Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 18I / e / 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 $\mathrm{lcm}(2,n)$. 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 $m\equiv 2(\mathrm{mod}4)$; containment in $\mathbb{Q}({\zeta }_n)$ forces this conductor to divide $n$, which gives $m\mid \mathrm{lcm}(2,n)$.

If $n$ is even, $\mathrm{lcm}(2,n)=n$, 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

Solved by gpt-5.6-sol high.