Decomposition of an ideal as a product of two ideals How to show $$5\mathbb{Z}[\sqrt[3]{2}] = (5, \sqrt[3]{2}+2)(5, (\sqrt[3]{2})^2+3\sqrt[3]{2}-1).$$
Firstly, I think that I can say that $$(5, \sqrt[3]{2}+2)(5, (\sqrt[3]{2})^2+3\sqrt[3]{2}-1)= (25,5(\sqrt[3]{2}+2),5((\sqrt[3]{2})^2+3\sqrt[3]{2}-1),5((\sqrt[3]{2})^2+\sqrt[3]{2})).$$ Since 5 divides each term, I have that $(5, \sqrt[3]{2}+2)(5, (\sqrt[3]{2})^2+3\sqrt[3]{2}-1) \subseteq (5)$. But now how do I show the other direction? Any help would be appreciated. Thanks in advance.

anon gave you the elementary answer, but if we want to use more machinery, we can use the Dedekind-Kummer theorem.
Namely, to factor $5\mathcal{O}_K$ (where $K=\mathbb{Q}(\sqrt[3]{2})$) it suffices to factor $x^3-2$ in $\mathbb{F}_5[x]$. But, this factors as $(x+2)(x^2+3x+4)$. Thus, we know that 
$$5\mathcal{O}_K=(5,\sqrt[3]{2}+2)(5,\sqrt[3]{2}^2+3\sqrt[3]{2}+4)$$
But, by subtracting $5$ from the second entry in the second ideal gives your factorization. 
