D$ and $N$ yields a six-term periodic exact sequence from a short exact sequence of modules. Taking alternating products of orders in that exact sequence provesFor a finite module, $|NM|=|M|/|\ker N|$ and $|DM|=|M|/|\ker D|$ show that the two Tate cohomology groups have equal order. Thus the Herbrand quotient of a finite module is one. For the trivial integral module $\mathbb{Z}$, the two groups are $\mathbb{Z}/n\mathbb{Z}$ and zero, giving $h_G(\mathbb{Z})=n$.

We also need cyclic cohomology of a regular lattice. In $R[G]$, fixed coefficient vectors are constant, and every such vector is a norm of a basis vector times its coefficient. The kernel of $N$ consists of vectors whose coefficient sum is zero. For such a vector $(a_i)$, solve $b_{i-1}-b_i=a_i$ successively around the cycle; the final equation holds precisely because ${\sum }_i{a}_i=0$. Thus $\ker N=DR[G]$. Both Tate cohomology groups vanish, so $h_G(R[G])=1$, for $R=\mathbb{Z}$ or ${\mathbb{Z}}_p$, and likewise for finite direct sums.

Now take a cyclic extension $L/K$ of p-adic fields of degree $n$. Write $U_L^r=1+{\mathfrak{p}}_L^r$. For $r>v_L(p)/(p-1)$, the convergent p-adic logarithm and p-adic exponential give mutually inverse $G$-equivariant group isomorphismswhere the right side is additive. The convergence assertion follows from the estimates $v_L(x^j/j)\ge j{v}_L(x)-v_L(p)v_p(j)$ and $v_L(x^j/j!)\ge j{v}_L(x)-v_L(p)j/(p-1)$; formal inverse identities therefore hold on this ball.

By the normal basis theorem, choose $\theta$ so that $\{{\sigma }^i\theta \}$ is a $K$-basis of $L$. The lattice $A={\sum }_i{\mathcal{O}}_K{\sigma }^i\theta$ is a regular ${\mathcal{O}}_K[G]$-module. It is a finite direct sum of regular ${\mathbb{Z}}_p[G]$-modules, hence its Tate cohomology vanishes. The lattices $A$ and ${\mathfrak{p}}_L^r$ are commensurable: their intersection has finite index in both. The exact sequences with these finite quotients establish finiteness of the Tate cohomology groups and show $h_G({\mathfrak{p}}_L^r)=1$. Thus $h_G(U_L^r)=1$. Since ${\mathcal{O}}_L^{\times }/U_L^r$ is finite, $h_G({\mathcal{O}}_L^{\times })=1$ as well.

The $G$-equivariant valuation exact sequence isIts last term has trivial $G$-action. The multiplication rule gives the Herbrand quotient of the local multiplicative group:All the Tate cohomology groups here are finite by the lattice argument and the exact sequence; finiteness of the norm index has not been assumed.

For completeness we prove the needed Hilbert theorem 90. If $a\in L^{\times }$ has field norm one, put $A_0=1$ and $A_i={\prod }_{j=0}^{i-1}{\sigma }^j(a)$. The linear map $c\mapsto b={\sum }_{i=0}^{n-1}{A}_i{\sigma }^i(c)$ is not zero by linear independence of distinct field embeddings. Choose $c$ with $b\ne 0$. The relation $A_n=1$ gives $a\sigma (b)=b$, so $a=b/\sigma (b)$. This proves ${\hat{H}}^{-1}(G,L^{\times })=0$. The zeroth Tate cohomology group is $K^{\times }/N_{L/K}{L}^{\times }$. Consequently the cyclic local norm index isWriting $n=ef$, the valuation of a field norm is $v_K(Nx)=f{v}_L(x)$. The valuation quotient has order $f$, and a norm with valuation zero comes from a unit. ThereforeIn particular every unit is a norm in an unramified extension. These results give the norm assertions used in the quadratic Hilbert symbol computation.

