145 lines
4.8 KiB
Markdown
145 lines
4.8 KiB
Markdown
# Finite Field Arithmetic
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- A **finite field** is a set containing a finite number of elements
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- This is sometimes called a *Galois Field*
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- In a Galois field you can:
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- Add
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- Subtract
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- Multiply
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- Invert (divide)
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- Fields are an extension of *groups* and related to *rings*
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### Groups
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A group is a set of elements $G$ together with an operation $\circ$ that combines two elements of $G$
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> 1. The operation $\circ$ is **closed**
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> - i.e. for all $a,b \in G$ then $a\circ b=c\in G$
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> 2. The operation is associative
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> - i.e. $a\circ(b\circ c) = (a\circ b)\circ c$ for all $a,b,c \in G$
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> 3. There is an element $1\in G$ called a **neutral element** such that $a\circ 1 = 1\circ a = a$ for all $a\in G$
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> 4. For each $a \in G$ there exists an element $a^{-1}\in G$ called the **inverse** of $a$ such that $a\circ a^{-1} = a^{-1}\circ a = 1$
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> 5. A group $G$ is **abelian** (commutative) if $a\circ b = b \circ a$ for all $a,b\in G$
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##### Example Group
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- 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
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- Every element would have an inverse where $a + (-a) = 0$ mod m
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- This group would not form a group with multiplication, as not all elements would have an inverse
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- We wouldn’t have an inverse, we would need $5\times \frac15=1$ however $\frac15 \notin \mathbb{Z}$
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### Fields
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A field $F$ is a set of elements with the following properties
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> 1. All elements of $F$ form an **additive group** with the group operation $+$ and the neutral element 0
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> 2. All elements of $F$ except 0 form a multiplicative group with the group operation $\times$ and the neutral element 1
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> 3. When the two group operations are mixed, the distributivity law holds.
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> - i.e. for all $a,b,c \in F, a\cdot(b+c) = (a\cdot b) + (a\cdot c)$
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##### Example Field
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- The set of real numbers $\mathbb{R}$ is a field with neutral element 0 for addition and 1 for multiplication
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- Every real number $a$ has a additive inverse $-a$
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- Every non-zero number $a$ has a multiplicative inverse $\frac{1}{a}$
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#### Finite Fields
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> A finite field only exists if it has $p^m$ elements
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>
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> Where:
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>
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> - $p$ is a prime
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> - $m$ is a positive integer
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###### Examples
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- There is a field with 11 elements: $GF(11)$
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- There is a field with 256 elements: $GF(256)$ or $GF(2^8)$
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- $GF(12)$ is not a finite field $(2^2 \cdot3)$
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###### Prime and Extension Fields
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When $m=1$ it creates a **prime field**
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When $m>1$ it creates an **extension field**
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### Prime Fields
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- A prime field $GF(p)$ contains the integers $\{0,1,...p-1\}$
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- These operations satisfy the properties of fields (*closure*)
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##### Inversion in Prime Fields
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$a \cdot a^{-1} \equiv 1 \space (mod \space p)$
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- A modular inverse exists when $gcd(a,p) = 1$
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- Because $p$ is prime, every number has a multiplicative inverse
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- $gcd(a,p) = 1, \forall a \neq0 \in GF(p)$
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- $a^{-1}$ can be calculated using the **extended Euclidean algorithm**
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#### Extension Fields
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- In prime fields, the elements are integers
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- Elements in extension fields $GF(2^m)$ are polynomials of degree $m$
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$a_{m-1}x^{m-1}, ..., a_1x + a_0 = A(x) \in GF(2^m)$
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where $a_i \in GF(2) = \{0,1\}$
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The coefficients of the polynomial are elements in $GF(2)$ the **sub-field**
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##### Example $GF(2^3)$
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- The field $GF(2^3)$, sometimes called $GF(8)$ is an extension field containing elements of the form: $A(x) = a_2 x^2 + a_1x^1 + a_0$
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- Its often easier to simply write the coefficients $(a_2, a_1, a_0)$ e.g. 001 or 101
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- $GF(2^3) = \{0, 1, x, x+1, x^2, x^2+1, x^2 + x, x^2 + x + 1\}$
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- $|GF(2^3)| = 8$
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#### Arithmetic in $GF(2^3)$
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- Adding or subtracting two polynomials happens as expected, but adding the coefficients
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- $A(x) = x^2 + x + 1$
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- $B(x) = x^2 + 1$
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- $A(x) + B(x) = (1+1)x^2 + (1)x + (1+1) = x$
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- mod 2 is simply `xor`
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- Addition and subtraction are identical
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#### Multiplication in $GF(2^3)$
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- $A(x) = x^2 + x + 1$
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- $B(x) = x^2 + 1$
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- $A(x) \cdot B(x) = (x^2 + x + 1)(x^2 + 1) = x^4 + x^3 + (1+1)x^2 + x + 1$
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- $x^4 + x^3 + x + 1$ however this is **not in the field**
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- The result must be reduced by the result modulo an **irreducible polynomial**
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$$
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A(x) \cdot B(x) = x^4 + x^3 + x + 1\space (mod \space x^3 + x + 1)
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$$
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- This means we have to do polynomial long division
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##### Inversion
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- Inversion is performed in a similar way to prime fields, we find:
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- $A(x) \cdot A^{-1}(x) \equiv 1 \space (mod \space P(x))$
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- $A^{-1}(x)$ is calculated using the extended euclidean algorithm
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### AES’ Finite Field
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- AES uses the extension field $GF(2^8)$ for many of its operations
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- Operations are the same as those in other $GF(2^m)$ fields, using the irreducible polynomial
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$$
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P(x) = x^8 + x^4 + x^3 + x + 1
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$$
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- As you might expect, these polynomials are typically represented as single bytes
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