高级算法 (Fall 2017)/Finite Field Basics

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Field

Let [math]\displaystyle{ S }[/math] be a set, closed under two binary operations [math]\displaystyle{ + }[/math] (addition) and [math]\displaystyle{ \cdot }[/math] (multiplication). It gives us the following algebraic structures if the corresponding set of axioms are satisfied.

Structures Axioms Operations
field commutative
ring
ring abelian
group
group monoid semigroup 1. Addition is associative: [math]\displaystyle{ \forall x,y,z\in S, (x+y)+z= x+(y+z). }[/math] [math]\displaystyle{ + }[/math]
2. Existence of additive identity 0: [math]\displaystyle{ \forall x\in S, x+0= 0+x=x. }[/math]
3. Everyone has an additive inverse: [math]\displaystyle{ \forall x\in S, \exists -x\in S, \text{ s.t. } x+(-x)= (-x)+x=0. }[/math]
4. Addition is commutative: [math]\displaystyle{ \forall x,y\in S, x+y= y+x. }[/math]
5. Multiplication distributes over addition: [math]\displaystyle{ \forall x,y,z\in S, x\cdot(y+z)= x\cdot y+x\cdot z }[/math] and [math]\displaystyle{ (y+z)\cdot x= y\cdot x+z\cdot x. }[/math] [math]\displaystyle{ +,\cdot }[/math]
6. Multiplication is associative: [math]\displaystyle{ \forall x,y,z\in S, (x\cdot y)\cdot z= x\cdot (y\cdot z). }[/math] [math]\displaystyle{ \cdot }[/math]
7. Existence of multiplicative identity 1: [math]\displaystyle{ \forall x\in S, x\cdot 1= 1\cdot x=x. }[/math]
8. Multiplication is commutative: [math]\displaystyle{ \forall x,y\in S, x\cdot y= y\cdot x. }[/math]
9. Every non-zero element has a multiplicative inverse: [math]\displaystyle{ \forall x\in S\setminus\{0\}, \exists x^{-1}\in S, \text{ s.t. } x\cdot x^{-1}= x^{-1}\cdot x=1. }[/math]

The semigroup, monoid, group and abelian group are given by [math]\displaystyle{ (S,+) }[/math], and the ring, commutative ring, and field are given by [math]\displaystyle{ (S,+,\cdot) }[/math].

Examples:

  • Infinite fields: [math]\displaystyle{ \mathbb{Q} }[/math], [math]\displaystyle{ \mathbb{R} }[/math], [math]\displaystyle{ \mathbb{C} }[/math] are fields. The integer set [math]\displaystyle{ \mathbb{Z} }[/math] is a commutative ring but is not a field.
  • Finite fields: Finite fields are called Galois fields. A Galois field of order [math]\displaystyle{ q }[/math], denoted as [math]\displaystyle{ \mathsf{GF}(q) }[/math], is a finite field containing [math]\displaystyle{ q }[/math] elements.
    • [math]\displaystyle{ {\mathbb{Z}_p} }[/math]: For any integer [math]\displaystyle{ n\gt 1 }[/math], [math]\displaystyle{ \mathbb{Z}_n=\{0,1,\ldots,n-1\} }[/math] (where addition [math]\displaystyle{ + }[/math] and multiplication [math]\displaystyle{ \cdot }[/math] are defined modulo [math]\displaystyle{ n }[/math]) is a commutative ring. Sometimes, this ring is denoted as [math]\displaystyle{ \mathbb{Z}/p\mathbb{Z} }[/math] and is called the quotient ring. In particular, for prime [math]\displaystyle{ p }[/math], [math]\displaystyle{ \mathbb{Z}_p }[/math] is a field. This can be verified by Fermat's little theorem.
    • [math]\displaystyle{ \mathsf{GF}(2^n) }[/math]:
    • [math]\displaystyle{ \mathsf{GF}(p^n) }[/math]:
    • Other examples: