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

The ideal-form correspondence for imaginary quadratic fields and the reduction bound $a\le \sqrt{31/3}$ give exactly the forms $[1,1,8]$, $[2,1,4]$ and $[2,-1,4]$. Thus the class number is three and the ideal class group is cyclic of order three.

Every field $\mathbb{Q}({\zeta }_{59^r})$ contains $\mathbb{Q}(\sqrt{-59})$ and is totally ramified over it at the prime above $59$. The class number divisibility under total ramification therefore shows that its class number is divisible by three. Its ring of integers is not a principal ideal domain. The degrees $58\cdot 59^{r-1}$ strictly increase, giving infinitely many distinct examples.

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 $\omega =(1+\sqrt{229})/2$. Since $229\equiv 1(\mathrm{mod}4)$ is squarefree,We first compute the ideal class group, rather than assume the proposed extension has the full Hilbert-class-field degree.

The real quadratic Minkowski bound for ideal classes is $\sqrt{229}/2<8$. Thus every ideal class has an integral representative of norm at most seven. The prime two is inert because $229\equiv 5(\mathrm{mod}8)$, while seven is inert because $229\equiv 5(\mathrm{mod}7)$ is a nonsquare. The inert ideal $(2)$ has norm four and is principal; a prime above seven has norm forty-nine. The only possible nonprincipal prime factors of these small representatives are therefore primes above three and five.

Reduction of $X^2-X-57$ gives roots $0,1$ modulo three and $2,4$ modulo five. WriteTheir conjugate pairs multiply to $(3)$ and $(5)$. Direct norm calculations giveThe element $6+\omega$ vanishes at the residue roots $0$ modulo three and $4$ modulo five, and not at the conjugate roots. Its absolute norm forces each exponent to be one. Likewise $15+2\omega$ vanishes only at the root zero modulo three, and its norm forces exponent three. ThusThese also give the factorizations suggested by the hint: $(13+\sqrt{229})=(2)PQ$ and $(16+\sqrt{229})=P^3$. The classes of $P,Q$ generate the ideal class group; the first relation makes $[Q]=[P{]}^{-1}$ and the second makes $[P{]}^3=1$. Hence the class number divides three.

It is not one. The algebraic integerso it is a unit. If $P$ were principal, its generator $a=x+y\omega$ would have norm $\pm 3$. Multiply $a$ by a suitable integer power of $ϵ$ so thatIts conjugate then also has absolute value at most $\sqrt{3ϵ}$ because $|a{a}^{\prime }|=3$. ThereforeThe final inequality is $12ϵ<229$, immediate from the displayed value of $ϵ$. Thus the integer $y$ is zero, which makes $|N(a)|=x^2=3$ impossible. ConsequentlyThis proves the ideal class group of Q of square root 229. No claim that $ϵ$ is a fundamental unit is needed. Its norm minus one also realizes all real sign patterns after multiplication by $-1$, so the ordinary and narrow class numbers agree.

Now let $f(X)=X^3-4X-1$ and let $H$ be its splitting field. The possible rational roots $\pm 1$ are not roots, so $f$ is irreducible. Its discriminant isThus its Galois group is $S_3$, since an irreducible cubic has group $A_3$ precisely when its discriminant is a square. Its quadratic subfield is $\mathbb{Q}(\sqrt{229})=K$, and $H/K$ is cyclic of degree three. If $\alpha$ is one root, then $H=K(\alpha )$.

For every rational prime $\ell \ne 229$, the reduction of $f$ is separable. Its roots lift in an unramified extension containing its residue splitting field, by Hensel's lemma. Hence its splitting field is unramified over $\mathbb{Q}$ at all these primes, and consequently over $K$ there.

