# Prove the following lemma(dealing with polynomials)

If $p$ is a prime number and $a_0,a_1,\ldots , a_{p - 1}$ are rational numbers satisfying $$a_0 + a_1 \alpha + a_2 \alpha ^2 + \ldots + a_{ p - 1} \alpha ^{ p - 1} = 0$$ where $$\alpha = \cos(\frac{2 \pi}{p}) + i \cdot \sin(\frac{2 \pi}{p} ) = e^{\frac{2i\pi}{p}}$$ then $a_0 = a_1 = \ldots = a_{p -1}$.

I think that it is helpful to consider:

(i) $g(x) = 1+ X^1 + \ldots + X^{p - 1}$ roots(particularly, $\alpha$ and its conjugate) and the fact that it is irreducible.

(ii) $f(x) = a_0+ a_1X^1 + \ldots + a_{p - 1}X^{p - 1}$.

If $p$ is prime, $1+X+\dots+X^{p-1}$ is the $p$-th cyclotomic polynomial and it is irreducible. Actually it is the minimal polynomial of $\alpha$ and its conjugates.
What can you deduce for $f(X)$?
• Well, we know that $(x-\alpha)(x-\beta) \mid \gcd(f(X),g(X))$ since both $f$ and $g$ are zero for $X=\alpha,\beta$, where $\beta$ is the conjugate of $\alpha$ Jul 13, 2018 at 15:48
• You forget $f(X)$ is the generator of $\{g(X)\in\mathbf Q[X]\mid g(\alpha)=0\}$. Jul 13, 2018 at 15:55
• I am not sure I understand correctly. What do you mean by $f(X)$ generator of $h(X)$, where $h(\alpha)=0$? Jul 13, 2018 at 16:08
• Not of $g(X)$ – a generator of the set of $g(X)$s which vanish at $\alpha$ (this is an ideal in $\mathbf Q[X]$). Jul 13, 2018 at 16:12