$\sqrt{2+\sqrt{2}}+\frac{1}{2}\sqrt[3]{3}$ is not integral over $\mathbb{Z}$ - solution check

Question

I want to check if $\sqrt{2+\sqrt{2}}+\frac{1}{2}\sqrt[3]{3}$ is integral over $\mathbb{Z}$ or not. I tried to show that it's not by doing this:

Since $x^3-3=0$ is a polynomial in $\mathbb{Z}[x]$ with $\sqrt[3]{3}$ as a root, this number is integral over $\mathbb{Z}$.

Same for $(x^2-2)^2-2$ and $\sqrt{2+\sqrt{2}}$.

From Atiyah-MacDonald's Proposition 5.1, $\mathbb{Z}[\sqrt[3]{3},\sqrt{2+\sqrt{2}}]$ is a finitely generated $\mathbb{Z}$-module, but $\sqrt{2+\sqrt{2}}+\frac{1}{2}\sqrt[3]{3}$ is not an element of this module (due to the $\frac{1}{2}$ coefficient), and so it's not integral over $\mathbb{Z}$.

Can someone tell me if this argument is correct or not?

Answer 1

Hint: If $\alpha,\beta$ are integral over $\mathbb Z$, then $\alpha\pm \beta$ are integral over $\mathbb Z$. Now let $$\alpha=\sqrt{2+\sqrt{2}}+\frac 12\sqrt[3]{3}$$ and $$\beta=\sqrt{2+\sqrt{2}}$$ and try to prove by contradiction.