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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.

Let $j(\mathfrak{b})$ be the Minkowski embedding of a number field applied to a nonzero fractional ideal. Under the metric $|x{|}^2={\sum }_{v\text{ real}}{x}_v^2+2{\sum }_{v\text{ complex}}|x_v{|}^2$, its covolume is $\sqrt{|D_K|}N(\mathfrak{b})$. For ordinary coordinate Lebesgue measure $\prod d{x}_v\prod d\mathrm{Re}{z}_{v}d\mathrm{Im}{z}_v$, its covolume is instead $2^{-r_2}\sqrt{|D_K|}N(\mathfrak{b})$. Here $D_K$ is the field discriminant, and the absolute norm of a fractional ideal is positive and multiplicative. The formula follows by taking the embedding determinant of an integral basis, then using the index of an integral ideal and scaling to handle a fractional ideal.

For the cyclotomic field $\mathbb{Q}({\zeta }_n)$, arithmetic Frobenius automorphisms at primes $\ell \nmid n$ send ${\zeta }_n$ to ${\zeta }_n^{\ell }$: reduction distinguishes the nth roots of unity, and Frobenius raises residues to their $\ell$th powers. Multiplicativity therefore sends a fractional ideal with positive rational generator $a$, prime to $n$, to the exponent $a\mathrm{mod}n$. Its kernel consists exactly of positive generators congruent to one modulo $n$. This proves Artin reciprocity for the ray class group of modulus $n\infty$ without assuming the general reciprocity theorem.

An ideal class is an element of the ideal class group, represented by a nonzero fractional ideal. Two fractional ideals represent the same class exactly when their quotient is a principal ideal.

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.

An integral ideal is a fractional ideal of an integral domain $A$ that is contained in $A$. For a number field, take $A$ to be its ring of integers ${\mathcal{O}}_K$.

The narrow ideal class group of a number field is the group of nonzero fractional ideals modulo principal fractional ideals generated by totally positive elements. It is the ray class group for the modulus consisting of all real Archimedean places and trivial finite part. The forgetful map to the ordinary ideal class group is surjective, with kernel $\{\pm 1{\}}^{r_1}/\mathrm{sgn}({\mathcal{O}}_K^{\times })$ under the unit signature map.

Write $R={\mathcal{O}}_K$. We prove unique factorization of ideals in a number field, rather than assuming this conclusion as a property of a Dedekind domain. Three preliminary facts about $R$ will be useful. Its additive group is a finite-rank free abelian group, so every ideal is finitely generated and the ascending chain condition holds. It is an integrally closed domain, by integral dependence is transitive. Finally, every nonzero prime ideal is a maximal ideal. For this last fact, choose $0\ne a\in P$. Multiplication by $a$ is an injective endomorphism of the finite-rank free abelian group $R$ with nonzero determinant, so $R/(a)$ is finite. Consequently $R/P$ is a finite integral domain and hence a field.

First, every nonzero proper ideal contains a product of nonzero prime ideals. Otherwise choose a maximal counterexample $J$ under inclusion, using the ascending chain condition. It cannot be a prime ideal. Choose $x,y\notin J$ with $xy\in J$. The strictly larger ideals $J+(x)$ and $J+(y)$ each contain a product of nonzero prime ideals, allowing the empty product $R$ when necessary. Multiplying those products gives a product contained ina contradiction.

Next fix a nonzero prime ideal $P$, choose $0\ne a\in P$, and choose a product of nonzero prime idealswith $r$ minimal. Because $(a)\subseteq P$, primality and maximality imply $P_j=P$ for some $j$; reorder so that $P_1=P$. Minimality provides $b\in P_2\cdots P_r$ with $b\notin (a)$, taking the remaining product to be $R$ if $r=1$. ThusDefine the fractional idealIt contains $R$ and $x$. The ideal $P{P}^{-1}$ lies between $P$ and $R$, so it is $P$ or $R$. If it were $P$, then $xP\subseteq P$. Applying the determinant trick to a finite generating set of an ideal for $P$ gives a monic polynomial over $R$ satisfied by $x$. Since $R$ is an integrally closed domain, this would put $x$ in $R$, a contradiction. ThereforeThis proves that each nonzero prime ideal is an invertible fractional ideal, independently of the desired factorization theorem.

