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Every ideal class of this real quadratic number field contains an integral ideal of ideal norm at most $\sqrt{14}<4$, by the diamond form of the Minkowski convex body theorem. The only possible norms are one, two and three. The sole prime ideal above two has norm two and is principal, generated by $4+\sqrt{14}$, whose field norm is two. The prime three is inert, since $X^2-14$ is irreducible modulo three, so there is no ideal of norm three. Every class therefore has a principal small representative, and the ideal class group is trivial.

This illustrates why the small-ideal bound can prove a class number without listing arbitrary ideals: only prime ideals of a few possible norms need to be tested. Real quadratic fields require the absolute field norm; the generator here happens to have positive norm.

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.

Let $d_K$ denote the field discriminant, and put $n=[K:\mathbb{Q}]\ge 2$. Let $r_1$ be the number of real field embeddings and $r_2$ the number of conjugate pairs of nonreal field embeddings, so $n=r_1+2r_2$. The Minkowski bound for ideal classes provides a nonzero integral ideal $A$ satisfyingRearranging gives the Minkowski lower bound for a number-field discriminantBecause $2r_2\le n$ and $\pi /4<1$, this is at leastApply the arithmetic-geometric mean inequality to $1,2,\dots ,n$:It follows thatwhere $n/(n+1)\ge 2/3$ for $n\ge 2$ was used. ThusThe source writes $|\Delta (K{)}^2|>1$. If $\Delta (K)$ denotes the embedding determinant, its square is $d_K$; if it denotes the field discriminant, the displayed result also implies the stated inequality. The conclusion therefore does not depend on that notational convention.

To compute the Ideal class group of Q of square root thirty-five, put $\theta =\sqrt{35}$. The ring of integers of a quadratic field formula gives $R=\mathbb{Z}[\theta ]$, and the field discriminant is $4\cdot 35=140$. Both field embeddings are real, so the Minkowski bound for ideal classes isEvery ideal class therefore has an integral ideal representative of ideal norm at most five. By prime ideal factorization and multiplicativity of the ideal norm, the ideal class group is generated by prime ideals of ideal norm at most five.

Apply the Dedekind factorization theorem to $X^2-35$. At two and five it has a repeated linear factor, givingAt three, $X^2-35\equiv X^2+1$ has no root, so $(3)$ itself is a prime ideal of ideal norm nine, outside the bound. These are all the relevant prime ideals: a prime ideal of ideal norm four would have to lie above two with residue degree two, but the factorization above two has only the single factor of degree one. Consequently $[P_2]$ and $[P_5]$ generate, and each has order of a group element dividing two.

The algebraic integer $5+\theta$ has field norm $25-35=-10$. It belongs to both $P_2$ and $P_5$. Hence its principal ideal is contained in $P_2\cap P_5=P_2{P}_5$. Both have ideal norm ten, soIt follows that $[P_5]=[P_2{]}^{-1}=[P_2]$, and the ideal class group is generated by one element of order of a group element at most two.

Finally, $P_2$ is not a principal ideal. If $P_2=(a+b\theta )$ with $a,b\in \mathbb{Z}$, the equality of ideal norm and absolute field norm would implyModulo five this requires $a^2\equiv 2$ or $3$, whereas squares modulo five are $0,1,4$. Thus $[P_2]\ne 1$, provingThis computes the ordinary ideal class group, with no positivity restriction on generators.

Let $\alpha =\sqrt[3]{3}$. The polynomial $X^3-3$ is Eisenstein at three, so the number field has degree of a field extension three. Use the permitted ring of integers $R=\mathbb{Z}[\alpha ]$, whose integral basis is $1,\alpha ,{\alpha }^2$. The field traces $\mathrm{Tr}(1)=3$, $\mathrm{Tr}(\alpha )=\mathrm{Tr}({\alpha }^2)=0$, $\mathrm{Tr}({\alpha }^3)=9$, and $\mathrm{Tr}({\alpha }^4)=0$ giveThere is one real field embedding and one conjugate pair of nonreal field embeddings. The Minkowski bound for ideal classes is thereforeEvery ideal class has an integral ideal representative of ideal norm at most four, so it is enough to show that every prime ideal of ideal norm at most four is a principal ideal.

