# Roots of unity

After we’ve finally proved The Fundametal theorem of Galois Theory, we can officially say that we understand the correspondence, and we know exactly when it behaves as we want it to behave.

So, it’s time to see some results! The first topic I want to discuss is related to the roots of unity. We’ve met them throughout our journey several times, but I didn’t really care about them, or, at least I didn’t want to mention anything special about them.

However, there are a lot of things to say about them and they are very important. Moreover, their minimal polynomials over $\mathbb{Q}$ are very improtant as well, and they have some really great properties.

So, in this post, we’ll understand better what the roots of unity are, and what properties they satisfy. I am not going to use the tools from galois theory in this post, though in the next post – when I’ll present the minimal polynomials of them, galois theory will play it’s role…

## What is a root of unity

Just in case you are not familiar with this term – a root of unity of order $n$ is some number (can be complex of course) $\rho$ such that:

\rho^n=1

In other words, the roots of unity of order $n$ are the roots of the polynomial:

x^n-1

For example, $i=\sqrt{-1}$ is a root of unity of order 4, since:

i^4=(i^2)^2=(-1)^2=1

### How to find the roots of unity of order $n$$n$

We know that there are total of $n$ roots of unity – since the polynomial $x^n-1$ has degree $n$. So if we will be able to find $n$ roots, we can stop searching for more.

Moreover, their magnitude is 1 since:

x^n-1 =0 \Rightarrow x^n=1 \Rightarrow |x|^n=1\overset{|x|\geq0}{\Rightarrow} |x|=1

Note that 1 is a root of unity for every $n$, but thats a ‘boring’ root, we want to find more… How? We can use the well known fact:

e^{2\pi i }=\cos 2\pi +i\sin 2\pi = 1

So the number $e^{\frac{2\pi i }{n}}$ is going to be a root of unity of order $n$:

(e^{\frac{2\pi i }{n}})^n=e^{\frac{2\pi i \cdot n}{n}}=e^{2\pi i}=1

Great, but this one reveals all of them: for every integer $0\leq k\leq n-1$, the number $e^{\frac{2\pi i \cdot k}{n}}$ is also a root of unity:

(e^{\frac{2\pi i \cdot k}{n}})^n=e^{\frac{2\pi i \cdot k\cdot n}{n}}=e^{2\pi i\cdot k}=1

And they are all different (why?).

So I’ll denote $\rho_n=e^{2\pi i n}$ from now on, and we know that the roots of unity are:

\rho_n^0=1\ ,\rho_n\ , \rho_n^2\ , \rho_n^3\ \dots,\ \rho_n^{n-1}

In fact, the set of the roots of order $n$, $\Omega_n$ is a group under multiplication:

\Omega_n=\{1,\rho_n,\rho_n^{2},\dots,\rho_n^{n-1}\}

Let’s see some properties of it:

## The Group

First things first, this group is of order $n$ – it has $n$ elements. Moreover, it is cyclic$\rho_n$ is it’s generator.

Therefore, by Lagrange’s theorem – the order of an element in the group divides $n$. And we can take it one step forward: If $d$ is a divisor of $n$, say $n=dm$ for some $n\in\mathbb{N}$, then for every $\rho\in\Omega_{d}$:

\rho^n=\rho^{dm}=(\rho^{d})^m\overset{\rho\in\Omega_d}{=}1^m=1 \\ \Downarrow \\
\rho\in\Omega_n

Therefore:

\Omega_d\sube \Omega_n

And that’s true for every divisor of $n$, so:

\bigcup_{d|n}\Omega_d\sube \Omega_n

Note that this union is not disjoint and it is in fact the whole group – since $n$ is a divisor of itself… However, this discussion gave us some sort of an understanding on the situation:

Instead of adding to the union the groups of units of order $n$ – it would be better if will only consider their generators, those are exactly the elements $\rho\in\Omega_d$ such that:

\text{for every } k< d : \rho^k\neq 1 \text{ and } \rho^d=1

and they are exaclty the elements of order $d$ in $\Omega_n$. Such roots of unity have a special name – those are the primitive roots of unity of order $d$. So if we will denote by $\Omega_d^*$ the set of the primitive roots of unity of order $d$ we’ll get that:

\bigcup_{d|n}\Omega_d^*=\Omega_n

And this is indeed a disjoint union.

