Adjoining an nth root of unity to $\mathbb{Q}$

Question

$\zeta_n$ is the nth root of unity (or $e^{\frac{2\pi i}{n}}).$ How to show $\zeta_n^k+(\zeta _n^{-1})^k$ an element of $\mathbb{Q}(\zeta _n+\zeta _n^{-1})$?

$\zeta_n^{-1}=\zeta_n^{n-1}$.

Since $(\zeta_n+\zeta_n^{-1})^k\in \mathbb{Q}(\zeta _n+\zeta _n^{-1})$, $(\zeta_n+\zeta_n^{-1})^k=\sum_{i=0}^k(\zeta_n)^i(\zeta_n ^{-1})^{k-i}=\sum_{i=0}^k(\zeta_n)^{2i-k}$, but how do I show $\zeta_n^k+(\zeta _n^{-1})^k$ is a term linear combination of these sums?

EDITED: For example, $\zeta^2+\zeta^{-2}=(\zeta + \zeta^{-1})^2-2,$ $\zeta^3+ \zeta^{-3}=(\zeta+\zeta^{-1})^3-3(\zeta+\zeta^{-1})+6$

Answer 1

Hint: Induction on $k$.

It's a relatively uncomplicated induction.

Answer 2

Here is another approach.

Note that $\zeta_n+\zeta_n^{-1}$ is a real number (why?). Also, you have that $[\mathbb{Q}(\zeta_n):\mathbb{Q}(\zeta_n+\zeta_n^{-1})]=2$ (why?). But, since $\mathbb{R}\cap \mathbb{Q}(\zeta_n)$ is another subextension of $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ with $[\mathbb{Q}(\zeta_n):\mathbb{Q}(\zeta_n)\cap \mathbb{R}]=2$ (why?) this implies from what we've said that $\mathbb{Q}(\zeta_n+\zeta_n^{-1})=\mathbb{Q}(\zeta_n)\cap\mathbb{R}$.

Now, to prove that $\zeta_n^k+\zeta_n^{-k}\in \mathbb{Q}(\zeta_n+\zeta_n^{-1})$, the above shows it suffices to prove it's real. But,

$$\overline{\zeta_n^k+\zeta_n^{-k}}=\overline{\zeta_n}^k+\overline{\zeta_n}^{-k}=(\zeta_n^{-1})^k+(\zeta_n^{-1})^{-k}=\zeta_n^k+\zeta_n^{-k}$$

Answer 3

Basic idea from palindromic polynomials. Easy to show that $x^n+x^{-n}$ can be expressed as a polynomial in $x^1+x^{-1}$.