Now take a nonzero proper ideal $A$ and a maximal ideal $P$ containing it. The ideal $B=P^{-1}A$ satisfiesThe first inclusion is strict: if $A=B$, the same $x\in P^{-1}\setminus R$ would stabilize $A$, and the determinant trick would again force $x\in R$. Repeating gives a strictly ascending chain until it reaches $R$; the ascending chain condition ensures this happens after finitely many steps. The identities $A=PB$ then express $A$ as a product of nonzero prime ideals.

For uniqueness, suppose $P_1\cdots P_r=Q_1\cdots Q_s$. Since the right-hand side is contained in $P_1$, some $Q_j$ is contained in $P_1$. Both are nonzero maximal ideals, so $Q_j=P_1$. Multiplication by $P_1^{-1}$ cancels this factor. Repeat. A nonempty product of proper prime ideals cannot equal $R$, so both products finish together. This proves uniqueness up to reordering, including repeated factors.

A Dedekind domain is a Noetherian integrally closed domain in which every nonzero prime ideal is maximal. The equivalent local characterization is that its localizations at nonzero prime ideals are discrete valuation rings. A field is harmless as the dimension-zero case.

Write $R$ for the domain, $F$ for its fraction field and $I$ for the nonzero fractional ideal. Its proposed inverse is an $R$-submodule of $F$. Choose $0\ne d\in R$ with $dI\subseteq R$, so $d\in I^{-1}$ and the inverse is nonzero. Choose also $0\ne a\in I$. Every $x\in I^{-1}$ satisfies $xa\in R$, hence $I^{-1}\subseteq a^{-1}R$. Writing $a=r/s$ with $r,s\in R$ gives $r{I}^{-1}\subseteq sR\subseteq R$. Thus $I^{-1}$ is a fractional ideal; the Noetherian hypothesis also makes it finitely generated.

For a nonzero maximal prime ideal $\mathfrak{p}$, the discrete valuation ring $R_{\mathfrak{p}}$ makes $I_{\mathfrak{p}}$ principal, say $I_{\mathfrak{p}}=a_{\mathfrak{p}}{R}_{\mathfrak{p}}$. We need to justify that inversion commutes with this localization. One inclusion is immediate. Conversely, if $x{I}_{\mathfrak{p}}\subseteq R_{\mathfrak{p}}$, finite generation of $I$ supplies a common $s\notin \mathfrak{p}$ such that $sxI\subseteq R$. Therefore $sx\in I^{-1}$, provingIt follows that $(I{I}^{-1}{)}_{\mathfrak{p}}=R_{\mathfrak{p}}$ at every such prime ideal. Globally $I{I}^{-1}\subseteq R$. If this were a proper ideal, it would be contained in a maximal ideal and would remain proper after localizing there, a contradiction. In the field case every nonzero fractional ideal is the whole field. Hence in all casesThis proves that every nonzero fractional ideal of a Dedekind domain is an invertible fractional ideal, rather than presupposing its inverse exists.

