Ideal class group

The ideal class group is the group of nonzero fractional ideals modulo principal fractional ideals.

Let $d$ be a positive square-free integer congruent to two modulo four and $p\ge 3$ a prime. If $p$ does not divide $h(\mathbb{Q}(\sqrt{-d}))$ and $p\mid d$, then $x^2+d=y^p$ has no positive integer solution. More generally, if $\gcd (x,a)=1$ and $x^2+a^{2}d=y^p$, the two conjugate principal ideals are coprime. Parity excludes a shared prime above two; for an odd common prime, divisibility of both conjugates would force the underlying rational prime to divide both $x$ and $d$, giving valuation one on the left side, incompatible with a $p$th power.

Unique prime ideal factorization consequently gives $(x+a\sqrt{-d})=A^p$. The class-number condition forces $A$ principal; the only units are $\pm 1$, both odd powers. Hence $x+a\sqrt{-d}=(u+v\sqrt{-d}{)}^p$. When $a=1$ and $p\mid d$, the coefficient of $\sqrt{-d}$ on the right is divisible by $p$: its linear term has coefficient $p$, and every higher odd term contains $d$. It cannot equal one. Conversely, an actual solution under these conditions forces divisibility of the class number by $p$. For example $13^2+174=7^3$ proves $3\mid h(\mathbb{Q}(\sqrt{-174}))$.

Let $d>1$ be a positive square-free integer congruent to one modulo four, with distinct prime factors $p_1,\dots ,p_k$. The ring of integers is $\mathbb{Z}[\sqrt{-d}]$. Each ramified prime ideal $P_i=(p_i,\sqrt{-d})$ satisfies $P_i^2=(p_i)$, so its ideal class has order dividing two. Their product is the principal ideal $(\sqrt{-d})$.

There are no further subset relations. For a nonempty proper subset $S$, put $m={\prod }_{i\in S}{p}_i$. If ${\prod }_{i\in S}{P}_i=(u+v\sqrt{-d})$, the ideal norm gives $m=u^2+d{v}^2$. But $1<m<d$ forces $v=0$ and makes the square-free integer $m$ a square, a contradiction. Thus the homomorphism ${\mathbb{F}}_2^k\to \mathrm{Cl}(K)$ defined by the ramified classes has exactly the all-ones relation, and its image is $C_2^{k-1}$. This identifies a subgroup, without asserting that these classes exhaust the ideal class group.

Write $\eta =\sqrt{-34}$. The ring of integers is $\mathbb{Z}[\eta ]$ with discriminant $-136$. The Minkowski bound for ideal classes is less than eight, so classes of prime ideals of norm at most seven generate the ideal class group. These are the ramified prime $P_2=(2,\eta )$ and conjugate pairs at five and seven; three is inert with prime norm nine. Set $P_5=(5,\eta -1)$ and $P_7=(7,\eta +1)$. Conjugate prime classes are inverse, so these three classes suffice.

The field norm identities $N(4+\eta )=50$ and $N(1+\eta )=35$ give $(4+\eta )=P_2{P}_5^2$ and $(1+\eta )={\bar{P}}_5{P}_7$, by reduction modulo the indicated primes and equality of norms. Thus $[P_5{]}^2=[P_2]$ and $[P_7]=[P_5]$. Since $P_2^2=(2)$ but $a^2+34b^2=2$ has no integer solution, $[P_2]$ has order two. Therefore $[P_5]$ has order four and generates the entire group, proving the displayed isomorphism.

Put $\theta =\sqrt{-30}$. The ring of integers is $\mathbb{Z}[\theta ]$ and its discriminant is $-120$. The Minkowski bound for ideal classes $4\sqrt{30}/\pi <7$ shows that the ideal class group is generated by the classes of the ramified prime ideals $P_2=(2,\theta )$, $P_3=(3,\theta )$ and $P_5=(5,\theta )$. There is no additional prime ideal of norm four because two ramifies with residue degree one. The relations $P_i^2=(i)$ and $(\theta )=P_2{P}_3{P}_5$ show that these classes have order dividing two and their product is one. None is a principal ideal: the element field norm $a^2+30b^2$ cannot equal two, three or five. Thus there are exactly four classes, with the displayed group structure. This also supplies the class-number input for the Hilbert class field of Q of square root minus thirty.

