Exercise 10.2.3

Proof. Here we suppose that $p$ is an odd prime.

The splitting field of ${\Phi }_p$ over $\mathbb{Q}$ is $L=\mathbb{Q}({\zeta }_p,{\zeta }_p^2,\dots ,{\zeta }_p^{p-1})=\mathbb{Q}({\zeta }_p)$. As ${\Phi }_p({\zeta }_p)=0$, and as ${\Phi }_p(x)$ is irreducible over $\mathbb{Q}$ (Proposition 4.2.5), ${\Phi }_p$ is the minimal polynomial of ${\zeta }_p$ over $\mathbb{Q}$, hence

$\text{∙}$

Suppose that $p$ is a Fermat prime, so $p=2^n+1$, where $n=2^k,k\in \mathbb{N}$. Then $[L:\mathbb{Q}]=2^n$, where $L$ is the splitting field of ${\Phi }_p$ over $\mathbb{Q}$. By the authorized Theorem 10.1.12, ${\zeta }_p$ is constructible.

$\text{∙}$

Conversely, if we suppose that ${\zeta }_p$ is constructible, by the same Theorem 10.1.12, $[L:\mathbb{Q}]=2^n$ for some integer $n\ge 0$, so $p=2^n+1$. As $p$ is prime, by Exercise 1, $n=2^k,k\in \mathbb{N}$, so $=2^{2^k}+1$ is a Fermat prime.

Conclusion: if $p$ is an odd prime, ${\zeta }_p$ is constructible if and only if $p$ is a Fermat prime. □