Kronecker–Weber theorem

Remarks

The converse holds as well: $\Gal(\Q(\zeta_n)/\Q) \cong (\Z/n\Z)^\times$ is abelian, so every subfield of $\Q(\zeta_n)$ is abelian over $\Q$. One standard proof reduces to cyclic extensions of prime power degree and studies their ramification groups at the ramified primes. A key input is that $\Q$ has no nontrivial extension unramified at all primes, a consequence of the Hermite–Minkowski theorem. Another proof comes from class field theory: the ray class field of $\Q$ for the modulus $n\infty$ is $\Q(\zeta_n)$.

Example: every quadratic field is abelian over $\Q$, so it lies in a cyclotomic field. For instance $\sqrt{2} = \zeta_8 + \zeta_8^{-1}$ lies in $\Q(\zeta_8)$, and $\sqrt{5} = \zeta_5 - \zeta_5^2 - \zeta_5^3 + \zeta_5^4$ lies in $\Q(\zeta_5)$.

The abelian hypothesis is needed. The field $\Q(\sqrt[3]{2})$ is not contained in any $\Q(\zeta_n)$: it is not Galois over $\Q$, while every subfield of $\Q(\zeta_n)$ is Galois over $\Q$.

Going deeper

For a student

The maximal abelian extension of $\Q$ is the field generated by all roots of unity. Its Galois group is $\widehat{\Z}^\times$, the unit group of the profinite completion $\widehat{\Z}$ of $\Z$. Hilbert's twelfth problem asks for similar explicit generators for the abelian extensions of other number fields. For imaginary quadratic fields this is achieved by complex multiplication, using values of elliptic and modular functions.