Proving $5S= \langle 5,{\alpha} +2\rangle\,\langle 5, {\alpha}^2+3{\alpha}-1\rangle$ I need help in verifying the following equality:
$$ 5S = \langle 5, \alpha  +2\rangle\, \langle 5, {\alpha}^2+ 3 \alpha -1\rangle $$
where $S= \mathbb{Z}[{2}^{1/3}]$ and $\alpha = 2^{1/3}$.
It seems so easy to prove but I can prove only one inclusion that product of those two ideals above is in $5S$. Though I have a few ideas to prove the other inclusion:


*

*Most boring way- I'm not even sure if it will work because I left it mid-way. Any element of $5S$ is of the form $5(a_0+a_1{\alpha}+ a_2{\alpha}^2)$ so try to express it in as an element of $\langle 25, 5{\alpha}+10, 5{\alpha}^2+15{\alpha}-5, 5{\alpha}^2+5{\alpha}\rangle$ which is the product of the given two ideals.

*Try to prove that intersection of the given ideals is in $5S$. I know intersection contains $5S$ but I'm not sure how to find it.


It would be great if someone could provide me some hint/s on proving this equality. I don't want a full solution of this seemingly easy question.
Reference: Number Fields, Daniel Marcus
Thanks.
 A: Let $I$ denote the ideal $\langle 5,\alpha+2,\alpha^2+3\alpha-1,\alpha^2+\alpha\rangle$ of $\mathbb{Z}[\alpha]$.  Observe that $$5I=\langle 5,\alpha+2\rangle \,\langle 5,\alpha^2+3\alpha-1\rangle$$ (I did not check, but I put my faith in your calculations).  I claim that $1\in I$.  To show this, first note that $\alpha \in I$ because
$$\alpha =\alpha(\alpha+2)-(\alpha^2+\alpha)\,.$$
Thus, $1\in I$, as
$$1 =\alpha(\alpha+3)-(\alpha^2+3\alpha-1)\,.$$
A: You can use another method that I'm sure is discussed in Marcus' book. 
Consider $K=\mathbb{Q}(\alpha)$, where $\alpha = \sqrt[3]{2}$. First we'll prove that $\mathbb{Z}[\alpha]$ is the ring of integers of $K$, i.e. $\mathcal O_K = \mathbb{Z}[\alpha]$. To prove this first note that $\text{disc } \mathbb{Z}[\alpha] = -108$. Also we know the following:
$$\left | \frac{\mathcal O_K}{\mathbb{Z}[\alpha]} \right |^2 \big | \text{ disc } \mathbb{Z}[\alpha] = -3^3\cdot 2^2$$
However we get that $x^3-2$ is an Eisenstein polynomial for $p=2$ and hence $2 \nmid \left | \frac{\mathcal O_K}{\mathbb{Z}[\alpha]} \right |$. Similarly $(x-1)^3 - 2$ is an Eisenstein polynomial for $p=3$ and hence $3 \nmid \left | \frac{\mathcal O_K}{\mathbb{Z}[\alpha]} \right |$. Therefore $\mathcal O_K = \mathbb{Z}[\alpha]$
Now to find out how $5\mathcal O_K$ factors into ideals we factorize $x^3-2$ in $\mathbb{F}_5[x]$. We then get $x^3 - 2 = (x+2)(x^3 + 3x + 4)$ in $\mathbb{F}_5[x]$. Therefore we have that:
$$5\mathcal O_K = \langle 5,\alpha + 2 \rangle \langle 5, \alpha^2 + 3\alpha + 4 \rangle$$
Hence the proof.
