<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] Working in $\Bbb Z[x]/(x^3-2)$ I get
$$(5,x+2)(5,x^2+3x-1)=\big(25,5(x^2+3x-1),5(x+2),(x+2)(x^2+3x-1)\big).$$
Then $(x+2)(x^2+3x-1)=x^3+5x^2+5x-2=5x(x+1)$. Hence the above ideal is
$$=(5)(5,x^2+3x-1,x+2,x^2+x).$$
Use the Euclidean division algorithm (which applies here) to simplify the right-hand factor. For example $x^2+3x-1\equiv x-1$ mod $x+2$ so we may replace $x^2+3x-1$ with $x-1$.
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