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The imaginary quadratic field $K=\mathbb{Q}(\sqrt{-163})$ has discriminant $-163$, and its Minkowski bound for ideal classes isThe polynomial $X^2-X+41$ defining its ring of integers of a quadratic field has no root modulo $2,3,5,$ or $7$, so these rational primes are inert. Factoring an integral ideal of norm at most eight into prime ideals then shows that its only possible nontrivial factor is the principal ideal $(2)$, of norm four. Every ideal class is therefore principal, and

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 $4$.

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.

TakeWe justify that this is a Dedekind domain. It is an integral domain inside the number field $K=\mathbb{Q}(\sqrt{-5})$. As an additive group it is free of rank two over $\mathbb{Z}$. The subgroup theorem for finitely generated abelian groups says that every subgroup of such a group is finitely generated. Thus every ideal has finitely many additive generators, and those same elements generate it as a $D$-ideal. This proves that $D$ is Noetherian.

We use the standard formula for the ring of integers of a quadratic field: for squarefree $d$, it is $\mathbb{Z}[(1+\sqrt{d})/2]$ if $d\equiv 1(\mathrm{mod}4)$, and $\mathbb{Z}[\sqrt{d}]$ otherwise. Since $-5\equiv 3(\mathrm{mod}4)$, $D$ is the ring of integers of a number field $K$, meaning all elements of $K$ integral over $\mathbb{Z}$. We also use transitivity of integral extensions. If an element of $K$ is integral over $D$, it is integral over $\mathbb{Z}$ because $D$ is integral over $\mathbb{Z}$, and therefore belongs to $D$. Thus $D$ is an integrally closed domain.

Every nonzero prime ideal $P$ is maximal. Indeed, choose $0\ne \alpha =a+b\sqrt{-5}\in P$. Multiplying by its algebraic conjugate puts the nonzero integer $N(\alpha )=a^2+5b^2$ in $P$. Hence the integral domain $D/P$ is a quotient of the finite ring $D/N(\alpha )D$, and is finite. Every finite integral domain is a field, since multiplication by a nonzero element is injective and therefore surjective. The element $2$ has no inverse in $D$, since $1/2\notin \mathbb{Z}[\sqrt{-5}]$, so $D$ is not a field. Consequently $(0)$ is not maximal, and a maximal ideal above it is nonzero. Together with the preceding result, this gives Krull dimension exactly one.

Finally, consider the idealReduction modulo $2$ with $\sqrt{-5}\mapsto 1$ identifies $D/I$ with ${\mathbb{F}}_2$, so the additive index $[D:I]$ is $2$. If $I$ were the principal ideal $(a+b\sqrt{-5})$, multiplication by its nonzero generator would have integer matrixin the basis $1,\sqrt{-5}$. The index of its image is the absolute determinant, soThere are no integer solutions: $b\ne 0$ makes the right side at least $5$, while $b=0$ would require $a^2=2$. Thus $I$ is a nonprincipal ideal in the integers adjoined a square root of minus five, and $D$ is a Dedekind domain that is not a principal ideal domain.

LetIts minimal polynomial is $X^2-X+m$, and the algebraic norm givesSuppose that $N$ is composite and choose a prime divisor $q\le \sqrt{N}<m$. Since $X^2-X+m$ has the root $-n$ modulo $q$, the idealis a prime ideal of ideal norm $q$. The triviality of the ideal class group makes $\mathfrak{q}=(\beta )$ principal, so $|N_{L/\mathbb{Q}}(\beta )|=q$.

The ring of integers of a quadratic field consists of elementswhose norm is $(a^2+p{b}^2)/4$. If $b=0$, this norm is a square and cannot be the prime $q$. If $b\ne 0$, thenand integrality gives $q\ge m$, again a contradiction. Therefore the Euler prime-generating quadratic from class number one yields

For $K=\mathbb{Q}(\sqrt{-163})$, the fundamental discriminant is $-163$ and the Minkowski bound for ideal classes isThus every ideal class contains an integral ideal of norm at most eight.

The ring of integers of a quadratic field is $\mathbb{Z}[\theta ]$ for $\theta =(1+\sqrt{-163})/2$, whose minimal polynomial is $X^2-X+41$. Its discriminant $-163$ is a quadratic nonresidue modulo each of $3,5,$ and $7$; directly modulo two, $X^2+X+1$ has no root. Hence these primes are inert by the splitting of rational primes in a quadratic field. The only prime ideal over one of them with norm at most eight is $(2)$, which has norm four and is principal. Unique prime ideal factorization now makes every ideal of norm at most eight principal. The Minkowski bound therefore provesThis is the Ideal class group of Q of square root of minus one hundred sixty-three.

The ring of integers of a quadratic field has integral basis $1,\sqrt{d}$ when the square-free integer $d$ is not congruent to one modulo four. Since $d=2$,This is the ring of integers of Q of square root two.

The Minkowski convex body theorem states that if $\Lambda \subset {\mathbb{R}}^n$ is a full lattice and $C$ is a measurable, convex, centrally symmetric set withthen $C$ contains a nonzero point of $\Lambda$.

