text: # 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
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