We finish with the Hasse norm theorem: for a cyclic extension of number fields, an element is a global norm if and only if it is a norm at every place. More precisely the local norm is the norm from the product $L{\otimes }_K{K}_v$; in a Galois extension its factors have the same norm subgroup. The forward implication is immediate. For the converse associate to $a\in K^{\times }$ the cyclic algebra $(L/K,\sigma ,a)$. The splitting criterion for a cyclic algebra, proved below, identifies its splitting with $a\in N_{L/K}{L}^{\times }$. At a place where the tensor product of fields has several factors, this algebra is a matrix algebra over the cyclic algebra of one decomposition-field factor, with the generator restricted to its cyclic decomposition subgroup. Thus the local norm assumption makes its Brauer class zero at every place. Injectivity in the Albert-Brauer-Hasse-Noether theorem makes the global Brauer class zero, giving the desired global norm. This is the required proof outline; cyclicity is essential to this reduction.

A central simple algebra over $K$ is a finite-dimensional associative $K$-algebra with no nonzero proper two-sided ideals and with center $K$. The Artin–Wedderburn theorem writes it uniquely as $M_r(D)$ with $D$ a central division algebra. After a finite separable splitting field extension it becomes $M_d$, so its dimension is $d^2$, and $d$ is its degree of a central simple algebra. Passing from $D$ to a matrix algebra over $D$ does not change its Brauer class.

The Brauer group consists of these classes, with product induced by tensor product. Extending to a common splitting field shows that a tensor product of central simple algebras is again central simple; its two-sided ideals and center descend from the corresponding full matrix algebra. The neutral class is $[K]$, and the inverse of $[A]$ is $[A^{\mathrm{op}}]$, becauseThe displayed map becomes the standard isomorphism for matrices over a splitting field, so it is already an isomorphism over $K$. Thus Brauer classes measure precisely the obstruction to being a split matrix algebra.

An explicit family is supplied by cyclic algebras. Let $L/K$ be cyclic of degree $n$, with generator $\sigma$, and let $a\in K^{\times }$. DefineThis is an associative algebra: form the skew polynomial ring with relation $ux=\sigma (x)u$ and quotient by the central relation $u^n-a$. Reduction of powers gives the displayed $n^2$-dimensional $K$-basis. An element commuting with every $x\in L$ has no $u^i$ coefficient for $i\ne 0$, since ${\sigma }^i$ is not the identity. Commuting also with $u$ forces the remaining coefficient into $K$. Hence the center is $K$.

Here is an explicit splitting construction. On $L^n$, with basis $e_0,\dots ,e_{n-1}$, putThen $UD(x)=D(\sigma x)U$ and $U^n=aI$. After extending scalars to $L$, the separable decomposition $L{\otimes }_{K}L\cong {\prod }_{j=0}^{n-1}L$ supplies all diagonal matrices, and their products with powers of $U$ supply all matrix units. ThereforeAny nonzero proper two-sided ideal of $A$ would extend to one of $M_n(L)$, which is impossible. Thus this construction really gives a central simple algebra, with splitting field $L$.

Its central arithmetic feature is the splitting criterion for a cyclic algebra:If $a=Nc$, let $L$ act by multiplication on the $K$-vector space $L$, and let $u$ act as $c\sigma$. Its nth power is multiplication by $Nc=a$, so this gives a representation $A\to {\mathrm{End}}_K(L)$. It is injective by simplicity and an isomorphism by equal dimension. Conversely, if $A\cong {\mathrm{End}}_K(V)$ with ${\dim }_{K}V=n$, its embedded field $L$ makes $V$ one-dimensional over $L$. The relation for $u$ says that $u$ is $\sigma$-semilinear. In an $L$-basis it must have the form $c\sigma$, whose nth power is $Nc$. The relation $u^n=a$ forces $a=Nc$.

More generally the crossed-product algebra of a Galois extension attaches an algebra to a multiplicative two-cocycle. Changing the cocycle by a coboundary rescales its basis and preserves the Brauer class. Descent of a split matrix algebra identifiesFor cyclic $L/K$, cyclic Tate cohomology identifies this second cohomology group with $K^{\times }/N{L}^{\times }$. In a crossed product, rescale successive cyclic basis elements to powers of a single $u$; the remaining datum is $u^n=a$, changed by a field norm when $u$ is rescaled. This explains why cyclic algebras give every class split by a cyclic extension, and why multiplication of their parameters is addition of their Brauer classes.

