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A subgroup of the group of fractional ideals prime to a modulus of a number field, containing the principal ray subgroup. It is the inverse image of a subgroup of the ray class group. The existence theorem of global class field theory realizes it as the kernel of an Artin map, and identifies its quotient with the corresponding Galois group. This is the ideal-theoretic class-field meaning of congruence subgroup.

For a modulus of a number field $\mathfrak{c}={\mathfrak{c}}_0{\mathfrak{c}}_{\infty }$, the multiplicity at a finite place is the exponent of its prime ideal in ${\mathfrak{c}}_0$. At a real Archimedean place it is one if that place occurs in ${\mathfrak{c}}_{\infty }$ and zero otherwise. Complex places have multiplicity zero. These effective modulus multiplicities encode congruences and positivity requirements in the ray class group.

A modulus of a number field is $\mathfrak{m}={\mathfrak{m}}_0{\mathfrak{m}}_{\infty }$, with ${\mathfrak{m}}_0$ a nonzero integral ideal and ${\mathfrak{m}}_{\infty }$ a subset of the real places. Let $I_{\mathfrak{m}}$ be the group of fractional ideals prime to ${\mathfrak{m}}_0$, and let $P_{\mathfrak{m}}$ consist of principal ideals $(a)$ with $a\equiv 1$ at each prime power in ${\mathfrak{m}}_0$ and $a>0$ at each selected real place. The ray class group isThe ray class field $K_{\mathfrak{m}}$ is the finite abelian extension for which the Artin map has kernel $P_{\mathfrak{m}}$ and induces an isomorphism ${\mathrm{Cl}}_{\mathfrak{m}}(K)\cong \mathrm{Gal}(K_{\mathfrak{m}}/K)$. Equivalently it is the maximal abelian extension whose conductor divides $\mathfrak{m}$.

To compute its order use the ray class exact sequence, including signatures:The middle map lifts a residue and sign tuple to an element of $K$ by weak approximation theorem, then takes its principal ideal as a ray class. Different lifts differ by a principal ray factor. Its kernel consists exactly of tuples represented by global units. Every ordinary ideal class has a representative prime to the finite modulus, again by weak approximation theorem; this gives surjectivity of the last map. If $U_{\mathfrak{m}}$ denotes the ray units, the kernel of the unit map is $U_{\mathfrak{m}}$. The Chinese remainder theorem and counting units in each prime-power quotient give the ray class number formula

First take $K=\mathbb{Q}(\sqrt{-3})$. Its ring of integers is $\mathbb{Z}[\omega ]$, where $\omega =(-1+\sqrt{-3})/2$ and ${\omega }^2+\omega +1=0$. The Minkowski bound for ideal classes is $(2/\pi )\sqrt{3}<2$, so every ideal class has a representative of norm one and $h_K=1$. Its units are the six roots of unity. Modulo seven, the polynomial $X^2+X+1$ has distinct roots $2,4$, and thereforeThe unit group of this quotient has order $36$. The unit $1+\omega$ maps to $(3,5)$, which has order six; consequently unit reduction is injective and $U_{(7)}=\{1\}$. The ray class number formula gives $36/6=6$.

We also identify the quotient explicitly. Conjugation interchanges the two factors, so the residue norm is $(x,y)\mapsto xy$. Its kernel has order six and equals the image of the six units, since $3\cdot 5=1(\mathrm{mod}7)$. HenceThe field $M=K({\zeta }_7)=\mathbb{Q}({\zeta }_{21})$ has degree six over $K$, since the degrees over $\mathbb{Q}$ are twelve and two. It is unramified outside seven by the cyclotomic computation and the base-change argument above. At primes over seven, $K$ has completion ${\mathbb{Q}}_7$, and $M/K$ is the tame degree-six cyclotomic extension; its conductor exponent is one. Its relative arithmetic Artin symbol at an ideal $\mathfrak{a}$ prime to seven acts byIf $\mathfrak{a}=(a)$ with $a\equiv 1(\mathrm{mod}7{\mathcal{O}}_K)$, then $N_{K/\mathbb{Q}}(a)\equiv 1(\mathrm{mod}7)$. Thus the Artin map kills the principal ray subgroup. It factors through the order-six ray class group; the residue norm quotient computed above shows that it is onto the degree-six Galois group, so the fields coincide. This proves the explicit ray class field of Q of square root minus three modulo seven:

Now take $K=\mathbb{Q}(\sqrt{-6})$, and write $s=\sqrt{-6}$. Its ring of integers is $\mathbb{Z}[s]$, its discriminant is $-24$, and its units are $\pm 1$. The quotient by $(s)$ iswhose two units are both images of $\pm 1$. The ray class exact sequence therefore gives ${\mathrm{Cl}}_{(s)}(K)\cong \mathrm{Cl}(K)$.

For its class number, the Minkowski bound for ideal classes is $(2/\pi )\sqrt{24}<4$, so each ideal class contains an integral ideal of norm at most three. The primes of norms two and three areThese are the only nontrivial possibilities within the bound, and their ideal classes coincide and have order at most two. The ideal ${\mathfrak{p}}_2$ is not principal: its being principal would give an integer solution to $x^2+6y^2=2$, which is impossible. Thus $h_K=2$, the ideal class group is $C_2$, and the ray field is the Hilbert class field.

