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For a primitive $m$th root of unity ${\zeta }_m$, every Galois automorphism sends ${\zeta }_m$ to a unique ${\zeta }_m^a$ with $a$ invertible modulo $m$. The assignment $\sigma \mapsto a$ is a group homomorphism, since composition multiplies the exponents, and is injective because ${\zeta }_m$ generates the cyclotomic field. The irreducibility of cyclotomic polynomials gives degree $\phi (m)$, equal to the order of the target, so the map is an isomorphism. The cyclotomic character on an absolute Galois group factors through this finite quotient and generally has a nontrivial kernel.

Let ${\zeta }_n$ be a primitive nth root of unity. The cyclotomic polynomial isInduction on $n$ shows that ${\Phi }_n$ is monic with integer coefficients. Here is the essential irreducibility of cyclotomic polynomials argument. Let $f$ be the minimal polynomial of ${\zeta }_n$ and let $\ell \nmid n$ be a prime. If ${\zeta }_n^{\ell }$ had a different monic irreducible polynomial $g$, then $f\mid g(X^{\ell })$. Reducing modulo $\ell$ gives $\bar{f}\mid {\bar{g}}^{\ell }$, so $\bar{f}$ and $\bar{g}$ have a common factor. As $fg\mid {\Phi }_n$, this would give a repeated factor of $X^n-1$ modulo $\ell$, impossible because its derivative is $n{X}^{n-1}$. Thus ${\zeta }_n^{\ell }$ is conjugate to ${\zeta }_n$. Every positive integer prime to $n$ is a product of such primes, so all primitive nth roots of unity are conjugate. ConsequentlyIn particular the cyclotomic field is an abelian extension. Complex conjugation corresponds to $-1$; for $n>2$ its fixed field is $\mathbb{Q}({\zeta }_n+{\zeta }_n^{-1})$, the maximal real subfield, of degree $\phi (n)/2$.

The ring of integers is $\mathbb{Z}[{\zeta }_n]$. One way to see that there is no missing index is to work at each prime $p$. Write $n=p^{a}m$ with $p\nmid m$. The prime-to-$p$ roots of unity have separable reduction; their local factors generate unramified extensions with full integer rings. Over each such factor, adjoining ${\zeta }_{p^a}$ is generated by the Eisenstein polynomial ${\Phi }_{p^a}(1+T)$, with uniformizer ${\zeta }_{p^a}-1$, and hence gives the full integer ring of that totally ramified factor. The coprimality of $p^a$ and $m$ means that ${\zeta }_n$ generates both sets of roots of unity. Thus ${\mathbb{Z}}_p[{\zeta }_n]$ is maximal at every $p$, proving the integral assertion.

The same local description computes ramification and splitting. For $p\nmid n$, the ramification index is one, the residue degree is the multiplicative order of $p$ modulo $n$, and there are $\phi (n)/f$ primes above $p$. If $p\mid n$, put $m=n/p^a$; thenwhere ${\mathrm{ord}}_1(p)=1$. Indeed, the prime-to-$p$ part is unramified, and the $p$-power part is totally ramified by the preceding Eisenstein polynomial. Notice the exception $p=2,a=1$: its ramification index is one, consistently with $\mathbb{Q}({\zeta }_{2m})=\mathbb{Q}({\zeta }_m)$ for odd $m$. Thus finite ramification occurs only at the indicated conductor primes, with that exception.

We now prove cyclotomic Artin reciprocity over the rational numbers, using arithmetic Frobenius automorphisms. If $\ell \nmid n$ and $\mathfrak{L}$ lies over $\ell$, then Frobenius sends the reduction of ${\zeta }_n$ to its $\ell$th power. Distinct nth roots of unity remain distinct modulo $\mathfrak{L}$, since $X^n-1$ has separable reduction. Hence the Artin symbol is exactlyExtend multiplicatively to the group $I(n)$ of rational fractional ideals prime to $n$. Such an ideal has a unique positive rational generator $q$ whose numerator and denominator are prime to $n$. The formula isEvery unit residue modulo $n$ has a positive integer representative prime to $n$, so this Artin map is surjective. Its kernel is precisely the subgroup $P_{n\infty }$ generated by positive $q\equiv 1(\mathrm{mod}n)$. ThereforeThis proves the Artin reciprocity law for this extension, including its principal-ray kernel. The infinite component of the modulus records positivity and therefore the possible action of complex conjugation.

The cyclotomic polynomial of the primitive $p$th root $\omega$ isBy the irreducibility of cyclotomic polynomials, this is the minimal polynomial of $\omega$ over $\mathbb{Q}$. The prime cyclotomic field degree therefore gives

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.