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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.

For a number field $k$, its ordinary Hilbert class field $H_k$ is the maximal finite abelian extension unramified at every finite prime and with every real place splitting completely. The Artin reciprocity map inducesThe image of an unramified prime ideal is its Frobenius automorphism. Thus a prime splits completely in $H_k$ precisely when its ideal class is trivial. This field packages ideal-class obstructions into splitting behavior. Ideals become principal in $H_k$ by the principal ideal theorem. Specifying the real-place convention matters for real fields: omitting it leads instead to the narrow class-field problem. Both illustrative fields below are imaginary quadratic, so there are no real places to discuss.

First consider $k=\mathbb{Q}(\sqrt{-23})$, whose quadratic discriminant is $-23$. The correspondence between ideal classes and reduced primitive positive definite binary quadratic forms lets us compute its class number without guessing a class field. A reduced form $(a,b,c)$ obeys $|b|\le a\le c$, $b^2-4ac=-23$, and $a\le \sqrt{23/3}<3$; a boundary case uses $b\ge 0$. For $a=1$ the only class is $(1,1,6)$; for $a=2$ the two possibilities are $(2,1,3)$ and $(2,-1,3)$. Hence $h_k=3$ and its ideal class group is cyclic of order three.

Take $g(X)=X^3-X-1$. Its only possible rational roots are $\pm 1$, neither a root, so it is irreducible. Its polynomial discriminant is $4-27=-23$, which is nonsquare. Thus its splitting field $H$ has Galois group $S_3$ and the quadratic fixed field of $A_3$ is $k$, generated by the square root of its discriminant. Therefore $H/k$ is cyclic of degree three and $H=k(\theta )$ for a root $\theta$ of $g$.

At a prime other than $23$, the separable residue polynomial splits over a finite residue-field extension; Hensel's lemma puts all its roots in an unramified local extension. At $23$ there is the exact factorizationThe simple factor lifts by Hensel's lemma. The remaining quadratic factor has discriminant valuation one, because the cubic discriminant is $-23$ and the resultant with the simple factor is a unit. Its local splitting field is ramified quadratic, with inertia group of order two. Its intersection with $A_3$ is trivial, so the relative extension $H/k$ is unramified at $23$ as well. There are no real places. An unramified abelian extension of degree $h_k=3$ must be all of $H_k$, givingThis is the Hilbert class field of Q of square root minus twenty-three.

The other alternative, $k=\mathbb{Q}(\sqrt{-30})$, has quadratic discriminant $-120$. Reduced forms have $a\le \sqrt{40}<7$ and even $b$. Testing $1\le a\le 6$, $|b|\le a$, and $c=(b^2+120)/(4a)\ge a$ gives exactlyThus $h_k=4$. Each class is its own inverse, because replacing $b$ by $-b$ leaves these representatives unchanged; consequently the ideal class group is $C_2\times C_2$.

Set $H=\mathbb{Q}(\sqrt{2},\sqrt{-3},\sqrt{5})$. The square classes $2,-3,5$ are independent, so $[H:\mathbb{Q}]=8$. Their product gives $\sqrt{-30}$, and $H/k$ is abelian of degree four. Its seven quadratic subfields have discriminantsFor a multiquadratic abelian extension, the conductor-discriminant formula takes the product of these quadratic character conductors, namely the absolute quadratic discriminants. Hence $|d_H|=120^4$. The relative discriminant tower formula givesThe integral relative discriminant ideal therefore has norm one and is the unit ideal. Thus $H/k$ is everywhere unramified, and its degree equals the class number. This proves the other explicit answer:It agrees with the Hilbert class field of Q of square root minus thirty. Both alternatives have been computed independently from their class numbers and ramification.

