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For a finite abelian extension of number fields, local reciprocity maps into the local decomposition groups combine to a continuous map on the idele group: outside finitely many places they kill units, and an idele has nonunit components only finitely often. The Artin reciprocity law is equivalent to the compatibility that this product kills every principal idele. Conversely, restrict a global continuous Artin homomorphism to each single-place embedding of $K_v^{\ast }$. These maps reconstruct the global map, by density of finite-component ideles and triviality on a basic open subgroup of units. Merely having unrelated local maps omits the essential principal-idele compatibility.

Let ${\zeta }_n$ be a primitive nth root of unity. The cyclotomic polynomial isInduction on $n$ shows that ${\Phi }_n$ is monic with integer coefficients. Here is the essential irreducibility of cyclotomic polynomials argument. Let $f$ be the minimal polynomial of ${\zeta }_n$ and let $\ell \nmid n$ be a prime. If ${\zeta }_n^{\ell }$ had a different monic irreducible polynomial $g$, then $f\mid g(X^{\ell })$. Reducing modulo $\ell$ gives $\bar{f}\mid {\bar{g}}^{\ell }$, so $\bar{f}$ and $\bar{g}$ have a common factor. As $fg\mid {\Phi }_n$, this would give a repeated factor of $X^n-1$ modulo $\ell$, impossible because its derivative is $n{X}^{n-1}$. Thus ${\zeta }_n^{\ell }$ is conjugate to ${\zeta }_n$. Every positive integer prime to $n$ is a product of such primes, so all primitive nth roots of unity are conjugate. ConsequentlyIn particular the cyclotomic field is an abelian extension. Complex conjugation corresponds to $-1$; for $n>2$ its fixed field is $\mathbb{Q}({\zeta }_n+{\zeta }_n^{-1})$, the maximal real subfield, of degree $\phi (n)/2$.

The ring of integers is $\mathbb{Z}[{\zeta }_n]$. One way to see that there is no missing index is to work at each prime $p$. Write $n=p^{a}m$ with $p\nmid m$. The prime-to-$p$ roots of unity have separable reduction; their local factors generate unramified extensions with full integer rings. Over each such factor, adjoining ${\zeta }_{p^a}$ is generated by the Eisenstein polynomial ${\Phi }_{p^a}(1+T)$, with uniformizer ${\zeta }_{p^a}-1$, and hence gives the full integer ring of that totally ramified factor. The coprimality of $p^a$ and $m$ means that ${\zeta }_n$ generates both sets of roots of unity. Thus ${\mathbb{Z}}_p[{\zeta }_n]$ is maximal at every $p$, proving the integral assertion.

The same local description computes ramification and splitting. For $p\nmid n$, the ramification index is one, the residue degree is the multiplicative order of $p$ modulo $n$, and there are $\phi (n)/f$ primes above $p$. If $p\mid n$, put $m=n/p^a$; thenwhere ${\mathrm{ord}}_1(p)=1$. Indeed, the prime-to-$p$ part is unramified, and the $p$-power part is totally ramified by the preceding Eisenstein polynomial. Notice the exception $p=2,a=1$: its ramification index is one, consistently with $\mathbb{Q}({\zeta }_{2m})=\mathbb{Q}({\zeta }_m)$ for odd $m$. Thus finite ramification occurs only at the indicated conductor primes, with that exception.

We now prove cyclotomic Artin reciprocity over the rational numbers, using arithmetic Frobenius automorphisms. If $\ell \nmid n$ and $\mathfrak{L}$ lies over $\ell$, then Frobenius sends the reduction of ${\zeta }_n$ to its $\ell$th power. Distinct nth roots of unity remain distinct modulo $\mathfrak{L}$, since $X^n-1$ has separable reduction. Hence the Artin symbol is exactlyExtend multiplicatively to the group $I(n)$ of rational fractional ideals prime to $n$. Such an ideal has a unique positive rational generator $q$ whose numerator and denominator are prime to $n$. The formula isEvery unit residue modulo $n$ has a positive integer representative prime to $n$, so this Artin map is surjective. Its kernel is precisely the subgroup $P_{n\infty }$ generated by positive $q\equiv 1(\mathrm{mod}n)$. ThereforeThis proves the Artin reciprocity law for this extension, including its principal-ray kernel. The infinite component of the modulus records positivity and therefore the possible action of complex conjugation.

Use arithmetic Frobenius automorphisms: residues are raised to the size of the base residue field. Let $S$ contain the finite primes ramified in the finite abelian extension $L/K$. The Artin map is the group homomorphismwhere $I_K^S$ is the group of fractional ideals supported outside $S$. The Artin symbol ${\mathrm{Frob}}_{\mathfrak{p}}$ is independent of the prime above $\mathfrak{p}$, since the extension is abelian. Commutativity makes the displayed extension from prime ideals well defined. Defining this map does not yet assert the Artin reciprocity law, which describes its congruence group kernel.

For an intermediate field $K\subseteq E\subseteq L$, restriction givesTo prove it, start with an unramified prime $\mathfrak{p}$ and a prime of $L$ above it. Its Frobenius automorphism acts on the residue of every integral element of $E$ by raising it to $N\mathfrak{p}$. Its restriction is therefore the Frobenius automorphism for $E/K$ at the restricted prime. The identity for all allowed fractional ideals follows by the group homomorphism property. On Galois groups, restriction is the quotient by $\mathrm{Gal}(L/E)$, so passing to a subextension quotients the Artin map accordingly.

