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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.

For each finite place $v\in S$, choose $d_v\in K_v^{\times }$ such that $K_v(\sqrt{d_v})/K_v$ is the unramified quadratic extension. Such a choice exists even at residue characteristic two. At odd residue characteristic use a unit with nonsquare residue. At residue characteristic two, choose $\overline{c}$ for which $X^2-X-\overline{c}$ is irreducible over the finite residue field, lift $c$ to ${\mathcal{O}}_v$, and set $d_v=1+4c$. The lifted polynomial has unit derivative at every residue point and gives the unramified degree-two extension; its root is $(1+\sqrt{d_v})/2$.

The square classes are open in $K_v^{\times }$. Indeed the strong form of Hensel lemma applied to $X^2-u$ at $1$ shows that $u$ is a square whenever $v(u-1)>2v(2)$. Thus any sufficiently close approximation $a$ to $d_v$ has $a/d_v$ square. By weak approximation for number fields, choose $a\in K$ satisfying this at every finite $v\in S$. Then the local extensions $K_v(\sqrt{a})$ are all unramified of degree two. At least one such condition ensures $a$ is not a square in $K$. If $S$ contains no finite place, impose the same condition at one auxiliary finite place.

Set $L=K(\sqrt{a})$. At every specified finite prime there is one prime of $L$ above it, with ramification index one and residue degree two. Hence $\boxed{\mathfrak{p}{\mathcal{O}}_L\text{ is prime for every finite }\mathfrak{p}\in S}$. The dyadic case is covered by the unramified construction, rather than by a nonexistent nonsquare residue unit in characteristic two.

The idele group is the restricted productwith componentwise multiplication and the restricted product topology on the idele group relative to the unit groups at finite places. We need weak approximation for number fields only at the finitely many specified coordinates, not density in the full restricted-product topology.

Here is a proof including the Archimedean places. At each finite place in $S$, choose $a_v\in K$ sufficiently close to $x_v$, using the definition of the completion. Choose a nonzero $d\in {\mathcal{O}}_K$ clearing their denominators. Pick powers ${\mathfrak{p}}_v^{m_v}$ so that $|z/d{|}_v<\epsilon$ whenever $z\in {\mathfrak{p}}_v^{m_v}$. Put $I={\prod }_{v\in S,v\text{ finite}}{\mathfrak{p}}_v^{m_v}$. For each positive rational integer $M$ coprime to the rational primes below these places, the Chinese remainder theorem gives $b_M\in {\mathcal{O}}_K$ with $b_M\equiv Md{a}_v(\mathrm{mod}{\mathfrak{p}}_v^{m_v})$. All of $b_M+I$ satisfy the same congruences.

Under the Minkowski embedding of a number field, $I$ is a full lattice in ${\mathbb{R}}^{r_1}\times {\mathbb{C}}^{r_2}$. A fundamental parallelepiped is bounded, so every real vector is within a fixed distance of this lattice, uniformly over its translates. Choose a target vector whose coordinates at the specified Archimedean places are those of $Md{x}_v$, assigning arbitrary values at the other Archimedean places. There is $b\in b_M+I$ within that fixed distance of the target. Dividing by $Md$ makes all specified Archimedean errors tend to zero as $M\to \infty$ through the allowed integers. At finite places $M$ is a unit, so the earlier congruence bounds persist. By taking the initial approximations sufficiently close, the ultrametric inequality gives the requested errors there too. Thus some $\boxed{y=b/(Md)\in K}$ satisfies every required inequality. If there are no finite places, set $I={\mathcal{O}}_K$ and omit the congruences.

The idele group is the multiplicative restricted productIts restricted product topology has basic open sets ${\prod }_{v\in S}{U}_v\times {\prod }_{v\notin S}{\mathcal{O}}_v^{\times }$, where $S$ is finite and contains every infinite place, and $U_v$ is open in $K_v^{\times }$. In particular, the restriction to almost all local unit groups is part of the topology, not merely the unrestricted product topology.

The diagonal map $K^{\times }\to J_K$ is injective and well-defined by finite support of the principal ideal's valuations; its image consists of principal ideles. To prove discreteness of principal ideles, take the neighbourhood of $1$ which requires $x_v\in {\mathcal{O}}_v^{\times }$ at every finite place and $|x_v-1|<1/2$ in the ordinary real or complex modulus at every infinite place. A diagonal element in it is a global unit, so $x-1\in {\mathcal{O}}_K$. If $x\ne 1$, the nonzero integer $N_{K/\mathbb{Q}}(x-1)$ has absolute value at least one, but the archimedean bounds make its absolute value less than $2^{-[K:\mathbb{Q}]}$. This contradiction proves $K^{\times }$ is a discrete subgroup of $J_K$.