For a finite field $K={\mathbb{F}}_q$, every finite splitting field $L={\mathbb{F}}_{q^n}$ is cyclic over $K$. The field norm is the power map $x\mapsto x^{(q^n-1)/(q-1)}$ on its cyclic multiplicative group, and is onto ${\mathbb{F}}_q^{\times }$. Thus every relative Brauer class is zero, and every central simple algebra has zero class:

For a non-Archimedean local field $K$, the valuation classification of central division algebras shows that every Brauer class splits over a finite unramified extension. For the degree-$n$ unramified extension $L/K$ with arithmetic Frobenius $\sigma$, units are norms and norm valuations are multiples of $n$. Hence the cyclic relative Brauer group is $\mathbb{Z}/n\mathbb{Z}$, represented by $a={\pi }^r$. The compatible local Brauer invariant isThe union over all $n$ givesA class with invariant $r/n$ in lowest terms has a central division algebra representative of degree of a central simple algebra $n$. This is the local valuation classification used above; the explicit cyclic algebras exhibit all of its invariants. At the Archimedean places, $\mathrm{Br}(\mathbb{C})=0$, while $\mathrm{Br}(\mathbb{R})=\mathbb{Z}/2\mathbb{Z}$: its nontrivial class is the quaternion algebra $(\mathbb{C}/\mathbb{R},\text{conjugation},-1)$, since complex norms are positive.

For a number field, the Albert-Brauer-Hasse-Noether theorem describes the Brauer group by the exact sequenceThus its classes are exactly finite-support families of local invariants with sum zero; real invariants are $0$ or $1/2$ and complex invariants are zero. Every global Brauer class has a cyclic splitting extension, so cyclic algebras also realize all global classes. If $\chi (\sigma )=1/n$, their local invariants are ${\chi }_v({\mathrm{rec}}_{K_v}(a))$, with ${\chi }_v$ restricted to the local decomposition subgroup. The sum-zero relation for a principal parameter is the invariant form of Artin reciprocity. Injectivity supplies the local-to-global splitting test, which yields the Hasse norm theorem for cyclic extensions as explained in the preceding solution. Here the local valuation classification and the global invariant theorem are the structural theorems used to describe the Brauer groups; the construction and norm-splitting calculation have been proved explicitly.

Normalize each finite-place valuation to have image $\mathbb{Z}$. For a radical class $[a]$, the residue of $v(a)$ modulo $n$ is well-defined, because replacing $a$ by $a{b}^n$ adds $nv(b)$. A suitable finite set isOne can compute the last union from any finite generating set of $\Delta$. Each representative has only finitely many nonzero valuations, so this indeed defines a finite set depending only on $n$ and the classes, not on a choice of representatives. Including all infinite places safely allows possible real-to-complex ramification.

Fix $v\notin S$ and work over the completion $K_v$. A representative has $v(a)=nk$, so $a={\pi }^{nk}u$ with $u$ a unit. Adjoining an $n$th root of $a$ is the same as adjoining an $n$th root of $u$. Since $n$ is invertible in the residue field, $X^n-\bar{u}$ is separable: its derivative $n{X}^{n-1}$ has no common zero with it.

Take a finite extension of the finite residue field containing all roots of all these finitely many reduced polynomials. The corresponding finite unramified extension of $K_v$ exists by the classification proved in Question 2. Hensel's lemma lifts each simple residue root to a root of the original polynomial in that unramified extension. Thus all the required radicals, and every conjugate root, lie in an unramified local extension. Every completion of the radical field at $v$ is therefore unramified. This proves finite ramification support for radical extensions.

The local argument does not assume ${\mu }_n\subset K$: adjoining all roots proves the stronger unramified assertion for the splitting field, and hence for any field obtained by choosing the radicals. For $n=1$ the radical extension is trivial; the displayed set is still valid.

Use the following simple-root Hensel lemma. Let $R$ be a complete discrete valuation ring, with maximal ideal $\mathfrak{m}$, and let $f\in R[X]$. If $f(a_0)\in \mathfrak{m}$ and $f^{\prime }(a_0)\in R^{\ast }$, there is a unique $a\in R$ with $a\equiv a_0(\mathrm{mod}\mathfrak{m})$ and $f(a)=0$.

