More About Computing Class Groups

If $ \mathfrak{p}$ is a prime of $ \O_K$, then the intersection $ \mathfrak{p}\cap \mathbf{Z}=p\mathbf{Z}$ is a prime ideal of $ \mathbf {Z}$. We say that $ \mathfrak{p}$ lies over $ p\in\mathbf{Z}$. Note $ \mathfrak{p}$ lies over $ p\in\mathbf{Z}$ if and only if $ \mathfrak{p}$ is one of the prime factors in the factorization of the ideal $ p\O_K$. Geometrically, $ \mathfrak{p}$ is a point of $ \Spec (\O_K)$ that lies over the point $ p\mathbf{Z}$ of $ \Spec (\mathbf{Z})$ under the map induced by the inclusion $ \mathbf{Z}\hookrightarrow \O_K$.

Lemma 7.3.1   Let $ K$ be a number field with ring of integers $ \O_K$. Then the class group $ \Cl (K)$ is generated by the prime ideals $ \mathfrak{p}$ of $ \O_K$ lying over primes $ p\in\mathbf{Z}$ with $ p\leq B_K = \sqrt{\vert d_K\vert}\cdot
\left(\frac{4}{\pi}\right)^s\cdot \frac{n!}{n^n}$, where $ s$ is the number of complex conjugate pairs of embeddings $ K\hookrightarrow \mathbf{C}$.

Proof. Theorem 7.1.2 asserts that every ideal class in $ \Cl (K)$ is represented by an ideal $ I$ with $ \Norm (I)\leq B_K$. Write $ I=\prod_{i=1}^m
\mathfrak{p}_i^{e_i}$, with each $ e_i\geq 1$. Then by multiplicativity of the norm, each $ \mathfrak{p}_i$ also satisfies $ \Norm (\mathfrak{p}_i)\leq B_K$. If $ \mathfrak{p}_i \cap
\mathbf{Z}=p\mathbf{Z}$, then $ p\mid \Norm (\mathfrak{p}_i)$, since $ p$ is the residue characteristic of $ \O_K/\mathfrak{p}$, so $ p\leq B_K$. Thus $ I$ is a product of primes  $ \mathfrak{p}$ that satisfies the norm bound of the lemma. $ \qedsymbol$

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

  1. Use the algorithms of Chapter 4 to list all prime ideals $ \mathfrak{p}$ of $ \O_K$ that appear in the factorization of a prime $ p\in\mathbf{Z}$ with $ p\leq B_K$.
  2. Find the group generated by the ideal classes $ [\mathfrak{p}]$, where the $ \mathfrak{p}$ are the prime ideals found in step 1. (In general, this step can become fairly complicated.)
The following three examples illustrate computation of $ \Cl (K)$ for $ K=\mathbf{Q}(i), \mathbf{Q}(\sqrt{5})$ and $ \mathbf{Q}(\sqrt{-6})$.

Example 7.3.2   We compute the class group of $ K=\mathbf{Q}(i)$. We have

$\displaystyle n = 2, \quad r=0, \quad s=1, \quad d_K = -4,
$

so

$\displaystyle B_K = \sqrt{4}\cdot \left(\frac{4}{\pi}\right)^1
\cdot\left(\frac{2!}{2^2}\right) = \frac{8}{\pi} <3.
$

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

$\displaystyle 2\O_K = (1+i)^2,
$

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

Example 7.3.3   We compute the class group of $ K=\mathbf{Q}(\sqrt{5})$. We have

$\displaystyle n = 2, \quad r = 2, \quad s=0, \quad d_K = 5,
$

so

$\displaystyle B = \sqrt{5}\cdot \left(\frac{4}{\pi}\right)^0\cdot
\left(\frac{2!}{2^2}\right) < 3.$

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

Example 7.3.4   In this example, we compute the class group of $ K=\mathbf{Q}(\sqrt{-6})$. We have

$\displaystyle n = 2, \quad r=0, \quad s=1, \quad d_K = -24,
$

so

$\displaystyle B = \sqrt{24} \cdot \frac{4}{\pi} \cdot
\left(\frac{2!}{2^2}\right)\sim 3.1.
$

Thus $ \Cl (K)$ is generated by the prime ideals lying over $ 2$ and $ 3$. We have $ \O_K=\mathbf{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

$\displaystyle \mathfrak{p}_2 = (2,\sqrt{-6})$   and$\displaystyle \qquad
\mathfrak{p}_3 = (3,\sqrt{-6}).
$

Also, $ \mathfrak{p}_2^2 = 2\O_K$ and $ \mathfrak{p}_3^2=3\O_K$, so $ \mathfrak{p}_2$ and $ \mathfrak{p}_3$ define elements of order dividing $ 2$ in $ \Cl (K)$.

Is either $ \mathfrak{p}_2$ or $ \mathfrak{p}_3$ principal? Fortunately, there is an easier norm trick that allows us to decide. Suppose $ \mathfrak{p}_2 = (\alpha)$, where $ \alpha=a+b\sqrt{-6}$. Then

$\displaystyle 2=\Norm (\mathfrak{p}_2) = \vert\Norm (\alpha)\vert = (a+b\sqrt{-6})(a-b\sqrt{-6})
= a^2 + 6b^2.$

Trying the first few values of $ a,b\in\mathbf{Z}$, we see that this equation has no solutions, so $ \mathfrak{p}_2$ can not be principal. By a similar argument, we see that $ \mathfrak{p}_3$ is not principal either. Thus $ \mathfrak{p}_2$ and $ \mathfrak{p}_3$ define elements of order $ 2$ in $ \Cl (K)$.

Does the class of $ \mathfrak{p}_2$ equal the class of $ \mathfrak{p}_3$? Since $ \mathfrak{p}_2$ and $ \mathfrak{p}_3$ define classes of order $ 2$, we can decide this by finding the class of $ \mathfrak{p}_2\cdot \mathfrak{p}_3$. We have

$\displaystyle \mathfrak{p}_2\cdot \mathfrak{p}_3 = (2,\sqrt{-6})\cdot (3,\sqrt{-6})
= (6,2\sqrt{-6}, 3\sqrt{-6}) \subset (\sqrt{-6}).$

The ideals on both sides of the inclusion have norm $ 6$, so by multiplicativity of the norm, they must be the same ideal. Thus $ \mathfrak{p}_2\cdot \mathfrak{p}_3=(\sqrt{-6})$ is principal, so $ \mathfrak{p}_2$ and $ \mathfrak{p}_3$ represent the same element of $ \Cl (K)$. We conclude that

$\displaystyle \Cl (K) = \langle \mathfrak{p}_2 \rangle = \mathbf{Z}/2\mathbf{Z}.$

William Stein 2012-09-24