 Solution

Fix a primitive $n$th root of unity ${\zeta }_n$. The $n$th cyclotomic polynomial isIts roots are exactly the roots of unity of order $n$.

Partitioning all $n$th roots of unity by exact order gives the cyclotomic factorizationWe prove ${\Phi }_n\in \mathbb{Z}[X]$ by induction on $n$. The assertion is clear for $n=1$. If it holds for every proper divisor of $n$, thenThe denominator is monic and belongs to $\mathbb{Z}[X]$. Exact polynomial division shows that the quotient lies in $\mathbb{Q}[X]$, and Gauss's lemma makes this monic quotient an element of $\mathbb{Z}[X]$.

Now let $K\subseteq \mathbb{C}$. The minimal polynomial of ${\zeta }_n$ over $K$ divides ${\Phi }_n$, so every $K$-conjugate of ${\zeta }_n$ is another primitive root ${\zeta }_n^a$ and lies in $K({\zeta }_n)$. The extension is therefore normal. It is separable because the characteristic is zero, and hence it is Galois.

Every $\sigma \in \mathrm{Gal}(K({\zeta }_n)/K)$ hasfor some $a\in (\mathbb{Z}/n\mathbb{Z}{)}^{\times }$. Since ${\zeta }_n$ generates the extension, this gives an injective homomorphismThe group on the right is abelian, so the Galois group is abelian, as summarized by the cyclotomic field construction.