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Let $L$ be a splitting field of $f(X)=X^m-1$ over $k$. Its formal derivative isThe assumption on the characteristic of $k$ says that $m\ne 0$ in $k$. Any root $\alpha$ of $f$ is nonzero, and hence $f^{\prime }(\alpha )=m{\alpha }^{m-1}\ne 0$. Thus $f$ and $f^{\prime }$ have no common root, so $f$ is a separable polynomial. Since it has degree $m$ and splits over $L$, it has exactly $m$ distinct roots. This is the separability of roots of unity.

Fix a primitive root of unity ${\zeta }_m=e^{2\pi i/m}$. Define the $m$th cyclotomic polynomial byIt is monic and has degree given by the Euler totient function $\phi (m)$. Partitioning all $m$th roots of unity according to their exact multiplicative order gives the cyclotomic factorizationWe now prove by mathematical induction that ${\Phi }_m\in \mathbb{Z}[X]$. The base case is ${\Phi }_1=X-1$. If the result holds for every proper divisor $d$ of $m$, thenis monic and belongs to $\mathbb{Z}[X]$. Polynomial division of $X^m-1$ by the monic integer polynomial $G$ produces a quotient and remainder in $\mathbb{Z}[X]$. The factorization over $\mathbb{C}$ says that the remainder is zero and the quotient is ${\Phi }_m$, proving the claim.

It remains to prove irreducibility of cyclotomic polynomials. Let $f\in \mathbb{Z}[X]$ be a monic irreducible polynomial dividing ${\Phi }_m$, let $\epsilon$ be a root of $f$, and write ${\Phi }_m=fg$. We claim that ${\epsilon }^p$ is a root of $f$ for every prime number $p\nmid m$. Otherwise ${\epsilon }^p$ is a root of $g$, so $\epsilon$ is a root of $g(X^p)$. Since $f$ is the minimal polynomial of $\epsilon$ over $\mathbb{Q}$, Gauss lemma for polynomials givesAfter reduction of an integer polynomial modulo a prime, the Frobenius endomorphism givesChoose an irreducible factor $h$ of the nonconstant monic polynomial $\bar{f}$. Then $h\mid \bar{g}$, so $h^2\mid \bar{f}\bar{g}={\bar{\Phi }}_m$. The cyclotomic factorization would then make $h^2$ divide $X^m-1$ over ${\mathbb{F}}_p$. This is impossible because its derivative $m{X}^{m-1}$ is coprime to it when $p\nmid m$.

Thus $f({\epsilon }^p)=0$ for every prime $p\nmid m$. Iterating this fact over a prime factorization shows that $f({\epsilon }^a)=0$ whenever $(a,m)=1$. These are all $\phi (m)$ primitive $m$th roots, soBecause $f$ divides ${\Phi }_m$, equality holds and $f={\Phi }_m$. Hence every cyclotomic polynomial is irreducible over $\mathbb{Q}$.

Finally, direct multiplication givesAlsoApplying the cyclotomic factorization to the two factors on the right givesand thereforeSince $X^2-X+1={\Phi }_6(X)$, cancellation yieldsThe thirtieth cyclotomic polynomial is irreducible over $\mathbb{Q}$ by the theorem just proved, so $F$ is irreducible.

The cyclotomic polynomial iswhere ${\zeta }_n$ is primitive. Partitioning all $n$th roots by exact order gives the cyclotomic factorizationInduct on $n$. The quotient of the monic integer polynomial $X^n-1$ by the product of the already constructed monic ${\Phi }_d$, $d<n$, lies in $\mathbb{Q}[X]$ and is monic. Gauss lemma then shows that this quotient ${\Phi }_n$ lies in $\mathbb{Z}[X]$.

If the prime $p$ does not divide $n$, then over ${\mathbb{F}}_p$,Thus $X^n-1$ has no repeated root. Every factor, including the reduction of ${\Phi }_n$, is square-free, proving the Separability of a cyclotomic polynomial modulo p.

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.