高级算法 (Fall 2017)/Finite Field Basics: Difference between revisions
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* '''Infinite fields''': <math>\mathbb{Q}</math>, <math>\mathbb{R}</math>, <math>\mathbb{C}</math> are fields. The integer set <math>\mathbb{Z}</math> is a commutative ring but is not a field. | * '''Infinite fields''': <math>\mathbb{Q}</math>, <math>\mathbb{R}</math>, <math>\mathbb{C}</math> are fields. The integer set <math>\mathbb{Z}</math> is a commutative ring but is not a field. | ||
* '''Finite fields''': Finite fields are called '''Galois fields'''. The number of elements of a finite field is called its '''order'''. A finite field of order <math>q</math>, is denoted as <math>\mathsf{GF}(q)</math>. | * '''Finite fields''': Finite fields are called '''Galois fields'''. The number of elements of a finite field is called its '''order'''. A finite field of order <math>q</math>, is usually denoted as <math>\mathsf{GF}(q)</math> or <math>\mathbb{F}_q</math>. | ||
**<math>{\mathbb{Z}_p}</math>: For any integer <math>n>1</math>, <math>\mathbb{Z}_n=\{0,1,\ldots,n-1\}</math> under modulo- | ** '''Prime field''' <math>{\mathbb{Z}_p}</math>: For any integer <math>n>1</math>, <math>\mathbb{Z}_n=\{0,1,\ldots,n-1\}</math> under modulo-<math>p</math> addition <math>+</math> and multiplication <math>\cdot</math> forms a commutative ring. It is called '''quotient ring''', and is sometimes denoted as <math>\mathbb{Z}/n\mathbb{Z}</math>. In particular, for '''prime''' <math>p</math>, <math>\mathbb{Z}_p</math> is a field. This can be verified by [http://en.wikipedia.org/wiki/Fermat%27s_little_theorem Fermat's little theorem]. | ||
** '''Boolean arithmetics''' <math>\mathsf{GF}(2)</math>: The finite field of order 2 <math>\mathsf{GF}(2)</math> contains only two elements 0 and 1, with bit-wise XOR as addition and bit-wise AND as multiplication. <math>\mathsf{GF}(2^n)</math> | |||
** Other examples: There are other examples of finite fields, for instance <math>\{a+bi\mid a,b\in \mathbb{Z}_3\}</math> where <math>I=\sqrt{-1}</math>. This field is isomorphic to <math>\mathsf{GF}(9)</math>. In fact, the following theorem holds for finite fields of given order. | |||
{{Theorem|Theorem| | |||
:A finite field of order <math>q</math> exists if and only if <math>q=p^k</math> for some prime number <math>p</math> and positive integer <math>k</math>. Moreover, all fields of a given order are isomorphic. | |||
}} | |||
=Polynomial over a field= | |||
Given a field <math>\mathbb{F}</math>, the '''polynomial ring''' <math>\mathbb{F}[x]</math> consists of all polynomials in the variable <math>x</math> with coefficients in <math>\mathbb{F}</math>. Addition and multiplication of polynomials are naturally defined by applying the distributive law and combining like terms. The '''degree''' <math>\mathrm{deg}(f)</math> of a polynomial <math>f\in \mathbb{F}[x]</math> is the exponent on the '''leading term''', the term with a nonzero coefficient that has the largest exponent. | |||
Because <math>\mathbb{F}[x]</math> is a ring, we cannot do division the way we do it in a field like <math>\mathbb{R}</math>, but we can do division the way we do it in a ring like <math>\mathbb{Z}</math>, leaving a '''remainder'''. The equivalent of the '''integer division''' for <math>\mathbb{Z}</math> is as follows. | |||
{{Theorem|Proposition (division for polynomials)| | |||
:Given a polynomial <math>f</math> and a nonzero polynomial <math>g</math> in <math>\mathbb{F}[x]</math>, there are unique polynomials <math>q</math> and <math>r</math> such that <math>f =q\cdot g+r</math> and <math>\mathrm{deg}(r)<\mathrm{deg}(g)</math>. | |||
}} | |||
The proof of this is by induction on <math>\mathrm{deg}(f)</math>, with the basis <math>\mathrm{deg}(f)<\mathrm{deg}(g)</math>, in which case the theorem holds trivially by letting <math>q=0</math> and <math>r=f</math>. | |||
As we turn <math>\mathbb{Z}</math> (a ring) into a finite field <math>\mathbb{Z}_p</math> by taking quotients <math>\bmod p</math>, we can turn a polynomial ring <math>\mathbb{F}[x]</math> into a finite field by taking <math>\mathbb{F}[x]</math> modulo a "prime-like" polynomial, using the division of polynomials above. | |||
{{Theorem|Definition (irreducible polynomial)| | |||
:An '''irreducible polynomial''', or a '''prime polynomial''', is a non-constant polynomial <math>f</math> that ''cannot'' be factored as <math>f=g\cdot h</math> for any non-constant polynomials <math>g</math> and <math>h</math>. | |||
}} |
Latest revision as of 05:59, 19 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.