We verify its actual identity rather than stopping at the degree. PutThe two quadratic fields $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{-3})$ are distinct, so $[H:\mathbb{Q}]=4$ and $[H:K]=2$. The integral elements $1,\sqrt{2},b,\sqrt{2}b$ are a rational basis. Their trace matrix is the Kronecker product of the two quadratic trace matrices, with discriminants $8$ and $-3$. Its determinant is therefore $8^2(-3{)}^2=576$. The discriminant-index formula for an integral lattice shows that the field discriminant divides $576$. On the other hand the relative discriminant ideal formula givesThe norm on the right is a positive integer, so equality is forced. Thus the relative discriminant ideal is the unit ideal, and $H/K$ is unramified at every finite prime. There are no real places of $K$. The extension is quadratic and therefore Abelian; its degree equals $h_K$, identifying it with the Hilbert class field of Q of square root minus six. Hence the ray class field of Q of square root minus six modulo square root minus six is

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 a number field $K$ and a modulus of a number field $\mathfrak{m}$, let $I_K(\mathfrak{m})$ be the group of fractional ideals prime to its finite part. Let $P_K(\mathfrak{m})$ consist of principal ideals generated by elements congruent to one modulo its finite part and positive at each real place in its infinite part. The ray class group is ${\mathrm{Cl}}_{\mathfrak{m}}(K)=I_K(\mathfrak{m})/P_K(\mathfrak{m})$.

The classification theorem of class field theory asserts the existence of a ray class field $K_{\mathfrak{m}}$ and an Artin map isomorphismIts intermediate fields correspond contravariantly to subgroups of this ray class group. Equivalently, for each congruence subgroup of fractional ideals $P_K(\mathfrak{m})\subseteq H\subseteq I_K(\mathfrak{m})$ there is a unique finite abelian extension with Artin group kernel $H$ and group $I_K(\mathfrak{m})/H$. Every finite abelian extension occurs for some modulus. For a fixed modulus, these are exactly the extensions whose conductor divides that modulus; the same field may be presented using larger moduli, with the appropriate pulled-back group kernel.

We quote the conductor facts needed to specialize this classification. The finite conductor exponent is zero exactly at the unramified primes. A real place occurs in the conductor exactly when it becomes complex. Thus conductor one means unramified at every finite prime and split at every real place. With $\mathfrak{m}=1$ the ray class group is the ordinary ideal class group, so classification gives an extension $H_K/K$ withIt is unramified at all finite primes, and its real places split. Conversely, every finite abelian extension with those properties has conductor one and lies in $K_1=H_K$. This proves the existence and maximality of the ordinary Hilbert class field. Its degree is finite by finiteness of the ideal class group. The real-place convention matters: allowing real places to become complex gives the narrow class field instead; for the imaginary quadratic field in part (ii) this distinction disappears.

For a number field $K$ and a modulus of a number field $\mathfrak{m}={\mathfrak{m}}_0{\mathfrak{m}}_{\infty }$, let $I_K(\mathfrak{m})$ be the group of fractional ideals prime to ${\mathfrak{m}}_0$, and let $P_{K,1}(\mathfrak{m})$ consist of principal fractional ideals $(a)$ with $a\equiv 1$ at the finite modulus and $a>0$ at the real places in ${\mathfrak{m}}_{\infty }$. The ray class group is ${\mathrm{Cl}}_{\mathfrak{m}}(K)=I_K(\mathfrak{m})/P_{K,1}(\mathfrak{m})$.

The Main theorem of class field theory states that there is a unique finite abelian ray class field $K_{\mathfrak{m}}/K$ for which the Artin reciprocity map inducesIt is unramified outside the finite modulus and has no ramification at real places omitted from ${\mathfrak{m}}_{\infty }$. More generally, subgroups $H$ with $P_{K,1}(\mathfrak{m})\subseteq H\subseteq I_K(\mathfrak{m})$ correspond, with inclusions reversed, to intermediate abelian fields, with Galois group $I_K(\mathfrak{m})/H$. Every finite abelian extension occurs for some modulus. Equivalently, $K_{\mathfrak{m}}$ is the maximal abelian extension with conductor of an abelian extension dividing $\mathfrak{m}$. This states both reciprocity and the existence theorem of global class field theory.

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 $4$.

A modulus of a number field is a formal productwhere ${\mathfrak{m}}_0={\prod }_{\mathfrak{p}}{\mathfrak{p}}^{n_{\mathfrak{p}}}$ is a nonzero integral ideal and ${\mathfrak{m}}_{\infty }$ is a product of distinct real places. Only finitely many exponents $n_{\mathfrak{p}}$ are nonzero.

A modulus of a number field is a formal productwhere ${\mathfrak{m}}_0={\prod }_{\mathfrak{p}}{\mathfrak{p}}^{n_{\mathfrak{p}}}$ is a nonzero integral ideal and ${\mathfrak{m}}_{\infty }$ is a product of distinct real embeddings of $K$. Only finitely many $n_{\mathfrak{p}}$ are nonzero; complex places do not occur.

For a positive integer $m$, the finite ray class group of $\mathbb{Q}$ is $(\mathbb{Z}/m\mathbb{Z}{)}^{\times }/⟨-1⟩$: the unit group permits either sign of a rational generator. Its ray class field is the maximal real subfield of the cyclotomic field $\mathbb{Q}({\zeta }_m)$. Including the real place requires positive principal generators and gives the full group $(\mathbb{Z}/m\mathbb{Z}{)}^{\times }$ and full cyclotomic field. Thus modulus $(5)$ gives $\mathbb{Q}(\sqrt{5})$, while $(5)\infty$ gives $\mathbb{Q}({\zeta }_5)$. The Artin reciprocity map sends a prime $\ell \nmid m$ to the exponent $\ell$, taken modulo $\{\pm 1\}$ in the finite case. Ramification at real places is possible only when those places are included in the modulus of a number field.

A ray unit for a modulus of a number field is a unit congruent to one modulo every finite prime power in the modulus and positive at every real place in it. The group of ray units is the kernel of the unit group map to the residue and signature group of a modulus.

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.