# More About Computing Class Groups

If is a prime of , then the intersection is a prime ideal of . We say that lies over . Note lies over if and only if is one of the prime factors in the factorization of the ideal . Geometrically, is a point of that lies over the point of under the map induced by the inclusion .

Lemma 7.3.1   Let be a number field with ring of integers . Then the class group is generated by the prime ideals of lying over primes with , where is the number of complex conjugate pairs of embeddings .

Proof. Theorem 7.1.2 asserts that every ideal class in is represented by an ideal  with . Write , with each . Then by multiplicativity of the norm, each also satisfies . If , then , since is the residue characteristic of , so . Thus  is a product of primes  that satisfies the norm bound of the lemma.

This is a sketch of how to compute :

1. Use the algorithms of Chapter 4 to list all prime ideals of that appear in the factorization of a prime with .
2. Find the group generated by the ideal classes , where the are the prime ideals found in step 1. (In general, this step can become fairly complicated.)
The following three examples illustrate computation of for and .

Example 7.3.2   We compute the class group of . We have

so

Thus is generated by the prime divisors of . We have

so is generated by the principal prime ideal . Thus is trivial.

Example 7.3.3   We compute the class group of . We have

so

Thus is generated by the primes that divide . We have , where satisfies . The polynomial is irreducible mod , so is prime. Since it is principal, we see that is trivial.

Example 7.3.4   In this example, we compute the class group of . We have

so

Thus is generated by the prime ideals lying over and . We have , and satisfies . Factoring modulo and we see that the class group is generated by the prime ideals

and

Also, and , so and define elements of order dividing in .

Is either or principal? Fortunately, there is an easier norm trick that allows us to decide. Suppose , where . Then

Trying the first few values of , we see that this equation has no solutions, so can not be principal. By a similar argument, we see that is not principal either. Thus and define elements of order in .

Does the class of equal the class of ? Since and define classes of order , we can decide this by finding the class of . We have

The ideals on both sides of the inclusion have norm , so by multiplicativity of the norm, they must be the same ideal. Thus is principal, so and represent the same element of . We conclude that

William Stein 2012-09-24