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?
 A: 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 $2$.
A: 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.
