Algebraic Numbers, Global Fields Have Finite Classgroups

Global Fields Have Finite Classgroups

1. Every ideal has finite index. This is true for Z and Z_p[t], and it propagates upward to the integral ring. We'll need this for the proof of step 3 below.

2. There are finitely many prime ideals with index below c, for any integer c. This is true for Z and Z_p[t], and it propagates upward to the integral ring.

3. Every ideal H has an x such that the index of x in H is no larger than a fixed integer l. this becomes a sufficient condition for a finite class group. We'll prove this for Z, and then for Z_p[t].

Let a number field E/Q have dimension n, and let S be the algebraic integers therein. Since Z is a pid, S is a free Z module. Give S a basis b₁ through b_n.

let H be an ideal in S. Choose m so that mⁿ ≤ |H| < (m+1)ⁿ. This can be done because H has finite index, as shown in step 1.

Consider the set of elements in S spanned by b₁ through b_n, with coefficients between 0 and m inclusive. Its cardinality exceeds the number of cosets of H, so two elements lie in the same coset. Their difference x is an element in H, with coefficients ≤ m in absolute value.

Recall from an earlier section that the index of x equals its norm. And the norm of x is the product of the norms of the images of x under all possible embeddings. Let c() be an embedding of S in the complex plane. In fact there are n embeddings, c₁(S) through c_n(S). Concentrate on a particular embedding c_i.

The norm of a sum of vectors in the complex plane is at most the sum of the norms of the individual vectors. Thus the norm of c_i(x) is at most m times the sum of|c_i(b_j)|.

Taking the product over all embeddings pulls out n factors of m, giving mⁿ. What remains is a product over all embeddings, of the sum of the norms of the images of b₁ through b_n. Call this expression l. Now |x| ≤ mⁿl.

We know |H| ≥ mⁿ, so l×|H| ≥ |x|. That completes the proof for number fields.

Let's look at our favorite example, adjoining the square root of -5. (We'll prove this is dedekind elsewhere.) Use 1 and sqrt(5)i as a basis. The embeddings are the identity map and complex conjugation. Apply the first embedding and find the sum of the norms, which is |1|+|sqrt(5)i|, or 1+sqrt(5). The other embedding produces the same norms, and the same sum. Take the square of 1+sqrt(5) and get (approximately) 10.472, which is our value of l.

It may seem strange to think of the index of x in H as a decimal, like 10.472. After all, there are an integer number of cosets of x in H. However, the value of l, derived from the images of our basis in the complex plane, acts as an upper bound. There are 10 or fewer cosets of x in H.

When looking for the generators of the class group (which is nontrivial since the ring is not a ufd), one can restrict attention to primes with index < 11. Such a prime Q in S contracts to a prime P in Z, and S/Q is at least as large as Z/P. Thus we can restrict attention to the primes in S lying over 2, 3, 5, and 7.

Next let E be an n dimensional extension of Z_p(t). Let S be the integral ring, and assume a basis b₁ through b_n. Given an ideal H, choose m as before, but restrict m to successive powers of p. Thus mⁿ ≤ |H| < (pm)ⁿ. for notational convenience, let m = p^j.

Let polynomials of degree j or less act as coefficients on b₁ through b_n. There are pm of these, and when raised to the n^th, the cardinality of S/H is exceeded. some x, using these coefficients, lies in H. Let's find the valuation of |x|. (This is a bit like taking logs.) The norm, or index, is p raised to this valuation.

Instead of any old basis for E, that hapens to live in S, I'm going to select a particular basis. Let Q be a prime ideal in S that lies over t. Then replace S with the localization S_Q, which is part of the larger fraction field E. This is integral over the base ring localized about t, which is the same base ring. Since S is integrally closed, and S_Q is integral and contains S, S_q = S.

Remember that S_Q is a local dedekind domain, and a dvr. Instead of talking about Q the prime ideal, let's say q is an element of this dvr such that q^m is an associate of t. Now we are ready to build the basis. The residue field over q is a finite extension of Z_p. Build a basis for the field extension, say the powers of y up to y^l. Then cross this with the powers of q from 0 to k-1. This forms a basis for S.

As you recall, we wanted to place a bound on the valuation of a linear combination of basis elements, with j degree polynomials as coefficients. Start with the constant terms on the polynomials. This gives something in Z_p times b₁, plus something in Z_p times b₂, and so on through b_n. The valuations of each of these terms are 0, 1/k, 2/k, 3/k, and so on up to (k-1)/k.

Now do the same for the linear terms. The result is the same as above, except everything is multiplied by t. Thus the valuations are 1, 1+1/k, 1+2/k, and so on up to 1+(k-1)/k. Repeat this process through t^j, giving a maximum valuation of j+(k-1)/k. This is the largest possible valuation for x.

There are n embeddings of x to consider, but each keeps the finite field extension of Z_p in place, and moves q to one of its conjugates. The valuations are unchanged. One can multiply the maximum valuation by n, for all n embeddings. The result is less than n times (j+1).

Remember that the index of H is at least p^jn. The index of {x} is less than p^n×(j+1). The quotient is strictly less than pⁿ. This bound proves the class group is finite, and as we saw above, it constrains the primes that can act as generators for the class group when the extension is not a ufd. In particular, we only need look at primes over primes of degree less than n.