高级算法 (Fall 2017)/Finite Field Basics: Difference between revisions
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The semigroup, monoid, group and abelian group are given by <math>(S,+)</math>, and the ring, commutative ring, and field are given by <math>(S,+,\cdot)</math>. | The semigroup, monoid, group and abelian group are given by <math>(S,+)</math>, and the ring, commutative ring, and field are given by <math>(S,+,\cdot)</math>. | ||
Examples: | |||
* Infinite fields: <math>\mathbb{Q}</math>, <math>\mathbb{R}</math>, <math>\mathbb{C}</math> are fileds. And <math>\mathbb{Z}</math> is a commutative ring but is not a field. | |||
* Finite fields: For any integer <math>n>1</math>, the modulo ring <math>\mathbb{Z}_n=\{0,1,\ldots,n-1\}</math> (where addition <math>+</math> and multiplication <math>\cdot</math> are defined modulo <math>n</math>) is a commutative ring. In particular, for '''prime''' <math>p</math>, <math>\mathbb{Z}_p</math> is a field. |
Revision as of 14:34, 13 September 2017
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.
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] |
---|---|---|---|---|---|---|---|
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] | |||||||
6. Multiplication is associative: [math]\displaystyle{ \forall x,y,z\in S, (x\cdot y)\cdot z= x\cdot (y\cdot z). }[/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 fileds. And [math]\displaystyle{ \mathbb{Z} }[/math] is a commutative ring but is not a field.
- Finite fields: For any integer [math]\displaystyle{ n\gt 1 }[/math], the modulo ring [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. In particular, for prime [math]\displaystyle{ p }[/math], [math]\displaystyle{ \mathbb{Z}_p }[/math] is a field.