<s>[INST] 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.
 [/INST] 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. 
</s>