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.

The ring of integers is $\mathbb{Z}[\sqrt{35}]$ and the field discriminant is $140$. The Minkowski bound for ideal classes is $\sqrt{35}<6$. At three the defining polynomial is irreducible, so the only prime ideals needed to generate the ideal class group are $P_2=(2,\sqrt{35}-1)$ and $P_5=(5,\sqrt{35})$, of ideal norms two and five. They satisfy $P_2^2=(2)$, $P_5^2=(5)$ and $P_2{P}_5=(5+\sqrt{35})$, so one class of order of a group element at most two generates. The field norm equation $a^2-35b^2=\pm 2$ is impossible modulo five, proving $P_2$ is not a principal ideal. Hence the ideal class group has exactly two elements.

Put $\omega =(1+\sqrt{229})/2$, $P=(3,\omega )$ and $Q=(5,\omega -4)$. The Minkowski bound for ideal classes is $\sqrt{229}/2<8$, so the ideal class group is generated by primes above three and five; two and seven are inert. The prime ideal factorizations $(6+\omega )=PQ$ and $(15+2\omega )=P^3$ reduce it to a cyclic group of order dividing three. It is not trivial: the unit $ϵ=(15+\sqrt{229})/2$ has field norm minus one. If an algebraic integer had norm of absolute value three, multiplying by a power of this unit would make both real absolute values at most $\sqrt{3ϵ}$. For $a=x+y\omega$, this bounds $|y|\sqrt{229}=|a-a^{\prime }|\le 2\sqrt{3ϵ}<\sqrt{229}$, forcing $y=0$, which cannot have norm of absolute value three.

For $s=\sqrt{-21}$, the integer ring is $\mathbb{Z}[s]$. The Minkowski bound for ideal classes is less than six. The ramified primes $(2,1+s)$ and $(3,s)$ give independent classes of order two; their product is the class of $(5,s-2)$ because $(3+s)$ factors as the product of these three prime ideals. Element norms $a^2+21b^2$ cannot equal 2, 3 or 6, so these classes are distinct. Ideals of norm at most five supply no further classes, proving the four-element group.

Parity and valuations at 3 and 7 make the ideals $(y+\sqrt{-21})$ and $(y-\sqrt{-21})$ coprime in any hypothetical solution. Their product is a cube, so the first is an ideal cube. The ideal class group has exponent two, forcing its cube root ideal to be principal. Cubing $a+b\sqrt{-21}$ would then give coefficient $3b(a^2-7b^2)$ on $\sqrt{-21}$, which cannot equal one. This proves the nonexistence of integer solutions without assuming unique element factorization.

The narrow ideal class group of a number field is the group of nonzero fractional ideals modulo principal fractional ideals generated by totally positive elements. It is the ray class group for the modulus consisting of all real Archimedean places and trivial finite part. The forgetful map to the ordinary ideal class group is surjective, with kernel $\{\pm 1{\}}^{r_1}/\mathrm{sgn}({\mathcal{O}}_K^{\times })$ under the unit signature map.

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.

For a negative fundamental discriminant $D$, primitive positive definite integral binary quadratic forms of discriminant $D$, modulo proper equivalence, correspond bijectively to the ideal class group of $\mathbb{Q}(\sqrt{D})$. The displayed ideal gives one conventional choice of this correspondence. Reduced positive definite binary quadratic forms give unique class representatives with the usual boundary sign convention, making class-number computations finite.

In $K=\mathbb{Q}(\sqrt{-14})$, the Minkowski bound for ideal classes is less than five. The ideals $\mathfrak{p}=(2,\sqrt{-14})$ and $\mathfrak{q}=(3,1+\sqrt{-14})$ generate all classes, with ${\mathfrak{p}}^2=(2)$ and $(2-\sqrt{-14})=\mathfrak{p}{\mathfrak{q}}^2$. No element has field norm two, so $[\mathfrak{p}]$ has order two and $[\mathfrak{q}]$ has order four.

An ideal class is an element of the ideal class group, represented by a nonzero fractional ideal. Two fractional ideals represent the same class exactly when their quotient is a principal ideal.