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

Put $t=\sqrt{-30}$ and $K=\mathbb{Q}(t)$. The ring of integers is $\mathbb{Z}[t]$, and the fundamental discriminant is $D_K=-120$. For a reduced positive definite binary quadratic form $(a,b,c)$ of discriminant $-120$, reduction givesThe coefficient $b$ is even; checking $1\le a\le 6$ and $|b|\le a$, with the usual nonnegative sign choice at boundary equalities, leaves preciselyThe correspondence between proper quadratic forms and ideal classes therefore gives $h_K=4$. All four classes equal their inverses, since inversion replaces $b$ by $-b$. HenceThe three nontrivial classes may be represented by the ramified prime ideals ${\mathfrak{p}}_2=(2,t)$, ${\mathfrak{p}}_3=(3,t)$ and ${\mathfrak{p}}_5=(5,t)$. They square to principal ideals and satisfy ${\mathfrak{p}}_2{\mathfrak{p}}_3{\mathfrak{p}}_5=(t)$. None is principal: an integral generator of norm $2$, $3$ or $5$ would require $a^2+30b^2$ to equal that number, which is impossible. The relation then makes the three classes distinct, consistently with the four-form calculation.

ConsiderThe three rational square classes are independent, so $[H:\mathbb{Q}]=8$. Their radical product is $\sqrt{-30}$, so $K\subset H$ and $[H:K]=4$; the extension $H/K$ is abelian. To verify that it is unramified, rather than assuming that radical adjunctions are harmless at $2,3,5$, use the conductor-discriminant formula. The seven quadratic subfields have fundamental discriminantsThe formula givesThe discriminant tower identity now givesThus the relative discriminant is the unit ideal and no finite prime ramifies. The base has no real places, so there is no infinite ramification to check. This unramified abelian degree-four extension has degree $h_K$, and part (i) provesThis is the Hilbert class field of Q of square root minus thirty.

Next translate prime representation into an ideal class. For $p\nmid 30$, we claimIf a representation exists, $\alpha =2x+yt$ lies in ${\mathfrak{p}}_2$ and has $N_{K/\mathbb{Q}}\alpha =2p$. The integral ideal $(\alpha ){\mathfrak{p}}_2^{-1}$ has norm $p$, so it is a prime ideal $\mathfrak{P}$. Its class is $[{\mathfrak{p}}_2{]}^{-1}=[{\mathfrak{p}}_2]$. Conversely, if $[\mathfrak{P}]=[{\mathfrak{p}}_2]$, the product ${\mathfrak{p}}_2\mathfrak{P}$ is principal, say $(\alpha )$. Since this ideal lies in ${\mathfrak{p}}_2$, its generator has the form $\alpha =2x+yt$ with integers $x,y$. Taking norms gives $4x^2+30y^2=2p$, proving the reverse implication. This is prime representation via an ideal class, and it proves sufficiency as well as necessity.

Identify the required class in $\mathrm{Gal}(H/K)$ through the Artin map. We can use the explicit representation $17=2+15$: $2+t$ has norm $34$, so the prime above $17$ obtained from $(2+t){\mathfrak{p}}_2^{-1}$ has class $[{\mathfrak{p}}_2]$. Its Frobenius in the multiquadratic extension acts on the three radicals with signsTheir product is one, so this automorphism fixes $K$. Because $H/\mathbb{Q}$ is abelian, for any unramified rational prime $p$ that splits in $K$, the Frobenius at either prime of $K$ of norm $p$ is its rational Frobenius restricted to $H/K$. The Artin isomorphism therefore makes the required class condition equivalent toConversely, these signs already imply splitting in $K$, since their product is $(\frac{-30}{p})=1$. No extra splitting condition is missing.

