Integral Domain

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

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

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

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

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

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

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

D2.  (R,.) be a semi group.

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

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

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

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

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

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

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

D4. D is without zero divisor :- i.e. a.b = 0 => a=0 or b=0 or both zero.

So (D,+,.) is a Integral Domain…………..

Integral Domain:- A ring (R,+,.) is called integral domain if it satisfies three more exioms

(1)   Commutative.

(2)   It have a unit element.

(3)   Without zero divisor.

But it have atleast two element.

Examples :- set of rational numbers is a integral domain which have a zero element and unit element.

2) set of real number is a integral domain .

Theoram:- Ring with modulo p of integers be a integral domain iff p be a prime number.

Proof :- suppose  (Ip,+p , .p) be a ring of integers with modulo p.

To show:- p be a prime number.

Assume contrary , p is not a prime number and let p =m . n where 1<m<p ,1<n<p.

Therefore         [n] .p [m] = [n.m] 

     =[p] = [0] (mod p)

                        i.e.     [n] .p [m] =[0] (mod p)

            but       [m] …[0] and [n]≠ [0].

So, (Ip,+p , .p) is not a integral domain.

Therefore  p is a prime number.

 Conversaly :-  let p be a prime number and [m],[n]∈Ip such that [m].p[n]=[0]
 =>  r.s = 0 (modp)
 =>    r.s is divided by p
   =>either r divided by p or s divided s divided by p, since p is prime number.
 =>Either r=0(mod p) or s = 0(mod P)
    =>[r] = [0] or [s] = [0]
Therefore (Ip,+p , .p) is a integral domain.

Ordered Integral Domain:-  A Integral domain (D,+,.) is called ordered integral domain if a subset D+ of D such that

1)      D+ ,will be closure corresponding to D with respect to ‘+’ and ‘.’ i.e. if a,b∈D+ => a+b∈D+ and a.b∈D+.

2)      For all a∈D+ its true for one and only one, a=0 , a∈D+,  -a∈D+ .

Theoram:- prove that set of complex numbers is not a ordered integral domain.

Proof:- let C be a set of complex numbers.we know that (C,+,.) be a integral domain. suppose C+ be a set of all positive complex numbers of C. where 0+i0 is a zero element of C+.

since i≠0 therefore for trichotomy law it will either i∈C+ or -i∈C+.

now C+ is always closure for '.' and '+'. 

let i∈C+ => i.i ∈C+=> -1∈C+

so i∈C+ , -1∈C+ => i(-1)∈C+ => -i∈C+

so i∈C+ => -i∈C+ this is contradiction of trichonotomy law. 

similarly -i∈C+ => i∈C+  ,this is again contradiction .

therefore (C+,+,.) is not a ordered integral domain.

                                                                    Hence proved.....