For a modulus of a number field, write $\mathfrak{m}$ for its finite ideal part together with any selected real places. Let $I_{\mathfrak{m}}$ be the group of fractional ideals coprime to the finite part. Let $P_{\mathfrak{m}}$ consist of principal ideals $(a)$ with $a\equiv 1(\mathrm{mod}{\mathfrak{p}}_v^{m_v})$ locally at each finite place in the modulus, and $a>0$ at each selected real place. The ray class group isEquivalently it is the quotient $J_K/(K^{\times }{U}_{\mathfrak{m}})$, with the corresponding principal-unit factors at finite places and positive multiplicative factors at selected real places. The question now assumes that no infinite places occur.

Put $R_{\mathfrak{m}}={\prod }_{v\in S}{\mathcal{O}}_v^{\times }/(1+{\pi }_v^{m_v}{\mathcal{O}}_v)$. Global units map to $R_{\mathfrak{m}}$ by reduction. To map $R_{\mathfrak{m}}$ to the ray class group, choose, by the Chinese remainder theorem, an integral $a$ having the prescribed unit residues at all $v\in S$, and send the tuple to the ray class of $(a)$. A different choice changes the ratio by an element congruent to one at every such place, so the map is well-defined and multiplicative.

The forgetful map to the ordinary ideal class group is surjective: weak approximation for number fields multiplies an ideal by a principal ideal so that all its valuations in $S$ vanish. Its kernel consists exactly of classes represented by principal ideals coprime to $S$, and their generators supply the local unit residues. Finally, a residue tuple maps to the identity ray class precisely when its chosen $a$ differs from a congruence-one generator by a global unit. Thus the ray class exact sequence isThe leftmost arrow need not be injective, explaining the absence of an initial zero.

For $K=\mathbb{Q}(\sqrt{2})$, the ring of integers of Q of square root two is $\mathbb{Z}[\sqrt{2}]$. Its absolute field norm is a Euclidean function. Indeed, for $a+b\sqrt{2}\in K$, choose nearest integers $m,n$; writing $u=a-m$, $w=b-n$, both bounded by $1/2$, gives $|u^2-2w^2|\le 1/2<1$. Applying this to a quotient proves Euclidean division, hence the ring is a principal ideal domain and $\mathrm{Cl}(K)=0$. This is norm-Euclideanity of the integers adjoined square root two.

Write $\mathfrak{p}=(\sqrt{2})$ and $\mathfrak{q}=(1+2\sqrt{2})$. Their ideal norms are $2$ and $7$, respectively; ${\mathfrak{p}}^2=(2)$ and the residue field at $\mathfrak{q}$ is ${\mathbb{F}}_7$, with $\sqrt{2}\equiv 3$ because $1+2\sqrt{2}\equiv 0$. The finite local quotient product for the requested modulus isThe first factor consists of $1$ and $1+t$.

The given fundamental unit $ϵ=1+\sqrt{2}$ and $-1$ generate the global units. Their images areThe first image has order six; its cube is $(1+t,1)$ and its square is $(1,2)$, generating the first factor and the order-three subgroup of the second. The second image supplies the missing order-two element of ${\mathbb{F}}_7^{\times }$. Thus the global unit map is surjective onto all $12$ local unit residues. By the exact sequence, or the unit-surjectivity criterion for a trivial finite ray class group,There is no positivity condition here: the modulus has no infinite part, so using the unit $-1$ is legitimate and essential in the surjectivity calculation.

Here a divisor is a modulus of a number field, rather than an arbitrary real-weighted divisor. Writewhere ${\mathfrak{c}}_0$ is a nonzero integral ideal, the finite multiplicities $m_{\mathfrak{p}}$ are nonnegative integers, and ${\mathfrak{c}}_{\infty }$ is a set of real Archimedean places. The multiplicity of a place in a modulus is $m_v(\mathfrak{c})=m_{\mathfrak{p}}$ at the finite place corresponding to $\mathfrak{p}$, is $1$ at a real place in ${\mathfrak{c}}_{\infty }$, and is $0$ at every other infinite place. In particular complex places have multiplicity $0$.

Let $I_{\mathfrak{c}}$ be the group of nonzero fractional ideals prime to ${\mathfrak{c}}_0$. PutLet $P_{\mathfrak{c}}$ consist of the principal ideals $(a)$ with $a\in K_{\mathfrak{c},1}^{\times }$. The generalized ideal class group is the ray class groupLet $U={\mathcal{O}}_K^{\times }$ be the unit group, let $U_{\mathfrak{c}}=U\cap K_{\mathfrak{c},1}^{\times }$ be the group of ray units, and define the residue and signature group of a modulusThe first factor is omitted when ${\mathfrak{c}}_0={\mathcal{O}}_K$. There is a group homomorphism $\rho :U\to G_{\mathfrak{c}}$ recording the unit's finite residues and its signs. Its kernel is $U_{\mathfrak{c}}$.