The supplementary law for $2$, quadratic reciprocity for $-3$ and $5$, and then the Chinese remainder theorem turn these sign conditions intoTheir four simultaneous residue classes modulo $120$ are $17,23,47,113$. Finally, $2$ is represented by $(x,y)=(\pm 1,0)$, whereas $3$ and $5$ cannot be represented: $y\ne 0$ gives a value at least $15$, and $y=0$ gives twice a square. We have proved the complete criterion

The ring of integers is ${\mathcal{O}}_k=\mathbb{Z}[\sqrt{7}]$, since $7\equiv 3(\mathrm{mod}4)$, and its discriminant is $28$. Rationalizing givesBoth are algebraic integers, so $\epsilon$ is a unit. Set $\mathfrak{p}=(3+\sqrt{7})$. Since $(3+\sqrt{7}{)}^2=2\epsilon$,Also $N(\mathfrak{p})=|N(3+\sqrt{7})|=2$. It is the unique ideal of ideal norm two: any quotient of size two is ${\mathbb{F}}_2$, and the image of $\sqrt{7}$ must satisfy $z^2=1$, hence be one. The kernel of a ring homomorphism onto ${\mathbb{F}}_2$ is uniquely $(2,1+\sqrt{7})$, which is $\mathfrak{p}$.

The given Minkowski constant means every ideal class has an integral representative of ideal norm at most $\frac{1}{2}\sqrt{28}=\sqrt{7}<3$. Such a ideal norm is one or two. The norm-one ideal is the whole ring, and the sole norm-two ideal is principal. Thus the class number is one.

If $x^2-7y^2=2$, the principal ideal $(x+y\sqrt{7})$ has ideal norm two and hence equals $\mathfrak{p}$. Its generator differs from $3+\sqrt{7}$ by a unit. Conversely multiplication by a norm-one unit preserves the field norm two. With the assumed fundamental unit, all units are $\pm {\epsilon }^n$; in fact a norm-minus-one unit is impossible since $x^2\equiv -1(\mathrm{mod}7)$ has no solution. Therefore the complete answer isFor $n=1$ and the positive sign, the product is $45+17\sqrt{7}$, giving $(x,y)=(45,17)$; $45^2-7\cdot 17^2=2$ checks the result.

Since $26\equiv 2(\mathrm{mod}4)$, the ring of integers is $\mathbb{Z}[\sqrt{26}]$ with discriminant $104$. Write $P=(2,ϵ+1)=(2,\sqrt{26})$ and $Q_{\pm }=(5,ϵ\pm 1)$. Reduction of $X^2-26$ modulo two gives a double root, and modulo five gives the roots $\pm 1$. Thus $N(P)=2$, $N(Q_{\pm })=5$, and $(5)=Q_{+}{Q}_{-}$. Directly $P^2=(4,2\sqrt{26},26)=(2)$, since $26-6\cdot 4=2$. The element $ϵ+1=6+\sqrt{26}$ belongs to $P$ and $Q_{+}$ and has field norm ten, so equality of ideal norms givesThe Minkowski bound is $\sqrt{104}/2=\sqrt{26}<6$, so every ideal class contains an integral ideal of ideal norm at most five. Three is inert and gives no prime ideal of ideal norm three; ideals of ideal norm four are $P^2$, principal. Thus the class group is generated by $P,Q_{+},Q_{-}$, and the displayed relations make every class either trivial or $[P]$. The class $[P]$ is nontrivial: a generator would have field norm $\pm 2$, but $x^2-26y^2=\pm 2$ would give $x^2\equiv 2$ or $11(\mathrm{mod}13)$, neither a quadratic residue. Hence the class number is two.

The unit $ϵ=5+\sqrt{26}$ has field norm $-1$. To show it is fundamental, suppose $1<u<ϵ$ is a positive real embedding of a unit $x+y\sqrt{26}$. Its conjugate is $\pm 1/u$, so $y=(u-u^{\prime })/(2\sqrt{26})$ is a positive integer less than two. Thus $y=1$, and $x^2-26=\pm 1$ forces $x=\pm 5$. Only $5+\sqrt{26}$ is greater than one, contrary to the strict inequality. Therefore no smaller unit exceeds one. Multiplying any positive unit by a suitable power of ${ϵ}^{-1}$ puts it in $[1,ϵ)$, proving that all units are $\pm {ϵ}^n$.