Here $d\not\equiv 3(\mathrm{mod}4)$, so the ring of integers of a quadratic field isUnder its complex embedding it has real-lattice covolume $\sqrt{d}$. An integral ideal $I$ has index $N(I)$ in ${\mathcal{O}}_K$, so its covolume is $\sqrt{d}N(I)$. Apply Minkowski's theorem to the disk $|z|\le R$, whose area is $\pi R^2$. Taking $R^2$ down to $4\sqrt{d}N(I)/\pi$ gives a nonzero $\alpha \in I$ withSince the field norm is $N_{K/\mathbb{Q}}(\alpha )=|\alpha {|}^2$ in this imaginary quadratic field, this proves the imaginary quadratic Minkowski disk bound

Because $(\alpha )\subseteq I$, the idealis integral, represents the inverse class of $I$, and hasThus every ideal class has a bounded-norm representative. Only finitely many integral ideals have bounded norm, proving the finiteness of the ideal class group.

Now take $K=\mathbb{Q}(\sqrt{-22})$. Its ring of integers is $\mathbb{Z}[\sqrt{-22}]$, its discriminant is $-88$, andEvery class therefore has an integral representative of norm at most five. The polynomial $X^2+22$ is irreducible modulo three and modulo five, so both primes are inert and there are no ideals of norm three or five. At two it reduces to $X^2$, and the splitting of rational primes in a quadratic field givesAn ideal of norm four is ${\mathfrak{p}}_2^2=(2)$ and is principal. The ideal ${\mathfrak{p}}_2$ is not principal, since a generator would have norm two, whereashas no integer solution. Consequently the Ideal class group of Q of square root of minus twenty-two is

Suppose finally that $y^3=x^2+22$. If $x$ were even, the right-hand side would be $2$ modulo four, which is not a cube modulo four. Hence $x$ and $y$ are odd. In ${\mathcal{O}}_K$,The two ideals on the left are coprime. Indeed, a common prime ideal would divide both $2x$ and $2\sqrt{-22}$. It cannot lie above two because their norms are odd. It would therefore lie above eleven and force $11\mid x$; but then $x^2+22$ has $11$-adic valuation exactly one, impossible for the cube $y^3$.

By unique factorization of ideals in a number field, coprimality impliesfor some ideal $\mathfrak{a}$. The class group has order two, so it has no element of order three; $[\mathfrak{a}{]}^3=1$ therefore forces $\mathfrak{a}$ to be principal. The only units are $\pm 1$, both of which can be absorbed into a cube, and hencefor integers $a,b$. Comparing the coefficient of $\sqrt{-22}$ givesThus $b=\pm 1$. For $b=1$ this requires $3a^2=23$, and for $b=-1$ it requires $3a^2=21$; neither is possible. Therefore the result on no integral solutions to y cubed equals x squared plus twenty-two is

First suppose $|{\sigma }_i(\alpha )|=1$ for every complex embedding. The coefficients of the monic polynomialare rational algebraic integers and hence integers. They are uniformly bounded in $m$, because each is an elementary symmetric sum of numbers of modulus one. Only finitely many such integer polynomials can occur, so only finitely many algebraic integers ${\alpha }^m$ occur. Two powers coincide; since $\alpha \ne 0$, this gives ${\alpha }^{r-s}=1$. Thus $\alpha$ is a root of unity. This is Kronecker theorem on algebraic integers in the unit disk.

The second printed assertion is literally false for $\alpha =0$. For the intended statement with $\alpha \ne 0$, use the Minkowski embedding: algebraic integers of $K$ form a lattice, so only finitely many have all conjugates of modulus at most $2$. A nonzero algebraic integer with every conjugate of modulus at most one hasso every conjugate has modulus one and the first part applies. Consequently every non-root in that finite box has ${\max }_i|{\sigma }_i(\alpha )|>1$. Choose $c>1$ no larger than the least of these finitely many maxima, taking $c=2$ if there are no non-roots. Then every nonzero $\alpha \in {\mathcal{O}}_K$ satisfying $|{\sigma }_i(\alpha )|<c$ for all $i$ is a root of unity.

Dirichlet unit theorem states that for signature $(r_1,r_2)$,and the logarithmic embedding of number field units maps the free part to a full lattice in the hyperplane where the weighted coordinate sum is zero.

