The Ideal Class Group is Finite

In this post I’m going to prove that every number field K has only finitely many ideal classes. Next post, I’m going to compute some explicit class groups and show how this whole thing is applicable to Diophantine equations where we must use rings of integers that don’t have unique factorizations.

First, if X = {x1, x2, …, x(n)} is an integral basis of O(K), and the set of ring homomorphisms from K to C is {w1, w2, …, w(n)}, then let x(i, j) be w(i)(x(j)), that is that ith conjugate of x(j). Each x(i, j) has an absolute value inherited from C, which may or may not be equal to N(x(j)). Now, define m(i) to be the sum of |x(i, j)| over all j, and m to be product of m(i) over all i. Then m is a positive real number depending on K and X.

Second, every ideal I of O(K) has some element a such that |N(a)| <= m*N(I). To see why, define k to be the integer such that k^n <= N(I) < k^(n+1) (see errata below), where n is the degree of K over Q. Now, consider the set S of elements of the form a1x1 + a2x2 + … + a(n)x(n), where every a(i) is an integer between 0 and k inclusive. S has k^(n+1) elements, so the natural function from S to O(K)/I can’t be injective.

Now, we can find two distinct elements in S with the same image in O(K)/I, defined by, say, b(j) and c(j). Then b–c = (b1-c1)x1 + … + (b(n)-c(n))x(n) is in I, and for each j, –k <= b(j)-c(j) <= k. |N(b–c)| is the product of the absolute values of the conjugates of (b1-c1)x1 + … + (b(n)-c(n))x(n); each absolute value is at most the absolute value of the sum of that conjugate of each (b(j)-c(j))x(j) over all j, which is at most kx(j). The |N(b–c)| is at most the product of the sum of kx(i, j), which is just (k^n)*m <= N(I)*m.

The bound here, m, is really bad. Using some theorems on lattices in space, it’s possible to get a far lower bound; for instance, if K = Q(SQRT(-5)), the lowest m is derived from the integral basis {1, SQRT(-5)}, for which m = 10.47 to two decimal places. The theorem on lattices gives m = 1.42.

The third step is showing that every ideal I is equivalent in the class group to some ideal of norm less than m. To see why that holds, we use fractional ideals. If I^(-1) is the inverse of I, then we can find some c such that cI^(-1) is an integral ideal of O(K). Then there exists an a in I with |N(a)| <= m*N(cI^(-1)). Since (a) is contained in cI^(-1), it is divisible by it, so their quotient, (a/c)I, is an integral ideal. But now N((a/c)I) = |N(a)|/N(cI^(-1)) <= m.

Recall that the definition of the class group is H = F/P, where F is the multiplicative group of fractional ideals and P is the multiplicative group of principal fractional ideals. Two ideals I and J have the same representative in H if we can find some element b in the field such that I = bJ. But that shows that (a/c)I is equivalent to I in H.

Finally, there are only finitely many ideals of norm at most m. But this means that every ideal is equivalent in H to an ideal drawn from a finite set. In particular, every element of H has a representative in a finite set. So H is finite, and we’re done.

Errata: the bound on Q(SQRT(-5)) is not 1.42 by the theorem on lattices. The bound is never less than 2. The actual bound is ((4/pi)^s)*(n!/n^n)*SQRT(|d(K)|), where n is the degree of K and n = r + 2s, where r is the number of real conjugates of K and s is the number of complex conjugate pairs of conjugates of K. For K = SQRT(-5), n = 2, r = 0, and s = 1, so the bound is (4/pi)*(1/2)*SQRT(20) = 2.85. The estimate of 1.42 was obtained by erroneously letting d(K) = 5.

Also, thanks to commenter Bob: k^(n+1) is a mistake. The set S has (k+1)^n elements. There are k+1 choices for each a(i), and n different i‘s. We can’t define k to satisfy k^n <= N(I) < k^(n+1), because we’re not guaranteed to have such a k, let alone a unique one. What we need is k^n <= N(I) < (k+1)^n.