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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 $p\in \mathbb{N}_{primes}$)

user5721565
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  • A general way to prove that a monic polynomial $f$ is irreducible over $\Bbb Z$ is, for example, Eisenstein, or reduction criterion. The minimal polynomial, however, of $\zeta_p+\zeta_p^{-1}$ is unique, so you know the proof from the above post. – Dietrich Burde Jan 15 '19 at 19:16
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    In JoseBrox answer that $G$ is isomorphic to a subgroup of $\mathbb{Z}_p^*$ is obvious, it is less obvious that it is the whole of it, at first you need to admit it (that $[\mathbb{Q}(\zeta):\mathbb{Q}] = p-1$) so $[\mathbb{Q}(\zeta):\mathbb{Q}(\zeta+\zeta^{-1})] =2$ implies $[\mathbb{Q}(\zeta+\zeta^{-1}):\mathbb{Q}] =(p-1)/2$ and $\deg(f) = (p-1)/2$ means $f$ is irreducible – reuns Jan 16 '19 at 00:58
  • @reuns , That's great! The cyclotomic polynomial of deg $p-1$ is irreducible thus $[\mathbb{Q}(\zeta_p)] = p-1$ and $x^2 - (\zeta_p + \zeta_p ^{-1})x + 1 |_{\zeta_p} = 0 $ ,Also there is not polynomial with $\zeta_p$ has a root from deg 1, due to the fact $\zeta_p\in \mathbb{C}$ so, $\zeta_p$ is a root $\Rightarrow \bar{\zeta_p}$ is a root. Apply $[\mathbb{Q}(\zeta_p):\mathbb{Q} ] =[\mathbb{Q}(\zeta_p):\mathbb{Q}(\zeta_p + \zeta_p^{-1})]\cdot [\mathbb{Q}(\zeta_p+\zeta_p^{-1}):\mathbb{Q} ] \Rightarrow [\mathbb{Q}(\zeta_p+\zeta_p^{-1}):\mathbb{Q} ] = \frac{p-1}{2}$ – user5721565 Jan 16 '19 at 06:01

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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 $(p-1)/2$.

Jose Brox
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