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Let $K$ have degree $n$, and suppose a rational prime $p$ factors into $n$ distinct prime ideals of norm $p$. If $n>p$, then $K$ is not monogenic. Otherwise the Dedekind factorization theorem would force the degree-$n$ minimal polynomial of a power-basis generator to have $n$ distinct linear factors over ${\mathbb{F}}_p$, although only the $p$ polynomials $X-a$ are available.

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

For $p\equiv 1(\mathrm{mod}20)$, put $\theta =\sqrt{-p}$. Since $-p\equiv 3(\mathrm{mod}4)$, the ring of integers of a quadratic field formula gives $R=\mathbb{Z}[\theta ]$. Modulo five,The two roots are distinct, so the Dedekind factorization theorem givesIn particular $P\ne \overline{P}$.

We prove that $[P]$ has order of a group element greater than $n$. Suppose instead that $P^k$ is a principal ideal for some $1\le k\le n$. Since $P^k\subseteq R$, its generator is in $R$: write $P^k=(a+b\theta )$ with $a,b\in \mathbb{Z}$. Taking ideal norms gives the positive field normBut $5^k\le 5^n<p$, so $b=0$. Thus $a^2=5^k$, forcing $k$ to be even and $a=\pm 5^{k/2}$. This would givecontrary to unique factorization of ideals in a number field and $P\ne \overline{P}$. Consequently no such $k$ exists. The finiteness of the ideal class group ensures that $[P]$ has finite order of a group element, soFor every positive integer $n$, the allowed infinitude of prime numbers in the progression $1(\mathrm{mod}20)$ supplies one larger than $5^n$. Applying the construction proves unbounded element orders in imaginary quadratic ideal class groups. In particular, this establishes elements of arbitrarily large order of a group element, rather than merely unbounded class numbers.

Since $\mathcal{O}=\mathbb{Z}[\theta ]$, reduction gives $\mathcal{O}/p\mathcal{O}\cong {\mathbb{F}}_p[X]/(\bar{g})$. The maximal ideals of this quotient are precisely those generated by the distinct irreducible factors ${\bar{g}}_i$. Their inverse images are thereforethe distinct prime ideals above $p$. In fact $\mathcal{O}/{\mathfrak{p}}_i\cong {\mathbb{F}}_p[X]/({\bar{g}}_i)$ is a field of size $p^{\mathrm{deg}{\bar{g}}_i}$. Every prime containing $p$ arises this way; independence of the choice of lift $g_i$ follows because another lift differs by a multiple of $p$.

To prove the ideal factorization, set $J={\prod }_i{\mathfrak{p}}_i^{e_i}$. A generator of the product either contains a factor $p$, and hence lies in $p\mathcal{O}$, or is the product ${\prod }_i{g}_i(\theta {)}^{e_i}$. The latter also lies in $p\mathcal{O}$, since its polynomial reduces to $\bar{g}$ and $g(\theta )=0$. Thus $J\subset p\mathcal{O}$. The ideal norm is multiplicative for nonzero ideals in the ring of integers; henceEqual finite indices together with the containment imply equality. ThereforeThis proves the Dedekind factorization theorem in the stated monogenic case by containment and norms, rather than assuming the desired factorization.

The polynomial is an Eisenstein polynomial at $5$, so it is irreducible and $[K:\mathbb{Q}]=3$. Its polynomial discriminant isOne must remove the index at $2$ before using this as the field discriminant. Put $\beta =({\alpha }^2+\alpha )/2$. Direct reduction using ${\alpha }^3=5\alpha -20$ givesThus $B=\mathbb{Z}+\mathbb{Z}\alpha +\mathbb{Z}\beta$ is a ring finite over $\mathbb{Z}$, so its elements are integral. Its basis has discriminant of elements of a number field $-2575=-5^2\cdot 103$. The discriminant-index formula for an integral lattice says $\mathrm{disc}(B)=[{\mathcal{O}}_K:B{]}^2{\Delta }_K$, so only $5$ could divide the remaining index.

We use these standard different ideal facts: the norm of the different is the absolute field discriminant; its localization is the local different; an unramified prime has different exponent zero, and a tamely ramified prime of ramification index $e$ has different exponent $e-1$. The Eisenstein polynomial at $5$ gives total ramification of degree three, which is tame because $5\nmid 3$. Its contribution to $v_5({\Delta }_K)$ is therefore two. This excludes a further index factor $5$, proving ${\mathcal{O}}_K=B$ and $\boxed{{\Delta }_K=-2575}$.

Since the index $[{\mathcal{O}}_K:\mathbb{Z}[\alpha ]]=2$ is prime to $5$ and $103$, Dedekind factorization theorem applies at both primes. Modulo $5$, $f=X^3$, giving a single prime $\mathfrak{p}=(5,\alpha )$ with residue degree one and ramification index three. Modulo $103$,Thus $\mathfrak{q}=(103,\alpha -6)$ has residue degree one and ramification index two; the other prime over $103$ is unramified. Both ramified primes are tame. No other prime ramifies because none divides ${\Delta }_K$. Consequently

Evaluation at $\alpha$ gives the ring isomorphismand reducing modulo $p$ givesFor each $i$,This quotient is a field because ${\overline{g}}_i$ is irreducible. Hence $P_i$ is a nonzero prime ideal; it is nonzero because it contains $p$. Its ideal norm is

The ideals are distinct. Indeed, if $P_i=P_j$ with $i\ne j$, then both ${\overline{g}}_i$ and ${\overline{g}}_j$ would vanish in the same quotient. Since distinct monic irreducibles are coprime, a Bezout identity between them would make $1$ vanish there, a contradiction.

