I want to prove that $\alpha=\frac{\sqrt[3]{2}}{2}$ is not integral over $\mathbb{Z}$ (i.e. not an algebraic integer). Is there a way to modify the following reasoning to make it work?
A quick check shows that $P(x)=4x^3-1$ is the minimal polynomial of this $\alpha$ over $\mathbb Q$. So I set $I=\{p(x)\in \mathbb Q[x]:p(\alpha)=0\} =(P(x)) \triangleleft \mathbb Q[x]$. I claim that if $q(x)$ is a monic polynomial in $\mathbb Z[x]\cap I$ then $P(x)|q(x)$ in $\mathbb Z[x],$ not just in $\mathbb Q[x]$. I.e. I claim that if $P(x)p(x)=q(x)$ for some $p(x)\in \mathbb Q[x]$, we also have $p(x) \in \mathbb Z[x].$ But $p(x)$ is a root of the polynomial $P(x)y-q(x) \in \mathbb Z[x][y]$ and so $p(x)$ is integral over $\mathbb Z[x]$. But $\mathbb Z[x]$ is integrally closed, since it's a UFD and therefore $p(x)\in \mathbb Q[x] \implies p(x) \in \mathbb Z[x]$.
I have realised that after writing this, $p(x)$ actually isn't integral as the polynomial $P(x)y-q(x)$ is not monic. Is there any way of salvaging this argument, or proving the result a different way? I want to say something like "If the monic minimal polynomial of $\alpha \in\mathbb C$ over $\mathbb Q$ does not have integer coefficients (In my above example, this polynomial would be $ x^3-\frac{1}{4} \notin\mathbb Z[x]$) then this $\alpha$ is not an algebraic integer."