At 229 there is exactly one double residue root and one simple root:The simple root lifts to $r\in {\mathbb{Q}}_{229}$, and division gives $f=(X-r)q$ with $q$ monic quadratic. The discriminant-product identity isHere $q(r)=f^{\prime }(r)$ is a unit, since $r$ has the simple residue root. Therefore the quadratic discriminant has valuation one, and its root field is the ramified quadratic field ${\mathbb{Q}}_{229}(\sqrt{229})$. This is already the completion of $K$. The local splitting field has no further extension over it: the simple root lies in ${\mathbb{Q}}_{229}$ and the other two lie in that quadratic field. Thus $H/K$ is unramified even at 229. Equivalently its rational inertia is a transposition and has trivial intersection with $A_3$.

The positive cubic discriminant gives three distinct real roots, so $H$ is totally real and there is no infinite ramification over $K$. We have exhibited an everywhere-unramified cyclic extension of degree $h_K=3$. The Hilbert-class-field characterization in part (i) now proves

Finally let $p\ne 229$ be rational prime. A prime ideal of norm $p$ is principal if and only if its Artin symbol in $H_K/K$ is trivial. If it is principal, a generator has norm $\pm p$; multiplying by the norm-minus-one unit if necessary gives norm $+p$. Conversely an element of norm $p$ generates a prime ideal of norm $p$. Hence representation by $x^2+xy-57y^2$ is equivalent to $p$ splitting in $K$ with trivial Artin symbol in $H_K/K$, or to complete splitting of $p$ in $H_K/\mathbb{Q}$.

Let ${\tau }_p\in S_3$ be its rational Frobenius, defined up to conjugacy. Its restriction to the discriminant quadratic field is its sign. The splitting criterion for that quadratic field isFor odd $p$ this follows by quadratic reciprocity from the character of discriminant 229; for $p=2$ both criteria are negative since $229\equiv 5(\mathrm{mod}8)$. The polynomial is separable modulo $p$, and the degrees of its factors are the cycle lengths of ${\tau }_p$. Thus a root modulo $p$ means that ${\tau }_p$ fixes a root. An even permutation of three letters with a fixed point is the identity: a nonidentity three-cycle has none. Therefore the two stated conditions together are exactly complete splitting in $H_K$, andThis is prime representation by the norm form of Q of square root 229, with the norm sign and the prime two explicitly covered.

The preceding constructions lead to the Iwasawa theorem on local cyclotomic units. Its significance is that an explicitly calculated local unit quotient is governed by a p-adic L-function; class field theory then suggests that the same analytic object should govern global arithmetic. The latter prediction is the Iwasawa main conjecture, rather than a consequence of the local calculation alone.

Fix compatible roots of unity ${\zeta }_{p^n}$ and put $K_n={\mathbb{Q}}_p({\zeta }_{p^n})$, ${\pi }_n={\zeta }_{p^n}-1$. Let $V_{\infty }$ be the norm limit of the local principal units. Take its even part under complex conjugation to obtain $U_{\infty }^1$, the local principal-unit limit in the real tower. Let $C_{\infty }^1$ be the closed subgroup generated by norm-compatible real cyclotomic units, with their finite-order residue factors removed. WriteHere the last map is augmentation. Normalize the p-adic zeta pseudomeasure ${\zeta }_p$ byThe expression $x^k$ is well defined on $G$ for even $k$. A pseudomeasure lies in the total fraction ring of the Iwasawa algebra; the pole on its trivial finite-character component is removed by the augmentation ideal, so $I(G){\zeta }_p\subset {\Lambda }_G$. In this normalization the local theorem isThe following steps outline its proof and explain where the earlier questions enter.