Let $P^{\mathfrak{c}}$ consist of all principal fractional ideals prime to ${\mathfrak{c}}_0$, and put $R_{\mathfrak{c}}=P^{\mathfrak{c}}/P_{\mathfrak{c}}$. The two short exact sequences areandFor the first short exact sequence, weak approximation for number fields realizes every choice of finite unit residues and real signs by some $a\in K^{\times }$ prime to ${\mathfrak{c}}_0$. Send that data to the class of $(a)$ in $R_{\mathfrak{c}}$. Changing $a$ without changing its residues and signs multiplies it by an element of $K_{\mathfrak{c},1}^{\times }$, so the map is well-defined. Its kernel consists exactly of data arising from units: if $(a)=(b)$ with $b\in K_{\mathfrak{c},1}^{\times }$, then $a/b\in U$. This gives $R_{\mathfrak{c}}\cong G_{\mathfrak{c}}/\rho (U)$. For the second short exact sequence, forget the ray conditions. Its kernel is $R_{\mathfrak{c}}$, and weak approximation for number fields gives a representative prime to ${\mathfrak{c}}_0$ for every ideal class. These are the two parts of the ray class exact sequence.

For $K=\mathbb{Q}(\sqrt{15})$, the ring of integers of a quadratic field is ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{15}]$. The finite modulus is trivial and both real places occur, so $H_{\mathfrak{c}}$ is the narrow ideal class group and $G_{\mathfrak{c}}=\{\pm 1{\}}^2$. The two field embeddings send $\sqrt{15}$ to $\sqrt{15}$ and $-\sqrt{15}$. The given unit $ϵ=4+\sqrt{15}$ is positive at both places, since $4-\sqrt{15}>0$; the unit $-1$ is negative at both. Thus the unit signature map has imageThe given ideal class group has order $2$. The second short exact sequence therefore givesTo determine the group structure, retain the ideal $\mathfrak{p}=(2,1+\sqrt{15})$ that generates the ordinary ideal class group. We have ${\mathfrak{p}}^2=(2)$: all generators $4$, $2(1+\sqrt{15})$ and $(1+\sqrt{15}{)}^2$ lie in $(2)$, whileputs $2$ in ${\mathfrak{p}}^2$. Since $2$ is totally positive, $[\mathfrak{p}{]}^2=1$ in the narrow ideal class group. Its image in the ordinary ideal class group is nontrivial, so $[\mathfrak{p}]$ has order exactly $2$.

The class of the principal ideal $(\sqrt{15})$ is a nontrivial element of $R_{\mathfrak{c}}$. Its two signs are $(+,-)$, and multiplying by a unit can only reverse both signs or neither, so no generator of this ideal is totally positive. Its square is $(15)$, which does have a totally positive generator. Thus $[(\sqrt{15})]$ is another element of order $2$, distinct from $[\mathfrak{p}]$ because its ordinary ideal class is trivial. These two elements are independent and generate all four classes. ConsequentlyThe nontrivial ordinary ideal class already has a lift of order $2$, so the extension in the second short exact sequence splits; it cannot be cyclic of order $4$.

At finitely many finite places of a number field, choose square-class parameters for the local unramified quadratic extensions. Local square classes are open, including at dyadic places by the strong form of Hensel lemma. Weak approximation for number fields gives a global parameter in all the prescribed local classes. Its quadratic extension is unramified of residue degree two at those places, so their extended prime ideals remain prime. If no finite place is prescribed, an auxiliary local nonsquare condition ensures the global extension is nontrivial.

This group records finite unit residues and signs at the selected real places of a modulus of a number field. Omit the residue factor for trivial finite modulus. Its quotient by the image of the unit group is naturally the kernel of the map from the ray class group to the ordinary ideal class group. The weak approximation for number fields theorem supplies elements realizing the specified data, and units account for changing a generator of a principal fractional ideal.

For finitely many pairwise inequivalent nontrivial absolute values on a field, an element of the field can approximate independently prescribed targets in the associated completion of a valued field at each place to arbitrary positive accuracies. Equivalently the diagonal map has dense image in the finite product. Requiring finitely many inequivalent places is essential; the conclusion controls finitely many inequalities, not infinitely many independently prescribed bounds at once, or incompatible targets at one place. Weak approximation for number fields is the class-field-theoretic instance.