The class number of a number field is the order $h_K=|\mathrm{Cl}(K)|$ of its ideal class group.

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.

If a number field of degree $n$ and signature $(r_1,r_2)$ has discriminant $d_K$, every ideal class contains an integral ideal $I$ with

For ${\alpha }^3=6$, the ring of integers is $\mathbb{Z}[\alpha ]$ and its field discriminant is $-972$. The Minkowski bound for ideal classes is $16\sqrt{3}/\pi <9$. Prime ideals of norm at most eight lie above $2,3,5,7$. The primes above two, three and the degree-one prime above five are generated by $\alpha -2$, $3+2\alpha +{\alpha }^2$ and $\alpha -1$, of norm magnitudes $2,3,5$. At seven the residue roots are $3,5,6$; the primes at roots six and three are generated by $\alpha +1$ and ${\alpha }^2+\alpha -5$, both of norm seven. Their product with the remaining prime is $(7)$, making that prime principal too. Every possible Minkowski representative is therefore principal.

Since every nonzero integral ideal has norm at least one, the Minkowski bound for ideal classes impliesfor every number field of degree $n$ and signature $(r_1,r_2)$.

For $K=\mathbb{Q}(\sqrt[3]{3})$ with ${\mathcal{O}}_K=\mathbb{Z}[\sqrt[3]{3}]$, the field discriminant is $-243$ and the Minkowski bound is $8\sqrt{3}/\pi <5$. The prime ideals above $2$ and $3$ are principal, so every ideal class has a principal representative and

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

The field has ring of integers $\mathbb{Z}[\sqrt{-33}]$, discriminant $-132$, and Minkowski bound below eight. The nonprincipal ramified ideals above two and three generate four classes, and the split primes above seven introduce no new class. Hence

The ring of integers is $\mathbb{Z}[\sqrt{-22}]$, its discriminant is $-88$, and its Minkowski bound for ideal classes is $4\sqrt{22}/\pi <6$. The primes three and five are inert, whileThe ideal ${\mathfrak{p}}_2$ is nonprincipal because $a^2+22b^2=2$ has no integer solution. Every class is represented by an ideal of norm at most five, so $[{\mathfrak{p}}_2]$ is the only possible nontrivial class and the class group is $C_2$.

The equation $y^3=x^2+22$ has no integer solutions. Factoring it in $\mathbb{Z}[\sqrt{-22}]$, the two conjugate factors are coprime. Unique factorization of ideals in a number field and the absence of three-torsion in the class group force $x+\sqrt{-22}$ to be a cube up to a unit. Writing it as $(a+b\sqrt{-22}{)}^3$ makes its $\sqrt{-22}$ coefficientwhich has no integer solution.

Let $m\ge 2$, let $p=4m-1$ be prime, and suppose $\mathbb{Q}(\sqrt{-p})$ has class number one. If $n\ge 0$ and $n^2+n+m<m^2$, then $n^2+n+m$ is prime. The key identity is the algebraic normA hypothetical prime divisor below $m$ would split and yield an algebraic integer of that norm, but the positive norm form $(a^2+p{b}^2)/4$ represents no prime below $m$.

The order of the class of a fractional ideal $I$ is the least positive integer $n$ for which $I^n$ is principal.

Let $p\equiv 11(\mathrm{mod}12)$, $K=\mathbb{Q}(\sqrt{-p})$, and let $\mathfrak{p}$ be a prime-ideal factor of $(3)$. If $p>3^{n+2}$ and $n$ is odd, then $[\mathfrak{p}]$ does not have order $n$. Otherwise ${\mathfrak{p}}^n=(\alpha )$ would givefor $\alpha =(a+b\sqrt{-p})/2$. The bound forces $b=0$, after which $3^n=N(\alpha )$ is an integer square, impossible for odd $n$.

If every integral ideal $I$ contains nonzero $\alpha$ with $|N(\alpha )|\le CN(I)$, then $(\alpha )I^{-1}$ is an integral representative of $[I{]}^{-1}$ with norm at most $C$.

Bounded-norm representatives prove class-group finiteness because only finitely many integral ideals have norm at most a fixed constant.

Every element of a finite ideal class group has finite order. Therefore every nonzero ideal has a positive power that is principal.