Because the chosen lifts satisfyfor some $H\in \mathbb{Z}[X]$, evaluation at $\alpha$ givesEvery generator of the product ideal ${\prod }_i{P}_i^{e_i}$ either contains a factor $p$ or is the displayed product, soOn the other hand, multiplicativity of ideal norms and ${\sum }_i{e}_i\mathrm{deg}{\overline{g}}_i=\mathrm{deg}f=[L:\mathbb{Q}]$ giveTwo finite-index ideals with one contained in the other and equal index are equal. Thus the Dedekind factorization theorem is proved in this monogenic case:

The minimal polynomial of $\beta$ isand part (b) showed that ${\mathcal{O}}_K=\mathbb{Z}[\beta ]$. Modulo $3$,The quadratic has no root in ${\mathbb{F}}_3$, so it is irreducible. The Dedekind factorization theorem therefore givesThe two factors are distinct prime ideal, with norms $3$ and $3^2=9$, respectively.

Modulo $2$,and the quadratic is irreducible over ${\mathbb{F}}_2$. HenceThis is a product of two distinct proper prime ideals, so $(2)$ is not prime. These decompositions are collected in prime ideals above two and three in the cubic field of discriminant minus 307.

The Dedekind factorization theorem says that, when a rational prime $p$ does not divide the index of an order $\mathbb{Z}[\theta ]\subseteq {\mathcal{O}}_K$, a factorizationof the minimal polynomial modulo $p$ gives

Let $K=\mathbb{Q}(\sqrt{d})$ with square-free $d$, and let $p$ be odd. The resulting splitting of rational primes in a quadratic field is:

Finally suppose $d>0$ and
$\alpha =A/C+(B/C)\sqrt{d}$ has norm $-1$, with integers
$A,B,C$ chosen coprime. ThenIf an odd $p\equiv 3(\mathrm{mod}4)$ ramified, then $p\mid d$. Reduction modulo $p$ gives
$A^2\equiv -C^2(\mathrm{mod}p)$. Since $-1$ is not a square modulo such a prime, $p\mid A,C$. Dividing the equation's divisibility shows $p\mid B$ as well (use that square-free $d$ has $p$-adic valuation one), contradicting coprimality. Hence no prime $p\equiv 3(\mathrm{mod}4)$ ramifies.

Put $\theta =\sqrt[3]{3}$. The given ring of integers has integral basis $1,\theta ,{\theta }^2$. The discriminant of the power basis, equivalently of $T^3-3$, isThe field has signature $(1,1)$: one real embedding and one conjugate pair of complex embeddings. The Minkowski bound for ideal classes is thereforeEvery ideal class consequently has an integral representative of norm at most four.

It remains to inspect prime ideals above $2$ and $3$. Modulo $2$,The Dedekind factorization theorem giveswhereButso the principal ideal $(\theta -1)$ is contained in ${\mathfrak{p}}_2$ and has the same norm; henceAlsosoBoth primes above $2$ are principal.

Finally,Thus the unique prime above $3$ is the principal ideal $(\theta )$. By unique factorization of ideals in a number field, every integral ideal of norm at most four is built from these principal prime ideals. Every ideal class is therefore trivial, and the Class group of Q of cube root of three is

Let $\alpha =\sqrt[3]{2}$. Its minimal polynomial is $f(X)=X^3-2$, whose discriminant is $-108$. Since $5\nmid 108$, the prime $5$ does not divide the index $[{\mathcal{O}}_K:\mathbb{Z}[\alpha ]]$, so the Dedekind factorization theorem applies at $5$. In ${\mathbb{F}}_5[X]$,The quadratic factor has discriminant $3$, which is a quadratic nonresidue modulo $5$, so it is irreducible. ThereforewhereTheir residue-field degrees are one and two. Both factors of $f$ occur with multiplicity one, so both prime-ideal exponents are one. Hence neither ${\mathfrak{p}}_1$ nor ${\mathfrak{p}}_2$ is ramified.

Dedekind factorization theorem states that if $K=\mathbb{Q}(\theta )$ with integral $\theta$ and $p$ does not divide $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$, then a factorization $\overline{f}={\prod }_i{\overline{f}}_i^{e_i}$ of its minimal polynomial of an algebraic element into distinct monic irreducible polynomials over ${\mathbb{F}}_p$ givesThe factors are distinct prime ideals; the residue degrees and ramification indices satisfy ${\sum }_i{e}_i\mathrm{deg}{f}_i=[K:\mathbb{Q}]$. Unique prime ideal factorization holds in the ring of integers.