(Note how lagrange’s theorem approves the equality – if $\rho\in\Omega_n$ is a root of unity and it has order $d$ in the group, then it is in fact a primitive root of unity of order $d$, hence belongs to $\Omega_d^*$)

Let’s see an example:

#### Order 12

The group $\Omega_{12}$ is made of the elements:

\color{purple}\rho_{12}^0=\rho_1=1\ (\text{order 1})
\\
\text{}\\
\color{blue} \rho_{12}^6=\rho_2=-1 \ (\text{order 2})
\\
\text{}\\
\color{green} \rho_{12}^{4}=\rho_3, \rho_{12}^{8}=\rho_3^2 \ (\text{order 3})
\\
\text{}\\
\color{orange} \rho_{12}^{3}=\rho_4, \rho_{12}^{9}=\rho_4^3 \ (\text{order 4})
\\
\text{}\\
\color{red} \rho_{12}^2=\rho_6, \rho_{12}^{10}=\rho_6^5 \ (\text{order 6})
\\
\text{}\\
\color{black}\rho_{12},\rho_{12}^5,\rho_{12}^7,\rho_{12}^{11} \ (\text{order 12})
\\

Note how each color here represents a primitive root of unity of some order.

Do you think that you can determine if a root is primitive of some order? I’ll give you hint – focus on the powers of $\rho_{12}$ and try to find some connection between them to 12

## Who are the primitive roots of unity?

Since we have found a decomposition of the group $\Omega_n$ to disjoint sets of primitive roots of unity of different order, we would like to know who exactly are they – how can we describe a primitive root of unity of order $n$?

Let’s just play with and see where it leads us: Suppose that $\rho_n^k$ is a primitive root of unity of order 12, then for every $1\leq m< n$:

(\rho_n^{k})^m=\rho_n^{k\cdot m}\neq 1

But we know that we can express 1 as:

1=\rho_n^n=(\rho_n^n)^{r}=\rho_n^{n\cdot r}

for every $r\in\mathbb{N}$. Therefore, we are left with the inequality:

\rho_n^{k\cdot m}\neq \rho_n^{n\cdot r}

Which implies that for every $1\leq m< n$ and for every $r\in\mathbb{N}$:

k\cdot m\neq n\cdot r

Do you see what we just got? This fact says exactly that $k$ and $n$ are co-prime, that is, their greatest common divisor is 1:

\text{gcd}(k,n)=1

Let’s quickly prove it:

Suppose that $\text{gcd}(k,n)=d>1$, then there exist some $m_1,m_2$ such that:

k=d\cdot m_1\\
n=d\cdot m_2 \\


So we can multiply the first equation by $m_2$ and the second by $m_1$ to get:

k\cdot m_2=d\cdot m_1m_2\\
n\cdot m_1=d\cdot m_2m_1 \\

Note that $m_2< n$, and we can now compare the expressions from both equations to get:

k\cdot m_2=
n\cdot m_1 \ \ \ \ (m_2< n, m_1\in\mathbb{N})

And that’s a contradiction to what we’ve just showed!

Great! Now we know exactly what the primitive roots of order $n$ are: Those are the roots with powers that are co-prime with $n$:

\rho_n^k\text{ is primitive}  \iff (k,n)=1

We are going to describe primitive roots using this fact, so make sure that’s clear.

So now we know the size of the set $\Omega_n^*$:

|\Omega_n^*|=|\{\rho_n^k: 1\leq k< n\land  (k,n)=1\}|=|\{1\leq k< n:(k,n)=1\}|=:\varphi(n)

Where $\varphi(n)$ is Euler totient function!

Recall that we were able to express the group $\Omega_n$ as the disjoint union:

\Omega_n=\bigcup_{d|n}\Omega_d^*

And we can compare sizes to get:

n=|\Omega_n|=|\bigcup_{d|n}\Omega_d^*|\overset{\text{disjoint union}}{=}\sum_{d|n}|\Omega_d^*|=\sum_{d|n}\varphi(d)

And that’s just a beautiful identity! It relates a natural number with it’s divisors in such an elegant way!

## Summary

So that’s what I wanted to say about roots of unity, in the next post I am going to present the cyclotomic polynomial of order $n$.