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. The number of elements of a finite field is called its order. A finite field of order [math]\displaystyle{ q }[/math], is usually denoted as [math]\displaystyle{ \mathsf{GF}(q) }[/math] or [math]\displaystyle{ \mathbb{F}_q }[/math].
- Prime field [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] under modulo-[math]\displaystyle{ p }[/math] addition [math]\displaystyle{ + }[/math] and multiplication [math]\displaystyle{ \cdot }[/math] forms a commutative ring. It is called quotient ring, and is sometimes denoted as [math]\displaystyle{ \mathbb{Z}/n\mathbb{Z} }[/math]. 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.
- Boolean arithmetics [math]\displaystyle{ \mathsf{GF}(2) }[/math]: The finite field of order 2 [math]\displaystyle{ \mathsf{GF}(2) }[/math] contains only two elements 0 and 1, with bit-wise XOR as addition and bit-wise AND as multiplication. [math]\displaystyle{ \mathsf{GF}(2^n) }[/math]
- Other examples: There are other examples of finite fields, for instance [math]\displaystyle{ \{a+bi\mid a,b\in \mathbb{Z}_3\} }[/math] where [math]\displaystyle{ I=\sqrt{-1} }[/math]. This field is isomorphic to [math]\displaystyle{ \mathsf{GF}(9) }[/math]. In fact, the following theorem holds for finite fields of given order.
Theorem - A finite field of order [math]\displaystyle{ q }[/math] exists if and only if [math]\displaystyle{ q=p^k }[/math] for some prime number [math]\displaystyle{ p }[/math] and positive integer [math]\displaystyle{ k }[/math]. Moreover, all fields of a given order are isomorphic.
Polynomial over a field
Given a field [math]\displaystyle{ \mathbb{F} }[/math], the polynomial ring [math]\displaystyle{ \mathbb{F}[x] }[/math] consists of all polynomials in the variable [math]\displaystyle{ x }[/math] with coefficients in [math]\displaystyle{ \mathbb{F} }[/math]. Addition and multiplication of polynomials are naturally defined by applying the distributive law and combining like terms. The degree [math]\displaystyle{ \mathrm{deg}(f) }[/math] of a polynomial [math]\displaystyle{ f\in \mathbb{F}[x] }[/math] is the exponent on the leading term, the term with a nonzero coefficient that has the largest exponent.
Because [math]\displaystyle{ \mathbb{F}[x] }[/math] is a ring, we cannot do division the way we do it in a field like [math]\displaystyle{ \mathbb{R} }[/math], but we can do division the way we do it in a ring like [math]\displaystyle{ \mathbb{Z} }[/math], leaving a remainder. The equivalent of the integer division for [math]\displaystyle{ \mathbb{Z} }[/math] is as follows.
Proposition (division for polynomials) - Given a polynomial [math]\displaystyle{ f }[/math] and a nonzero polynomial [math]\displaystyle{ g }[/math] in [math]\displaystyle{ \mathbb{F}[x] }[/math], there are unique polynomials [math]\displaystyle{ q }[/math] and [math]\displaystyle{ r }[/math] such that [math]\displaystyle{ f =q\cdot g+r }[/math] and [math]\displaystyle{ \mathrm{deg}(r)\lt \mathrm{deg}(g) }[/math].
The proof of this is by induction on [math]\displaystyle{ \mathrm{deg}(f) }[/math], with the basis [math]\displaystyle{ \mathrm{deg}(f)\lt \mathrm{deg}(g) }[/math], in which case the theorem holds trivially by letting [math]\displaystyle{ q=0 }[/math] and [math]\displaystyle{ r=f }[/math].
As we turn [math]\displaystyle{ \mathbb{Z} }[/math] (a ring) into a finite field [math]\displaystyle{ \mathbb{Z}_p }[/math] by taking quotients [math]\displaystyle{ \bmod p }[/math], we can turn a polynomial ring [math]\displaystyle{ \mathbb{F}[x] }[/math] into a finite field by taking [math]\displaystyle{ \mathbb{F}[x] }[/math] modulo a "prime-like" polynomial, using the division of polynomials above.
Definition (irreducible polynomial) - An irreducible polynomial, or a prime polynomial, is a non-constant polynomial [math]\displaystyle{ f }[/math] that cannot be factored as [math]\displaystyle{ f=g\cdot h }[/math] for any non-constant polynomials [math]\displaystyle{ g }[/math] and [math]\displaystyle{ h }[/math].