An example showing $\mathbb{Z}[\sqrt[3]{7}]$ is not a UFD

Question

It cannot be a UFD because it's the ring of integers of $\mathbb{Q}(\sqrt[3]{7})$ and has class number 3. How can we give an example showing this?

Answer 1

• Take $\mathfrak{p}$ a non-principal inversible prime ideal.

• Take $a\in \mathfrak{p},\not \in \mathfrak{p}^2$ (this can be done in the finite ring $O_K/(p^2)$ where $p=char(O_K/\mathfrak{p})$)

• Let $m$ be the order of $\mathfrak{p}$ in the class group, so $\mathfrak{p}^m=(b)$.

• $b$ is irreducible but $(b)$ is not a prime ideal.

• $a^m \in (b)$

• Neither $a$ nor $a^{m-1}$ is in $(b)$ so $a^m$ has more than one factorization.