Class field theory

Global class field theory classifies finite abelian extensions of a number field using ideal-class or idele-class quotients and the Artin reciprocity map.

For a cyclic extension of number fields, an element of the base field is a global field norm exactly when it is a norm at every finite and infinite completion. The splitting criterion for a cyclic algebra encodes the norm obstruction as a Brauer group class. The Albert-Brauer-Hasse-Noether theorem makes a class that splits at every completion split globally, giving the result. Cyclicity is essential; arbitrary noncyclic extensions can have additional local-global norm obstructions.

An Artin L-function encodes traces of Frobenius elements in a Galois representation and converts character sums over primes into analytic information.

Only the inertia group invariants enter an Artin L-function local factor. A Frobenius automorphism lift acts unambiguously on this subspace, since other lifts differ by inertia. For an unramified one-dimensional group representation, the factor is $(1-\chi ({\mathrm{Frob}}_{\mathfrak{p}})(N\mathfrak{p}{)}^{-s}{)}^{-1}$; for a ramified one-dimensional group representation, the invariant subspace is zero and the factor is one. This explains the zero values used at conductor primes by quadratic Dirichlet characters.

For the standard representation of the symmetric group $S_3$, the unramified local factor of an Artin L-function, with $t=p^{-s}$, is respectively the displayed factor for identity, transposition, and three-cycle Frobenius cycle types. These follow by removing the trivial summand from the three-point permutation representation. If the inertia group contains a three-cycle, its invariant subspace in the standard representation is zero, so the ramified local factor is one.

Assume that every nontrivial one-dimensional Artin L-function over the intermediate fields is holomorphic and nonzero at one, while the trivial one is the base Dedekind zeta function with a simple pole. By Brauer induction, $\chi ={\sum }_j{n}_j{\mathrm{Ind}}_{H_j}^G{\psi }_j$ with integers $n_j$ and linear characters ${\psi }_j$. Compatibility of Artin L-functions with induced representations expresses $L_K(\chi ,s)$ as the corresponding product of $L_{L^{H_j}}({\psi }_j,s{)}^{n_j}$. Its order at one is $-{\sum }_j{n}_j⟨{\psi }_j,1_{H_j}⟩$. Frobenius reciprocity turns this sum into $-⟨\chi ,1_G⟩$. Hence every nontrivial irreducible character gives order zero and a holomorphic nonzero function near one. This local conclusion does not assert global Artin holomorphy.

The ray class field realizes the ray class group through the Artin reciprocity map and is the maximal abelian extension with bounded conductor.

A modulus is a formal product $\mathfrak{m}={\mathfrak{m}}_0{\mathfrak{m}}_{\infty }$, where ${\mathfrak{m}}_0$ is a nonzero integral ideal and ${\mathfrak{m}}_{\infty }$ is a product of distinct real places.

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.

The ray class group modulo $\mathfrak{m}$ is $I_K(\mathfrak{m})/P_K(\mathfrak{m})$, where the numerator consists of fractional ideals prime to ${\mathfrak{m}}_0$ and the denominator consists of $(\alpha )$ with $\alpha \equiv 1$ at the finite modulus and positive at every real place in ${\mathfrak{m}}_{\infty }$.

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.

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.

For a finite modulus $\mathfrak{m}$, there is an exact sequenceWith real places in the modulus, the residue-unit term and the unit map also record the required signs.

For a modulus with no infinite component, suppose the ordinary ideal class group is trivial. Then the ray class group is trivial exactly when the global unit group maps surjectively onto the product of local unit residue groups at the finite modulus. Negative real units are allowed: positivity restrictions occur only when the modulus includes the corresponding real places.

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.

If $U_{\mathfrak{m},1}$ consists of units congruent to one modulo ${\mathfrak{m}}_0$ and positive at the selected real places, then

For an unramified prime $\mathfrak{p}$ in an abelian extension $L/K$, the Artin symbol is the unique Galois element acting on residue fields by $x\mapsto x^{N\mathfrak{p}}$.

For an unramified prime $\mathfrak{p}$ in a finite abelian extension $L/K$, the residue degree equals the order of its Artin symbol. Consequently the number of primes above $\mathfrak{p}$ is $[L:K]$ divided by that order.

