Exercise 6.5.4

Proof. The roots of $f=x^n-1$ in $ℂ$ are ${\zeta }^k,0\le k<n$, where $\zeta =e^{2\mathit{i\pi }∕n}$.

The splitting field of $f$ over $\mathbb{Q}$ is $Q(1,\zeta ,\cdots ,{\zeta }^{n-1})=\mathbb{Q}(\zeta )$. Moreover, every root ${\zeta }^k$ is of the form ${\zeta }^k={𝜃}_k(\zeta )$, where ${𝜃}_k=x^k,0\le k\le n-1$.

so by definition $x^n-1=0$ is an Abelian equation. □