An element of field norm $\pm 10$ has principal ideal $P{Q}_{+}$ or $P{Q}_{-}$ by prime ideal factorization. The first is generated by $ϵ+1$; the second by $ϵ-1=4+\sqrt{26}$, of field norm $-10$. Two generators of the same principal ideal differ by a unit. Therefore all, and only, the integer solutions areThe two displayed signs are independent; multiplication by a norm-minus-one unit exchanges the two field norm signs.

The ideal ${\mathfrak{p}}_2=(2,\alpha )$ has ideal norm two. Its square is contained in $(2)$ and has norm four, so $(2)={\mathfrak{p}}_2^2$. A generator would have field norm $u^2+26v^2=2$, impossible for integers $u,v$. Thus its ideal class has order two.

Modulo three, $X^2+26=(X-1)(X+1)$, giving $(3)={\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$ with ${\mathfrak{p}}_3=(3,1-\alpha )$. A generator of ${\mathfrak{p}}_3$ would have norm three, again impossible. If ${\mathfrak{p}}_3^2$ were principal, its generator would have norm nine. The only possibilities are $\pm 3$, whose ideal is ${\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$, not ${\mathfrak{p}}_3^2$.

Finally $N(1-\alpha )=27$. This element lies in ${\mathfrak{p}}_3$ but not ${\overline{\mathfrak{p}}}_3$, because $\alpha \equiv -1$ in the latter quotient. Unique unique factorization of ideals in a number field and norms therefore give

Modulo five, $X^2+26=(X-2)(X+2)$. Choose ${\mathfrak{p}}_5=(5,2+\alpha )$; then $(5)={\mathfrak{p}}_5{\overline{\mathfrak{p}}}_5$, and $N(2+\alpha )=4+26=30$. The element $2+\alpha$ belongs to ${\mathfrak{p}}_2$, ${\mathfrak{p}}_3$ and ${\mathfrak{p}}_5$. Their product has norm $2\cdot 3\cdot 5=30$, soThe imaginary-quadratic Minkowski bound for ideal classes is $(2/\pi )\sqrt{104}<7$. Every ideal class has an integral representative of norm at most six, whose prime ideal factors lie over two, three or five. Thus these prime ideals generate the ideal class group. Conjugate classes are inverses, and the displayed relation eliminates the class over five. The remaining generators have orders two and three, so their product has order six and generates the whole group:

Write $r=\sqrt{10}$. The ideals are ${\mathfrak{p}}_2=(2,r)$, ${\mathfrak{p}}_3=(3,r+1)$ and ${\overline{\mathfrak{p}}}_3=(3,r-1)$, with residue fields of orders 2, 3 and 3 respectively. Thus their ideal norms are 2, 3 and 3.

The generators of ${\mathfrak{p}}_2^2$ are $4,2r,10$. They are multiples of 2, and the integer combination of 4 and 10 giving 2 proves ${\mathfrak{p}}_2^2=(2)$. Similarly ${\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$ is generated by $9,3r+3,3r-3,r^2-1=9$. The difference of the middle generators is 6; together with 9 this generates 3. Therefore the product is $(3)$.

The ideals ${\mathfrak{p}}_2$ and ${\mathfrak{p}}_3$ are coprime, so their product equals their intersection. The element $4+r$ belongs to both, hence $(4+r)\subseteq {\mathfrak{p}}_2{\mathfrak{p}}_3$. Both ideals have norm 6, so equality holds. Likewise $2+r$ belongs to ${\mathfrak{p}}_2$ and ${\overline{\mathfrak{p}}}_3$, and its norm is $-6$. ThusThe discriminant is 40. The Minkowski bound for ideal classes in a real quadratic field is $\sqrt{40}/2=\sqrt{10}<4$, so every ideal class has an integral representative of norm 1, 2 or 3. The only possible prime ideals with those norms are the ones just listed. Their classes satisfy $[{\mathfrak{p}}_2{]}^2=1$ and $[{\mathfrak{p}}_3]=[{\overline{\mathfrak{p}}}_3]=[{\mathfrak{p}}_2]$. Thus there are at most two classes. The ideal ${\mathfrak{p}}_2$ is not principal, since a generator would have norm $\pm 2$, already ruled out. Therefore the class number is exactly two.

