Leting $\delta:=\sqrt[3]{2}$, prove that $\mathbb{Z}[\delta]$ is the ring of integers of $\mathbb{Q}(\delta)$.
Here's what I've done: let $K:=\mathbb{Q}(\delta)$. Since $[K:\mathbb{Q}]=3$, then $1, \delta, \delta^2$ are $\mathbb{Q}$-linearly independent (in particular, they are $\mathbb{Z}$-independent). Since $1, \delta, \delta^2$ are integers of $K$, we have that $\mathbb{Z}[\delta]=<1, \delta, \delta^2>_{\mathbb{Z}}\subset\mathcal{O}_{K}$.
To prove $\supset$, I did the following: if $z=a+b\delta+c\delta^2$ is an integer of $K$, its minimal polynomial $p(x)=x^3+c_2x^2+c_1x+c_0$ lies in $\mathbb{Z}[x]$ and has roots $\sigma_1(z),\sigma_2(z)$ and $\sigma_3(z)$, where
\begin{align*} \sigma_1 &:\delta\mapsto\delta\\ \sigma_2 &:\delta\mapsto\omega\delta\\ \sigma_3 &:\delta\mapsto\omega^2\delta \end{align*}
are the embeddings fixing $\mathbb{Q}$ and $\omega^3=1$. Using Viète's relations we get:
\begin{align*} c_2&=-\sum_{i}\sigma_i(z)= -3a\\ c_1&= \sum_{i<j}\sigma_i(z)\sigma_j(z)=3a^2-6bc\\ c_0&= -\prod_{i}\sigma_i(z)=-(a^3+2b^3+4c^3-6abc) \end{align*}
which gives us restrictions for $a, b, c$, since $c_0, c_1, c_2\in\mathbb{Z}$. Since $\delta z$ and $\delta^2 z$ are also integers of $K$, we calculate its respective minimal polynomials and find new restrictions for $a, b, c$.
After a very tiresome case analysis, I could prove that $a, b, c\in\mathbb{Z}$, concluding the proof.
This solution was given as a hint by the professor, and seems to me not very natural and, above all, way too long. My question is: is there a more natural and/or shorter solution for this? Thanks!
