# Finiteness of the Class Group via Geometry of Numbers

We have seen examples in which is not a unique factorization domain. If is a principal ideal domain, then it is a unique factorization domain, so it is of interest to understand how badly fails to be a principal ideal domain. The class group of measures this failure. As one sees in a course on Class Field Theory, the class group and its generalizations also yield deep insight into the possible abelian Galois extensions of .

Definition 10.4.1 (Class Group)   Let be the ring of integers of a number field . The of  is the group of nonzero fractional ideals modulo the sugroup of principal fractional ideals , for .

Note that if we let denote the group of nonzero fractional ideals, then there is an exact sequence

A basic theorem in algebraic number theory is that the class group is finite, which follows from the first part of the following theorem and the fact that there are only finitely many ideals of norm less than a given integer.

Theorem 10.4.2 (Finiteness of the Class Group)   Let be a number field. There is a constant that depends only on the number , of real and pairs of complex conjugate embeddings of  such that every ideal class of contains an integral ideal of norm at most , where . Thus by Lemma 10.3.5 the class group of  is finite. One can choose such that every ideal class in contains an integral ideal of norm at most

The explicit bound in the theorem is called the Minkowski bound, and I think it is the best known unconditional general bound (though there are better bounds in certain special cases).

Before proving Theorem 10.4.2, we prove a few lemmas. The strategy of the proof will be to start with any nonzero ideal , and prove that there is some nonzero , with very small norm, such that is an integral ideal. Then will be small, since is small. The trick is to determine precisely how small an we can choose subject to the condition that be an integral ideal, i.e., that .

Let be a subset of . Then is if whenever then the line connecting and lies entirely in . We say that is if whenever then also. If is a lattice in , then the of is the volume of the compact real manifold , which is the same thing as the absolute value of the determinant of any matrix whose rows form a basis for .

Lemma 10.4.3 (Blichfeld)   Let be a lattice in , and let be a bounded closed convex subset of that is symmetric about the origin. Assume that . Then  contains a nonzero element of .

Proof. First assume that . If the map is injective, then

a contradiction. Thus is not injective, so there exist such that . By symmetry . By convexity, the average of and is also in . Thus , as claimed.

Next assume that . Then for all there is , since . If then the are all in , which is finite since is bounded and is discrete. Hence there exists for arbitrarily small . Since is closed, .

Lemma 10.4.4   If and are lattices in , then

Proof. Let be an automorphism of  such that . Then defines an isomorphism of real manifolds that changes volume by a factor of . The claimed formula then follows.

Fix a number field with ring of integers . Let be the embedding

 Re   Re   Im   Im

where are the real embeddings of and are half the complex embeddings of , with one representative of each pair of complex conjugate embeddings. Note that this is not exactly the same as the one at the beginning of Section 10.2.

Lemma 10.4.5

Proof. Let . From a basis for we obtain a matrix whose th row is

Re   Re   Im   Im

and whose determinant has absolute value equal to the volume of . By doing the following three column operations, we obtain a matrix whose rows are exactly the images of the under all embeddings of into , which is the matrix that came up when we defined .
1. Add times each column with entries Im to the column with entries Re.
2. Multiply all columns Im by , thus changing the determinant by .
3. Add each columns with entries Re to the the column with entries Im.
Recalling the definition of discriminant, we see that if  is the matrix constructed by the above three operations, then . Thus

Lemma 10.4.6   If is a nonzero fractional ideal for , then is a lattice in , and

Proof. We know that is a nonzero rational number. Lemma 10.4.5 implies that is a lattice in , since has rank as abelian group and spans , so is also a lattice in . For the volume formula, combine Lemmas 10.4.4-10.4.5 to get

Proof. [Proof of Theorem 10.4.2] Let be a number field with ring of integers , let be as above, and let be the function defined by

Notice that if then .

Let be any closed, bounded, convex, subset that is symmetric with respect to the origin and has positive volume. Since  is closed and bounded,

exists.

Suppose  is any nonzero fractional ideal of . Our goal is to prove there is an integral ideal with small norm. We will do this by finding an appropriate . By Lemma 10.4.6,

Let , where . Then

so by Lemma 10.4.3 there exists . Since is the largest norm of an element of , the largest norm of an element of is at most , so

Since , we have , so is an integral ideal of that is equivalent to , and

Notice that the right hand side is independent of . It depends only on , , , and our choice of . This completes the proof of the theorem, except for the assertion that can be chosen to give the claim at the end of the theorem, which we leave as an exercise.

Corollary 10.4.7   Suppose that is a number field. Then .

Proof. Applying Theorem 10.4.2 to the unit ideal, we get the bound

Thus

and the right hand quantity is strictly bigger than for any and any (exercise).

Subsections
William Stein 2004-05-06