Math 129: Algebraic Number Theory Lecture 9

William Stein

Thursday, March 4, 2004

Today we prove that the class group of the ring of integers of a number field is finite, discuss how to compute it in some examples, then introduce the group of units. We will prove the main structure theorem for the group of units on Tuesday. The notes about proof of finiteness of the class group are not given below, since they are in the handout for Lecture 8.

1 Remarks on Computing the Class Group

If 𝔭 is a prime of O_{K}, then the intersection 𝔭 ∩ Z = pZ is a prime ideal of Z. We say that 𝔭 lies over p ∈ Z. Note 𝔭 lies over p ∈ Z if and only if 𝔭 is one of the prime factors in the factorization of the ideal pO_{K}. Geometrically, 𝔭 is a point of Spec(O_{K}) that lies over the point pZ of Spec(Z) under the map induced by the inclusion Z ↪ O_{K}.

Lemma 1.1. Let K be a number field with ring of integers O_{K}. Then the class group Cl(K) is generated by the prime ideals 𝔭 of O_{K} lying over primes p ∈ Z with p ≤ B_{K} = \sqrt{|d_{K}|} · (\frac{4}{π})^{s} · \frac{n!}{n^{n}}, where s is the number of complex conjugate pairs of embeddings K ↪ C.

Proof. We proved before that every ideal class in Cl(K) is represented by an ideal I with Norm(I) ≤ B_{K}. Write I = ∏_{i=1}^{m} 𝔭_{i}^{e_{i}}, with each e_{i} ≥ 1. Then by multiplicativity of the norm, each 𝔭_{i} also satisfies Norm(𝔭_{i}) ≤ B_{K}. If 𝔭_{i} ∩Z = pZ, then p | Norm(𝔭_{i}), since p is the residue characteristic of O_{K}/𝔭, so p ≤ B_{K}. Thus I is a product of primes 𝔭 that satisfies the norm bound of the lemma, whcih proves the lemma.

This is a sketch of how to compute Cl(K):

1. Use the “factoring primes” algorithm to list all prime ideals 𝔭 of O_{K} that appear in the factorization of a prime p ∈ Z with p ≤ B_{K}.

2. Find the group generated by the ideal classes [𝔭], where the 𝔭 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 Cl(K) for K = Q(i), Q(\sqrt{5}) and Q(\sqrt{−6}).

Example 1.2. We compute the class group of K = Q(i). We have

n = 2, r = 0, s = 1, d_{K} = −4,

so

B_{K} = \sqrt{4} · (\frac{4}{π} )^{1} · (\frac{2!}{2^{2}} ) = \frac{8}{π} < 3.

Thus Cl(K) is generated by the prime divisors of 2. We have

2O_{K} = (1 + i)^{2},

so Cl(K) is generated by the principal prime ideal 𝔭 = (1 + i). Thus Cl(K) = 0 is trivial.

Example 1.3. We compute the class group of K = Q(\sqrt{5}). We have

n = 2, r = 2, s = 0, d_{K} = 5,

so

B = \sqrt{5} · (\frac{4}{π} )^{0} · (\frac{2!}{2^{2}} ) < 3.

Thus Cl(K) is generated by the primes that divide 2. We have O_{K} = Z[γ], where γ = \frac{1+\sqrt{5}}{2} satisfies x^{2} − x − 1. The polynomial x^{2} − x − 1 is irreducible mod 2, so 2O_{K} is prime. Since it is principal, we see that Cl(K) = 1 is trivial.

Example 1.4. In this example, we compute the class group of K = Q(\sqrt{−6}). We have

n = 2, r = 0, s = 1, d_{K} = −6,

so

B = \sqrt{6} · \frac{4}{π} ∼ 3.1.

Thus Cl(K) is generated by the prime ideals lying over 2 and 3. We have O_{K} = Z[\sqrt{−6}], and \sqrt{−6} satisfies x^{2} + 6 = 0. Factoring x^{2} + 6 modulo 2 and 3 we see that the class group is generated by the prime ideals

𝔭_{2} = (2, \sqrt{−6}) and 𝔭_{3} = (3, \sqrt{−6}).

Also, 𝔭_{2}^{2} = 2O_{K} and 𝔭_{3}^{2} = 3O_{K}, so 𝔭_{2} and 𝔭_{3} define elements of order dividing 2 in Cl(K).

Is either 𝔭_{2} or 𝔭_{3} principal? Fortunately, there is an easier norm trick that allows us to decide. Suppose 𝔭_{2} = (α), where α = a + b\sqrt{−6}. Then

2 = Norm(𝔭_{2}) = | Norm(α)| = (a + b\sqrt{−6})(a − b\sqrt{−6}) = a^{2} + 6b^{2}.

Trying the first few values of a, b ∈ Z, we see that this equation has no solutions, so 𝔭_{2} can not be principal. By a similar argument, we see that 𝔭_{3} is not principal either. Thus 𝔭_{2} and 𝔭_{3} define elements of order 2 in Cl(K).

Does the class of 𝔭_{2} equal the class of 𝔭_{3}? Since 𝔭_{2} and 𝔭_{3} define classes of order 2, we can decide this by finding the class of 𝔭_{2} · 𝔭_{3}. We have

𝔭_{2} · 𝔭_{3} = (2, \sqrt{−6}) · (3, \sqrt{−6}) = (6, 2\sqrt{−6}, 3\sqrt{−6}) ⊂ (\sqrt{−6}).

The ideals on both sides of the inclusion have norm 6, so by multiplicativity of the norm, they must be the same ideal. Thus 𝔭_{2} · 𝔭_{3} = (\sqrt{−6}) is principal, so 𝔭_{2} and 𝔭_{3} represent the same element of Cl(K). We conclude that

Cl(K) = ⟨𝔭_{2}⟩ = Z/2Z.

2 The Group of Units

Definition 2.1 (Unit Group). The group of units U_{K} associated to a number field K is the group of elements of O_{K} that have an inverse in O_{K}.

Proposition 2.2. An element a ∈ K is a unit if and only if Norm(a) = ±1.

Theorem 2.3. The group U_{K} of units of O_{K} is the product of a finite cyclic group of roots of unity with a free abelian group of rank r + s − 1, where r is the number of real embeddings of K and s is the number of complex conjugate pairs of embeddings.

Example 2.4 (Pell’s Equation). The classical Pell’s equation is, given square-free d > 0, to find all positive integer solutions (x, y) to the equation x^{2} − dy^{2} = 1. Note that if x + y\sqrt{d} ∈ Q(\sqrt{d}), then

Norm(x + y\sqrt{d}) = (x + y\sqrt{d})(x − y\sqrt{d}) = x^{2} − dy^{2}.

The solutions to Pell’s equation thus form a finite-index subgroup of the group of units in the ring of integers of Q(\sqrt{d}). Theorem 2.3 implies that for any d the solutions to Pell’s equation form an infinite cyclic group, a fact that takes substantial work to prove using only elementary number theory (for example, using continued fractions).