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Let's define the sequence $x_0=0$ and $ x_{i+1} = \sqrt{x_i+2}$ taking always the positive root. Prove that the field extension $\Bbb Q \subset \Bbb Q(x_i) $ is cyclic with degree $2^i$

Well.. at least it's easy to find a tower of fields $ \Bbb Q = E_0 \subset E_1 \subset ... E_i $ such that $ x_i \in E_i$ and $[E_{n+1}:E_n]=2$. So $[\Bbb Q(x_i) : \Bbb Q] \le 2^i $ and divides $2^i$ so is of the form $2^j$ with $ 1\le j \le i$. But I don't know how to continue :S And also I don't know how to prove that the extension is cyclic (it's group of \Bbb Q automorphism is cyclic )

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I'm guessing that your problem includes not being sure that $E_n = \mathbb{Q}(x_n)$? It may help to observe that $\mathbb{Q}(x_i) = \mathbb{Q}(x_i, x_{i-1})$.... – Hurkyl Nov 21 '12 at 9:53
up vote 7 down vote accepted

The easy way to solve this is to notice that $x_n = 2\cos(\pi 2^{-(n+1)})$, and first study the tower of fields $F_n = \Bbb Q(e^{2i\pi/2^{n+2}})$. It is easy to show that the Galois group of $F_n$ over $\Bbb Q$ is isomorphic to $(\Bbb Z / 2^{n+2} \Bbb Z)^*$ : pick a primitive $2^{n+2}$th root of unity $\zeta$. Then for $\alpha \in (\Bbb Z / 2^{n+2} \Bbb Z)^*$ there is a unique automorphism $\sigma_\alpha$ of $F_n$ sending $\zeta$ to $\zeta^\alpha$. $\sigma_{\alpha \beta} = \sigma_\alpha \circ \sigma_\beta$, and $\sigma_1 = id$.

Next, $E_n$ is the real subfield of $F_n$ : the subfield of $F_n$ fixed by complex conjugation. Since conjugation is $\sigma_{-1}$, the Galois group of $E_n$ is the quotient $(\Bbb Z / 2^{n+2} \Bbb Z)^*/\langle -1 \rangle$.

Finally, we need more on the structure of $(\Bbb Z / 2^{n+2} \Bbb Z)^*$. It turns out that for $n\ge 1$, $3$, of order $2^n$ and $-1$, of order $2$, generate it : they give an isomorphism $\Bbb Z/2^n\Bbb Z \times \{\pm 1\} \to (\Bbb Z / 2^{n+2} \Bbb Z)^* $. Since we are quotienting by $\{\pm 1\}$, the resulting group is isomorphic to $\Bbb Z/2^n\Bbb Z$

As a bonus, we get an explicit generator of $Aut(E)$ (the one corresponding to $ \pm 3$ in $(\Bbb Z/2^{n+2} \Bbb Z)^*/\langle -1 \rangle $): let $P(X) = X^3-3X$ and $Q(X) = X^2-2$ (so that $Q(x_{n+1}) = x_n$), and define $\sigma_n$ on $E_n$ by $\sigma_n(x_n) = P(x_n)$. We check that it is a compatible family of field automorphisms :

$P \circ Q = (X^2-2)^3-3(X^2-2) = X^6-6X^4+9X²-2 = (X^3-3X)^2-2 = Q \circ P$

So we have : $\sigma_0(0) = P(x_0) = 0$.
$\sigma_{n+1}(x_n) = \sigma_{n+1}(Q(x_{n+1})) = Q(P(x_{n+1})) = P(Q(x_{n+1})) = P(x_n) = \sigma_n(x_n)$ .

Thus they give field automorphisms and $\sigma_{n+1}|_{E_n} = \sigma_n$ so we can glue them together to define $\sigma$ on $E_\infty = \cup E_n$.

The fact that $\sigma$ is a generator of the Galois group of $E_n$ translates to the fact that $P^{\circ 2^n} = X \pmod {Q^{\circ n}}$, and $P^{\circ 2^{n-1}} = -X \pmod {Q^{\circ n}}$

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