Exercise 9.1.16

Proof. Here we write ${\zeta }_k=e^{2\mathit{i\pi }∕k}$ for all subscript $k$.

(a)

${\zeta }_n={({\zeta }_{\mathit{nm}})}^m\in \mathbb{Q}({\zeta }_{\mathit{nm}})$, and ${\zeta }_m={({\zeta }_{\mathit{nm}})}^n\in \mathbb{Q}({\zeta }_{\mathit{nm}})$,therefore $$\mathbb{Q}({\zeta }_n,{\zeta }_m)\subset \mathbb{Q}({\zeta }_{\mathit{nm}}).$$

As $n\wedge m=1$, there exists integers $u,v$ such that $1=\mathit{un}+\mathit{vm}$.

Therefore ${\zeta }_{\mathit{nm}}={({\zeta }_{\mathit{nm}}^n)}^u{({\zeta }_{\mathit{nm}}^m)}^v={\zeta }_m^u{\zeta }_n^v\in \mathbb{Q}({\zeta }_n,{\zeta }_m)$, hence

$$\mathbb{Q}({\zeta }_{\mathit{nm}})\subset \mathbb{Q}({\zeta }_n,{\zeta }_m)$$

We have proved

$$\mathbb{Q}({\zeta }_{\mathit{nm}})=\mathbb{Q}({\zeta }_n,{\zeta }_m)$$

(b)

By Corollary 9.1.10, $[\mathbb{Q}({\zeta }_{\mathit{nm}}):\mathbb{Q}]=\varphi (\mathit{nm})$. As $n\wedge m=1,\varphi (\mathit{nm})=\varphi (n)\varphi (m)$ (Lemma 9.1.1), so $[\mathbb{Q}({\zeta }_{\mathit{nm}}):\mathbb{Q}]=\varphi (n)\varphi (m)$, and by part (a), this is equivalent to $$[\mathbb{Q}({\zeta }_n,{\zeta }_m):\mathbb{Q}]=\varphi (n)\varphi (m).$$

Using the Tower Theorem,

$$\varphi (n)\varphi (m)=[\mathbb{Q}({\zeta }_n,{\zeta }_m):\mathbb{Q}]=[\mathbb{Q}({\zeta }_n,{\zeta }_m):\mathbb{Q}({\zeta }_m)][\mathbb{Q}({\zeta }_m):\mathbb{Q}]=\varphi (m)[\mathbb{Q}({\zeta }_n,{\zeta }_m):\mathbb{Q}({\zeta }_m)],$$

thus

$$\varphi (n)=[\mathbb{Q}({\zeta }_m)({\zeta }_n):\mathbb{Q}({\zeta }_m)].$$

Let $f$ be the minimal polynomial of ${\zeta }_n$ over $\mathbb{Q}({\zeta }_m)$. Then

$$\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=[\mathbb{Q}({\zeta }_m)({\zeta }_n):\mathbb{Q}({\zeta }_m)]=\varphi (n).$$

${\zeta }_n$ is a root of ${\Phi }_n(x)\in \mathbb{Q}[x]\subset \mathbb{Q}({\zeta }_m)[x]$, therefore $f\mid {\Phi }_n$ in $\mathbb{Q}({\zeta }_m)[x]$. Moreover these two polynomials are monic of same degree $\varphi (n)$, so they are identical. ${\Phi }_n=f$ is so irreducible over $\mathbb{Q}({\zeta }_m)$.