If $ f(x) $ is irreducible, odd degree, and $ \alpha $ is a root of $ f(x) $, then $ \mathbb{Q}(\alpha) = \mathbb{Q}(\alpha^{2^{k}}) $ for $k \geq 0$ Let $ f(x) \in \mathbb{Q}[x] $ be irreducible of odd degree, and let $ F = \mathbb{Q}(\alpha) $ for some root $ \alpha $ of $ f(x) $. Prove that $ F = \mathbb{Q}(\alpha^{2^k}) $ for all $ k \geq 0 $.
I was totally stuck on this question until I found a very similar question and solution
here for the case of $k = 1$.
I thought I could use that proof and generalize it for arbitrary $k$, but all my attempts have failed. I also tried induction but I don't think using induction helped at all, because it seems like we are back at the original problem when trying to prove $ \alpha^{2^k} \in \mathbb{Q}  ( \alpha^{2^{k+1}}  ) $. Maybe I'm missing something; any help is appreciated.
 A: *

*Since $\alpha$ is a root of an irreducible $f(x) \in \mathbb{Q}[x]$, it follows that $[F : \mathbb{Q}] = [\mathbb{Q}(\alpha) : \mathbb{Q}] = \deg f,$ which is odd.


*Now, for any $k \geq 0$
$$[F : \mathbb{Q}] = \left[F : \mathbb{Q}\left(\alpha^{2^k}\right)\right]\left[\mathbb{Q}\left(\alpha^{2^k}\right) : \mathbb{Q}\left(\alpha^{2^{k+1}}\right)\right]\left[\mathbb{Q}\left(\alpha^{2^{k+1}}\right) : \mathbb{Q}\right]$$
so $\left[\mathbb{Q}\left(\alpha^{2^k}\right) : \mathbb{Q}\left(\alpha^{2^{k+1}}\right)\right]$ divides an odd number, so it is odd.


*Since $g(x) = x^2 - \alpha^{2^{k+1}} \in \left(\mathbb{Q}\left(\alpha^{2^{k+1}}\right)\right)[x]$ has $g\left(\alpha^{2^k}\right) = 0$, then $$\left[\mathbb{Q}\left(\alpha^{2^k}\right) : \mathbb{Q}\left(\alpha^{2^{k+1}}\right)\right] = \left[\left(\mathbb{Q}\left(\alpha^{2^{k+1}}\right)\right)\left(\alpha^{2^k}\right) : \mathbb{Q}\left(\alpha^{2^{k+1}}\right)\right] \leq 2,$$ but it cannot be $2$, as we've already seen it's odd, so it must equal $1$.
Therefore $\mathbb{Q}\left(\alpha^{2^{k+1}}\right) = \mathbb{Q}\left(\alpha^{2^k}\right)$.  By induction, it follows that $$\mathbb{Q}(\alpha) = \mathbb{Q}\left(\alpha^{2^0}\right) = \cdots = \mathbb{Q}\left(\alpha^{2^k}\right)$$ for any $k \geq 0$.
A: You can also do the following, and get an idea of explicit rational expressions for $\alpha^{2^k}$ in terms of $\alpha^{2^{k+1}}$.

Write $f(x)=xp(x^2)+q(x^2)$, with $p,q\in\mathbb{Q}[x]$.
[This is a simple idea that is useful to remember for many other problems.]

Since $f$ has odd degree, then $p\neq0$. Moreover, from $f$ being irreducible you get that $p(\alpha^2)\neq0$, since otherwise $p(x^2)$ would be a polynomial of smaller degree that has $\alpha$ as a root.
From $0=f(\alpha)=\alpha p(\alpha^2)+q(\alpha^2)$ we get that $\alpha\in\mathbb{Q}(\alpha^2)$. Therefore $F\subset\mathbb{Q}(\alpha^2)$. The reverse inclusion is clear.
Observe that $q^2(x^2)-x^2p^2(x^2)=f(x)f(-x)=f_2(x^2)$, with $f_2$ of odd degree and $f_2(\alpha^2)=0$. We can do the same trick from the beginning, but with $f_2(x)=xp_2(x^2)+g_2(x^2)$. Evaluating at $x=\alpha^2$ we get that $\alpha^2\in\mathbb{Q}(\alpha^4)$. This shows that $\mathbb{Q}(\alpha^2)\subset\mathbb{Q}(\alpha^4)$. The reverse inclusion is clear.
The rest is just induction, repeating this idea with $f_{k+1}$ defined by $f_k(x)f_k(-x)=f_{k+1}(x^2)$.

The other idea used here, that is useful to remember, is that if $\alpha$ is a root of $g(x)$, $g(x)g(-x)=G(x^2)$, then $\alpha^2$ is a root of $G(x)$.