For the proof, define Newton's method iterates $a_{r+1}=a_r-f(a_r)/f^{\prime }(a_r)$. Each denominator stays a unit, because all iterates are congruent to $a_0$. Taylor's polynomial identity givesIf an iterate is already a root, the process stops. Otherwise these valuations tend to infinity, so the iterates form a Cauchy sequence in the complete ring. Its limit $a$ has the required residue and satisfies $f(a)=0$. If $a,b$ were two such roots, thenThe parenthesis is a unit, forcing $a=b$.

Now let $K/{\mathbb{Q}}_p$ be finite, with residue field $k={\mathbb{F}}_q$. For every $m\ge 1$ choose a monic irreducible polynomial $\bar{f}\in k[X]$ of degree $m$ and lift it to a monic $f\in {\mathcal{O}}_K[X]$. The lift is irreducible over $K$, since a factorization would reduce to a factorization of $\bar{f}$, by the monic form of Gauss's lemma over the valuation ring. If $L=K(\alpha )$ for a root, its residue field contains $k(\bar{\alpha })$ of degree $m$. The inequality $e(L/K)f(L/K)\le [L:K]=m$ forces residue degree $m$ and ramification index one. Thus it is an unramified extension of degree $m$.

Conversely, in any unramified extension of degree $m$, choose a generator of its residue field and lift its separable minimal polynomial. Hensel's lemma supplies a root with that residue. Its degree is at least $m$ and hence it generates the extension. The same unique lifting of all residue conjugates shows that the extension is Galois andOne can see uniqueness inside a fixed algebraic closure explicitly: the roots of $X^{q^m-1}-1$ lift all nonzero residue elements, and a lifted primitive residue element generates the extension. Hencewith arithmetic Frobenius acting by $\zeta \mapsto {\zeta }^q$. The case $q^m-1=1$ simply gives $K_1=K$. These are all finite unramified extensions; $K_m\subseteq K_r$ exactly when $m\mid r$. Their union is the maximal unramified extension, whose Galois group is $\hat{\mathbb{Z}}$.

Finally prove norm-surjectivity on units using the Herbrand quotient. For a cyclic group $G$ of order $m$, writeThe quotient is multiplicative on short exact sequences, equals one for finite modules, equals $m$ on the trivial module $\mathbb{Z}$, and equals one on induced regular modules. For a finite module, the equality follows directly from $|M^G||(\sigma -1)M|=|M|=|\ker N||NM|$.

Let $U_L={\mathcal{O}}_L^{\ast }$. For sufficiently large $r$, the equivariant p-adic logarithm identifies $U_L^r=1+{\mathfrak{p}}_L^r$ with ${\mathfrak{p}}_L^r$ additively. In the unramified case this is an induced regular lattice. Indeed choose a normal basis of the finite residue extension and lift its generating element. Its $m$ Galois conjugates reduce to a basis and hence form an ${\mathcal{O}}_K$-basis of ${\mathcal{O}}_L$; their coordinate determinant is a unit. Since a uniformizer of $K$ is also one of $L$, multiplication by its $r$th power gives the same regular-module structure on ${\mathfrak{p}}_L^r$. For a regular module, invariants are norm images and a vector of coefficient sum zero is a cyclic difference, so both indicated Tate cohomology groups vanish. Consequently $h_G(U_L^r)=1$. The quotient $U_L/U_L^r$ is finite, so $h_G(U_L)=1$.

Apply multiplicativity toIt gives $h_G(L^{\ast })=m$. Hilbert 90 makes ${\hat{H}}^{-1}(G,L^{\ast })$ trivial, so $[K^{\ast }:N_{L/K}{L}^{\ast }]=m$. On the other handso the quotient maps onto $\mathbb{Z}/m\mathbb{Z}$. As its order is already $m$, its unit kernel is trivial. A norm which is a unit comes from a unit, by the same valuation formula. ThereforeThe standard normal-basis theorem for finite fields and the convergent logarithm/exponential on sufficiently deep principal units are the ingredients used in evaluating the Herbrand quotient; the norm conclusion itself has been deduced here.