Modulo two,where the quadratic factor is an irreducible polynomial over the finite field ${\mathbb{F}}_2$. The Dedekind factorization theorem givesNowThus $(\alpha -1)=P_2$, by containment and equality of ideal norms, and cancellation in the second identity gives $({\alpha }^2+\alpha +1)=Q_2$. This explicitly handles the prime ideal of ideal norm four, as well as the one of ideal norm two.

Modulo three the polynomial is $X^3$, soThese exhaust the prime ideals of ideal norm at most four. Every bounded representative factors into the listed principal ideals, and hence is a principal ideal. Therefore the Class group of Q of cube root of three 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

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 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.

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.

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 $4$.

Write $n=[L:K]$. The field trace gives a nondegenerate $K$-bilinear trace pairingThe inverse different, or codifferent, is the trace-dual latticeIt contains $S$, because the trace of an element integral over $R$ belongs to the integrally closed ring $R$. It is stable under multiplication by $S$: if $s,t\in S$ and $x\in {\mathfrak{D}}_{S/R}^{-1}$, then $\mathrm{Tr}(sxt)=\mathrm{Tr}(x(st))\in R$. Nondegeneracy of the trace pairing and finite generation of $S$ show that this trace dual is a finitely generated $R$-module spanning $L$. Therefore it is a fractional ideal of $S$.

Its inverseis the different ideal. Since $S\subseteq {\mathfrak{D}}_{S/R}^{-1}$, every such $x$ lies in $S$; hence ${\mathfrak{D}}_{S/R}$ is an integral ideal of $S$.

The discriminant ideal $\mathrm{disc}(S/R)$ is locally generated bywhere ${\alpha }_1,\dots ,{\alpha }_n$ is a local $R$-basis of $S$. Equivalently, it is the image of the determinant of the trace pairingThis formulation makes the definition independent of a basis, since changing a basis multiplies its discriminant by the square of the determinant of the change-of-basis matrix.

The asserted identity of ideals can be checked after localization at every nonzero prime ideal of $R$. We may therefore assume that $R$ is a discrete valuation ring and choose a basis ${\alpha }_1,\dots ,{\alpha }_n$ of $S$. Let ${\beta }_1,\dots ,{\beta }_n$ be its trace-dual basis, so $\mathrm{Tr}({\alpha }_i{\beta }_j)={\delta }_{ij}$; this is a basis of ${\mathfrak{D}}_{S/R}^{-1}$. If $T=(\mathrm{Tr}({\alpha }_i{\alpha }_j))$, thenThus the determinant measuring the inclusion $S\subseteq {\mathfrak{D}}_{S/R}^{-1}$ is $\det T$. The determinant description of the norm of a fractional ideal consequently givesLocalization then proves the equality over the original Dedekind domain.

Finally, the determinant-of-pairing map identifies the invertible $R$-module $({\det }_{R}S{)}^{\otimes 2}$ with the discriminant ideal. Hence in the ideal class groupup to the harmless inverse caused by the convention used to identify invertible modules with fractional ideals. In either convention the class is a square.

Distinct prime-power ideals are pairwise comaximal. For comaximal ideals, product equals intersection, so induction givesThe Chinese remainder theorem therefore gives the ring isomorphism

The ideal approximation theorem, proved by applying this Chinese-remainder map one prime power deeper, suppliesEquivalently, $v_{P_i}(\alpha )=m_i$ at every prime dividing $I$.

Since $(\alpha )\subseteq I$, the fractional idealis an integral ideal and $I{I}^{\prime }=(\alpha )$ is a principal ideal. At every $P_i\mid I$,
$v_{P_i}(I^{\prime })=v_{P_i}(\alpha )-m_i=0$, so $I+I^{\prime }={\mathcal{O}}_K$.

Choose $\beta \in I$ and $\gamma \in I^{\prime }$ with $\beta +\gamma =1$. Then
$\alpha ,\beta \in I$. For any $x\in I$,Thusso every ideal of ${\mathcal{O}}_K$ has a generating set of two elements.

Minkowski convex body theorem states that a centrally symmetric convex measurable subset of ${\mathbb{R}}^n$ with volume greater than $2^n$ times the covolume of a full Euclidean lattice contains a nonzero lattice point. The equivalent closed-boundary form follows by compactness for compact convex bodies.

