How to see if $ x = \sqrt{2 +\sqrt{2}}+\dfrac{1}{2} \sqrt[3]{3}$ is integral over $\mathbb{Z}$ or not?? I don't think it is, as $\dfrac{1}{2}$ is not . But I am having trouble writing out an explicit proof.
Should I use any of the equivalent conditions like trying to show that $Z[x]$ is not a finitely generated $\mathbb{Z}$ module, etc??
On
$\sqrt{2+\sqrt{2}}$ is integral over $\mathbb{Z}$, as its monic irreducible polynomial is $x^4-4x^2+2$. Hence if $x$ is integral over $\mathbb{Z}$ then so is $\frac{1}{2}\sqrt[3]{3}=x-\sqrt{2+\sqrt{2}}$.
Similarly, $\sqrt[3]{9}$ is integral over $\mathbb{Z}$, hence if $\frac{1}{2}\sqrt[3]{3}$ is integral over $\mathbb{Z}$ then so is $$ \frac{1}{2}\sqrt[3]{3}\cdot\sqrt[3]{9}=\frac{3}{2} $$ But the only rational numbers which are integral over $\mathbb{Z}$ are the integers, so $\frac{3}{2}$ isn't integral over $\mathbb{Z}$, which implies that $x$ isn't integral either.
The algebraic integers are closed under addition, subtraction, multiplication, and $n$-th roots, for any positive integer $n$.
It follows that $\sqrt{2 +\sqrt{2}}$ is an algebraic integer.
Let $x = \sqrt{2 +\sqrt{2}}+\frac{1}{2} \sqrt[3]{3}$.
If $x$ is an algebraic integer, then
$${\small{\frac{1}{2}}} \sqrt[3]{3} = x - \sqrt{2 +\sqrt{2}}$$
is an algebraic integer, but then $\left( {\small{\frac{1}{2}}} \sqrt[3]{3}\right)^3 $ is an algebraic integer, contradiction, since $\frac{3}{8}$ is not an algebraic integer.