Exercise 4.3.25 (Primitive roots of unity)

Proof.

(a)

We recall this property of the function $\exp <!--\; FUNCTION\; APPLICATION\; -->$: $$\begin{array}{lll}\hfill \forall <!--\; FUNCTION\; APPLICATION\; -->z\in ℂ,\forall <!--\; FUNCTION\; APPLICATION\; -->z^{\prime }\in ℂ,(e^z=e^{z^{\prime }}⟺\exists <!--\; FUNCTION\; APPLICATION\; -->k\in \mathbb{Z},z^{\prime }=z+2\mathit{ik\pi }).& & \hfill \text{(1)}\end{array}$$

We can write every complex number $\zeta$ in the form $\zeta =\rho e^{\mathit{ix}}$, where $x\in ℝ$, and $\rho \ge 0$. Let $n$ be a positive integer. Then

$$\begin{array}{llll}\hfill {\zeta }^n=1& ⟺{\rho }^n{e}^{\mathit{nix}}=1& \hfill & \\ \hfill & ⟺\{\begin{array}{c}{\rho }^n=1\hfill \\ e^{\mathit{nix}}=1\hfill \end{array}& \hfill & \\ \hfill & ⟺\{\begin{array}{c}\rho =1\hfill \\ \exists <!--\; FUNCTION\; APPLICATION\; -->a\in \mathbb{Z},\mathit{nx}=2\mathit{\pi a},(\text{by (1)})\hfill \end{array}& \hfill & \\ \hfill & ⟺\exists <!--\; FUNCTION\; APPLICATION\; -->a\in \mathbb{Z},\zeta =e^{\frac{2\mathit{i\pi a}}{n}}.& \hfill & \end{array}$$

Since $e^{\frac{2\mathit{i\pi }(a+\mathit{kn})a}{n}}=e^{\frac{2\mathit{i\pi a}}{n}}$,

$$\exists <!--\; FUNCTION\; APPLICATION\; -->a\in \mathbb{Z},\zeta =e^{\frac{2\mathit{i\pi a}}{n}}⟺\exists <!--\; FUNCTION\; APPLICATION\; -->a\in [[1,n]],\zeta =e^{\frac{2\mathit{i\pi a}}{n}}.$$

In conclusion, $\zeta$ is a $n$th root of unity if and only if $\zeta$ is one of the $n$ numbers $e^{2\mathit{\pi ia}∕n}$ where $a=1,2,\dots ,n$ (for my part, I prefer $a=0,1,\dots ,n-1$ ).

(b)

We compute the order of $\zeta =e^{2\mathit{\pi ia}∕n}$, where $a\in \mathbb{Z}$. For every $l\in \mathbb{Z}$, $$\begin{array}{llllll}\hfill {\zeta }^l=1& ⟺e^{2\mathit{\pi ila}∕n}=1& \hfill & & \hfill & \\ \hfill & ⟺\exists <!--\; FUNCTION\; APPLICATION\; -->k\in \mathbb{Z},2\mathit{\pi ila}∕n=2\mathit{\pi ik}& \hfill (\mathit{by}(1))& & \hfill & & \hfill \\ \hfill & ⟺\exists <!--\; FUNCTION\; APPLICATION\; -->k\in \mathbb{Z},\mathit{la}=\mathit{nk}& \hfill & & \hfill & \\ \hfill & ⟺n\mid \mathit{la}& \hfill & & \hfill & \\ \hfill & ⟺\frac{n}{n\wedge a}\mid l\frac{a}{n\wedge a}& \hfill & & \hfill & \\ \hfill & ⟺\frac{n}{n\wedge a}\mid l& \hfill (\text{since }\frac{n}{n\wedge a}\wedge \frac{a}{n\wedge a}=1)& & \hfill & & \hfill \end{array}$$

Thus the least positive integer $l$ such that ${\zeta }^l=1$ (i.e. the order of $\zeta$) is

$$\mathrm{ord}(\zeta )=\frac{n}{n\wedge a}.$$

By definition, $\zeta$ is a primitive root if and only if $\mathrm{ord}(\zeta )=n$. This is equivalent to $n=\frac{n}{n\wedge a}$, so $\zeta$ is a primitive root if and only if $n\wedge a=1$.