Mathspics: Field theory

Feild ;-

An algebric expression (F,+,.) where F is a non-empty set with two operations “addition and dot” is called ring if it is satisfies following axioms..

D1. (F, + ) be an abelian group.

1) closure for addition:- if a,b ∈F => a+b∈F for all a,b∈F

2) associate for addition:-if a,b,c ∈F then (a+b)+c=a+(b+c) for all a,b,c∈F

3) identity for addition:- if for all a∈F there are exist 0∈F such that a+0=a for all a∈F

4) inverse for addition:- if for all a∈F then there are exist –a∈F such that a+(-a) ∈F for all a∈F

5) commutative low :- for all a,b∈F such that a+b=b+a for all a,b∈F

D2. for (R,.).

1) closure for multifi. :- for all a,b∈F => a.b∈F for all a,b∈F

2) associate for dot :- for all a,b,c∈F st (a.b).c=a.(b.c) for all a,b,c∈F

3) Commutative for dot:- for all a,b∈F => a.b =b.a

4) unit element :- 1 ∈F st a.1 = a =1.a for all a∈F

5) Inverse for dot:- for all a(≠0) ∈F there are exist a^-1 ∈F st a a^-1= 1

D3. distribution law :- for all a,b,c∈F

1) left distribution:- a.(b+c)=a.b+a.c for all a,b,c∈F

2) Right distribution :- (b+c).a=b.a+c.a for all a,b,c∈F.

So (F,+,.) is a Field…………..

Subfield:- A subset F' of F is called subfield if it is satisfies all conditions of a field.i.e. if F' shall be a Feild itself then it is called subfield.
Prime field :- A field ( F,+,.) is called prime field if it have not any proper subfield .
Theoram:-Prove that Every field always be a integral domain.
proof :- :- let (F,+,.) be a field. Since we know that every field be a commutative ring with a unit element.

To show:- (F,+,.) will a integral domain.

For this we will only to show that every field does not contains a zero divisor.

Let a,b∈F be orbitary element of F .and let a≠0 such that ab≠0

Since a≠0 => there are exist a^-1∈F .

So, ab=0 => a^-1(ab) = a^-1.0 [a.0=0]

=>( a^-1a) b = 0 [associate law]

=>1.b = 0 [a^-1a = 1]

=> b=0 [1.b=b]

Similarly, suppose ab = 0 and b≠0

b≠0 => b^-1∈F.

therefore ab = 0 => (ab) b^-1= b^-1.0

=> a(b b^-1) = 0

=> a.1 = 0

=> a =0

Therefore in a field ab=0 => a=0 or b=0

Therefore there are no zero divisor in a field and hence every field is a integral domain.

Hence proved………………

{note : - but its conversaly is not true. We can show it like below}

Proof :- let (F,+,.) be a field and let it contains distinct unit and zero elements i.e. 1≠0 .

Let a be a non-zero element of F then

a^-1 = 0 => a a^-1= a.0 [left multiple with a]

=> 1 = 0 [a a^-1=1 and a.0 =0]

=> a.1 = a.0 [left multiplel with a]

=> 1 = 0 [ cencellation law]

This is a contradiction therefore integral domain is not a field.

Hence proved………

Theorem :- prove that every finite integral domain will be a field.

Proof :- let (D,+,.) be a commutative ring with zero divisor which have n finite element a₁, a₂ ……. a_n.i.e.

D = {a₁, a₂ ……. a_n.}.

To prove : - D be a Field. For this we should show that there are exist an element 1∈D such that 1a =a for all a∈D and it contains inverse with respect to dot of all elements.

Let a≠ 0 ∈ D. therefore aa₁, aa₂ ……. aa_nwill be all elements of D and they will be distinct from each others.if not then because aai = aaj where i≠j

aai = aaj a≠0 => a(ai-aj) = 0

=> ai –aj = 0

=> ai =aj

=> i = j

This is a contradiction because i≠j so we can say aa₁, aa₂ ……. aa_nall are distinct elements of D .

Now, a≠0 and a∈ D => there are exist aa_i∈ D such that aai = a where 1≤i≤n.

ð aai = a.1 where ai ∈ D

ð ai = 1

=>1∈ D

Now to show 1 is the unit element of D .

Let y be a an element of D and x∈ D then ax=y=xa