The Artin map sends an ideal prime to the ramified primes to the product of its Frobenius elements. For a suitable modulus, its kernel is the norm or congruence subgroup associated with the abelian extension.

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.

For an abelian extension $L/K$ of number fields, restriction to an intermediate field takes each Artin symbol to that of the smaller extension. For a finite base extension $E/K$, restriction identifies $\mathrm{Gal}(LE/E)$ with $\mathrm{Gal}(L/(L\cap E))$. At an unramified prime $\mathfrak{q}$ of $E$ above $\mathfrak{p}$, its residue-field Frobenius restricts to the $f(\mathfrak{q}/\mathfrak{p})$th power of the Frobenius automorphism at $\mathfrak{p}$. Since $N_{E/K}\mathfrak{q}={\mathfrak{p}}^{f(\mathfrak{q}/\mathfrak{p})}$, the group homomorphism property proves the displayed identity for fractional ideals supported over unramified primes.

The conductor ${\mathfrak{f}}_{L/K}$ is the smallest modulus through which the Artin reciprocity map for a finite abelian extension $L/K$ factors. Its finite prime divisors are exactly the ramified primes, with exponents measuring the depth of ramification.

For a finite abelian extension of number fields, its relative discriminant ideal is the product of the finite conductor ideals of the characters of its Galois group, with the trivial character contributing one. In a multiquadratic extension of $\mathbb{Q}$, the nontrivial characters correspond to its quadratic subfields, and their conductors are the absolute values of the corresponding fundamental discriminants. This gives an efficient discriminant computation and can prove that a compositum is unramified over a quadratic subfield.

Finite abelian extensions of $K$ correspond contravariantly to congruence subgroups $H$ with $P_K(\mathfrak{m})\subseteq H\subseteq I_K(\mathfrak{m})$, and the Artin map identifies $I_K(\mathfrak{m})/H$ with the Galois group.

The ray class field modulo $\mathfrak{m}$ is the maximal abelian extension whose Artin map factors through the ray class group; its degree is the ray class number.

For $K=\mathbb{Q}(\sqrt{-6})$, reduction of $\mathbb{Z}[\sqrt{-6}]$ modulo $(\sqrt{-6})$ gives $\mathbb{Z}/6\mathbb{Z}$. Its unit group is covered by the units $\{1,-1\}$ of $K$. The ray class exact sequence therefore identifies the ray class group with the ordinary ideal class group, of order two. Consequently this ray class field is the Hilbert class field of Q of square root minus six. To verify its lack of ramification, the integral basis candidate $1,\sqrt{2},b,\sqrt{2}b$, where $b=(1+\sqrt{-3})/2$, has discriminant $8^2(-3{)}^2=576$. The relative discriminant formula also forces the field discriminant to be divisible by $(-24{)}^2=576$. Equality proves that the relative discriminant ideal is trivial.

The ring of integers is $\mathbb{Z}[\omega ]$, where ${\omega }^2+\omega +1=0$, and its class number is one by the Minkowski bound for ideal classes. Modulo seven it is ${\mathbb{F}}_7\times {\mathbb{F}}_7$. The six roots of unity reduce to the kernel of the product map on the two residue unit groups, so the ray class group is ${\mathbb{F}}_7^{\times }$, of order six. The relative Artin symbol on $K({\zeta }_7)$ is exponentiation by the norm of a fractional ideal modulo seven. It kills the principal ray subgroup, and $[K({\zeta }_7):K]=6$, proving that this is the ray class field.

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.

The adele ring is the restricted product of the completions $K_v$ with respect to the valuation rings ${\mathcal{O}}_v$ at finite places.

Given spaces $X_i$ with distinguished open subsets $U_i$, their restricted product consists of tuples $(x_i)$ for which $x_i\in U_i$ at all but finitely many indices.

A basic open set in a restricted product ${\prod }_v^{\prime }(X_v,U_v)$ is a product of open local sets that equals the distinguished subset $U_v$ at all but finitely many indices.

The idele group is the restricted product ${\prod }_v^{\prime }{K}_v^{\times }$ with respect to ${\mathcal{O}}_v^{\times }$ at finite places.

An idele obtained by mapping the same nonzero element of a number field into every completion. Only finitely many local valuations are nonzero, so this diagonal tuple belongs to the idele group.

