linear algebra - How do I find the minimal polynomial for 11th roots of unity sum $zeta+zeta^{-1}$. Non-trivial seemingly

Let $\zeta$ be a primitive 11th root of unity, how does one show that $\alpha= \zeta+\zeta^{-1}$ has minimal polynomial $x^5+x^4-4x^3-3x^2+3x+1$?

I have been unable to solve this problem that I would have assumed was simple for hours... It gives a hint that we can evaluate the minimal polynomial of $\zeta$ at $\zeta$ and divide each side by $\zeta^5$, and rewrite in terms of $\alpha$, but I can't get it:

$$\zeta=z\text{ for saving type it 200 times}$$
$$z^{10}+z^9+\cdots+z+1=0$$
$$z^5+z^4+\cdots+z^{-4}+z^{-5}=0$$
$$z^5+z^4+z^3+z^2+1+z^{-2}+z^{-3}+z^{-4}+z^{-5}+\alpha=0$$

How is this meant to give me the minimal polynomial?

Answer

It depends how much Galois theory you know. I would do it as follows, which is not so easily seen to be related to the hint you were given. The Galois group of the splitting field of $x^{11}-1$ is cyclic of order $10$, and is generated by $\sigma: \sigma(\zeta) = \overline{\zeta}$ and $\tau : \tau(\zeta) = \zeta^{4}$. The element $\alpha$ is already fixed by $\sigma$. Now $\tau^{5}$ is the identity,
and it follows that the coefficients of the polynomial
$(x - \alpha)(x - \tau(\alpha))(x-\tau^{2}(\alpha))(x-\tau^{3}(\alpha))(x- \tau^{4}(\alpha))$ are fixed by both $\sigma$ and $\tau$, so are fixed by the full Galois group, and lie in $\mathbb{Q}$.

Hence if you expand out $(x - (\zeta+\zeta^{-1}))(x - (\zeta^{4}+\zeta^{-4}))(x - (\zeta^{5}+\zeta^{-5}))(x - (\zeta^{2}+\zeta^{-2}))(x - (\zeta^{3}+\zeta^{-3}))$, you will obtain a polynomial of degree $5$ in $\mathbb{Q}[x]$ satisfied by $\alpha$, and it is easy to check that this has to be the minimum polynomial of $\alpha$. By the way, I have made use of the fact that $\zeta^{11}= 1$, so that $\zeta^{16} = \zeta^{5}$ and $\zeta^{20} = \zeta^{9} = \zeta^{-2}$. Likewise $\zeta^{8} = \zeta^{-3}$.