Degree of $\sqrt[5]{2+\sqrt[3]{5+\sqrt{2}}} \cdot e^{2\pi i /3}$ as algebraic number

Question

In this post, ‘degree’ means ‘degree as an algebraic integer’. Let $\alpha = \sqrt[5]{2+\sqrt[3]{5+\sqrt{2}}}$ I see that $f(\alpha)=0$ for the polynomial $f(x) = \big((x^{5}-2)^{3}-5)\big)^{2}-2$, which is of degree $30$. I have two questions:

1. I want to show that the degree of $\alpha$ is $30$. How can one argue that $\alpha$ indeed has degree $30$? Since $\sqrt[5]{\sqrt[3]{\sqrt{2}}}$ has degree $30$, so I feel like it should not be difficult to prove that $\alpha$ has degree $30$ but I am struggling here.

2. To show that $\beta =\sqrt[5]{2+\sqrt[3]{5+\sqrt{2}}} \cdot e^{2\pi i /3}$ has degree equal to $$\text{degree }(\alpha) \times \text{degree } (e^{2\pi i/3}) = 30 \times 2 = 60$$ seems way harder to me. I do not even have much intuition into it and computations seem awful.

I did enter both of these questions into WolframAlpha and my guesses are correct. But I have no clue why. I would love to know reasons behind it. I know some Galois theory.

Answer 1

If you insist on doing the calculations by hand, here is one possible approach using algebraic number theory.

1. $\sqrt{2}$ generates a quadratic extension $K_2$ of ${\mathbb Q}$.
2. $5 + \sqrt{2}$ is not a cube in $K_2$ since its norm is $23$. Alternatively, if it is the cube of an element of $K_2$, it must be the cube of an algebraic integer $a + b \sqrt{2}$, which it is not. Thus $K_6 = K_2(\sqrt[3]{5 + \sqrt{2}})$ has degree $6$.
3. $2 + \sqrt[3]{5 + \sqrt{2}}$ is not a fifth power in $K_6$ because its norm down to $K_2$ is not (it is $2^3 + 5 + \sqrt{2} = 13 + \sqrt{2}$). This implies that $K_{30} = {\mathbb Q}(\alpha)$ has degree $30$.
4. This is Jyrki Lahtonen's argument: Clearly $K_{30}(\beta^3) = K_{30}$ and $\beta \not \in K_{30}$, hence $K_{30}(\beta) \ne K_{30}$, and this implies that $K_{30}(\beta) = K_{30}(e^{2\pi i/3})$ is a quadratic extension of $K_{30}$.