The Class Group

Definition 7.1.1 (Class Group)   Let be the ring of integers of a number field . The class group of  is the group of fractional ideals modulo the sugroup of principal fractional ideals , for .

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

That the class group is finite 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 (Proposition 6.3.6).

Theorem 7.1.2 (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 Proposition 6.3.6 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. There are other better bounds, but they depend on unproven conjectures.

The following two examples illustrate how to apply Theorem 7.1.2 to compute in simple cases.

Example 7.1.3   Let . Then , , and , so the Minkowski bound is

Thus every fractional ideal is equivalent to an ideal of norm . Since is the only ideal of norm , every ideal is principal, so is trivial.

Example 7.1.4   Let . We have , so , , , and the Minkowski bound is

We compute the Minkowski bound in SAGE as follows:
sage: K = QQ[sqrt(10)]; K
Number Field in sqrt10 with defining polynomial x^2 - 10
sage: B = K.minkowski_bound(); B
sqrt(10)
sage: B.n()
3.16227766016838

Theorem 7.1.2 implies that every ideal class has a representative that is an integral ideal of norm , or . The ideal is ramified in , so

If were principal, say , then would have norm . Then the equation

 (7.1)

would have an integer solution. But the squares mod  are , so (7.1.1) has no solutions. Thus defines a nontrivial element of the class group, and it has order  since its square is the principal ideal . Thus .

To find the integral ideals of norm , we factor modulo , and see that

If either of the prime divisors of were principal, then the equation would have an integer solution. Since it does not have one mod , the prime divisors of are both nontrivial elements of the class group. Let

Then

so the classes over  are equal.

In summary, we now know that every element of is equivalent to one of

or

Thus the class group is a group of order at most that contains an element of order . Thus it must have order . We verify this in SAGE below, where we also check that generates the class group.
sage: K.<sqrt10> = QQ[sqrt(10)]; K
Number Field in sqrt10 with defining polynomial x^2 - 10
sage: G = K.class_group(); G
Class group of order 2 with structure C2 of Number Field ...
sage: G.0
Fractional ideal class (3, sqrt10 + 1)
sage: G.0^2
Trivial principal fractional ideal class
sage: G.0 == G( (3, 2 + sqrt10) )
True


Before proving Theorem 7.1.2, we prove a few lemmas. The strategy of the proof is 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 is an integral ideal, i.e., that .

Let be a subset of . Then is convex if whenever then the line connecting and lies entirely in . We say that is symmetric about the origin if whenever then also. If is a lattice in the real vector space , then the volume 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 7.1.5 (Blichfeld)   Let be a lattice in , and let be a bounded closed convex subset of that is symmetric about the origin. If 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 . Because is symmetric about the origin, . 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 nonzero for arbitrarily small . Since is closed, .

Lemma 7.1.6   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, since , by definition.

Fix a number field with ring of integers .

Let be the real embeddings of and be half the complex embeddings of , with one representative of each pair of complex conjugate embeddings. Let be the embedding

 Re   Re   Im   Im

Note that this is not exactly the same as the one at the beginning of Section 6.2 if .

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 in Section 6.2.
1. Add times each column with entries Im to the column with entries Re.
2. Multiply all columns with entries Im by , thus changing the determinant by .
3. Add each columns that now has entries ReIm to the the column with entries Im to obtain columns ReIm.
Recalling the definition of discriminant, we see that if  is the matrix constructed by doing the above three operations to , then . Thus

Lemma 7.1.8   If  is a fractional -ideal, then is a lattice in  and

Proof. Since has rank as an abelian group, and Lemma 7.1.7 implies that also spans , it follows that is a lattice in . For some nonzero integer we have , so is also a lattice in . To prove the displayed volume formula, combine Lemmas 7.1.6-7.1.7 to get

Proof. [Proof of Theorem 7.1.2] Let be a number field with ring of integers , let be as above, and let be the function defined by

Notice that if then , and for any ,

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

exists.

Suppose  is any fractional ideal of . Our goal is to prove that there is an integral ideal with small norm. We will do this by finding an appropriate . By Lemma 7.1.8,

Let , where . Then

so by Lemma 7.1.5 there exists . Let be such that . 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 7.1.9   Suppose that is a number field. Then .

Proof. Applying Theorem 7.1.2 to the unit ideal, we get the bound

Thus

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

A prime ramifies in if and only if , so the corollary implies that every nontrivial extension of is ramified at some prime.

William Stein 2012-09-24