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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

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    $\begingroup$ $-1$ is the only primitive root of order $2$, so the case $n=2$ is vacuously true. Anyway, what do you know about the minimal polynomials of the primitive roots? There is a sleek argument using the piece of information that all primitive roots of unity of a given order share the same minimal polynomial. But I dare not use that, if you haven't heard about cyclotomic polynomials. $\endgroup$ – Jyrki Lahtonen Dec 28 '18 at 21:11
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    $\begingroup$ To give you a taste, the primitive roots of unity of order four are zeros of the polynomial $p(x)=x^2+1$. More precisely $p(x)=(x-\omega)(x-\eta)$ where $\omega$ and $\eta$ are the two primitive roots. Now, if $\eta=\omega+q$ for some rational number $q$, this means that $\omega$ is also a zero of the polynomial $f(x)=p(x+q)$ because $$f(\omega)=p(\omega+q)=p(\eta)=0.$$ Furthermore, $f(x)$ obviously also has rational coefficients. Meaning that $\omega$ is a zero of the greates common divisor of $p(x)$ and $f(x)$. But $p(x)$ is irreducible, so.... $\endgroup$ – Jyrki Lahtonen Dec 28 '18 at 21:17
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    $\begingroup$ But the irreducibility of cyclotomic polynomials is somewhat non-trivial in general. Therefore I needed to ask whether you are familiar with that. $\endgroup$ – Jyrki Lahtonen Dec 28 '18 at 21:21
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    $\begingroup$ For irreducibility of the cyclotomic polynomials over $\mathbb{Q}$ could use Eisenstein's criterion, using $\frac{x^{n} -1}{x-1}$, right ? @JyrkiLahtonen $\endgroup$ – jacopoburelli Dec 28 '18 at 21:23
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    $\begingroup$ Only when the order of those primitive roots is a prime (or a power of a prime), I think. And when $n$ is odd, the various primitve roots of unity of order $n$ have distinct imaginary parts, so their differences are trivially not rational because they are not even real. $\endgroup$ – Jyrki Lahtonen Dec 28 '18 at 21:28
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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 $\eta-\omega\notin\Bbb{Q}$.

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  • $\begingroup$ I knew I was missing something simple :-) $\endgroup$ – Jyrki Lahtonen Dec 29 '18 at 6:13
  • $\begingroup$ Just two things that are not entirely clear to me, why $\eta - \omega \in \mathbb{Z}$ ? And why can you remove the absolute value so easily ? $\endgroup$ – jacopoburelli Dec 29 '18 at 7:18
  • $\begingroup$ To the second point; if $\eta-\omega<0$ then $\omega-\eta>0$, and the question is symmetric in $\omega$ and $\eta$. $\endgroup$ – Servaes Dec 29 '18 at 9:25
  • $\begingroup$ To the first point; both $\eta$ and $\omega$ are elements of $\Bbb{Z}[\zeta_n]$, hence so is $\eta-\omega$. And $\Bbb{Z}[\zeta_n]\cap\Bbb{Q}=\Bbb{Z}$. $\endgroup$ – Servaes Dec 29 '18 at 9:26

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