minimum polynomial of $e^{2\pi i/p}$ for $p$ a prime number

Question

I wanna find the minimum polynomial of $e^{2\pi i/p}$ for $p$ a prime number.

I tried to fix a prime number $p$ and so we can write it as $x^p-1=(x-1)(x^{p-1}+x^{p-2}+...+1$). But how to continue now?

Answer 1

what you basically want to find is an extension of the rational numbers that contains the n-th root of unity $\alpha$ over $\mathbb{Q}$. Generally, for a positive integer $n$, $$T^n-1= \prod_{d\mid n} \Phi_d(T)$$ where $ \Phi_d(T)$ is the d-th cyclotomic polynomial of order $\phi(d)$, which is irreducible over $\mathbb{Q}$and gives you the field extension of $\mathbb{Q}$ containing precisely the d-th roots of unity.

$$\Phi_d(T) = \!\! \prod_{1 \leq k \leq d\atop \operatorname{ggT}(k, d) = 1} \left(T - e^{2 \pi \cdot \mathrm i k / d}\right)$$

Let $m_\alpha(T)$ be the minimal polynomial of $\alpha$. In your case $n=p$ and $\alpha = e^{2\pi i/p}$, thus here $m_\alpha(T)=\Phi_p(T)$.
The extension is then given by the quotient $\mathbb{Q}(\alpha) \cong\mathbb{Q}[T]/m_\alpha(T)$.
Now you have to convince yourself that $m_\alpha(T)=\frac{X^p-1}{X-1}=1+X+\dots X^{p-1}$ and that this polynomial is irreducible. Hint: Consider the automorphism of rings $\mathbb{Q}[T] \rightarrow \mathbb{Q}[T]$ sending $f(T) \mapsto f(T+1)$, then use Eisenstein criterium and argue that irreducibilty is persevered under isomorphisms.