The idele group is the restricted productwith respect to ${\mathcal{O}}_v^{\ast }$ at the finite places. Thus $a=(a_v)$ is an idele if all components are nonzero and $a_v$ is a unit at all but finitely many finite places. Its topology has basic open sets which equal the full unit subgroup outside finitely many places; it is not just the subspace topology of the unrestricted product. The group is locally compact, and the diagonal $K^{\ast }$ consists of the principal ideles. The idèle class group is $C_K=J_K/K^{\ast }$.

The finite valuations give the idele-to-ideal homomorphismIts kernel is the product of the finite unit groups and the infinite multiplicative groups; principal ideles map to principal fractional ideals. Thus further quotienting gives the ideal class group. The product of normalized local moduli defines $|a{|}_J$; at complex places the normalized modulus is the square of the ordinary absolute value. The product formula says that principal ideles have modulus one. These constructions package ideals, units and archimedean conditions into one group.

Here is the precise reciprocity equivalence. For a finite abelian extension $L/K$, the global law is a continuous Artin reciprocity mapwhich kills $K^{\ast }$ and sends a uniformizer idele at an unramified place to arithmetic Frobenius. Its full class-field-theoretic norm statement is $C_K/N_{L/K}{C}_L\cong \mathrm{Gal}(L/K)$. In the local formulation, the mapsare continuous, kill local units when unramified, send a local uniformizer to the same Frobenius, are compatible with restriction in towers, and obeyHere $D_v$ is the decomposition group of any prime above $v$, independent of that choice because the extension is abelian. The usual local norm theorem identifies $K_v^{\ast }/N_{L_w/K_v}{L}_w^{\ast }$ with $D_v$. The principal-idele condition is the essential global compatibility; existence of unrelated local norm isomorphisms alone would not establish global reciprocity.

Given the compatible local maps, defineOnly finitely many factors are nontrivial: there are finitely many ramified places, and an idele is a unit almost everywhere. The product is a homomorphism. Continuity follows by choosing local kernel neighbourhoods at those exceptional places and the full unit groups elsewhere. The boxed compatibility kills $K^{\ast }$, so the map descends to $C_K$. At every unramified place it has the required Frobenius value. The local norm maps also show that idele norms are killed, since the local component of an idele norm is ${\prod }_{w\mid v}{N}_{L_w/K_v}(b_w)$. Together with the class-field norm-index statement this gives the global norm-kernel formulation.

Conversely, embed $K_v^{\ast }$ in $J_K$ with every other component one, and set $r_v=\theta \circ {\iota }_v$. Each restriction is continuous and has the required unramified normalization. An open kernel of the finite-valued global map contains a basic neighbourhood whose unrestricted finite-place factors are full unit groups outside a finite set. Hence those local restrictions kill units. For an arbitrary idele, remove the finitely many exceptional and nonunit components; the remainder lies in that kernel neighbourhood. It follows directly that $\theta$ is exactly the product of its local restrictions, not merely a map agreeing on single-component ideles. Its triviality on $K^{\ast }$ is exactly the boxed product relation. With the decomposition and norm assertions of the reciprocity theorem, these restrictions are the local reciprocity maps into $D_v$ with kernel $N_{L_w/K_v}{L}_w^{\ast }$.

One can explicitly connect this idelic formulation with the classical ideal law without assuming an idelic law in advance. A modulus of a number field records finitely many nonnegative exponents at finite places and a selected subset of real places. For a modulus $\mathfrak{m}$, let $U_{\mathfrak{m}}$ have factors $1+{\mathfrak{p}}_v^{m_v}$ at finite places dividing the modulus, full units at other finite places, positive real numbers at specified real places, and full infinite multiplicative groups otherwise. Weak approximation for number fields supplies a global $a\in K^{\ast }$ making $a^{-1}j$ lie in the specified local factors at the finitely many modulus places. Its valuations outside the modulus define an ideal prime to $\mathfrak{m}$. A different such $a$ changes that ideal by a ray-principal ideal, and every ideal prime to $\mathfrak{m}$ is obtained. HenceAn ideal Artin map killing $P_K(\mathfrak{m})$ therefore extends to $J_K$ via this isomorphism; restricting it gives the local maps just constructed. Conversely, continuous local maps satisfying the product condition have an open product kernel containing some $U_{\mathfrak{m}}$ and hence make the ideal Artin map kill $P_K(\mathfrak{m})$. This proves the idelic local-global reciprocity equivalence at the reciprocity-law level as well as the formal conversion of maps. The arithmetic Frobenius normalization and product relation agree with Milne's local-global reciprocity formulation.