Let $R={\mathcal{O}}_k$, $S={\mathcal{O}}_K$ and $n=[K:k]$. For each prime ideal $\mathfrak{P}$ of $S$ above $\mathfrak{p}$ of $R$, set $f_{\mathfrak{P}}=[S/\mathfrak{P}:R/\mathfrak{p}]$. The relative norm of a fractional ideal can be specified on the unique prime ideal factorization byHere is why this definition is the intrinsic relative ideal norm from norms of elements, and why the residue weights occur. Localize at $\mathfrak{p}$. The finite torsion-free $R_{\mathfrak{p}}$-module $S_{\mathfrak{p}}$ is free of rank $n$, because $R_{\mathfrak{p}}$ is a discrete valuation ring. Multiplication by a nonzero integral $a$ has determinant $N_{K/k}(a)$. The Smith normal form givesDecompose the quotient at the prime ideals above $\mathfrak{p}$. In $S_{\mathfrak{P}}$ each quotient ${\mathfrak{P}}^j/{\mathfrak{P}}^{j+1}$ is one-dimensional over $S/\mathfrak{P}$, so has length $f_{\mathfrak{P}}$ over $R_{\mathfrak{p}}$. ThereforeFor an integral ideal $I$, all norms of elements of $I$ have valuations at least $\sum f_{\mathfrak{P}}{v}_{\mathfrak{P}}(I)$. These minima can be attained simultaneously at the finitely many prime ideals above a fixed $\mathfrak{p}$. In fact the Chinese remainder theorem for ideals, applied inside the finitely generated module $I$, chooses an $a\in I$ with nonzero image in each $I/\mathfrak{P}I$. Equivalently $v_{\mathfrak{P}}(a)=v_{\mathfrak{P}}(I)$ at all these primes. The generated ideal of the element norms consequently has exactly the displayed valuations. Clearing a denominator extends this description to fractional ideals.

Multiplication of fractional ideals adds their prime valuations. The displayed formula therefore provesIt also gives $N_{K/k}((a))=(N_{K/k}(a))$.

Finally, reduction of the free rank-$n$ module $S_{\mathfrak{p}}$ modulo $\mathfrak{p}$ has dimension $n$ over $R/\mathfrak{p}$. The factorization $\mathfrak{p}S={\prod }_i{\mathfrak{P}}_i^{e_i}$ and the Chinese remainder theorem for ideals identify this reduction with the product of the quotients by ${\mathfrak{P}}_i^{e_i}$. Their filtrations have $e_i$ successive quotients of dimension $f_i$ over $R/\mathfrak{p}$. Adding dimensions proves the fundamental identity for prime decomposition:

Use $R={\mathcal{O}}_k$, $S={\mathcal{O}}_L$, $T={\mathcal{O}}_K$, and denote their relative codifferents by $C_{L/k}$, $C_{K/L}$ and $C_{K/k}$. Transitivity of the field trace givesFor the forward direction, testing an element $t\in T$ against every $s\in S$ is legitimate because $st\in T$. For the reverse direction, take $s=1$ and use trace transitivity.

Localize at a nonzero base prime ideal. The intermediate ring is semilocal Dedekind domain, so its invertible fractional ideal $C_{L/k}$ is principal there, say $C_{L/k}=d^{-1}S$ locally. This principality follows by choosing, with the Chinese remainder theorem for ideals, a generator at each of its finitely many maximal ideals. The displayed criterion now becomesInvert these fractional ideals using Solution 1. Equality at every base localization gives the global different in a tower formulaThe second factor is extended to the upper ring of integers; that extension is implicit in the printed formula. At a local prime the exponents consequently satisfy $d(K/k)=d(K/L)+e(K/L)d(L/k)$.

For $R={\mathcal{O}}_k$ and $S={\mathcal{O}}_K$, define the inverse different, or codifferent, by the trace-dual latticeThe trace pairing is nondegenerate in this separable field extension, so this is a full fractional ideal of $S$. The relative different ideal is its inverse:Every integral $x$ has integral traces against $S$, so $S\subseteq {\mathfrak{D}}^{-1}$ and $\mathfrak{D}\subseteq S$.

For $k=\mathbb{Q}$, take an integral basis $b_1,\dots ,b_n$ and its trace pairing matrix $T=(\mathrm{Tr}(b_i{b}_j))$. In these coordinates the codifferent is $T^{-1}{\mathbb{Z}}^n$, andThe ideal norm of $\mathfrak{D}$ equals this index, by inversion of fractional ideal norms. Therefore the norm of the different ideal relation isHere $r_2$ is the number of conjugate pairs of complex embeddings. The different is an ideal of the upper field, whereas the relative discriminant ideal is its ideal norm in the base field; over $\mathbb{Q}$ the ordinary signed field discriminant also records the signature.

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.

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

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.

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.

