Determining whether or not an element is integral over $\mathbb Z$ 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." 
 A: If $\alpha$ is integral over $\mathbb{Z}$, then the $\mathbb{Z}$-module generated by 1, $\alpha$, $\alpha^2$, $\alpha^3$ ... is a finitely generated $\mathbb{Z}$-module, but this is obviously false for your $\alpha$ (the denominators get arbitrarily large).
A: Hint: If $\alpha$ is an integral over $\mathbb Z$, then $\alpha^3$ is also integral over $\mathbb Z$. 
Now if a number is rational and integral over $\mathbb Z$, then....
A: In general I think this follows, more or less straightly, from Gauss's Lemma, yet , in your case we can try as follows:
Suppose $\;\alpha\;$ is an alg. integer, so that there exists a monic polynomial $\;f(x)\in\Bbb Z[x]\;$ s.t. $\;f(\alpha)=0\;$ . But $\;p(x)=x^3-\dfrac14\;$ is the minimal polynomial of $\;\alpha\;$ over $\;\Bbb Q\;$, and of course $\;\Bbb Z\subset\Bbb Q\;$ , so that we have that $\;p\mid f\;$ . Yet in order to have $\;f\;$ as a multiple of $\;p(x)\;$ we must get rid of that $\;\frac14\;$ there, and this can't be done without violating the fact that $\;f\;$ is monic.
Fill in details.
A: Another perspective is to use valuations.  The $2$-adic valuation $\nu_2$ extends (in general nonuniquely) to a valuation on $\mathbb{Q}(\sqrt[3]{2})$.  Actually, you can show that there is only one prime lying over $2$ here with ramification index $3$, so the extension is unique. You have $\nu_{2}(\sqrt[3]{2}) = \frac{1}{3} \nu_2(2)$, with $\nu_2(2) = 1$.  Thus $$\nu_2(\frac{\sqrt[3]{2}}{2}) = \nu_2(\sqrt[3]{2}) - \nu_2(2)  < 0$$ so $\frac{\sqrt[3]{2}}{2}$ can't be integral over $\mathbb{Z}$ (to be integral, it has to have nonnegative value at every prime).
