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Let $p$ be a prime and suppose $e_1,e_2,\cdots,e_p$ denote some $p$-th roots of unity in $\mathbb{C}$, not necessarily distinct.

If
$$e_1 + e_2 + \cdots + e_p=0,$$ then, can we always conclude that all $e_i$'s are the distinct?


Note that for $p=2$ this is true, and so one may assume $p>2$.

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If $p$ is a prime, you can indeed only get $0$ as a sum of $p$ such $p$-th roots of unity by taking the $p$ distinct roots once each.

I don't know how much field theory you have done, but here is an explanation of something much more general: Let $\omega \neq 1$ be a complex number satisfying $\omega^{p} = 1.$ Then the $p$-th roots of unity are $\{\omega^{i}: 0 \leq i \leq p-1 \}.$ Furthermore, the polynomial $x^{p}-1 \in \mathbb{Q}[x]$ factors as $(x-1)( 1 + x + \ldots + x^{p-2}+x^{p-1}),$ and the second factor (which we now call $f(x)$) is irreducible in $\mathbb{Q}[x]$ by Eisenstein's criterion. Note that $\omega$ is a root of $f(x).$.

This implies by general theory that $\{1,\omega,\omega^{2},\ldots,\omega^{p-3}, \omega^{p-2} \}$ is linearly independent over $\mathbb{Q}.$ So every non-zero element which is expressible as a $\mathbb{Q}$-linear combination of these elements has a unique such expression. Since $-\omega^{p-1} = 1 + \omega+ \omega^{2}+ \ldots + \omega^{p-3} + \omega^{p-2}$ is the unique way to express $-\omega^{p-1}$ this way, we see that $0 = q(1 + \omega + \omega^{2} + \ldots + \omega^{p-1})$ is the only way to write $0$ as a $\mathbb{Q}$-linear combination of $\{1,\omega,\omega^{2},\ldots,\omega^{p-1} \},$ where $q$ can be any rational number. Hence the only way to write $0$ as a sum of $p$ such $p$-th roots of unity is the case $q = 1.$

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  • $\begingroup$ Thanks; simple, but I didn't thought with this perspective. Nice answer, and thanks for the transparent clarification. $\endgroup$ – Beginner May 4 '17 at 8:32
  • $\begingroup$ I was writing my answer when ancientmathematician's answer appeared. That answer is correct, though written more tersely. $\endgroup$ – Geoff Robinson May 4 '17 at 8:34
  • $\begingroup$ No offense, but this is the SAME proof as in the other answer with the difference being, that the other answer does it in a short and elegant way AND was given earlier. I cannot believe that the other answer has two downvotes, while this one is accepted. Of course the OP is free to choose the answer he wants to accept, but at least also upvote the other answer... $\endgroup$ – MooS May 4 '17 at 8:34
  • $\begingroup$ @Moos: No offence taken: I was writing my answer before I saw ancientmathematician's as I said above. $\endgroup$ – Geoff Robinson May 4 '17 at 8:36
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    $\begingroup$ @Moos:I think (s)he may have been referring to an earlier comment that someone made, which is now deleted. In any case, I have also pointed out above that the answer by ancientmathematician is correct. $\endgroup$ – Geoff Robinson May 4 '17 at 8:42
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Let $p$ be prime and let $\zeta:=\exp\frac{2\pi}{p}$. Then the $p$-th rootsof unity are $1,\zeta,\zeta^2,\dots,\zeta^{p-1}$.

Moreover, $$\sum_{0}^{p-1} a_i \zeta^{i}=0 \mbox{ for some } a_i\in\mathbb{Z} \Rightarrow a_i=c\mbox{ for all }.i$$ This is a consequence of the irreducibility of $X^{p-1}+\dots+X+1$, easily seen by Eisenstein.)

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  • $\begingroup$ I am not, the $a_i$ allow for multiple appearances. $\endgroup$ – ancientmathematician May 4 '17 at 8:23
  • $\begingroup$ I think the question asked if $p$ many (not necessarily distinct) $p^{\operatorname{th}}$ root of unity sums up to zero, does it followed that they are distinct? $\endgroup$ – Alex Vong May 4 '17 at 8:23
  • $\begingroup$ And my answer shows that the answer is "Yes". $\endgroup$ – ancientmathematician May 4 '17 at 8:24
  • $\begingroup$ @AlexVong He shows that the condition $\sum a_i=p$ (which is the hypothesis of the OP) implies all $a_i=1$. So he answered the question affirmatively. $\endgroup$ – MooS May 4 '17 at 8:25
  • $\begingroup$ @Beginner You are not thinking through the answer properly. It is 100% correct. $\endgroup$ – MooS May 4 '17 at 8:26

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