Choose a nonzero integral ideal $I$ in the inverse of any desired ideal class. For the nonzero element $x\in I$ supplied by the norm bound, $J=(x)I^{-1}$ is integral, belongs to the desired class and satisfiesThere are only finitely many integral ideals of bounded norm: identify ${\mathcal{O}}_K$ with ${\mathbb{Z}}^n$ as an additive group, so ideals of norm $m$ are among its subgroups of index $m$, of which there are finitely many. Since every ideal class has one of these bounded-norm representatives, the ideal class group is finite.

Put $w=\sqrt{-14}$. The ring of integers is $\mathbb{Z}[w]$, with discriminant $-56$. The Minkowski bound for ideal classes for an imaginary quadratic field says that every ideal class has an integral representative of norm at most $(2/\pi )\sqrt{56}<5$. Therefore the ideal class group is generated by primes above two and three: norm-four representatives contribute only products of primes above two.

The prime two ramifies: ${\mathfrak{p}}_2=(2,w)$ has norm two and ${\mathfrak{p}}_2^2=(2)$. It is not principal, since an element of norm two would require $a^2+14b^2=2$, which has no integral solution. Thus $[{\mathfrak{p}}_2]$ has order two. Since $X^2+14=(X-1)(X+1)$ modulo three, three splits into ${\mathfrak{p}}_3=(3,w-1)$ and its conjugate, whose classes are inverses.

The element $2+w$ has norm $18$. It belongs to ${\mathfrak{p}}_3$ but not its conjugate, and its norm has exactly one factor two. Unique prime ideal factorization therefore givesHence $[{\mathfrak{p}}_3{]}^2=[{\mathfrak{p}}_2{]}^{-1}=[{\mathfrak{p}}_2]\ne 1$. Since all generating classes are powers of $[{\mathfrak{p}}_3]$, $\boxed{\mathrm{Cl}(\mathbb{Q}(\sqrt{-14}))\cong \mathbb{Z}/4\mathbb{Z}}$.

Put $s=\sqrt{-17}$. The ring of integers of a number field is $\mathbb{Z}[s]$ and the field discriminant is $-68$. The Minkowski bound for ideal classes says that each ideal class has an integral ideal representative of norm at most $(2/\pi )\sqrt{68}<6$.

The prime two ramifies as $(2)={\mathfrak{p}}_2^2$, where ${\mathfrak{p}}_2=(2,1+s)$. Three splits as $(3)={\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$, choosing ${\mathfrak{p}}_3=(3,1+s)$, whereas five is inert because $-17\equiv 3(\mathrm{mod}5)$ is not a square. The norm bound therefore leaves only the classes of $1$, ${\mathfrak{p}}_2$, ${\mathfrak{p}}_3$ and ${\overline{\mathfrak{p}}}_3$; a norm-four ideal is ${\mathfrak{p}}_2^2$ and is principal.

The ideal ${\mathfrak{p}}_2$ is not principal, since an element of norm two would solve $a^2+17b^2=2$. On the other hand $N(1+s)=18$, and $1+s$ is divisible by ${\mathfrak{p}}_2$ and ${\mathfrak{p}}_3$ but not ${\overline{\mathfrak{p}}}_3$. Unique prime ideal factorization and the norm give $(1+s)={\mathfrak{p}}_2{\mathfrak{p}}_3^2$. Thus $[{\mathfrak{p}}_3{]}^2=[{\mathfrak{p}}_2]$, which has order two, and

For the Diophantine equation, $x$ must be positive. If $x$ were even, $y^2\equiv 15(\mathrm{mod}32)$, impossible for a square. Thus $x$ is odd and $y$ even. If $17\mid x$, then $17\mid y$, but $y^2+17$ would have 17-adic valuation one while $x^5$ has valuation divisible by five, also impossible.

