irreducibility of $(\zeta + \zeta ^{-1})$'s minimal polynomial over $\mathbb{Z}[x]$ This question relates to the next post, especially to Did's, answer (second answer in the post). 
Minimal Polynomial of $\zeta+\zeta^{-1}$
The answer gives a method to construct a monic polynomial $f\in\mathbb{Z}[x], $ of degree $\frac{p-1}{2}$ when $p\in \mathbb{N}_{primes}$, such that $f(\zeta + \zeta^{-1})=0$ , when $\zeta = e^{\frac{2\pi i}{p}}$.
However, I wish to know whether is there a general way to prove that a monic polynomial $f$ (as above) in $\mathbb{Z}[x]$ with a free coefficient $a_0 = 1$ , such that $f(e^{\frac{2\pi i}{p}}+e^{-\frac{2\pi i}{p}})=0$ is irreducible? (for $p\in \mathbb{N}_{primes}$) 
 A: The minimal polynomial $f$ of $\alpha:=\zeta+\zeta^{-1}$ over $\mathbb{Z}$ is the only monic irreducible polynomial having $\alpha$ as a zero; if $\alpha$ is a zero of another polynomial $g$, then $g$ is a multiple of $f$. Therefore, among monic polynomials having $\alpha$ as a zero, the only one which is irreducible is the one with minimal degree.
On the other hand, by Galois theory the minimal polynomial of $\alpha$ is the product of the linear factors produced by the conjugates of $\alpha$ by its Galois group of automorphisms. The Galois group $G$ of $\zeta$ is isomorphic to $\mathbb{Z}_p^*$, and since $\sigma\in G$ implies $\sigma(\zeta^{-1})=\sigma(\zeta)^{-1}$, and $\alpha=\zeta+\zeta^{-1}$, we have that $\sigma(\alpha)=\sigma^{-1}(\alpha)$; and if $\sigma,\tau\in G$ satisfy $\sigma(\alpha)=\tau(\alpha)$ then $\tau=\sigma^{\pm1}$. So the Galois group of $\alpha$ has half the elements than $G$, $(p-1)/2$, and hence the degree of the minimal polynomial of $\alpha$ is $(p-1)/2$.
In other words, $f$ monic with $\alpha$ as a root will be irreducible iff its degree is $(p-1)/2$.
