Sum of roots of unity with linear coefficients

Question

Given that $n$ is a positive integer and $\omega=e^{2\pi i/n}$, how do you prove that $a_0+a_1\omega+a_2\omega^2+...+a_{n-1}\omega^{n-1}=0$ is only true when all the $a$'s are equal?

Answer 1

Amended result. If $n$ is prime and $\omega=e^{2\pi i/n}$ and $a_0,\ldots,a_{n-1}$ are integers and $$a_0+a_1\omega+a_2\omega^2+\cdots+a_{n-1}\omega^{n-1}=0\ ,$$ then $a_0=a_1=a_2=\cdots=a_{n-1}$.

Proof. Under the stated conditions, $\omega$ is a root of the integer polynomial $$f(z)=a_0+a_1z+a_2z^2+\cdots+a_{n-1}z^{n-1}\ .$$ Therefore $f(z)$ is a (polynomial) multiple of the minimal polynomal of $\omega$, which is $$m(z)=1+z+z^2+\cdots+z^{n-1}\ .$$ Since $f(z)$ has the same degree as $m(z)$, it must in fact be a constant multiple of $m(z)$, and the result follows.

See also this book, problem 3.26 (shameless free plug).