Remarks on Computing the Class Group

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 $\mathfrak{p}$ $p\in\mathbf{Z}$. 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 11.1.1   Let $K$ be a number field with ring of integers $\O _K$. Then the class group $\Cl (K)$ is generated by the prime ideals $\mathfrak{p}$ 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. We proved before 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, whcih proves the lemma. $\qedsymbol$

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

1. Use the ``factoring primes'' algorithm 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, one must think more carefully about how to do this step.)

The following three examples illustrate computation of $\Cl (K)$ for $K=\mathbf{Q}(i), \mathbf{Q}(\sqrt{5})$ and $\mathbf{Q}(\sqrt{-6})$.

Example 11.1.2   We compute the class group of $K=\mathbf{Q}(i)$. We have

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

so

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

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

$$2\O _K = (1+i)^2,$$

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

Example 11.1.3   We compute the class group of $K=\mathbf{Q}(\sqrt{5})$. We have

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

so

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

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

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

so

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

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

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

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

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

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

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

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

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