# Calculating class numbers

It is a general question about simple examples of calculating class numbers in quadratic fields. Here are an excerpt from Frazer Jarvis' book Algebraic Number Theory:

"Example 7.20 For $K=\mathbb{Q}(\sqrt[3]{2} )$, the discriminant is 108, and $r_{2}=1$. So the Minkowski bound is $\approx 2.940$. So every ideal is equivalent to one whose norm is at most 2. The only ideal of norm 1 is the full ring of integers, which is principal; the ideal $(2)=\mathcal{p}_{2}^{3}$, where $\mathcal{p}_{2}=(\sqrt[3]{2})$ is also principal. Thus every ideal is equivalent to a principal ideal, so the class group is trivial."

The question is why does it suffice to look at the principal ideal generated by 2?

Since the Minkowski bound is less than $3$, we need to show that the only integral ideals of $\mathcal{O}_K$ with norm $1$ or $2$ are $I=(1)$ and $I=(\sqrt[3]{2})$. Both are principal, so the class number is $1$. The prime $2$ in $\mathcal{O}_K$ has the principal prime factorization $(2)=(\sqrt[3]{2})^3$; for more details and a proof see Keith Conrad's notes here, after Theorem $1$, and in Theorem $2$.

Because the Minkowski's bound is $\approx 2.940$ we only need to check the prime ideals of norm $2$ or less, as each ideal class group contains such ideal.

Consider a prime ideal $\mathfrak p$ in $\mathcal O_K$ of norm $2$. We know it must lie above a ideal $(p)$ of $\mathbb{Z}$. Now we have the following:

$$N(\mathfrak p) = \left | \frac{\mathcal O_K}{\mathfrak p} \right| = \left | \frac{\mathbb Z}{(p)} \right|^{\left[\frac{\mathcal O_K}{\mathfrak p} : \frac{\mathbb Z}{(p)} \right]} = N(p)^{f(\mathfrak p|p)}$$

Now note that $N(p) = p$ and so we must have that $p=2$. In particular $\mathfrak p$ is a prime ideal of $\mathcal O_K$, lying above $2$. So therefore we need to consider the factorization of $(2)$ in $\mathcal O_K$. It's not hard to notice that:

$$2 \mathcal O_K = (2,\sqrt[3]{2})^2 = (\sqrt[3]{2})^3$$

Finally the ideal class group of $\mathcal O_K$ is generated by $(1)$ and $(\sqrt[3]{2})$, so it must be the trivial group of one element and hence $\mathbb{Z}[\sqrt[3]{2}]$ is a PID.