Rational independence of higher golden ratios

Question

The higher Golden Ratios $(\phi_n)_{n=0,1,2,3,\dots}$ are the real numbers defined recursively by setting $\phi_0=0$ and for $n\ge1$, $\phi_n$ is the positive (real) root of $X^2-\phi_{n-1}X-1$. We find that $\phi_1=1$ and $\phi_2=\frac{1+\sqrt5}{2}$. In particular, $\phi_2$ is equal to the classical Golden Ratio. Moreover the sequence $(\phi_n)_{n=0,1,2,\dots}$ is strictly increasing. The function $\mathbb R^\times\buildrel f\over\to\mathbb R$ defined by $x\mapsto x-\frac1x$ sends $\phi_n$ to $\phi_{n-1}$ for each $n\ge1$ and so there is a nested chain of algebraic number fields $$\mathbb Q=\mathbb Q[\phi_1]\subseteq\mathbb Q[\phi_2]\subseteq\mathbb Q[\phi_3]\subseteq\dotsb.$$ For each $n\ge1$ we have $$|\mathbb Q[\phi_{n+1}]:\mathbb Q[\phi_n]|= 1\textrm{ or }2.$$ I guess it is always $2$. How to prove this guess?

Added following R.P.'s comment: I don't ask for a solution to a homework problem. More useful would be a recommendation of a text book or article where this kind of problem is addressed. There are some results with similar flavor in Kaplansky's book Fields and Rings (Chicago, 2nd ed. 1972) which look superficially promising such as Theorems 53 and 54 of the Miscellany that is section 12 of Part I. Some additional references would be a huge help and enough for me to consider the question closed.

Answer 1

Possibly this particular question is elementary enough to merit being closed, but it's a case of a widely studied problem in arithmetic dynamics. Let $f:\mathbb P^1\to\mathbb P^1$ be a rational map defined over $\mathbb Q$, let $\alpha\in\mathbb P^1(\mathbb Q)$ be an initial point, and consider the tower of fields given by iterated inverse images, $$ K_n := \mathbb Q\bigl(f^{-n}(\alpha)\bigr).$$ The question deals with a particular subtower within this tower for the map $f([x,y])=[x^2-y^2,xy]$ and point $\alpha=[1,1]$. There's a large literature studying the fields $K_n$ and their Galois groups that goes by the name arboreal representation, since $G(K_n/\mathbb Q)$ (usually) naturally embeds into the automorphism group of a $d$-branched tree (where $d=\deg(f)$).

Answer 2

Here's a sketch of a proof of the stronger statement over $\mathbb F_2$:

Define $\varphi_1 = 1$ and, for $n \ge 2$, let $\varphi_n$ be a root of $x^2 - \varphi_{n-1}x - 1 \in \mathbb F_2(\varphi_{n-1})[x]$. We claim that $[\mathbb F_2(\varphi_n) : \mathbb F_2(\varphi_{n-1})] = 2$ for all $n \ge 2$.

For small values of $n$, this can be checked directly. Now suppose for contradiction that the conclusion fails, and let $n$ be the smallest positive integer for which $\varphi_n \in \mathbb F_2(\varphi_{n-1})$ for some choice of $\varphi_{n-1}$ and $\varphi_n$ as above. Since $[\mathbb F_2(\varphi_{n-1}) : \mathbb F_2(\varphi_{n-2})] = 2$, we can write \begin{equation*} \varphi_n = a\varphi_{n-1} + b \end{equation*} with $a,b \in \mathbb F_2(\varphi_{n-2})$. Now, \begin{align*} 0 &= \varphi_n^2 + \varphi_{n-1}\varphi_n + 1\\ &= a^2\varphi_{n-1}^2 + b^2 + \varphi_{n-1}(a\varphi_{n-1} + b) + 1\\ &= (a^2 + a)\varphi_{n-1}^2 + b^2 + b\varphi_{n-1} + 1\\ &= (a^2 + a)(\varphi_{n-1}\varphi_{n-2} + 1) + b^2 + b\varphi_{n-1} + 1\\ &= \big((a^2 + a)\varphi_{n-2} + b\big)\varphi_{n-1} + a^2 + a + b^2 + 1. \end{align*} This implies that \begin{equation*} (a^2 + a)\varphi_{n-2} + b = 0 = a^2 + a + b^2 + 1, \end{equation*} hence \begin{equation*} b\varphi_{n-2}' = \frac{b}{\varphi_{n-2}} = a^2 + a = b^2 + 1, \end{equation*} where $\varphi_{n-2}'$ is the root of $x^2 + \varphi_{n-3}x + 1$ different from $\varphi_{n-2}$. This means $b \in \mathbb F_2(\varphi_{n-2}) = \mathbb F_2(\varphi_{n-2}')$ is a root of $x^2 + \varphi_{n-2}'x + 1$, but this contradicts our assumption that $x^2 + \varphi_{n-2}'x + 1$ is irreducible over $\mathbb F_2(\varphi_{n-2}')$.