# Finite Field Arithmetic

- A **finite field** is a set containing a finite number of elements
  - This is sometimes called a *Galois Field*
- In a Galois field you can:
  - Add
  - Subtract
  - Multiply
  - Invert (divide)
- Fields are an extension of *groups* and related to *rings*

### Groups

A group is a set of elements $G$ together with an operation $\circ$ that combines two elements of $G$

> 1. The operation $\circ$ is **closed**
>    - i.e. for all $a,b \in G$ then $a\circ b=c\in G$
> 2. The operation is associative
>    - i.e. $a\circ(b\circ c) = (a\circ b)\circ c$ for all $a,b,c \in G$
> 3. There is an element $1\in G$ called a **neutral element** such that $a\circ 1 = 1\circ a = a$ for all $a\in G$
> 4. For each $a \in G$ there exists an element $a^{-1}\in G$ called the **inverse** of $a$ such that $a\circ a^{-1} = a^{-1}\circ a = 1$
> 5. A group $G$ is **abelian** (commutative) if $a\circ b = b \circ a$ for all $a,b\in G$

##### Example Group

- The set of integers $\mathbb{Z}_m = \{0,1,...m-1\}$ with the operation addition modulo m form a group with the neutral element 0
- Every element would have an inverse where $a + (-a) = 0$ mod m
- This group would not form a group with multiplication, as not all elements would have an inverse
  - We wouldn’t have an inverse, we would need $5\times \frac15=1$ however $\frac15 \notin \mathbb{Z}$ 

### Fields

A field $F$ is a set of elements with the following properties

> 1. All elements of $F$ form an **additive group** with the group operation $+$ and the neutral element 0
> 2. All elements of $F$ except 0 form a multiplicative group with the group operation $\times$ and the neutral element 1
> 3. When the two group operations are mixed, the distributivity law holds.
>    - i.e. for all $a,b,c \in F, a\cdot(b+c) = (a\cdot b) + (a\cdot c)$

##### Example Field

- The set of real numbers $\mathbb{R}$ is a field with neutral element 0 for addition and 1 for multiplication
- Every real number $a$ has a additive inverse $-a$
- Every non-zero number $a$ has a multiplicative inverse $\frac{1}{a}$

![1646405235.png](img/1646405235.png)

#### Finite Fields

> A finite field only exists if it has $p^m$ elements
>
> Where:
>
> - $p$ is a prime 
> - $m$ is a positive integer

###### Examples

- There is a field with 11 elements: $GF(11)$
- There is a field with 256 elements: $GF(256)$ or $GF(2^8)$
- $GF(12)$ is not a finite field $(2^2 \cdot3)$

###### Prime and Extension Fields

When $m=1$ it creates a **prime field**

When $m>1$ it creates an **extension field**

### Prime Fields

- A prime field $GF(p)$ contains the integers $\{0,1,...p-1\}$

![1646405270.png](img/1646405270.png)

- These operations satisfy the properties of fields (*closure*)

##### Inversion in Prime Fields

$a \cdot a^{-1} \equiv 1 \space (mod \space p)$

- A modular inverse exists when $gcd(a,p) = 1$
- Because $p$ is prime, every number has a multiplicative inverse
  - $gcd(a,p) = 1, \forall a \neq0 \in GF(p)$
- $a^{-1}$ can be calculated using the **extended Euclidean algorithm**

#### Extension Fields

- In prime fields, the elements are integers
- Elements in extension fields $GF(2^m)$ are polynomials of degree $m$

$a_{m-1}x^{m-1}, ..., a_1x  + a_0 = A(x) \in GF(2^m)$

where $a_i \in GF(2) = \{0,1\}$

The coefficients of the polynomial are elements in $GF(2)$ the **sub-field**

##### Example $GF(2^3)$

- The field $GF(2^3)$, sometimes called $GF(8)$ is an extension field containing elements of the form: $A(x) = a_2 x^2 + a_1x^1 + a_0$
- Its often easier to simply write the coefficients $(a_2, a_1, a_0)$ e.g. 001 or 101
- $GF(2^3) = \{0, 1, x, x+1, x^2, x^2+1, x^2 + x, x^2 + x + 1\}$ 
  - $|GF(2^3)| = 8$

#### Arithmetic in $GF(2^3)$

- Adding or subtracting two polynomials happens as expected, but adding the coefficients
  - $A(x) = x^2 + x + 1$
  - $B(x) = x^2 + 1$
  - $A(x) + B(x) = (1+1)x^2 + (1)x + (1+1) = x$
- mod 2 is simply `xor`
- Addition and subtraction are identical

#### Multiplication in $GF(2^3)$

- $A(x) = x^2 + x + 1$
- $B(x) = x^2 + 1$
- $A(x) \cdot B(x) = (x^2 + x + 1)(x^2 + 1) = x^4 + x^3 + (1+1)x^2 + x + 1$
- $x^4 + x^3 + x + 1$ however this is **not in the field**
- The result must be reduced by the result modulo an **irreducible polynomial**

$$
A(x) \cdot B(x) = x^4 + x^3 + x + 1\space (mod \space x^3 + x + 1)
$$

- This means we have to do polynomial long division

![img](img/IMG_0230.jpg)

##### Inversion

- Inversion is performed in a similar way to prime fields, we find:
- $A(x) \cdot A^{-1}(x) \equiv 1 \space (mod \space P(x))$
  - $A^{-1}(x)$ is calculated using the extended euclidean algorithm

### AES’ Finite Field

- AES uses the extension field $GF(2^8)$ for many of its operations
- Operations are the same as those in other $GF(2^m)$ fields, using the irreducible polynomial

$$
  P(x) = x^8 + x^4 + x^3 + x + 1
$$

- As you might expect, these polynomials are typically represented as single bytes