The ideals $(y+s)$ and $(y-s)$ are consequently coprime. A common prime divides $2s$ and hence lies over two or seventeen, both excluded above. Their product is $(x{)}^5$, so unique prime ideal factorization gives $(y+s)={\mathfrak{a}}^5$. Since the ideal class group has order four, $[\mathfrak{a}{]}^5=1$ forces $[\mathfrak{a}]=1$. Write $\mathfrak{a}=(a+bs)$ with integers $a,b$. The only units are $\pm 1$ by their norm, and the sign can be absorbed into the fifth power. Hence $y+s=(a+bs{)}^5$.

Comparing the coefficient of $s$ givesFor $b=1$, the bracket would be one, but it is four modulo five. For $b=-1$, the bracket would be minus one, but it is one modulo three for either possible square residue of $a$. Both cases are impossible. Therefore there are no integer solutions.

Classical Kummer theory relates extraction of $n$th roots to Galois cohomology. In characteristic zero the exact sequenceand Hilbert theorem 90 identify $H^1(K,{\mu }_n)$ with $K^{\times }/K^{\times n}$. For an elliptic curve, the corresponding Kummer exact sequence of an elliptic curve isMultiplication by $n$ is surjective over the algebraic closure. If $nQ=P\in E(K)$, the cocycle $\sigma \mapsto \sigma Q-Q$ takes values in $E[n]$. Changing the choice of $Q$ changes it by a coboundary, and changing $P$ by an element of $nE(K)$ does not change its class. Conversely, a trivial cocycle class allows $Q$ to be adjusted by an $n$-torsion point to become $K$-rational. Thus the Kummer map of an elliptic curve is an injectionThis is the elliptic form of Kummer theory. The full cohomology group need not be finite; the arithmetic restriction on these classes is essential.

Choose a finite set $S$ of places containing the archimedean places, the primes over $n$, and all primes of bad reduction of an elliptic curve. At a finite place outside $S$, $E$ has good reduction and $n$ is a unit. A division point of the reduction of $P$ exists over the algebraic closure of the residue field. Smooth lifting gives a point over a finite unramified extension whose multiple differs from $P$ by an element of the kernel of reduction of an elliptic curve. In the formal group of an elliptic curve, $[n](T)=nT+O(T^2)$ is an isomorphism by the invertible morphism criterion for formal group laws, and its integral inverse converges on the maximal ideal. Correcting that difference produces a division point in the maximal unramified extension. Hence the Kummer map of an elliptic curve class is unramified outside $S$.

To prove finiteness explicitly, choose a finite Galois extension $L/K$ containing all $E[n]$ and all $n$th roots of unity, and enlarge $S$ to include its ramified places. A basis of $E[n]\cong (\mathbb{Z}/n\mathbb{Z}{)}^2$ identifies it over $L$ with ${\mu }_n^2$. Classical Kummer theory then identifiesThe restriction of every class in the image of $\delta$ belongs to $L(S_L,n{)}^2$, where the S-unramified power class group isIndeed an unramified local Kummer extension at residue characteristic prime to $n$ has valuation divisible by $n$: in an unramified field containing a root, $nv(a^{1/n})=v(a)$ with integral valuations.

The finiteness of S-unramified Kummer classes follows from the exact sequenceTo see the final map, write the ideal of $a$ away from $S_L$ as ${\mathfrak{a}}^n$ and take the ideal class of $\mathfrak{a}$. Its kernel is represented by an S-unit, after division by an $n$th power; conversely an $n$-torsion ideal class yields such an $a$. The S-unit group is finitely generated by the Dirichlet unit theorem together with the finitely many inverted primes. The ideal class group of the localized ring is a quotient of the finite ordinary ideal class group. Both outer groups are therefore finite.