The ideal class group of a number field is the group of nonzero fractional ideals of its ring of integers under multiplication, modulo the subgroup of principal fractional ideals. For degree $n$ and $r_2$ pairs of complex embeddings, the Minkowski bound for ideal classes states that each class contains an integral ideal $I$ withThere are only finitely many integral ideals of bounded ideal norm: as sublattices of the free abelian group of integers, they have bounded index, and finitely many subgroups have any fixed index. Thus the ideal class group is finite.

For the finite separable field extension define the inverse different, or codifferent, byIt is a fractional ideal of ${\mathcal{O}}_L$. The different ideal is its inverse fractional ideal.

Suppose the integer ring is generated by $\alpha$, with monic minimal polynomial of an algebraic element $g$ of degree $d$. We prove the derivative formula using the coefficient pairing for a monogenic algebra. If ${\alpha }_1,\dots ,{\alpha }_d$ are its distinct conjugate polynomial roots, Lagrange interpolation for a polynomial $h$ of degree less than $d$ givesComparing coefficients of $T^{d-1}$ yieldsFor any integral polynomial, first reduce modulo $g$; the coefficients stay integral because $g$ is monic over ${\mathcal{O}}_K$.

It follows that $g^{\prime }(\alpha {)}^{-1}{\mathcal{O}}_L$ lies in the codifferent. To prove equality, consider the matrixAll entries are integral. Entries with $i+j<d-1$ are zero, and those with $i+j=d-1$ are one. Reversing the column order makes the matrix triangular with diagonal ones; thus its determinant is a unit of ${\mathcal{O}}_K$.

Write an arbitrary $x\in L$ as $x=g^{\prime }(\alpha {)}^{-1}{\sum }_j{b}_j{\alpha }^j$ with $b_j\in K$. Membership in the codifferent says $A(b_j{)}_j\in {\mathcal{O}}_K^d$. Since $A$ is invertible over ${\mathcal{O}}_K$, this forces every $b_j$ integral. ThereforeSeparability ensures that $g^{\prime }(\alpha )\ne 0$; the trace pairing calculation also proves that no extra factor has been overlooked.

There is no infinite component in the modulus $(5)$. The ordinary ideal class group of $\mathbb{Q}$ is trivial, and its unit group is $\{\pm 1\}$. The ray class exact sequence therefore givesConcretely, a fractional ideal prime to $5$ has a rational generator $r=a/b$ with $5\nmid ab$. Reducing $r$ modulo $5$ is well-defined up to multiplication by the units $\pm 1$. Its ray class is trivial precisely when one of the two generators $r,-r$ is congruent to one. Thus the quotient has the two classes $\{1,4\}$ and $\{2,3\}$. No positivity restriction is imposed on a generator.

To identify the corresponding field, put $E=\mathbb{Q}({\zeta }_5+{\zeta }_5^{-1})=\mathbb{Q}(\sqrt{5})$. The maximal real subfield of the fifth cyclotomic field is fixed by complex conjugation, the element $-1$ of $(\mathbb{Z}/5\mathbb{Z}{)}^{\times }$. For a prime $\ell \ne 5$, the Frobenius automorphism of $\mathbb{Q}({\zeta }_5)/\mathbb{Q}$ is ${\zeta }_5\mapsto {\zeta }_5^{\ell }$. Its restriction to $E$ thus corresponds exactly to the residue class of $\ell$ modulo $\{\pm 1\}$. Multiplicativity extends this to the Artin reciprocity map on all fractional ideals prime to $5$, whose kernel is exactly $P_{\mathbb{Q},1}((5))$. Its field discriminant $5$ also confirms that $E$ is unramified outside $5$; moreover $E$ is a totally real number field. The uniqueness in the Main theorem of class field theory now gives

Including the real place changes the ray class group: a principal generator must then be positive as well as congruent to one. The residue and signature group of a modulus givesHere the isomorphism sends $(u,\epsilon )$ to $\epsilon u$. Consequently ${\mathbb{Q}}_{(5)\infty }=\mathbb{Q}({\zeta }_5)$, of degree four. The finite modulus gives the real quadratic field; the modulus with the real place gives the full cyclotomic field. The printed finite-modulus assertion needs no correction.

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.