# Remarks on Computing the Class Group

If is a prime of , then the intersection is a prime ideal of . We say that . 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 11.1.1   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. We proved before 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, whcih proves the lemma.

This is a sketch of how to compute :

1. Use the factoring primes'' algorithm 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, one must think more carefully about how to do this step.)
The following three examples illustrate computation of for and .

Example 11.1.2   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 11.1.3   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 11.1.4   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 2004-05-06