The set $S_t$ is centrally symmetric. It is convex because for $0\le \lambda \le 1$, the triangle inequality gives $\Vert \lambda u+(1-\lambda )v{\Vert }_1\le \lambda \Vert u{\Vert }_1+(1-\lambda )\Vert v{\Vert }_1\le t$. It is a diamond of area $2t^2$. For the embedded nonzero integral ideal $I$, the given covolume is $2N(I)\sqrt{d}$. Hence Minkowski convex body theorem applies whenFor the resulting nonzero $\alpha \in I$, write $x={\sigma }_1(\alpha )$, $y={\sigma }_2(\alpha )$. ThenDecrease $t$ to $2\sqrt{N(I)\sqrt{d}}$. A fixed slightly larger diamond contains only finitely many lattice points, so one nonzero $\alpha$ recurs along a sequence of such $t$ values. Passing to the limit provesGiven a target ideal class $C$, choose an integral ideal $I$ in $C^{-1}$. Since $\alpha \in I$, the ideal $J=(\alpha )I^{-1}$ is integral, belongs to $C$, and has ideal normFor $K=\mathbb{Q}(\sqrt{14})$, every ideal class thus has an integral representative of norm at most $\sqrt{14}<4$. The only possibilities are norms one, two or three. The prime two ramifies, giving the unique ideal of norm two $P_2=(2,\sqrt{14})$. The element $4+\sqrt{14}$ has field norm $16-14=2$, so $(4+\sqrt{14})=P_2$, making it principal. The prime three is inert because $X^2-14\equiv X^2+1(\mathrm{mod}3)$ is an irreducible polynomial; its prime ideal has norm nine, so there is no ideal of norm three. Every possible small representative is principal, proving class number one for Q square root of fourteen. Therefore

Work in ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{-d}]$, since $d\equiv 2(\mathrm{mod}4)$. First $x$ must be odd: if it were even, coprimality would make $a$ odd, so $x^2+a^{2}d\equiv 2(\mathrm{mod}4)$, which cannot be an odd power $y^p$ for $p\ge 3$. Thus $y$ is odd too. Also no odd prime $\ell$ can divide both $x$ and $d$. Such a prime does not divide $a$, and square-freeness of $d$ gives $v_{\ell }(x^2+a^{2}d)=1$, whereas $v_{\ell }(y^p)$ is divisible by $p$. Hence $\gcd (x,d)=1$.

Put $\alpha =x+a\sqrt{-d}$. Suppose a prime ideal $P$ divides both $(\alpha )$ and $(\overline{\alpha })$, and let $\ell$ be the rational prime below it. It then contains $2x$ and $2a\sqrt{-d}$. The case $\ell =2$ is impossible because $N((\alpha ))=y^p$ is odd. If $\ell$ is odd, the first inclusion gives $\ell \mid x$, and $\gcd (a,x)=1$ makes $a$ nonzero in the residue field. The second inclusion therefore gives $\sqrt{-d}\in P$, so $\ell \mid d$, contradicting the previous paragraph. Thus $(\alpha )$ and $(\overline{\alpha })$ are coprime ideals. In particular they generate the unit ideal, the precise meaning of coprimality here; no unique factorization of elements is assumed.

Their product is $(y{)}^p$. Unique prime ideal factorization and disjoint prime support imply $(\alpha )=A^p$ for an integral ideal $A$. In the ideal class group, $[A{]}^p=1$. Since $p$ does not divide the class number, Lagrange's theorem forces $[A]=1$. Write $A=(u+v\sqrt{-d})$. Equality of the principal ideals gives $\alpha =\epsilon (u+v\sqrt{-d}{)}^p$. The only units are $\pm 1$, since a unit satisfies $u^2+d{v}^2=1$ and $d\ge 2$. Both signs are $p$th powers for odd $p$, so they can be absorbed into the generator:Apply the class-number obstruction from a ramified odd exponent. For the equation with $a=1$ and $p\mid d$, compare the coefficients of $\sqrt{-d}$ in this identity. The coefficient on the right isIts $j=1$ term is divisible by $p$, and every $j\ge 3$ term contains a factor $d$, so the whole coefficient is divisible by $p$. The coefficient on the left is one, a contradiction. Thus there are no positive integer solutions under the stated class-number and divisibility assumptions.

