= Solution
Let $L=\mathbb Q(\zeta_n)$. We use the standard cyclotomic containment lemma: every root of unity in $\mathbb Q(\zeta_n)$ has order dividing $\operatorname{lcm}(2,n)$. One proof uses the conductor of a rational cyclotomic field. The conductor attached to a primitive $m$th root is $m$, except that it is $m/2$ when $m\equiv2\pmod4$; containment in $\mathbb Q(\zeta_n)$ forces this conductor to divide $n$, which gives $m\mid\operatorname{lcm}(2,n)$.
If $n$ is even, $\operatorname{lcm}(2,n)=n$, and the powers of $\zeta_n$ already give all $n$ possible roots of unity in $L$. If $n$ is odd, $-\zeta_n$ has order $2n$, so $L$ contains all $2n$th roots of unity, and the containment lemma shows there are no others. Thus the <roots of unity in a rational cyclotomic field> number
$$
\boxed{
\begin{cases}
n,&n\text{ even},\\
2n,&n\text{ odd}.
\end{cases}}
$$
Back to article page
