On the proof of: If $0<\frac{|χ(g)|}{χ(1)}<1$ then $\frac{χ(g)}{χ(1)}\notin\overline{\mathbb{Z}}$

Question

In class we proved the following theorem:

Let $σ:G\rightarrow GL(n,\mathbb{C})$ be a representation of $G$ and let $χ$ be the corresponding character.
Then $\forall g\in G$ we have: If $0<\frac{|χ(g)|}{χ(1)}<1$ then $\frac{χ(g)}{χ(1)}\notin\overline{\mathbb{Z}}$

The proof is pretty straight forward and goes as follows.

First fix $g\in G$ and set $a=\frac{χ(g)}{χ(1)}=\frac{χ(g)}{n}$.
We know that $χ(g)=\sum_{i=1}^{n}ω_i$, where each $ω_i$ is a root of unity.
We'll show that if $\frac{|χ(g)|}{χ(1)}<1$ and $\frac{χ(g)}{χ(1)}\in\overline{\mathbb{Z}}$ then $χ(g)=0$.

Let $p(x)=x^{k}+a_{k-1}x^{k-1}+\dots+a_{1}x+a_{0}$, $a_{i}\in\mathbb{Z}$ be the minimal polynomial of $a$.
Since $a=\frac{ω_{1}+\dots+ω_{n}}{n}$, then every other root of $p(x)$ is also of the form $\frac{ω_{1}'+\dots+ω_{n}'}{n}$ and hence we get that $|a_0|<1$, since $a_{0}=$product of all roots.

(Note: Actually to be exact, the proof we were given doesn't say "that since $a$ is of that form, then the rest of the roots must also be like that".
It rather says "Sicne every other root of $p(x)$ is also of that form, then...", but I assume it's the same thing.)

But $a_{0}\in\mathbb{Z}$ so $a_{0}=0$, and since $p(x)$ is irreducible we must have that $p(x)=x$ which implies that $a=0\Rightarrow χ(g)=0$.

Now I understand every part of it, apart from the implication that since $a$ is the sum of $n$ roots of unity divided by $n$, then the same must hold for the other roots.
How do we deduce such a claim?

Thank you in advance.

Answer 1

This has something to do with Galois theory. Since the minimal polynomial is irreducible, its roots are permuted transitively by the Galois group of the splitting field. But the Galois group sends roots of unity to roots of unity.