Finally,The integer $174=2\cdot 3\cdot 29$ is square-free, is two modulo four, and is divisible by three. If its imaginary quadratic class number were not divisible by three, the preceding result with $p=3$ would prohibit this solution. Consequently

Choose a nonzero integral ideal $B$ whose class is $[A{]}^{-1}$; one can multiply the fractional inverse $A^{-1}$ by a suitable nonzero integer to obtain such a $B$. By the given bound choose $0\ne \beta \in B$ with $|N(\beta )|\le c_{K}N(B)$. Put $I_A=(\beta )B^{-1}$. Since $(\beta )\subseteq B$, multiplication by $B^{-1}$ makes $I_A$ an integral ideal. Its class is $[B{]}^{-1}=[A]$, and multiplicativity of ideal norms givesThis proves the class representative bound without accidentally constructing a representative of the inverse class instead.

To determine the Ideal class group of Q of square root minus thirty-four, write $\eta =\sqrt{-34}$. The given integral basis has discriminant $-136$, so the imaginary-quadratic bound is $4\sqrt{34}/\pi <8$. Every class therefore has a representative of ideal norm at most seven, and factoring such representatives shows that the ideal class group is generated by prime ideals of ideal norm at most seven.

Two ramifies: let $P_2=(2,\eta )$, with $P_2^2=(2)$. Three is inert because $X^2+34\equiv X^2+1$ is irreducible modulo three, so its prime ideal ideal norm is nine and supplies no small generator. At five and seven the roots are $\pm 1$, so the primes of ideal norm five and seven are conjugate pairs. Set $P_5=(5,\eta -1)$ and $P_7=(7,\eta +1)$. Their conjugates have inverse ideal classes. No new prime ideal has ideal norm four, since the only prime above two has residue degree one; six is not a prime power. Thus $[P_2],[P_5],[P_7]$ generate the ideal class group.

The element field norms $N(4+\eta )=50$ and $N(1+\eta )=35$, together with reduction at the split primes, giveThe ideal norm products verify both exponents. Consequently $[P_5{]}^2=[P_2]$ and $[P_7]=[P_5]$, so the whole group is generated by $[P_5]$ and its order divides four. The class $[P_2]$ is not trivial: a generator would be an element of the ring of integers $a+b\eta$ of field norm two, but $a^2+34b^2=2$ has no integer solution. Since $[P_2{]}^2=1$, its order is exactly two. Hence $[P_5]$ has order four and

For integral solutions of x squared minus two equals y cubed, the equation implies $y\ge -1$, so $y+2>0$. If $y$ were even, then $x$ would be even and $x^2-2$ would be two or six modulo eight, whereas $y^3$ would be zero modulo eight. Thus $y$ and $x$ are odd. Also $3\nmid x$: otherwise $x^2-2\equiv 7(\mathrm{mod}9)$, which is not a cubic residue modulo nine. Reducing modulo three then gives $y\equiv 2(\mathrm{mod}3)$. ConsequentlyPut $\eta =\sqrt{-6}$ and $\alpha =x+(y+1)\eta$. Its field norm is $(y+2{)}^3$. The principal ideals $(\alpha )$ and $(\bar{\alpha })$ are coprime. Indeed a common prime ideal has underlying rational prime $p$ dividing $y+2$. The displayed gcd excludes two and three, so both two and $\eta$ are invertible modulo that prime ideal. The difference $\alpha -\bar{\alpha }=2(y+1)\eta$ therefore implies $p\mid y+1$, contradicting $p\mid y+2$.

Their product is $(y+2{)}^3$, so unique factorization of ideals in a number field and coprimality imply $(\alpha )=A^3$ for an integral ideal $A$. In the given ideal class group of order two, $[A{]}^3=1$ implies $[A]=1$. Hence $A=(u+v\eta )$ with integers $u,v$. Equality of principal ideals gives $\alpha =\epsilon (u+v\eta {)}^3$. The only units are $\pm 1$, and both are cubes, so the sign can be absorbed into $u,v$. Taking norms, and using $y+2>0$, givesThe coefficient of $\eta$ gives $y+1=3u^{2}v-6v^3$. Eliminating $y$ with the field norm identity yieldsThe denominator is nonzero for integral $v$. Polynomial division makes the integrality obstruction especially clear:If $u$ is integral, the reciprocal must be an integer, forcing $3v-1=\pm 1$. The only integral possibility is $v=0$. Then $u^2=1$, the field norm gives $y=-1$, and the real coefficient in the cube gives $x=u^3=\pm 1$. Both pairs directly satisfy the original equation. Therefore all integral solutions are

ideals are coprime when $A+B={\mathcal{O}}_K$, equivalently $1=a+b$ for some $a\in A$, $b\in B$. Always $AB\subseteq A\cap B$. If $x\in A\cap B$, then $x=xa+xb\in AB$, giving $\boxed{AB=A\cap B}$ in the coprime case. For nonzero integral ideals, the Chinese remainder theorem gives ${\mathcal{O}}_K/(AB)\cong {\mathcal{O}}_K/A\times {\mathcal{O}}_K/B$, and taking finite cardinalities yields $\boxed{N(AB)=N(A)N(B)}$.