First encode local units by Coleman power series. Evaluation at ${\pi }_n$ is surjective from $R$ to ${\mathcal{O}}_{K_n}$, and the product defining the Coleman norm operator givesThus norm-fixed series give norm-compatible sequences of local units. Conversely, lift $u_m$ from such a sequence to $f^{(m)}\in R^{\times }$ and take $g_m=N^{m-r_m}{f}^{(m)}$, where $r_m=\lfloor m/2\rfloor$. This has $g_m({\pi }_{r_m})=u_{r_m}$. The estimates in Question 2 give $N{g}_m/g_m\in 1+p^{m-r_m+1}R$. Consequently for fixed $n\le r_m$, the series $N^{r_m-n}{g}_m$ differs from $g_m$ modulo $p^{m-r_m+1}$ and its value at ${\pi }_n$ is $u_n$. The compactness of $R$ in the coarser of the topologies on integral formal power series supplies a convergent subsequence. Continuity of the Coleman norm operator and of evaluation shows that its limit is norm-fixed and takes every value $u_n$. It is a unit because the original principal units have residue $1$. Uniqueness follows from the Weierstrass preparation theorem: a nonzero series in $R$ has only finitely many zeros in the open unit disk, whereas the distinct ${\pi }_n$ give infinitely many interpolation points. This constructs the Coleman power series $f_u$.

Next turn its multiplicative information into an integral p-adic measure. For a norm-fixed unit $f$, use the Frobenius-corrected logarithmThe ratio belongs to $1+pR$ by reduction modulo $p$, and the logarithm divided by $p$ is integral because $p$ is odd. Equivalently $\mathcal{L}(f)=(1-\varphi /p)\log f$, interpreting the constant coefficient with the p-adic logarithm on units. The averaging identity for the logarithm is understood on the open $p$-adic unit disk, where the translated logarithmic series converge; the corrected result itself lies in $R$. The logarithmic product formula and Question 1 giveUnder the Amice transform, the root-of-unity average projects a p-adic measure onto its restriction to $p{\mathbb{Z}}_p$: the averaged factor $(1+T{)}^x$ survives exactly for $x\in p{\mathbb{Z}}_p$. Consequently $S\mathcal{L}(f)=0$ says that $\lambda (\mathcal{L}(f))$ is supported on ${\mathbb{Z}}_p^{\times }$. This gives a Galois group-equivariant homomorphism from local principal units to the Iwasawa algebra of ${\mathbb{Z}}_p^{\times }$.

The integral lifting step in this proof is the Coleman unit-measure exact sequenceHere $\mathrm{Col}(u)=\lambda (\mathcal{L}(f_u))$. The last map is surjective, since $m_1({\delta }_1)=1$, and it has the cyclotomic action on its target. Its composition with $\mathrm{Col}$ is zero: $D\varphi =p\varphi D$ implies $D\mathcal{L}(f)(0)=0$. The kernel on the unit side consists of norm-compatible $p$-power roots of unity. To see the shape of the kernel, write $z=\log (1+T)$; the equation $\mathcal{L}(f)=0$ becomes $h(pz)=ph(z)$ for $h(z)=\log f(e^z-1)$, so only its linear term can survive. Integral solutions are $f=(1+T{)}^c$, $c\in {\mathbb{Z}}_p$, after excluding finite residue factors by the principal unit condition. For the image, use the following explicit integral lifting of the corrected cyclotomic logarithm.

Take a p-adic measure $\mu$ on the units with first moment zero, and write $g$ for its Amice transform. Then $Sg=0$ and $g^{\prime }(0)=0$. Put $H=\exp (pg)\in 1+pR$ and solve $f^p=H\varphi (f)$ with $f(0)\in 1+p{\mathbb{Z}}_p$. The constant equation is solved by $f_0=\exp (p{g}_0/(p-1))$. In degree one the terms involving $f_1$ cancel, and the remaining condition is $H_1=0$, exactly $g_1=0$; choose any $f_1\in {\mathbb{Z}}_p$. For $n\ge 2$, after fixing the lower coefficients, the coefficient multiplying the unknown $f_n$ isThe other terms are divisible by $p$: for the already integral truncated series, $f^p\equiv \varphi (f)(\mathrm{mod}p)$, while $H\equiv 1(\mathrm{mod}p)$. Division therefore gives an integral $f_n$, since $f_0$ and $1-p^{n-1}$ are units. Recursion constructs $f\in R^{\times }$ with $\mathcal{L}(f)=g$. Now $pSg=\log (Nf/f)=0$, and $Nf/f\in 1+pR$ by Question 2. The p-adic logarithm is injective there, so $Nf=f$. This supplies the desired Coleman power series of a norm-compatible sequence of local units. The freely chosen linear coefficient accounts precisely for the kernel $(1+T{)}^c$. It proves the integral surjectivity step, including the integrality that rational inversion of $1-\varphi /p$ would leave unproved.

