Exercise 10.2.5

Proof.

(a)

The minimal polynomial of ${\zeta }_{2^s}$ over $\mathbb{Q}$ is ${\Phi }_{2^s}(x)$, and $$\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->({\Phi }_{2^s}(x))=\varphi (2^s)=2^{s-1}(2-1)=2^{s-1}.$$

Therefore $[\mathbb{Q}({\zeta }_{2^s}):\mathbb{Q}]=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->({\Phi }_{2^s}(x))=2^{s-1}$ is a power of 2. Since $\mathbb{Q}({\zeta }_{2^s})$ is the splitting field of $\Phi ({\zeta }_{2^s})$ over $\mathbb{Q}$, by Theorem 10.1.12, ${\zeta }_{2^s}$ is constructible.

Note 1. Since $x^{2^s}-1={\Phi }_1(x){\Phi }_2(x)\cdots {\Phi }_{2^{s-1}}(x){\Phi }_{2^s}(x)=(x^{2^{s-1}}-1){\Phi }_{2^s}(x)$, we see that ${\Phi }_{2^s}(x)=x^{2^{s-1}}+1$.

Note 2. Without Theorem 10.1.12, we can prove that ${\zeta }_{2^s}$ is constructible more geometrically by constructing a $2^s$-gon. ${\zeta }_2=-1$ is constructible. Reasoning by induction, suppose that ${\zeta }_{2^{s-1}}$ is constructible. Since ${({\zeta }_{2^s})}^2={\zeta }_{2^{s-1}}$, ${\zeta }_{2^s}$ is the intersection point of the circle $C(0,1)$ with the constructible bisector of the angle determined by $0,1,{\zeta }_{2^{s-1}}$, so is constructible.

(b)

Since $\gcd <!--\; FUNCTION\; APPLICATION\; -->(a,b)=1$, there exists $u,v\in \mathbb{Z}$ such that $\mathit{ua}+\mathit{vb}=1$.

Then $\frac{v}{a}+\frac{u}{b}=\frac{1}{\mathit{ab}}$, thus

$${\zeta }_a^v{\zeta }_b^u=e^{2\mathit{\pi i}\left(\frac{v}{a}+\frac{u}{b}\right)}=e^{\frac{2\mathit{\pi i}}{\mathit{ab}}}={\zeta }_{\mathit{ab}}.$$

So ${\zeta }_{\mathit{ab}}={\zeta }_a^v{\zeta }_b^u$, product of constructible numbers, is constructible.

(c)

We show first that two distinct Fermat primes $F_n=2^{2^n}+1,F_m=2^{2^m}+1,n<m$ are relatively prime. Suppose on the contrary that a prime $q$ divides $F_n$ and $F_m$. Then $2^{2^n}\equiv -1(\mathit{mod}q)$, and $$-1\equiv 2^{2^m}\equiv {(2^{2^n})}^{2^{m-n}}\equiv {(-1)}^{2^{m-n}}\equiv 1(\mathit{mod}q).$$

Therefore $q\mid 2$, so $q=2$, but this is impossible since $F_n$ is odd.

So $n\ne m\Rightarrow \gcd <!--\; FUNCTION\; APPLICATION\; -->(F_n,F_m)=1$.

Since ${\zeta }_{2^s},{\zeta }_{p_1},\dots ,{\zeta }_{p_r}$ are constructible, and $2^s,p_1,\dots ,p_r$ are relatively prime, by part (b), ${\zeta }_{2^s{p}_1},{\zeta }_{2^s{p}_1{p}_2},\dots ,{\zeta }_{2^s{p}_1\cdots p_r}={\zeta }_n$ are constructible.

Conclusion: if $p_1,\dots ,p_r$ are Fermat primes, and $n=2^s{p}_1\cdots p_r$, then ${\zeta }_n$ is constructible. □