For a nonzero prime ideal, the chain of finite-index inclusions givesby the given hypothesis. Distinct nonzero prime ideals are maximal and coprime, as are their positive powers. Unique factorization of ideals in a number field therefore gives $N(A)={\prod }_{P}N(P{)}^{v_P(A)}$. Exponents add under ideal multiplication, proving multiplicativity for arbitrary nonzero integral ideals. This is the standard domain of the finite ideal norm used here.

For the Dedekind factorization theorem, choose an integral primitive element $\theta$ of $K$ with monic minimal polynomial of an algebraic element $f\in \mathbb{Z}[X]$, and assume the rational prime $p$ does not divide the index $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$. Factorover ${\mathbb{F}}_p$, with distinct monic irreducible factors. For integer polynomial lifts $f_i$, the distinct prime ideals above $p$ are $P_i=(p,f_i(\theta ))$, andThus the residue degree is $\mathrm{deg}{\bar{f}}_i$, the ramification index is $e_i$, and ${\sum }_i{e}_i\mathrm{deg}{\bar{f}}_i=[K:\mathbb{Q}]$. The index restriction is essential to this direct factorization formula.

For $K=\mathbb{Q}(\sqrt{-2})$, the ring of integers is $\mathbb{Z}[\sqrt{-2}]$, so the index is one and the theorem applies to every prime. Modulo two, $X^2+2=X^2$, giving one prime ideal of ideal norm two. For an odd rational prime, a prime ideal has ideal norm $p$ if the polynomial splits, and ideal norm $p^2$ if it is irreducible. In the former case the supplied condition forces $p\equiv 1$ or $3(\mathrm{mod}8)$; in the latter case every odd square is one modulo eight. ConsequentlyThe odd part of any ideal norm is thus one or three modulo eight, because $3^2\equiv 1$. Multiplication by a power of two leaves possible residues $1,3$ for exponent zero, $2,6$ for exponent one, $4$ for exponent two, and $0$ for exponent at least three. Hence $\boxed{N(I)\not\equiv 5,7(\mathrm{mod}8)}$ for every nonzero integral ideal.

The product of ideals $AB$ consists of finite sums ${\sum }_j{a}_j{b}_j$ with $a_j\in A$, $b_j\in B$. Ideal divisibility $P\mid A$ means $A=PC$ for an integral ideal $C$; in this ring it is equivalent to $A\subseteq P$. The ideal norm is $N(A)=|{\mathcal{O}}_K/A|$, equivalently its additive lattice index.

If the prime ideal $P$ divides $AB$ but does not divide $A$, choose $a\in A\setminus P$. For every $b\in B$, $ab\in AB\subseteq P$, so primality gives $b\in P$. Thus $B\subseteq P$, proving $P\mid B$.

Every nonzero proper ideal $A$ is contained in a maximal ideal $P$, which is prime. Then $P\mid A$, so $A=PB$. Norm multiplicativity gives $N(B)=N(A)/N(P)<N(A)$ because $N(P)\ge 2$. Induction on the positive integer $N(A)$ gives a prime-ideal factorization, ending at norm one. For uniqueness, a prime in one factorization divides some prime in the other. Every nonzero prime ideal is maximal, since its quotient is a finite integral domain and hence a field. The two primes therefore coincide. Cancel that common factor using the permitted cancellation law, and continue. This proves uniqueness up to order.

The finite additive group ${\mathcal{O}}_K/A$ has order $N(A)$, so that integer annihilates the class of $1$. ThereforeIn particular an ideal of norm $m$ contains $(m)$. Such ideals correspond to certain additive subgroups of the finite group ${\mathcal{O}}_K/(m)$, so there are finitely many. Finiteness of these quotients follows from an integral basis; a nonzero ideal contains a nonzero principal ideal, whose index is the absolute field norm of its generator.