# Equivalence of Definitions of Primitive Root of Unity

## Theorem

Let $n \in \Z_{> 0}$ be a strictly positive integer.

Let $F$ be a field.

Let $U_n$ denote the set of all $n$-th roots of unity.

The following definitions of the concept of Primitive Root of Unity are equivalent:

### Definition 1

A primitive $n$th root of unity of $F$ is an element $\alpha \in U_n$ such that:

$U_n = \set {1, \alpha, \ldots, \alpha^{n - 1} }$

### Definition 2

A primitive $n$th root of unity of $F$ is an element $\alpha \in U_n$ such that:

$\forall m : 0 < m < n : \alpha^m \ne 1$

## Proof

### Definition 1 implies Definition 2

Let $\alpha \in U_n$ such that:

$U_n = \set{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}}$

By Definition of Power of Field Element:

$\alpha = \alpha^1$

Hence:

$U_n = \set{1, \alpha^1, \alpha^2, \ldots, \alpha^{n-1}}$

By Definition of Explicit Set Definition:

$\forall m \in \N: 0 < m < n : \alpha^m \ne 1$

$\Box$

### Definition 2 implies Definition 1

Let $\alpha \in U_n$ such that:

$\forall m \in \N : 0 < m < n: \alpha^m \ne 1$

Since $\alpha$ is a root of unity:

$\alpha \ne 0$

$\exists m, k \in \N : 0 \le m < k < n : \alpha^k = \alpha^m$

Then:

 $\ds \alpha^{\paren{k - m} }$ $=$ $\ds \alpha^k \circ \alpha^{-m}$ Sum of Indices Law for Field $\ds$ $=$ $\ds \alpha^k \circ \paren{\alpha^{m} }^{-1}$ Negative Index Law for Field $\ds$ $=$ $\ds \dfrac {\alpha^k} {\alpha^m}$ $\ds$ $=$ $\ds 1$ as $\alpha^k = \alpha^m$

Since $0 < k-m < n$, this contradicts the hypothesis:

$\forall m \in \N : 0 < m < n: \alpha^m \ne 1$

Hence:

$\forall m, k \in \N: 0 \le m < k < n : \alpha^k \ne \alpha^m$
$\set{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}} \subseteq U_n$
$n = \card {\set{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}}} \le \card {U_n}$

Consider the polynomial $X^n - 1$ over $F$.

By Definition of Root of Unity and Definition of Root of Polynomial:

$\beta \in F$ is a $n$-th root of unity if and only if $\beta$ is a root of the polynomial $X^n - 1$
The polynomial $X^n - 1$ has at most $n$ roots.

So:

$\card {U_n} \le n$

Hence:

$\card {U_n} = n = \card {\set{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}}}$
$U_n = \set{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}}$

$\blacksquare$