Finally restriction has finite kernel: inflation-restriction puts it in the finite group $H^1(\mathrm{Gal}(L/K),E[n])$. Thus the image of $\delta$ has finite restriction image and finite kernel, andThis proves the weak Mordell-Weil theorem. Combining it with the height descent lemma proves the full Mordell-Weil theorem. Local restrictions at every place refine the finite group used here to the Selmer group of an elliptic curve, which is useful for explicit descent calculations.

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 $4$.

Here ${\mathcal{O}}_K=\mathbb{Z}[s]$ with $s=\sqrt{-14}$ and field discriminant $-56$. The imaginary-quadratic Minkowski bound for ideal classes is $(2/\pi )\sqrt{56}<5$, so each ideal class has an integral representative of ideal norm at most four. Representatives therefore use prime ideals above two and three. PutThe prime two ramifies, with ${\mathfrak{p}}^2=(2)$, and three splits into $\mathfrak{q}\overline{\mathfrak{q}}=(3)$. No element has field norm two, because $N(a+bs)=a^2+14b^2$; thus $[\mathfrak{p}]$ is nontrivial of order two. Norm-four ideals represent $[\mathfrak{p}{]}^2=1$ and add no generator, so the class group is generated by $[\mathfrak{p}],[\mathfrak{q}]$.

The element $2-s$ has field norm 18. It belongs to $\mathfrak{p}$ and $\mathfrak{q}$, but not $\overline{\mathfrak{q}}$, since reduction there gives $s\equiv 1(\mathrm{mod}3)$. Its ideal therefore factors asThe exponents follow from the ideal norm 18, or from the root $s\equiv 2(\mathrm{mod}9)$ defining ${\mathfrak{q}}^2$. Hence $[\mathfrak{q}{]}^2=[\mathfrak{p}]$, so $[\mathfrak{q}]$ has order four and already generates all possible classes. ThereforeAs a separate check, the reduced forms of discriminant of a binary quadratic form $-56$ are $[1,0,14],[2,0,7],[3,2,5],[3,-2,5]$, giving exactly four classes.

Put $K=\mathbb{Q}(\sqrt{-6})$. Its fundamental discriminant is $-24$. The Minkowski bound for ideal classes isso every ideal class contains an integral ideal of norm $1$, $2$ or $3$. The primes above $2$ and $3$ are ramified, their classes have order at most two, andshows that they represent the same class. This class is nontrivial because the norm form $a^2+6b^2$ does not represent $2$. Hence

Now let $H=K(\sqrt{2})=\mathbb{Q}(\sqrt{2},\sqrt{-3})$. The discriminant of a biquadratic field is the product of the discriminants of its three quadratic subfields, soThe relative discriminant ${\mathfrak{D}}_{H/K}$ therefore has norm one, proving that $H/K$ is unramified at every finite prime. Since $[H:K]=2=|\mathrm{Cl}(K)|$, the Hilbert class field of Q of square root minus six is

Let $I$ be an integral ideal of an imaginary quadratic number field and $(e_1,e_2)$ an integral basis of $I$. Its normalized norm is a positive definite binary quadratic form. An unramified rational prime $p$ is represented by this form exactly when there is an integral prime ideal $P$ of norm $p$ with $[P]=[I{]}^{-1}$. Indeed, a representation gives an element $\alpha \in I$ with $N\alpha =pN(I)$, so $P=(\alpha )I^{-1}$ is integral of norm $p$. Conversely, $IP=(\alpha )$ with $\alpha \in I$ produces a representation by expressing $\alpha$ in the basis. Artin reciprocity then detects this ideal class by Frobenius in the Hilbert class field.

Given finitely many distinct places of a number field $v_1,\dots ,v_r$, the diagonal image of $K$ is dense in ${\prod }_i{K}_{v_i}$. One can therefore simultaneously prescribe finite congruences and real signs. This is useful in the ray class exact sequence and in finding representatives of an ideal class prime to a prescribed finite set of prime ideals.