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How do you characterize all the linear relations satisfied by $n$th roots of unity with real, integral and non-negative integral coefficients?

Here are two examples for 3rd and 4th root:

Let $\omega_{i}$ be the primitive $i$th root of unity, then

For $2$nd root, $1 + \omega_{2} = 0$;

For $3$rd root, $1+ \omega_{3} + \omega_{3}^{2}=0$

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I'm not sure what you're trying to say, but given the equation $\omega^n - 1 = 0$, we can factor it as $(\omega-1)(1+\omega + \omega^2 + \ldots + \omega^{n-1}) = 0$. If $\omega \neq 1$, then the latter factor must be zero. –  Ben Dec 7 '12 at 13:49
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Look up cyclotomic polynomials. In characteristic zero (presumably your main case of interest) that is all there is to is. In positive characteristic the cyclotomic polynomials are no longer irreducible, and there is more. –  Jyrki Lahtonen Dec 7 '12 at 13:53
    
@Ben Here for example for $3$rd root of unity, $1+\omega_{3} +\omega_{3}^{2}$ determine all linear relation means if $a_0 + a_1\omega_{3} + a_2\omega_{3}^{2} =0$ means $a_0 = a_1 = a_2$ –  kamal Dec 7 '12 at 16:52
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Do you insist on integer coefficients in your linear relations? –  Gerry Myerson Dec 8 '12 at 5:59
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@AndreasCaranti: Your vote of confidence is appreciated. Yet, the OP edited the question an added non-negativity of the coefficients. Yes, we can still describe the answer as the set of polynomials with non-neg coefficients in the ideal generated by the relevant cyclotomic polynomial, but such a description doesn't feel very useful at this point :-) –  Jyrki Lahtonen Feb 17 '13 at 6:38

1 Answer 1

From the comments, it looks like we're solving the equation $$\sum_{i=0}^{n-1} a_i \omega_n^i= 0$$ where $\omega_n$ is a primitive $n$th root of unity in $\mathbb{C}$ and the $a_i$s are real (if they can be complex, the problem seems pretty trivial; I'm not sure how to do it if they can only be rational). Take $\omega_n = e^{2\pi i/n}$. This splits into two equations: $$\sum_{i=0}^{n-1} a_i \cos(2\pi i/n) = 0$$ $$\sum_{i=0}^{n-1} a_i \sin(2\pi i/n) = 0.$$

$\sin 0 = 0$, so the second sum has no $a_0$ term. This means that $a_2, a_3, \ldots, a_{n-1}$ determine $a_1$: $$a_1 = -\frac{\sum_{i=2}^{n-1} a_i \sin(2\pi i/n)}{\sin(2\pi/n)}.$$ Now, in the same way, $a_1, a_2, \ldots, a_{n-1}$ determine $a_0$ from the first equation: $$a_0 = -\sum_{i=1}^{n-1} a_i \cos(2\pi i/n) = \frac{\sum_{i=2}^{n-1} a_i \sin(2\pi i/n)}{\sin(2\pi/n)} - \sum_{i=2}^{n-1} a_i \cos(2\pi i/n).$$

This means that $a_2, a_3, \ldots, a_{n-1}$ can be picked arbitrarily, and they determine $a_0$ and $a_1$ uniquely.

(Originally I had a lot of nonsense in this post, it has since been edited)

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I suspect OP wants integer coefficients, although I grant this is not stated in the question. –  Gerry Myerson Dec 8 '12 at 5:58
    
when I faced the question I thought about integer coefficient only, if you have answer for that will you like to share it. Also I like to know for the non negative integral coefficients. Currently, I do not know whether that follows from integral coefficients or not. –  kamal Dec 8 '12 at 6:34
    
OK, sorry. I'll leave this answer up anyway. For some reason the word linear made me think we're working over a field. –  Ben Dec 8 '12 at 14:19
    
Oh, unless you'd like me to delete it because it's more likely to attract new answers that way? –  Ben Dec 8 '12 at 14:23
    
@Ben Let the answer be there, it's useful. No need to be sorry, I am thankful for the answer. Let's see how interesting is the answer for integral and non-negative integral cases. –  kamal Dec 8 '12 at 15:27

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