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For each $n \in \mathbb{N}$, let $\zeta_n = e^{\frac{2\pi i}{n}}.$

For $n \geq 0$, let $K_n = \mathbb{Q}(\zeta_{2^{n+2}})$ (the cyclotomic field of $2^{n+2}$ root of unity). Let $a_n = \zeta_{2^{n+2}}+\zeta_{2^{n+2}}^{-1}$ and $K_n^+ = \mathbb{Q}(a_n).$

Fact : $K_n^+$ is the maximal real subfield of $K_n.$

I want to show that $[K_n : K_n^+] = 2$ and $[K_{n+1}^+ : K_n^+] = 2.$

I think $K_n = K_n^+(\zeta_{2^{n+2}})$, so I need to find an irreducible polynomail $p(x)$ over $K_n^+[x]$ such that $p(\zeta_{2^{n+2}}) = 0.$

Let $p(x) = (x - \zeta_{2^{n+2}})(x - \zeta_{2^{n+2}}^{-1}) = x^2 - a_nx + 1.$ I think this $p(x)$ works and yields that $$K_n\simeq K_n^+[x]\Big/(p(x)).$$

So $[K_n : K_n^+] = 2.$ Is this conclusion correct ?

I notice that $K_{n+1}$ contains $K_n$ as a subfield. So $K_n^+ \subseteq K_{n+1}^+$.

Therefore, it is valid to calculate $[K_{n+1}^+ : K_n^+].$

Since the degree of extension should be $2$, I try to construct a quadratic irreducible polynomail, but still struggling.

I also can show that $[K_n : \mathbb{Q}] = 2^{n+1}$ and $[K_n^+ : \mathbb{Q}] = 2$. Not sure fact this fact help showing $[K_{n+1}^+ : K_n^+] = 2.$

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For the first part, you have shown that $[K_n:K_n^+]\leq2$. To show equality, you must also check that $p(x)$ is irreducible, but that is trivial since the roots of $p(x)$ are nonreal.

For the second part note that, by transitivity,

$$\underbrace{[K_{n+1}:K_n]}_2\underbrace{[K_n:K_n^+]}_{2}=\underbrace{[K_{n+1}:K_{n+1}^+]}_2[K_{n+1}^+:K_n^+].$$

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Yes, this is fine. You have $\zeta=\zeta_{2^{n+2}}$ a root of $X^2-a_nX+1=0$, as this has real coefficients and non-real roots, it is irreducible over $\Bbb R$, and a fortiori over $K_n^+$.

One has $$[K_{n+1}:K_n^+]=[K_{n+1}:K_n][K_n:K_n^+]=4$$ and so $$4=[K_{n+1}:K_n^+]=[K_{n+1}:K_{n+1}^+][K_{n+1}^+:K_n^+]=2 [K_{n+1}^+:K_n^+]$$ so $[K_{n+1}^+:K_n^+]=2$.

If you want an equation for $a_{n+1}$, note that $$a_{n+1}^2=(\zeta_{2^{n+3}}+\zeta_{2^{n+3}})^2 =\zeta_{2^{n+2}}+2+\zeta_{2^{n+2}}^{-1}=a_n+2$$ so $a_{n+1}$ is a root of $X^2-a_n-2=0$.

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