The diagonal multiplicative group of a number field is discrete in its idele group. Require local units at every finite place and an archimedean neighbourhood of one. A diagonal element $x$ meeting these restrictions has integral $x-1$. If $x\ne 1$, its nonzero integer field norm cannot have absolute value less than one, while sufficiently small archimedean differences would force exactly that.

The map from an idele to its finite-place fractional ideal is surjective, has kernel the product of the infinite multiplicative groups and finite-place unit groups, and sends diagonal elements to principal fractional ideals. It realizes the ideal class group as the corresponding quotient of the idele group.

The idelic modulus is the product of normalized local moduli: ordinary real modulus, squared complex modulus and $(N\mathfrak{p}{)}^{-{\mathrm{ord}}_{\mathfrak{p}}(x)}$ at finite places. This continuous homomorphism to ${\mathbb{R}}_{>0}$ has the diagonal multiplicative group in its kernel by the product formula. It differs from the idele norm, which maps between idele groups of field extensions.

The norm-one idele group is the kernel of the idelic modulus. Its quotient by the diagonal $K^{\times }$ is compact for a number field and surjects onto the ideal class group.

An idele is an element of the idele group: a tuple of nonzero elements of the completion of a valued field at all places, with finite-place components in the local unit groups except at finitely many places.

The idele topology is the restricted product topology with respect to ${\mathcal{O}}_v^{\times }$. It is also the subspace topology induced by $x\mapsto (x,x^{-1})$ from ${\mathbb{A}}_K\times {\mathbb{A}}_K$.

The idele class group is $C_K={\mathbb{I}}_K/K^{\times }$, where $K^{\times }$ is embedded diagonally.

The ideles congruent to one modulo $\mathfrak{m}$ have finite components in $1+{\mathfrak{p}}^{n_{\mathfrak{p}}}$ at primes in the finite modulus and positive components at selected real places. Their image in $C_K$ is the idelic congruence subgroup.

For a finite abelian extension $L/K$, the norm group is the image of the idele-class norm $C_L\to C_K$. Class field theory gives $C_K/N_{L/K}{C}_L\cong \mathrm{Gal}(L/K)$.

The $v$-component of the idele norm is ${\prod }_{w\mid v}{N}_{L_w/K_v}(x_w)$. It descends to an idèle class group norm $C_L\to C_K$.

For a finite Galois extension $L/K$ with group $G$, the kernel of the global reciprocity map $C_K\to G^{\mathrm{ab}}$ is $N_{L/K}{C}_L$. It follows that this norm group equals the norm group from the maximal abelian subextension $L^{[G,G]}/K$.

Every finite abelian extension of $\mathbb{Q}$ is contained in a cyclotomic field.

For an odd prime $\ell$, the unique quadratic subfield of a prime cyclotomic field has no finite ramification outside $\ell$. The classification of quadratic discriminants identifies it with $\mathbb{Q}(\sqrt{{\ell }^{\ast }})$, where ${\ell }^{\ast }=(-1{)}^{(\ell -1)/2}\ell$. A fundamental discriminant factors as $D=D_2{\prod }_{\ell \mid D,\ell \text{ odd}}{\ell }^{\ast }$ with $D_2\in \{1,-4,8,-8\}$. The remaining quadratic number field is generated by $i$, $\sqrt{2}$, or $\sqrt{-2}$ and lies in $\mathbb{Q}({\zeta }_8)$. Taking the corresponding compositum proves the indicated quadratic case of the Kronecker–Weber theorem.

The Hilbert class field is the maximal unramified abelian extension of a number field. Its Galois group is canonically the ideal class group.

For a CM-field $K$ over its maximal real subfield $K^{+}$, the ordinary Hilbert class field $H_{K^{+}}$ is totally real because its real places split. Thus $H_{K^{+}}\cap K=K^{+}$. Its compositum with $K$ is an abelian extension of $K$ of degree $h_{K^{+}}$, and is everywhere unramified by unramified extensions remain unramified under base change. It lies in $H_K$, giving the divisibility. The argument does not require injectivity of extension of ideal classes.

