I know that this problem was asked before, but I'm just getting started at ring theory, and I'm having a hard time understanding it. Could someone give a detailed answer please?
My thoughts were to first determine whether $\mathbb{Z}[\sqrt[3]{2}]$ is a Euclidean Domain, a P.I.D., or a U.F.D. (hopefully at least P.I.D.), then determine whether $4+\sqrt[3]{4}$ is irreducible in $\mathbb{Z}[\sqrt[3]{2}]$, and do something about it.
The answered question didn't use any of these, and just proceeded by just simple calculations, involving some linear algebra. I would love to get some help without using these "Linear Algebraic" methods.
Hint: Let $\alpha=\sqrt[3]{2}$ and $I=(4+\alpha^2)$. Note that $(8+\alpha-2\alpha^2)(4+\alpha^2)=34$, and so $34 \mathbb{Z}[\sqrt[3]{2}] \subseteq I$.
($8+\alpha-2\alpha^2$ was found by searching for $a+b\alpha+c\alpha^2$ such that $(a+b\alpha+c\alpha^2)(4+\alpha^2) \in \mathbb Z$.)
This implies that $G=\mathbb{Z}[\sqrt[3]{2}]/I$ is an abelian group of exponent at most $34$. Since $\mathbb{Z}[\sqrt[3]{2}]$ is a free abelian group of rank $3$, $G$ generated by at most $3$ elements. Combining this with exponent $1,2,17,34$ leads to finitely many candidates. The possibilities can be reduced by further work on $I$.