Complex conjugation acts by $-1$ on both copies of ${\mathbb{Z}}_p(1)$. As $p$ is odd, taking the even part is exact and removes both end terms. It follows that $\mathrm{Col}:U_{\infty }^1\simeq {\Lambda }_G$. The integral exact sequence is essential here: the support calculation alone would not prove this isomorphism.

It remains to calculate the image of the cyclotomic units. For $a,b\in {\mathbb{Z}}_p^{\times }$, consider the symmetric cyclotomic unit seriesThe quotient has an invertible constant term $a/b$, because both numerator and denominator have a simple zero at $T=0$. The factor $X^{(b-a)/2}$ makes the expression unchanged under $X\mapsto X^{-1}$. The product over the roots of unity gives $N{f}_{a,b}=f_{a,b}$; one may verify this first for integer $a,b$ prime to $p$ and then extend by continuity. Dividing by the norm-compatible Teichmuller representative makes the series correspond to principal units and does not change its logarithm. The generating function for the Bernoulli numbers, with $z=\log X$, givesThe symmetric factor cancels the linear term. Applying $1-\varphi /p$, and then the moment formula of Question 3, yieldsfor even $k\ge 2$. These moments determine an even p-adic measure on the units: polynomials are dense by the Mahler theorem, even projection preserves density, and multiplication by $x^2$ is invertible on the units, so the positive even-degree polynomials are dense there as well. ThusThe closed ideal generated by all differences $[b]-[a]$ is $I(G)$. The closed subgroup generated by the corresponding cyclotomic units, including its Galois group translates, therefore has image exactly $I(G){\zeta }_p$. Taking the quotient of the isomorphism $U_{\infty }^1\simeq {\Lambda }_G$ proves the boxed local theorem.

Finally introduce global arithmetic. Let ${\overline{E}}_{\infty }^1$ be the closure of norm-compatible global units in the real tower, let $X_{\infty }^{+}$ be its p-ramified Iwasawa module, and let $Y_{\infty }^{+}$ be its unramified Iwasawa module. With compatible unit and splitting conventions, class field theory gives the class-field unit sequence; quotienting its unit terms by $C_{\infty }^1$ givesIn the cyclotomic tower the prime above $p$ is principal, so the unramified class-field convention also makes it split. On the torsion character components, multiplicativity of characteristic ideals impliesThe local theorem has already calculated the first factor analytically. Equality of the global unit and class-group factors would therefore identify the characteristic ideal of $X_{\infty }^{+}$ with the same ideal of the p-adic L-function. That is the content predicted by the Iwasawa main conjecture. More explicitly, for a nontrivial even finite character $\chi$, the character component of $I(G)$ is the whole component and the prediction isHere ${\Lambda }_{\chi }$ is the power-series algebra of the pro-$p$ part with coefficients in the character-value ring. The trivial component requires the augmentation correction for the zeta pole. Equivalently the characteristic ideals of ${\overline{E}}_{\infty }^1/C_{\infty }^1$ and $Y_{\infty }^{+}$ agree componentwise. A formulation using odd class-group characters must include the appropriate cyclotomic twist and inversion, as in Kummer reflection in Iwasawa theory. The motivation is an analytic formula for local units together with a global class-field exact sequence; the global equality is the additional arithmetic assertion.

