Ideal Class Group

Overview

Unique factorization of ideals survives in a number field's ring of integers even when unique factorization of elements fails. The ideal class group records the difference: it asks whether an ideal can be generated by a single element, while treating multiplication by any nonzero field element as a change of representative.

Its finiteness makes this obstruction computable. Its order, the class number, also occurs in analytic formulas and as the degree of the Hilbert class field.

Prerequisites

Use fractional ideals, their inverses and prime ideal factorizations. Know quotient groups and the distinction between units and nonzero field elements. For computations, know ideal norms and the discriminant of a number field.

Definition and conventions

Let $K$ be a number field, $I_K$ the multiplicative group of nonzero fractional $\mathcal O_K$-ideals, and $P_K$ its subgroup of principal fractional ideals. The ordinary ideal class group is

Thus $[I]=[J]$ exactly when $I=\alpha J$ for some $\alpha\in K^\times$. The identity class consists of all principal fractional ideals, not just the unit ideal. Multiplication is $[I][J]=[IJ]$.

Every class has an integral representative, since multiplication by a common denominator changes an ideal only by a principal factor. For real fields, requiring that factor to have a totally positive generator instead defines the narrow class group. This article uses the ordinary group.

Intuition

The ideal group keeps all prime ideal exponents. Passing to classes imposes every relation arising from a principal ideal. A nontrivial class is an obstruction to replacing ideal arithmetic by arithmetic of generators. An ideal can be nonprincipal while a power of it is principal; its class then has finite order.

This is separate from the unit group: units describe different generators of the same principal ideal, whereas classes describe ideals for which no generator exists.

Examples and non-examples

In $\mathbb Z[i]$, Euclidean division makes every ideal principal, so $\operatorname{Cl}(\mathbb Q(i))$ is trivial. The distinct ideals $(2)$ and $(3)$ therefore represent the same class. Equality of classes does not mean equality of ideals or equality of norms.

For a nontrivial calculation, let $K=\mathbb Q(\sqrt{-5})$, so $\mathcal O_K=\mathbb Z[\sqrt{-5}]$, and put $\mathfrak p=(2,1+\sqrt{-5})$. Its quotient ring is $\mathbb F_2$, so $N\mathfrak p=2$. If $\mathfrak p=(a+b\sqrt{-5})$, the principal norm formula would give $a^2+5b^2=2$, impossible for integers $a,b$. Hence $[\mathfrak p]\ne1$.

Writing $u=\sqrt{-5}$, the square is generated by $4$, $2+2u$ and $-4+2u$. These all lie in $(2)$, while their differences and an integer multiple of $4$ yield $2$. Therefore

The discriminant is $-20$. The Minkowski bound is $(2/\pi)\sqrt{20}<3$, so every ideal class has an integral representative of norm $1$ or $2$. There is only one ideal of norm $2$: a map onto $\mathbb F_2$ must send $u$ to the unique root $1$ of $x^2+5$ modulo $2$. Thus the two classes are $1$ and $[\mathfrak p]$, and the class group is cyclic of order two.

Important properties

For every number field of degree $n$, with $r_2$ pairs of complex embeddings and discriminant $D_K$, every class contains an integral ideal of norm at most

Only finitely many ideals have bounded norm, proving finiteness. Proof sketch: embed an inverse ideal as a lattice, use Minkowski's convex body theorem to find a sufficiently small nonzero element, then multiply by that element to obtain an integral representative of the original class.

Because $\mathcal O_K$ is Dedekind, its class group is trivial exactly when it is a principal ideal domain, equivalently a unique factorization domain. These equivalences should not be transferred to arbitrary commutative rings.

Connections and further reading

Ideal classes contribute equally to the leading asymptotic count of integral ideals; summing them introduces the factor $h_K$ in the residue of $\zeta_K$ at $1$. Reciprocity gives $\operatorname{Cl}(K)\cong\operatorname{Gal}(H_K/K)$ for the ordinary Hilbert class field. These are analytic and Galois interpretations of the same finite group. Sutherland, Sections 14.4–14.5 supplies the bounds used above.