For a square-free integer $d\equiv 1(\mathrm{mod}4)$, ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{-d}]$, so the index condition is automatic. If $p_i\mid d$, the polynomial $X^2+d$ reduces to $X^2$ modulo $p_i$. HenceTherefore $d{\mathcal{O}}_K=({\prod }_i{P}_i{)}^2=(\sqrt{-d}{)}^2$. By unique prime ideal factorization, ${\prod }_i{P}_i=(\sqrt{-d})$.

Let $S$ be a nonempty proper subset and put $m={\prod }_{i\in S}{p}_i$. For $d>1$, $1<m<d$ and $m$ is a square-free integer. If ${\prod }_{i\in S}{P}_i=(u+v\sqrt{-d})$, its ideal norm givesThe inequality $m<d$ forces $v=0$, leaving $m=u^2$, impossible for a square-free integer greater than one. Thus every such subproduct is nonprincipal.

These are ramified prime classes for discriminant minus four d. The ideal classes $[P_i]$ have order dividing two. Their product is the identity, and the preceding proof rules out every other nonempty subset relation. The homomorphism ${\mathbb{F}}_2^k\to \mathrm{Cl}(K)$ defined by these classes consequently has kernel exactly $\{0,(1,\dots ,1)\}$. Its image is thereforeThe boundary case $d=1$ has $k=0$ and makes the printed exponent $k-1$ meaningless; its field is $\mathbb{Q}(i)$ with trivial ideal class group. The subgroup statement is understood for $d>1$, while the prime-factor and proper-subset statements are empty for $d=1$.

The Dedekind factorization theorem says: if $K=\mathbb{Q}(\theta )$ with $\theta$ integral, minimal polynomial of an algebraic element $f\in \mathbb{Z}[X]$, and the rational prime $p$ does not divide $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$, factor $\bar{f}={\prod }_i{\bar{f}}_i^{e_i}$ into distinct monic irreducibles over ${\mathbb{F}}_p$. Then the distinct prime ideals above $p$ are $P_i=(p,f_i(\theta ))$, with integer lifts $f_i$, andThe index condition is part of the theorem and must be checked before applying it at a ramified prime.

For this cubic, $X^3-6$ is Eisenstein and has discriminant $-972$. We first verify ${\mathcal{O}}_K=\mathbb{Z}[\theta ]$. The square of the index divides 972, so only two and three can divide the index. If either prime $p$ divided it, the finite quotient group would contain an element of order $p$, giving an integral element $(a+b\theta +c{\theta }^2)/p$ with digits $0\le a,b,c<p$, not all zero. Its field norm isFor $p=2$, the numerator has valuation zero if $a=1$, valuation one if $a=0,b=1$, and valuation two if $a=b=0,c=1$. For $p=3$ the same three cases, according to the first nonzero digit, give valuations zero, one and two. None gives an integer field norm after dividing by $p^3$, contradicting integrality. Thus the index is one.

Multiplication by $a+b\theta$ in the power basis, or the product over the three conjugates, gives $\boxed{N(a+b\theta )=a^3+6b^3}$. DefineReduction of the cubic gives $X^3$ at two and three, $(X-1)(X^2+X+1)$ at five, and $(X-3)(X-5)(X-6)$ at seven. The quadratic factor at five is irreducible since its discriminant is two, a nonsquare modulo five. HenceThe norms are respectively $2,3,5,25,7,7,7$. Norms and reduction of the elements at those primes giveFor example their absolute element field norms are $6,5,2,21,70$; each indicated prime divides the element by substitution in its residue field, and the product of the indicated ideal norms already equals that absolute field norm. Thus no additional factor or exponent is possible.

In the ideal of the specified quotient every prime exponent cancels:Equality with the unit ideal implies that both $\mu$ and its inverse are integral, so $\boxed{\mu \text{ is a unit}}$. For the real root, $1<\theta <2$. The numerator is negative and has absolute value less than two, while the denominator is greater than five. Thus $-2/5<\mu <0$, proving $\boxed{\mu \ne \pm 1}$. Its field norm is $-1$, consistent with the ideal calculation.

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.

For every positive integer $n$, some imaginary quadratic field has an ideal class of order of a group element greater than $n$. Specifically, take a prime number $p\equiv 1(\mathrm{mod}20)$ with $p>5^n$. In $\mathbb{Z}[\sqrt{-p}]$, the Dedekind factorization theorem gives $(5)=P\overline{P}$, with distinct prime ideals $P=(5,\sqrt{-p}-2)$ and $\overline{P}=(5,\sqrt{-p}+2)$ of ideal norm five. If $P^k=(a+b\sqrt{-p})$ for $1\le k\le n$, the field norm equation $5^k=a^2+p{b}^2$ forces $b=0$. Then $k$ is even and $(a)=(5{)}^{k/2}=P^{k/2}{\overline{P}}^{k/2}$, contradicting unique factorization of ideals in a number field. Thus $[P]$ has order of a group element greater than $n$. The existence of suitable $p$ follows from Dirichlet's theorem on arithmetic progressions.