For a real quadratic field, the roots of unity are $\{\pm 1\}$ and the logarithms of positive units form a nonzero discrete subgroup $a\mathbb{Z}\subset \mathbb{R}$. Its smallest positive element is $\log ϵ$ for a unit $ϵ>1$. Given any positive unit $u>1$, choose $n$ so that $1\le u{ϵ}^{-n}<ϵ$; minimality forces the quotient to be $1$. Applying signs and inverses shows

Since $11\equiv 3(\mathrm{mod}4)$, the ring of integers of a quadratic field is $\mathbb{Z}[\sqrt{11}]$. A unit $a+b\sqrt{11}$ satisfies the Pell equation $a^2-11b^2=\pm 1$. The elementhas norm $1$. To prove it is the smallest unit above one without invoking the continued-fraction algorithm, suppose $1<a+b\sqrt{11}<10+3\sqrt{11}$. Replacing by its inverse or negative if needed makes $b>0$, and the conjugate relation implies $b\sqrt{11}<(u+u^{-1})/2<10$, so $b\le 3$. For $b=1,2,3$, the numbers $11b^2\pm 1$ are respectivelyOnly the last list contains a square, namely $100=10^2$, which yields the endpoint itself. Hence no smaller unit exists and

Let $n=[L:K]$ and write $g(X)={\prod }_{j=1}^n(X-{\alpha }_j)$. Lagrange interpolation, followed by summing over the conjugates, shows that the trace-dual of the power basis $1,\alpha ,\dots ,{\alpha }^{n-1}$ is contained in $g^{\prime }(\alpha {)}^{-1}{\mathcal{O}}_K[\alpha ]$ and has the same determinant. Hence

For $L=\mathbb{Q}(\sqrt{3})$, the ring of integers of a quadratic field is $\mathbb{Z}[\sqrt{3}]$. Taking $g=X^2-3$ givesFor $L=\mathbb{Q}(\sqrt{5})$, use $\alpha =(1+\sqrt{5})/2$ and $g=X^2-X-1$. Then

For $K=\mathbb{Q}(\sqrt{-3})$, putThen the ring of integers of a quadratic field is ${\mathcal{O}}_K=\mathbb{Z}[\omega ]$, andA unit must have norm $1$. Multiplying by four givesso $b\in \{0,\pm 1\}$, and checking these cases gives exactlyThese are precisely the sixth roots of unity, proving the description of the units of Q of square root minus three.

For $K=\mathbb{Q}(\sqrt{10})$, the ring of integers of a quadratic field isThe elementhas field norm $9-10=-1$, so it is a unit. We claim that

Let $u$ be any unit. Its norm is $\pm 1$. After changing its sign and replacing it by its inverse if necessary, its first real embedding satisfies $u\ge 1$. Choose $n\in \mathbb{Z}$ so thatWrite $v=a+b\sqrt{10}$ with $a,b\in \mathbb{Z}$. Its conjugate is $v^{\prime }=\pm v^{-1}$, so $|v^{\prime }|\le 1$. ThereforeThus $|a|\le 3$ and $|b|\le 1$. The equation $a^2-10b^2=\pm 1$ now shows directly that the only value in $[1,\epsilon )$ is $v=1$: if $b=0$ then $a=\pm 1$, while if $|b|=1$ then $a=\pm 3$ and the positive possibilities are either below $1$ or equal to $\epsilon$. Hence $u={\epsilon }^n$ after normalization, proving the claimed description of the units of Q of square root ten.

An algebraic number is an algebraic integer when it is a root of a monic polynomial in $\mathbb{Z}[X]$.

If $\alpha$ and $\beta$ are algebraic integers, the ring $\mathbb{Z}[\alpha ,\beta ]$ is a finitely generated $\mathbb{Z}$-module. Multiplication by $\alpha \beta$ preserves this module, so the algebraic integer module criterion shows that $\alpha \beta$ is an algebraic integer.

The ring of integers of a quadratic field givessince $2\equiv 2(\mathrm{mod}4)$. For a direct verification, if $x=a+b\sqrt{2}$ with $a,b\in \mathbb{Q}$ is integral, then its trace $2a$ and norm $a^2-2b^2$ are integers. Writing $2a=m\in \mathbb{Z}$ and reducing the norm condition in lowest terms shows first that $b$ can have denominator at most $2$; a denominator $2$ with odd numerator would require $m^2\equiv 2(\mathrm{mod}4)$, which is impossible. Thus $b\in \mathbb{Z}$, and then the norm condition forces $m$ even, so $a\in \mathbb{Z}$. Conversely, every $a+b\sqrt{2}$ with $a,b\in \mathbb{Z}$ is integral because $\sqrt{2}$ is integral and algebraic integers form a ring.

Now let $f$ have degree $d$ and roots ${\alpha }_1,\dots ,{\alpha }_d$. Sinceand $|{\alpha }_1|>1$, the integer $|a_d|$ cannot be $2$; hence $|a_d|=1$. Thus $f$ is monic up to sign and ${\alpha }_1$ is an algebraic integer.

Let $B$ be the product of all roots with modulus greater than one, counting multiplicity. Complex roots occur in conjugate pairs, so $B$ is real up to the signs of the real roots andAs a product of algebraic integers, $B$ is an algebraic integer; the equality forces $B=\pm 2$. Thereforeis the absolute value of the product of the remaining conjugates, is rational, and is an algebraic integer. It is consequently a nonzero rational integer. Every remaining factor has modulus at most one, so this integer is at most one. It must equal one, and henceThis is the mahler-measure norm argument at measure two.

The ring of integers of a quadratic field and its field discriminant are