In the absolute-value description, a place of a number field is an equivalence class of nontrivial absolute values, where equivalent absolute values give the same topology, or differ by a positive power. More generally this defines the rank-one absolute-value places used in weak approximation for a field. For a number field, finite places correspond to nonzero prime ideals of its ring of integers, while infinite places correspond to real embeddings and pairs of conjugate complex embeddings. The completion at $v$ is denoted $K_v$.

A valuation place can also be described by a valuation ring ${\mathcal{O}}_v\subset K$ and its residue map into $k_v$, extended by sending elements outside that ring to infinity. The finite places above are the discrete-valuation instances of this description.

The weak approximation theorem states that for finitely many pairwise inequivalent nontrivial absolute values $v_1,\dots ,v_r$, the diagonal image of $K$ is dense in ${\prod }_i{K}_{v_i}$. Equivalently, for arbitrary $a_i\in K_{v_i}$ and ${ϵ}_i>0$, there is $a\in K$ such thatThe places must be distinct: two requirements at the same completion need not be compatible. In the number-field setting this permits simultaneous prescribed finite congruences and archimedean approximations.

A central simple algebra over $K$ is a finite-dimensional associative unital $K$-algebra $A$ with no nonzero proper two-sided ideals and center exactly $K$. Its dimension is a square, ${\dim }_{K}A=n^2$. The Artin–Wedderburn theorem writes it as $M_r(D)$ for a uniquely determined central division algebra $D$ up to isomorphism. One obtains this structure by taking a minimal left ideal and its commuting division endomorphism ring; the simple algebra acts as all endomorphisms over that ring. Over a suitable finite splitting field it becomes a full matrix algebra.

The Brauer group identifies two central simple algebras if their division-algebra parts agree, equivalently if tensoring each with suitable matrix algebras makes them isomorphic. The product is tensor product, the identity is the class of $K$, and the inverse of $[A]$ is $[A^{\mathrm{op}}]$. Indeedis an isomorphism for a central simple algebra, so the product class is split. This also explains why matrix size is irrelevant to the Brauer class. The order of the class is its period, while the degree of its division-algebra representative is its index; these are different notions in general.

For the cyclic algebra construction, take a cyclic extension $L/K$ of degree $m$, a generator $\sigma$ and $a\in K^{\ast }$. SetThe multiplication is obtained by moving coefficients past $u$ and replacing every $u^m$ by $a$. It is associative: reducing a product of three monomials gives the same total exponent, the same power of $a$ and the same iterated automorphisms in either order. The scalar $a$ is fixed by $\sigma$ and ${\sigma }^m=1$, so reduction is consistent.

An element commuting with every $\ell \in L$ has only its $u^0$ coefficient nonzero, because ${\sigma }^i\ne 1$ for $0<i<m$. Commuting with $u$ then forces that coefficient into $K$, proving that the center is $K$. To prove simplicity, choose a nonzero element of a nonzero ideal with the fewest nonzero monomial coefficients. Since $u$ is invertible and each nonzero coefficient in $L$ is invertible, normalize one term to be the constant one. If another term remains, its automorphism differs from the identity; commuting with a suitable element of $L$ eliminates the constant term but not that other term. This gives a shorter nonzero element, a contradiction. The minimal element is therefore a unit, and the ideal is all of $A$. Thus $A$ is central simple of dimension $m^2$.

It splits over $L$. For example, over $L$ represent $\ell$ by the diagonal matrix with entries $\ell ,\sigma (\ell ),\dots ,{\sigma }^{m-1}(\ell )$, and let $u$ shift the standard basis down one place, sending the first vector to $a$ times the last. These matrices satisfy the defining relations. A primitive element of the separable extension $L/K$ has distinct conjugates; the Vandermonde matrix of their powers is invertible. Thus the diagonal images span all diagonal matrices after scalar extension to $L$. Multiplying the diagonal matrix units by powers of the shift gives every matrix unit, proving $A{\otimes }_{K}L\cong M_m(L)$.

