## 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.

### 10 Responses to The Ideal Class Group is Finite

1. DAS says:

Now you’re reminding me how much algebra I’ve forgotten. I used to be a hot shot math major. But in grad school I became a glorified statistician. Recently, I was working on a project involving some basic number theory (to deal with certain aspects of Fourier Transforms) and realized I’ve forgotten just about everything I learned as an undergrad.

Oy vey.

2. […] Recall from the previous post (now with errata) that for every number field K, there exists a constant m depending on K such that every ideal of K is equivalent in the class group H to an ideal of norm <= m. In addition, every ideal of norm <= m is generated by prime ideals of norm <= m; hence, it’s enough to look at the behavior of prime integers less than m under decomposition into ideals. […]

3. […] Recall from the proof of the finiteness of the class group that this shows that every ideal class of K contains a representative with norm <= SQRT(|d(K)|). […]

4. Bob says:

In paragraph 3, I think it should be

k^n

5. Bob says:

I don’t think my previous post appeared correctly: “In paragraph 3 I think it should be k^n

6. Bob says:

S has (k+1)^n elements… that’s all i’m trying to say. My comments keep getting truncated.

7. Alon Levy says:

Thanks for the correction.

I’m guessing your comment got truncated because you used a greater-than or lesser-than sign. Some HTML editing textboxes get overzealous and treat every lesser-than symbol as a tag, especially if you follow it with a character other than a space.

8. Shrenik Shah says:

Are you sure that the bound is never less than 2? Suppose you consider a fifth root of unity. The polynomial x^4+x^3+x^2+x+1 has discriminant 5^{5-2}=125, and s=2 in this case. We then have (4/\pi)^2\cdot (24/256) \cdot \sqrt{125}, which comes out to 1.699…, I think.

Shrenik

9. Alon Levy says:

Ah, good point. The bound is always strictly more than 1, but it can apparently be less than 2; actually, it’s 1.1 for the imaginary quadratic field Q(SQRT(-3)). But in that case the field is automatically a PID, since the only ideal of any number field that has norm 1 is the unit ideal.

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