The related class-number growth theorem illustrates why passage to an Iwasawa module is powerful. For the cyclotomic ${\mathbb{Z}}_p$-extension, let $p^{e_n}$ be the $p$-part of the class number at level $n$. The unramified Iwasawa torsion theorem, the Iwasawa module structure theorem and the class-group control relations give, for sufficiently large $n$,Here $\mu ,\lambda$ are the Iwasawa invariants. To see the two growth terms, use a topological generator $\gamma$ and, after a fixed sufficiently high level $n_0$, the norm polynomial ${\nu }_{n,n_0}=({\gamma }^{p^n}-1)/({\gamma }^{p^{n_0}}-1)$. An elementary summand $\Lambda /(p^a)$ contributes $a(p^n-p^{n_0})$ to the length of its quotient by this polynomial. A summand associated with a distinguished polynomial of degree $d$ contributes $dn$ plus a constant: at each root $\alpha$ the valuation of the norm polynomial is $n$ plus a fixed constant for large $n$. If $1+\alpha$ is a root of unity already killed at level $n_0$, its value is instead exactly $p^{n-n_0}$ and gives the same linear contribution. The finite errors in the structure and control maps stabilize and contribute to $\nu$. Thus module structure explains growth, while the Iwasawa main conjecture goes further by identifying the characteristic series of an Iwasawa module themselves with analytic ones.

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.

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.

Use the final printed restriction that $m$ is even. Square-freeness then gives $m\equiv 2(\mathrm{mod}4)$, and the integer ring is ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{-m}]$. An integer solution has $x>0$. If $y$ were even, $y^2+m$ would have 2-adic valuation one, impossible for a cube; hence $x,y$ are odd. For an odd prime $p\mid m$, $p\mid y$ would similarly give valuation one. Thus $\gcd (y,m)=1$.

The ideals generated by $y+\sqrt{-m}$ and $y-\sqrt{-m}$ are coprime. Any common prime ideal divides their sum and difference and therefore lies above a rational prime dividing $2m$ and $y$. The odd possibilities have just been excluded; above two, $\sqrt{-m}$ is zero modulo the ramified prime while $y$ is odd, so neither factor is divisible by it. Unique factorization of ideals in a number field innow yields $(y+\sqrt{-m})={\mathfrak{a}}^3$. Since the class number is coprime to three, the ideal class of $\mathfrak{a}$ is trivial. The ideal is principal, and the only units for this imaginary quadratic field with $m\ge 5$ are $\pm 1$, which can be absorbed into a cube. HenceComparing coefficients of $\sqrt{-m}$ givesThus $v=1$ with $3u^2=m+1$, or $v=-1$ with $3u^2=m-1$. Both alternatives cannot occur, since $m-1$ and $m+1$ cannot both be divisible by three. The eligible alternative permits at most the two choices $u=\pm \sqrt{(m\pm 1)/3}$, and thenConsequently there are at most two integer solutions $(x,y)$, which, when present, have the same $x$ and opposite $y$. The proof uses the evenness assumption; it does not silently claim the same coprimality argument for every odd $m$.

Let $K=\mathbb{Q}(\sqrt{-31})$. Its ring of integers of a quadratic field is $\mathbb{Z}[(1+\sqrt{-31})/2]$, and its fundamental field discriminant is $-31$.

First compute the class number. The ideal-form correspondence for imaginary quadratic fields identifies the ideal classes with proper equivalence classes of primitive positive definite binary quadratic forms of discriminant $-31$. Each has a unique reduced positive definite binary quadratic form representative under the convention $|b|\le a\le c$, with $b\ge 0$ when $|b|=a$ or $a=c$. From $31=4ac-b^2\ge 3a^2$ we get $1\le a\le \sqrt{31/3}<4$. Checking these three possible leading coefficients gives exactlyFor $a=3$, neither odd residue choice $b=\pm 1,\pm 3$ makes $(b^2+31)/12$ integral. The listed forms are primitive and satisfy the reduction convention, soThis computes the ideal class group of Q of square root minus thirty-one.

