= Solution
Let $\zeta_n$ be a primitive nth <root of unity>. The <cyclotomic polynomial> is
$$
\Phi_n(X)=\prod_{\substack{1\le a\le n\\(a,n)=1}}(X-\zeta_n^a),\qquad X^n-1=\prod_{d\mid n}\Phi_d(X).
$$
Induction 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 $nX^{n-1}$. 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. Consequently
$$
\boxed{[\mathbb Q(\zeta_n):\mathbb Q]=\varphi(n),\qquad\operatorname{Gal}(\mathbb Q(\zeta_n)/\mathbb Q)\cong(\mathbb Z/n\mathbb Z)^\times,\quad \sigma_a(\zeta_n)=\zeta_n^a.}
$$
In 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 $\varphi(n)/2$.
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 $\varphi(n)/f$ primes above $p$. If $p\mid n$, put $m=n/p^a$; then
$$
e_p=\varphi(p^a),\qquad f_p=\operatorname{ord}_m(p),\qquad g_p=\frac{\varphi(m)}{f_p},
$$
where $\operatorname{ord}_1(p)=1$. 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 exactly
$$
\operatorname{Art}((\ell))=\sigma_\ell.
$$
Extend 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 is
$$
\operatorname{Art}((q))=\sigma_{q\bmod n}.
$$
Every 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 $q\equiv1\pmod n$. Therefore
$$
\boxed{I(n)/P_{n\infty}\ \xrightarrow{\ \sim\ }\operatorname{Gal}(\mathbb Q(\zeta_n)/\mathbb Q).}
$$
This 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>.
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