Given an abelian extension $L/\mathbb{Q}$, such that $L=\mathbb{Q}(\alpha)$ for a root $\alpha$ of an irreducible polynomial $f(x)$. By Kronecker-Weber theorem we know that there exists $m \in \mathbb{N}$, such that $L \subseteq \mathbb{Q}(\zeta_m)$. How can we find explicitly the least integer $m \in \mathbb{N}$, with this property? Could you please introduce me to a reference for this?
I know that my question is very short, but I am very curious about this algorithm after this comment. (Also, this question was very informative for me.)
If $K/\Bbb{Q}$ is abelian then $K\subset \Bbb{Q}(\zeta_n)$. Clearly the $p$ ramified in $K$ must divide $n$.
Assuming that $K\subset \Bbb{Q}(\zeta_{rp^l}),p\nmid r$
Since $\Bbb{Q}(\zeta_{rp^l})/\Bbb{Q}(\zeta_r)$ is cyclic and totally ramified at $p$, then $K\subset \Bbb{Q}(\zeta_{rp^{l-1}})$ iff $$e(p,K/\Bbb{Q})=e(p,K(\zeta_r)/\Bbb{Q}(\zeta_r))= [K(\zeta_r):\Bbb{Q}(\zeta_r)] \le [\Bbb{Q}(\zeta_{rp^{l-1}}):\Bbb{Q}(\zeta_r)]= \phi(p^{l-1})$$
(there is a subtlety for $p=2$ here as $\Bbb{Q}(\zeta_8)/\Bbb{Q}$ is not cyclic)
(the given expression for $c_p$ is only valid for $p$ odd, we need a special case for $p=2$ as $\Bbb{Q}(\sqrt2)$ has ramification index $2$ at $p=2$ but is contained in $\Bbb{Q}(\zeta_8)$ not $\Bbb{Q}(i)$.