Why is $[\mathbb{Q}(e^{2\pi i/p}):\mathbb{Q}] = p-1 $?

Question

I know that $e^{2\pi i/p}$ is a root of $x^{p}-1$ and we can write:

$ x^{p}-1=\left(x-1\right)\left(\sum_{i=0}^{p-1}x^{i}\right) $

So $e^{2\pi i/p}$ is a root of $\sum_{i=0}^{p-1}x^{i}$, which means $[\mathbb{Q}(e^{2\pi i/p}):\mathbb{Q}] \leq p-1 $.

Can someone show the other direction?

Answer 1

You can get both directions from the fact that $e^{\frac{2 \pi i}{p}}$ is a root of $P(X)=X^{p-1}+X^{p-1}+..+X+1$, and that $P(X)$ is irreducible over $\mathbb Q[X]$.

You can prove the irreducibility by observing that $P(X)=\frac{X^p-1}{X-1}$ for $x\neq 1$, thus, for $X \neq 0$ we have

$$P(X+1)=\frac{(X+1)^p-1}{X}=\sum_{k=1}^p \binom{p}{k}X^{k-1}$$

use Eisenstein Criteria.