The particularly useful splitting criterion isIf $a=N(b)$, let the algebra act on $L$ by multiplication for $\ell$ and by the semilinear operator $b\sigma$ for $u$. Its $m$th power is multiplication by $N(b)=a$, so this gives an isomorphism with ${\mathrm{End}}_K(L)$, by simplicity and dimension. Conversely, a split algebra acts on an $m$-dimensional $K$-space $V$. Restriction to the subfield $L$ makes $V$ one-dimensional over $L$, and $u$ is a nonzero $\sigma$-semilinear operator. In an $L$-basis it is $b\sigma$, so $u^m=a$ forces $a=N(b)$. More generally the relative Brauer group split by a cyclic $L/K$ is $K^{\ast }/N_{L/K}{L}^{\ast }$, with multiplication of parameters corresponding to addition of classes.

For a finite field, a finite division algebra is commutative by Wedderburn's little theorem. Therefore every central simple algebra is a matrix algebra and $\boxed{\mathrm{Br}({\mathbb{F}}_q)=0}$. The cyclic-algebra viewpoint gives the same conclusion for cyclic classes because the norm between finite fields is surjective on their cyclic multiplicative groups.

For a nonarchimedean local field, the local Brauer invariant givesAn unramified cyclic extension of degree $m$, with arithmetic Frobenius generator, gives invariant $v_K(a)/m$ to its cyclic algebra. Question 2 shows that all units are norms; norm valuations are multiples of $m$. Thus the relative classes are $\mathbb{Z}/m\mathbb{Z}$, represented by powers of a uniformizer. Allowing all $m$ realizes all invariants. At the infinite local fields, $\mathrm{Br}(\mathbb{C})=0$, while $\mathrm{Br}(\mathbb{R})\cong \mathbb{Z}/2\mathbb{Z}$, with the nonzero class represented by Hamilton's quaternions and invariant $1/2$.

For a number field, the Albert-Brauer-Hasse-Noether theorem supplies the exact sequenceHence a global algebra is determined by its finitely many nonzero local invariants, whose sum is zero; it is split precisely when all its local classes vanish. Every division algebra over a number field is cyclic, so cyclic algebras provide representatives for the global classes as well. Their norm parameters translate splitting into local norm conditions and the reciprocity constraint on the sum of invariants. The local invariant theorem, Wedderburn's little theorem and the global invariant theorem are the standard structural results used here; their role in the requested computation is explicit.

Put $\omega =(1+\sqrt{229})/2$. Since $229\equiv 1(\mathrm{mod}4)$ is squarefree,We first compute the ideal class group, rather than assume the proposed extension has the full Hilbert-class-field degree.

The real quadratic Minkowski bound for ideal classes is $\sqrt{229}/2<8$. Thus every ideal class has an integral representative of norm at most seven. The prime two is inert because $229\equiv 5(\mathrm{mod}8)$, while seven is inert because $229\equiv 5(\mathrm{mod}7)$ is a nonsquare. The inert ideal $(2)$ has norm four and is principal; a prime above seven has norm forty-nine. The only possible nonprincipal prime factors of these small representatives are therefore primes above three and five.

Reduction of $X^2-X-57$ gives roots $0,1$ modulo three and $2,4$ modulo five. WriteTheir conjugate pairs multiply to $(3)$ and $(5)$. Direct norm calculations giveThe element $6+\omega$ vanishes at the residue roots $0$ modulo three and $4$ modulo five, and not at the conjugate roots. Its absolute norm forces each exponent to be one. Likewise $15+2\omega$ vanishes only at the root zero modulo three, and its norm forces exponent three. ThusThese also give the factorizations suggested by the hint: $(13+\sqrt{229})=(2)PQ$ and $(16+\sqrt{229})=P^3$. The classes of $P,Q$ generate the ideal class group; the first relation makes $[Q]=[P{]}^{-1}$ and the second makes $[P{]}^3=1$. Hence the class number divides three.