Now take a root $\theta$ of $F(T)=T^3+T-1$, and let $H$ be its splitting field over $\mathbb{Q}$. There is no rational root: the only candidates are $\pm 1$, and neither vanishes. Thus the cubic is irreducible. Its polynomial discriminant isSince this is nonsquare, the Galois group of an irreducible cubic is $S_3$. Its quadratic subfield is $K$: the product of pairwise root differences squares to $-31$ and changes sign under odd permutations. Thus $H/K$ is cyclic of degree three, and $H=K(\theta )$. Indeed the cubic field and $K$ have coprime degrees, so their compositum already has the full splitting-field degree six. The squarefree polynomial discriminant also shows, by the discriminant-index formula for an integral lattice, that $\mathbb{Z}[\theta ]$ is the full integer ring of the cubic field.

It remains to prove that $H/K$ is unramified everywhere. At any rational prime $\ell \ne 31$, the reduction of $F$ is separable. Over a finite residue extension in which it splits, all residue roots are simple and lift by Hensel lemma to the corresponding unramified extension of ${\mathbb{Q}}_{\ell }$. Therefore the local splitting field is unramified over ${\mathbb{Q}}_{\ell }$, and its extension over a completion of $K$ is unramified as well.

At the only possible ramified prime, direct reduction givesThe simple root lifts to $r\in {\mathbb{Q}}_{31}$. Write $F(T)=(T-r)Q(T)$, with $Q$ monic quadratic over ${\mathbb{Q}}_{31}$. The discriminant product formula givesHere $Q(r)=F^{\prime }(r)$ is a unit, so the quadratic discriminant has square class $-31$. Hence the entire local splitting field isIt is quadratic over ${\mathbb{Q}}_{31}$, but its extension over the completion of $K$ is trivial. Thus the prime over $31$ splits completely in $H/K$; in particular it is not ramified there. This is the local splitting-field argument behind an unramified cyclic cubic extension from a prime-discriminant cubic. There are no real places of the imaginary quadratic field $K$, so no infinite-place ramification remains to check.

The Hilbert class field theorem identifies the maximal everywhere unramified abelian extension with an extension of degree $h_K$. Our cyclic unramified extension already has that degree, so it must be the whole Hilbert class field. ThereforeEquivalently, it is the splitting field of $T^3+T-1$, of degree six over $\mathbb{Q}$ and degree three over $K$. Replacing $\theta$ by ${\theta }^{-1}$ gives the alternative defining cubic $T^3-T^2-1$. This is the Hilbert class field of Q of square root minus thirty-one.

The Hilbert class field $\tilde{K}$ is the maximal everywhere unramified abelian extension of the number field $K$, with complete splitting at real places if infinite places are included. Global class field theory gives the canonical Artin reciprocity mapso $[\tilde{K}:K]$ is the class number of $K$.

Put $s=\sqrt{-6}$, so ${\mathcal{O}}_K=\mathbb{Z}[s]$ and ${\mathcal{O}}_K^{\times }=\{\pm 1\}$. The primes $2$ and $3$ are ramified:Thus the finite modulus in the question is

The ray class exact sequence, with no real component in the modulus, givesThe Chinese remainder theorem for unit groups and the ramification relations yieldThe image of $-1$ kills the $C_2$ coming from ${\mathbb{F}}_3^{\times }$. Consequently the kernel of the map from the ray class group to the ordinary class group isSince the given class number is two, $|{\mathrm{Cl}}_{\mathfrak{m}}(K)|=12$.

It remains to distinguish $C_{12}$ from $C_6\times C_2$. LetThe prime $5$ splits in $K$, and $\mathfrak{q}$ represents the nontrivial ordinary ideal class because no element of $\mathbb{Z}[s]$ has norm $5$. Direct multiplication, or comparison of norms and valuations at the two primes over $5$, givesModulo $2$, the element $1+2s$ is $1$. Modulo $3$, its class $1+2ϵ$ is nontrivial and has order three because ${ϵ}^2=0$. Hence the ray class of $\mathfrak{q}$ has order six: its square is a nontrivial element of order three in the congruence kernel. The remaining order-two factor of that kernel, coming from $({\mathcal{O}}_K/(2){)}^{\times }$, is independent of $⟨[\mathfrak{q}]⟩$. Therefore

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