ideal class group is finite

We give two proofs of the finiteness of the class groupMathworldPlanetmath, one using the bound provided by Minkowski’s theorem ( and a second, more elementary, proof that does not provide the same computational benefits as Minkowski’s bound does. Both proofs rely on the following lemma:

Lemma 1.

If K is an algebraic extensionMathworldPlanetmath of Q and 0<dZ, then there are only a finite number of ideals of norm d.

The norm of a prime idealMathworldPlanetmathPlanetmathPlanetmath 𝔓 of K lying over a rational prime p is pf, where f is the residue fieldMathworldPlanetmath degree [𝒪K/𝔓𝒪K:/p], and there are at most [K:] prime ideals lying over any given rational prime. There are thus only a finite number of possibilities for ideals with norm d - simply factor d into a product of prime powers and note that each prime power must correspond to one of a finite number of possibilities.

The finiteness of the class group now follows trivially from Minkowski’s theorem:

Theorem 1.

If K is an algebraic extension of Q, then the class group of K, denoted Cl(K), is finite.

Minkowski’s theorem guarantees that each ideal class contains a representative integral ideal whose norm is bounded by a constant depending only on the field, and the lemma shows that there are only a finite number of integral ideals with norm less than that constant.

Minkowski’s theorem gives enough information about the size of the class group to be computationally useful in some cases (see the topic on using Minkowski’s constant to find a class numberMathworldPlanetmath). It does, however, require quite a bit of machinery. To see in a more elementary way that Cl(K) is finite, one can proceed as follows:

Proof. (alternate proof of theorem)
By the lemma, it suffices to show that there is some constant C, depending only on K, such that every class in Cl(K) has a representative I𝒪K with N(I)C. For uK, denote by Tu the linear map left multiplication by u, and let e1,,en be a basis of 𝒪K as a -module (where [K:]=n). Then if u=bi(u)ei, it follows that


is a polynomialPlanetmathPlanetmath of total degree at most n in the bi(u), whose coefficients are functions of the Tei and thus depend only on K (and not on u). Let C be the sum of the magnitude of those coefficients.

Let c be a class in Cl(K) and let J𝒪K be a representative of the class c-1. Consider S={i=1nriei 1riN(J)1/n+1}. The cardinality of S is strictly greater than N(J), while |𝒪K/J|=N(J). So by the pigenhole principle, two distinct elements of S are in the same J-coset of 𝒪K. Taking their difference, we get an element 0x=bi(x)eiJ, where not all the bi(x) are zero, and |bi(x)|N(J)1/n.

Now, since xJ, using unique factorizationMathworldPlanetmath of ideals in the Dedekind ring 𝒪K, we can construct an integral ideal I such that IJ=(x), so that I is in the class cCl(K). Finally,


so that N(I)C.

Title ideal class group is finite
Canonical name IdealClassGroupIsFinite
Date of creation 2013-03-22 17:57:23
Last modified on 2013-03-22 17:57:23
Owner rm50 (10146)
Last modified by rm50 (10146)
Numerical id 7
Author rm50 (10146)
Entry type Theorem
Classification msc 11R29
Related topic IdealNorm