For $a_{i}:=\sqrt{2+a_{i-1}}$, how to prove that $\mathbb{Q}\subset\mathbb{Q}(a_i)$ is cyclic of degree $2^i$?

Question

Let us define numbers $a_i\in\mathbb{R}_{\geq0}$ by $a_0=0$ and $a_{i+1}=\sqrt{2+a_i}$.

1. How do I prove that $\mathbb{Q}\subset\mathbb{Q}(a_i)$ is cyclic of degree $2^i$?
2. How do I prove that the root depth of $a_i$ is equal to $i$ for all $i\geq0$?

What I know:
1. I was thinking of finding a generator, since that would mean it is cyclic, but I don't know which to choose.
2. The quadratic closure of a field $F$ in $\overline{F}$ is $$F^q:=\bigcup_{j=0}^\infty F_j,$$ where $F_0=F$, and $F_j=F_{j-1}(\sqrt{F_{j-1}})$.
The root depth is the smallest number $j$ for which $x_i$ is in $F_j$. However I would not know how to prove the question with this information.

Answer 1

A plan of attack:

• Show that $a_n=2\cos\dfrac{\pi}{2^{n+1}}$.
• Let $\zeta$ be a complex primitive root of unity of order $2^\ell$, $\ell\ge3$. Recall (or do the exercise) that $\operatorname{Gal}(\Bbb{Q}(\zeta)/\Bbb{Q})$ is a direct product of the subgroups generated by $\sigma:\zeta\mapsto \zeta^5$ and $\tau:\zeta\mapsto \zeta^{-1}=\overline{\zeta}$. Here $\sigma$ has order $2^{\ell-2}$ and $\tau$ has order two.
• Conclude that the fixed field of $\tau$ is a cyclic extension of degree $2^{\ell-2}$.
• Prove that the fixed field of $\tau$ is $\Bbb{Q}(\cos(\pi/2^{\ell}))$ with Galois group cyclic generated by (a restriction) of $\sigma$.

Answer 2

All you have to do is show by induction that $\sqrt{2+a_i} \not \in \mathbb{Q}(a_i)$. The induction step is then to prove that $\sqrt{2+\sqrt{2+a_i}} \not \in \mathbb{Q}(\sqrt{2+a_i})$ That is that the equation $$2+\sqrt{2+a_i}=(a+b\sqrt{2+a_i})^2$$ is impossible. Squaring gives $$2+\sqrt{2+a_i}=a^2+b^2(2+a_i) + 2ab\sqrt{2+a_i}$$ and the induction hypothesis implies that $$2=a^2+b^2(2+a_i)$$ And using that $a_i>0$ we see this is impossible. So $[\mathbb{Q}(a_{i+1}): \mathbb{Q}(a_{i})]=2$