Now let $E/K$ be any finite extension and put $M=LE$. This is the translated extension: $M/E$ is abelian, and restriction identifies $\mathrm{Gal}(M/E)$ with $\mathrm{Gal}(L/(L\cap E))$. A prime $\mathfrak{q}$ of $E$ above a prime $\mathfrak{p}\notin S$ is unramified in $M/E$, because forming a compositum with $E$ preserves an unramified extension. Put $r=f(\mathfrak{q}/\mathfrak{p})$. The residue-field sizes satisfy $N\mathfrak{q}=(N\mathfrak{p}{)}^r$. Consequently the restriction to $L$ of Frobenius at $\mathfrak{q}$ acts on its integral residues by raising them to $(N\mathfrak{p}{)}^r$, givingBut the ideal norm is $N_{E/K}\mathfrak{q}={\mathfrak{p}}^r$. Multiplying this prime-by-prime identity provesThis is restriction and norm compatibility of the Artin map. The fractional ideal $\mathfrak{a}$ must be supported over primes unramified in $L/K$. No assumption that $E/K$ itself is unramified or Galois is needed. With geometric rather than arithmetic Frobenius, both maps are inverted and the compatibility identities remain unchanged.

The Brauer group of a field consists of equivalence classes of central simple algebras, with addition induced by tensor product, zero represented by matrix algebras, and inverse represented by the opposite algebra. Localization sends a global algebra to its scalar extensions at the completions. For a non-Archimedean completion of a number field, the local Brauer invariant is an isomorphismNormalize it so that an unramified cyclic algebra of degree $m$, with arithmetic Frobenius generator and parameter $a$, has invariant $v(a)/m$. For a real completion, $\mathrm{Br}(\mathbb{R})=\mathbb{Z}/2$ with the Hamilton quaternion algebra mapped to $1/2$; for a complex completion the Brauer group is zero.

The global computation is the exact sequence in the Albert-Brauer-Hasse-Noether theorem:Thus a global Brauer class is determined uniquely by a finite-support family of local invariants. Such a family occurs if and only if its sum is zero; at real places the only allowed entries are $0,1/2$, and at complex places the entry is zero. Injectivity is a local-global splitting principle; the middle exactness is the reciprocity constraint; surjectivity of the sum follows already from any one finite local summand. These are separate assertions, not merely different descriptions of the sum formula.

For an explicit group description, choose one finite place $v_0$. Assign arbitrary finite-support invariants at all other finite places and arbitrary allowed invariants at the real places, then set the invariant at $v_0$ equal to minus their sum. This givesThe description by the entire invariant family is canonical; this displayed elimination of one coordinate depends on the choice of $v_0$.

The connection with Artin reciprocity is especially transparent for cyclic algebras. Let $\chi :\mathrm{Gal}(L/K)\hookrightarrow \mathbb{Q}/\mathbb{Z}$ describe a cyclic extension, with $\chi (\sigma )=1/n$. Write $(\chi ,a)$ for the class of $(L/K,\sigma ,a)$. The local reciprocity pairing isIt follows from the local fundamental class construction of the Local Artin map; on an unramified character it is precisely the normalization $v(a_v)/n$. The group kernel in the parameter is the local norm group, and restriction/corestriction give the norm compatibility of this pairing.

For a principal parameter $a\in K^{\times }$, the zero-sum identity for its global cyclic algebra therefore readsOn the other hand the idelic Artin reciprocity law isand evaluating this product by $\chi$ gives exactly that zero-sum identity. Conversely, testing every cyclic character of a finite abelian Galois group separates its elements, so the zero-sum identities for cyclic algebras imply this principal-idele reciprocity law for every finite abelian extension. At almost all places a unit and an unramified character pair trivially, so the sums and products are finite.

The full idelic statement identifies the group kernel as $K^{\times }{N}_{L/K}{J}_L$ and induces $J_K/(K^{\times }{N}_{L/K}{J}_L)\cong \mathrm{Gal}(L/K)$. It explains why the local norm-residue pairings must glue globally. The global Brauer theorem supplies, in addition, the injectivity and realization statements needed to compute all algebra classes; the principal-product identity alone would not prove those statements. For cyclic algebras its injectivity yields the Hasse norm theorem, as in Question 3. In degree two, an invariant is either zero or $1/2$, so the zero-sum law becomes the Hilbert reciprocity law ${\prod }_v(a,b{)}_v=1$. For example, a quaternion algebra over $\mathbb{Q}$ can be ramified at precisely a prescribed finite set of places, including possibly the real place, if and only if that set has even cardinality; the invariants uniquely determine its Brauer class. This illustrates how arithmetic reciprocity becomes a concrete constraint on global algebra structure.

The Hilbert class field $H_K$ is the maximal unramified abelian extension of $K$, including the condition that real places do not become complex when infinite ramification is included. The Artin reciprocity law induces a canonical isomorphismConsequently $[H_K:K]=h_K$.

Let $G=\mathrm{Gal}(L/K)$ and letbe the fixed field of the commutator subgroup. Then $M/K$ is the maximal abelian subextension of $L/K$, withThe nonabelian norm-residue kernel theorem identifies the kernel of the global reciprocity mapwith $N_{L/K}{C}_L$. Applied to the abelian extension $M/K$, the Artin reciprocity law identifies the kernel of the same map with $N_{M/K}{C}_M$. Thereforeso the norm group of the Galois extension $L/K$ is the norm group of the abelian extension $M/K$.