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

  • 1
    $\begingroup$ $-1$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 3
    $\begingroup$ 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

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 $\eta-\omega\notin\Bbb{Q}$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.