TITLE: Prove that the degree of the splitting field of $x^p-1$ is $p-1$ if $p$ is prime
QUESTION [6 upvotes]: I came across this question I couldn't figure out how to solve:

If $p$ is a prime number, prove that the splitting field over $F$, the field of rational numbers, of the polynomial $x^p-1$ is of degree $p-1$.

Here's what I have so far. Denote $f(x) = x^p-1 \in F[x]$. Then, $f(x)$ can be written as the product of two irreducible polynomials, namely $(x-1)(x^{p-1}+\cdots+x+1)$. If we call this latter polynomial $q(x)$, we easily see that $\deg q(x) = p-1$. If $\alpha$ is a root of $q(x)$, then $\alpha$ is algebraic of degree $p-1$ over $F$. Thus, $[F(\alpha):F]=p-1$.
I think I now need to consider the splitting field of $q(x)$, but I'm not sure exactly how to proceed.
Thanks in advance.

REPLY [8 votes]: Exactly following the comments above, we have the following solution:
We know that $F(\alpha) \subset E$, the splitting field of $x^p-1$ over $F$. Since $[F(\alpha):F]=p-1$, and we want $[E:F]=p-1$, it is enough to show that $F(\alpha)=E$.
If we consider the roots of $q(x)$, we have $ e^{2\pi i \cdot 1/p}, e^{2\pi i  \cdot 2/p}, \cdots, e^{2 \pi i \cdot p-1/p}$. Suppose that we take $\alpha = e^{2\pi i \cdot 1/p}$ as a root. Then, we can represent $F(\alpha) = \alpha, \alpha^2, \cdots, \alpha^{p-1}$. Note that 1, the root of $x-1$, is trivially in $F(\alpha)$. Thus, all of the roots of $f(x)$ are contained in $F(\alpha)$, so it must be the case that the splitting field $E=F(\alpha)$.