The three reduced primitive binary quadratic forms of discriminant $-23$ are $(1,1,6)$ and $(2,\pm 1,3)$, so the class number is three. The displayed cubic has discriminant $-23$ and is irreducible. Its splitting field has Galois group $S_3$ and is cyclic cubic over $\mathbb{Q}(\sqrt{-23})$. Modulo $23$ it factors as $(X-10{)}^2(X-3)$. Lift the simple factor; the remaining quadratic has discriminant valuation one, so its local inertia group has order two. Its intersection with $A_3$ is trivial. Thus the relative cyclic cubic extension is everywhere unramified and equals the Hilbert class field.

The polynomial is irreducible with discriminant $229$, so its splitting field has Galois group $S_3$ and quadratic subfield $\mathbb{Q}(\sqrt{229})$. The splitting field is cyclic cubic over that subfield. It is unramified outside 229, and at 229 its inertia is a transposition, meeting the alternating cubic subgroup trivially. All roots are real. It is therefore everywhere unramified over the quadratic field. The ideal class group of Q of square root 229 has order three, so this extension is exactly the Hilbert class field.

The ideal class group of $\mathbb{Q}(\sqrt{-30})$ is $C_2\times C_2$. Its four reduced binary quadratic forms have coefficients $(1,0,30)$, $(2,0,15)$, $(3,0,10)$ and $(5,0,6)$. The displayed multiquadratic field has degree four over the base. The seven quadratic-subfield discriminants are $8,-3,5,-24,40,-15,-120$, whose product in absolute value is $120^4$. The conductor-discriminant formula and discriminant tower formula imply that the relative discriminant over $\mathbb{Q}(\sqrt{-30})$ is the unit ideal. This unramified abelian degree-four extension is therefore its Hilbert class field.

The Hilbert class field is the maximal everywhere-unramified abelian extension of a number field; its Galois group is canonically the ideal class group.

If a finite extension $B/A$ of number fields has a totally ramified finite prime, the ordinary Hilbert class field $H_A$ is disjoint from $B$. Its base change $B{H}_A/B$ is abelian and unramified at finite primes. Its real places also split, since a split-real unramified extension retains this property under base change. It therefore lies in the ordinary Hilbert class field of $B$, proving $h_A\mid h_B$ even when $A$ has real places.

The splitting field of $T^3+T-1$ has discriminant square class $-31$ and Galois group $S_3$. It is the Hilbert class field of $K=\mathbb{Q}(\sqrt{-31})$: the relative extension is unramified cyclic cubic, and the ideal class group of Q of square root minus thirty-one has order three. Locally at $31$, the simple residue root $28$ lifts and the remaining quadratic splits over ${\mathbb{Q}}_{31}(\sqrt{-31})$, so the prime of $K$ above $31$ splits completely in its Hilbert class field. The inverse generator satisfies $T^3-T^2-1=0$.

Suppose an irreducible monic integral cubic has discriminant $-p$ for a prime $p>3$, and let $L$ be its splitting field. Its Galois group is $S_3$ and its quadratic subfield is $\mathbb{Q}(\sqrt{-p})$. The extension over that subfield is cyclic cubic and everywhere unramified. Away from $p$, separable reduction and Hensel lemma put the roots in an unramified local extension. At $p$, discriminant valuation one gives a double residue root and a separate simple root; lift the simple root, and the remaining quadratic's discriminant has square class $-p$. Its splitting field is already the quadratic subfield's completion. The imaginary quadratic base has no real places.

If $M/L$ is an extension of number fields, $L$ has no real places, and one finite prime of $L$ is totally ramified in $M$, then $h_L\mid h_M$. The Hilbert class field $H_L$ has trivial intersection with $M$: any common intermediate extension would be both unramified and totally ramified at that prime. Thus $M{H}_L/M$ is an unramified abelian extension of degree $h_L$, and is contained in $H_M$.

The splitting field has Galois group $S_3$ and quadratic subfield $\mathbb{Q}(\sqrt{-59})$. The extension over that subfield is abelian of degree three and unramified everywhere. The only possible rational ramified prime is $59$; at that prime a simple residue root lifts, so the local splitting field is quadratic and coincides with the quadratic subfield's completion. It follows from the Hilbert class field degree that $3\mid h_{\mathbb{Q}(\sqrt{-59})}$.

The ideal class group of $K=\mathbb{Q}(\sqrt{-6})$ has order two, andThis quadratic extension of $K$ is unramified at every finite prime and therefore has the degree required of the Hilbert class field.