Polynomial divisibility and automorphisms of cyclotomic fields

Question

In my previous question Maximal real subfields of Cyclotomic subfields giving rise to identities I proved the following claim (I think):

CLAIM:If $\pi(x)$ is a monic irreducible polynomial such that $\pi(x)|\pi(x^2-2),$ then $\pi(x)$ is the minimal polynomial for $\zeta_n+\zeta_n^{-1}$ over $\mathbb Q$, where $n$ is an odd natural number.

Sketch of proof:If $\pi(\alpha)=0,$ then $\pi(\alpha^2-2)=0,$ so $\tau:\alpha \rightarrow \alpha^2-2$ is an automorphism of $\mathbb Q(\alpha)$.Construct the polynomial $x^2-\alpha.x+1$ with roots $\eta$ and $\eta^{-1}$.Extending $\tau$ to $\mathbb Q(\eta)$ and applying it to the quadratic, we will produce a quadratic with roots $\tau(\eta),\tau(\eta^{-1}),$ but a small calculation shows that they are also $\eta^2,\eta^{-2}$.So $\tau(\eta)=\eta^{\pm2},$ so $\eta$ is a root of unity $\zeta_n$ and $n$ is odd.

Let $f_{n+1}(x)=x.f_{n}(x)-f_{n-1}(x)$ be a sequence of polynomials starting at 2 and x.Then $f_m(u+u^{-1})=u^m+u^{-m}$.Thus one can generalise the problem as follows:

THEOREM:If $\pi(x)$ is a monic irreducible polynomial such that $\pi(x)|\pi(f_m(x)),m>1,$ then $\pi(x)$ is the minimal polynomial for $\zeta_n+\zeta_n^{-1}$ over $\mathbb Q$, where $(n,m)=1$.

All of this led me to consider $\pi(x)|\pi(g(x)),$ for an arbitrary $g(x)$.This is equivalent to asking for which $\alpha$ is $\tau:\alpha \rightarrow g(\alpha)$ an automorphism of $\mathbb Q(\alpha),$ since $\pi(x)$ is defined by $\alpha,$ being its minimal polynomial over $\mathbb Q$.

QUESTION 1: What really is going on that allows us to construct an auxiliary polynomial that works so well in the case of the theorem?Is there a more illuminating proof of the claim?My proof feels very contrived and unnatural.

QUESTION 2:Are these methods applicable to the general case?

Here are some of my thoughts: the case when $g(x)=x^2-2$ corresponds to $\mathbb Q(\zeta_n+\zeta_n^{-1})=\mathbb Q(\zeta_n)^H$,$H={(\pm1)}=G^{\frac{\varphi(n)}{2}}$,which is normal over Q, so all conjugates are expressible as "polynomials". What g is is simply one such polynomial.Expressing $\alpha=\eta+\eta^{-1}$ just allows us to apply g to it and arrive at $\eta^2+\eta^{-2},$ a conjugate.But constructing the auxiliary polynomial is crucial for the proof.When I tried to apply analogous reasoning for $H=G^{\frac{\varphi(n)}{3}}={(1,s,s^2)}

Answer 1

As an example in $\Bbb F_{13}$ with Galois group $G = C_{12}$, we chose a group with index $4$, so H = $\langle g^4 \rangle$ where $g$ is a generator of $G$. To find the invariant field corresponding to H I take the orbit of $\zeta$ under $H$ and take its sum giving $\eta = \zeta + \zeta^3 + \zeta^9$. The orbit of $\eta$ is $\{\eta, \zeta^2 + \zeta^5 + \zeta^6, \zeta^4 + \zeta^{10} + \zeta^{12}, \zeta^{7} + \zeta^8 + \zeta^{11}\}$, which I take as a basis of the subfield fixed by $H$ then I take the matrix $M$ whose rows are formed by the coefficients of the elements $1, \eta, \eta^2, \eta^3$ giving : $$ M = \begin{pmatrix} -1 & -1 & -1& -1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 2 & 0 \\ -5 & -3 & -6 & -3 \end{pmatrix} $$ To find out which combination of these rows gives the second (for example) element of the orbit I solve (I like column vectors, that's why the transpose) $$ (M^t)^{-1}\begin{pmatrix} 0\\ 1\\ 0\\ 0 \end{pmatrix} = \begin{pmatrix} -2\\ 4/3\\ 1\\ 2/3 \end{pmatrix} \implies \begin{pmatrix} 1 & x & x^2 & x^3\\ \end{pmatrix} \begin{pmatrix} -2\\ 4/3\\ 1\\ 2/3 \end{pmatrix} = 2/3x^3+x^2+4/3x-2 $$ HTH