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 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 ** is some number (can be complex of course) such that:

\rho^n=1

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

x^n-1

For example, 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

We know that there are total of roots of unity – since the polynomial has degree . So if we will be able to find 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 , 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 is going to be a root of unity of order :

(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 , the number 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 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 , 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 – it has elements. Moreover, it is **cyclic** – is it’s generator.

Therefore, by **Lagrange’s theorem** – the order of an element in the group divides . And we can take it one step forward: If is a divisor of , say for some , then for every :

\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 , 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 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 – it would be better if will only consider their **generators**, those are exactly the elements such that:

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

and they are exaclty the elements of order in . Such roots of unity have a special name – those are** the primitive roots of unity of order **. So if we will denote by the set of the primitive roots of unity of order 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 is a root of unity and it has order in the group, then it is in fact a primitive root of unity of order , hence belongs to )

Let’s see an example:

#### Order 12

The group 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 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 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 ?

Let’s just play with and see where it leads us: Suppose that is a primitive root of unity of order 12, then for every :

(\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 . Therefore, we are left with the inequality:

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

Which implies that for every and for every :

k\cdot m\neq n\cdot r

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

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

Let’s quickly prove it:

Suppose that , then there exist some such that:

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

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

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

Note that , 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 are: Those are the roots with powers that are **co-prime** with :

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

Where is **Euler totient function**!

Recall that we were able to express the group 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 .