Is $\mathbb{Z}[\sqrt[3]{2}]$ a principal ideal domain?
That is, is every ideal of $\mathbb{Z}[\sqrt[3]{2}]$ generated by a single element?
2025-01-13 00:11:35.1736727095
Is $\mathbb{Z}[\sqrt[3]{2}]$ a principal ideal domain?
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Yes, the number field $K=\mathbb{Q}(\sqrt[3]{2})$ has class number one, i.e., its ring of integers $\mathcal{O}_K=\mathbb{Z}[\sqrt[3]{2}]$ is a PID. This follows by using the Minkowski bound: we have $$ | \text{disc}(K)| = | \text{disc}(1, \sqrt[3]{2}, \sqrt[3]{4})| = 108 = 2^23^3, $$ so that the Minkowski bound for $K$ is given by $$ \frac{3!}{3^3}\cdot \frac{4}{\pi}\sqrt{\left|\text{disc}(K)\right|}=\frac{16\sqrt{3}}{3\pi}\sim 2.94. $$ But because $p=2$ has the principal prime factorisation $(2)=(\sqrt[3]{2})^3$, the class number of $K$ is equal to $1$.