Let $n \in \mathbb{N}$ be a natural number, and be $\omega$ and $\eta$ two differents n-th primitive roots in $\mathbb{C}$.
Prove that $\eta - \omega \notin \mathbb{Q}$
My attempt was to follow the false line of the following :
If i'd to prove that $\sqrt-2 - \sqrt-5 \notin \mathbb{Q}$, i'd try something by contradiction like : $$\sqrt-2 - \sqrt-5 = \alpha, \alpha \in \mathbb{Q}$$
$$\sqrt-2 = \sqrt-5 + \alpha$$ $$ -2 = \alpha^{2} + 2\alpha\sqrt-5 -5$$
But then $-2,\alpha^{2},-5 \in \mathbb{Q}$ which leads to $\sqrt-5 \in \mathbb{Q}$,false.
So here i'd like to re-write $$\eta = \omega + \alpha , \alpha \in \mathbb{Q} $$
And raise to the n-th power sothat $\eta \in \mathbb{Q}$, but then i'm unable to find some contradiction due to the difficulties in seeing the terms of the newton binomial $(\omega + \alpha )^{n}$.
Is this the right approach ?
Any help or tip would be appreciated,
Thanks a lot
Because $|\eta|=|\omega|=1$ and $\eta\neq\omega$ we have $0<|\eta-\omega|\leq2$, and switching $\eta$ and $\omega$ if necessary gives, without loss of generality, that $0<\eta-\omega\leq2$. Suppose now that $\eta-\omega\in\Bbb{Q}$. Because $\eta$ and $\omega$ are integral over $\Bbb{Z}$, so is $\eta-\omega$ and hence $\eta-\omega\in\Bbb{Z}$. This shows that $\eta-\omega\in\{1,2\}$.
If $\eta-\omega=1$ then $\eta=\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$ and $\omega=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$, the two $\pm$-signs being the same. But then one is a primitive third root of unity whereas the other is a primitive sixth root of unity, a contradiction.
If $\eta-\omega=2$ then $\eta=1$ and $\omega=-1$, but then one is a primitive first root of unity whereas the other is a primitive second root of unity, a contradiction.
We conclude that $\eta-\omega\notin\Bbb{Q}$.