These slides are consists on Integral Domain Characteristic of Ring Sub Ring Centre of Ring with its theorem's

1. Definition:
A commutative ring R is called an integral domain if it has no zero
i.e. a,b ϵ R if
ab=0 then either a=0 or b=0
Then set of integers Z is an integral domain
Lemma:
A commutative ring R is an integral domain if and only if cancellation
law holds in it with respect to multiplication
Proof:
Suppose cancellation law holds in R with respect to multiplication
Now for a,b ϵ R let

2. ab =0 where a ≠ o
also a.0 = 0
thus a . b = a . 0
By Cancellation law we have
b = 0
thus R has no zero divisor
There for R is an integral Domain
Conversely:
Suppose R is an integral domain i . e R has no zero divisor
Let ab = ac for a , b , c ϵ R with a ≠ o
ab – ac = 0
a(b – c ) = 0
Since R has no zero Divisor
thus b – c = 0 where a ≠ o
b = c i . e Cancellation law holds in R

3. If for a ring (R , + , .) there exists a least positive integer “n” such that
na = a + a + a + a + …… + a = 0 (n time)
i.e na = 0 Ɐ a ϵ R then n is called the characteristic of R
Def:
An element x ϵ R ( where R is ring ) is called an idempotent element if
X2 = x
Boolean Ring :
If each element of ring is idempotent then the ring is called Boolean
ring
Lemma :
If R is a Boolean ring then
(i) 2a = 0 where Ɐ a ϵ R
(ii) ab = ba R is Commutative
Characteristic of Ring

4. Proof :
( i ) 2a = a + a
= (a + a )2 R is Boolean X2 = x Ɐ x ϵ R
= (a + a) ( a + a)
= a2 + a2 + a2 + a2
= 4a2
2a = 4a a = a2 R is Boolean
4a – 2a = 0
2a = 0
(ii) now (a + a )2 = a + b R is Boolean
( a + b) ( a + b ) = a + b
a ( a + b) + b ( a + b ) = a + b Distributive law
a2 + ab + ba + b2 = a + b
a + ab + ba + b = a + b where a2 = a and b2 = b
ab + ba + ( a + b) = a + b
ab + ba = a + b – a – b
ab + ba = 0
ab = - ba

5. now
ab – ba = ab + (-ba)
= ab + ab where –ba = ab
= 2 (ab)
= 2a (b)
= 0 where 2a = 0 Ɐ a ϵ R by (
I )
this show that boolean ring is commutative

6. Def:
Let S be a non empty Subset of a Ring (R,+,.) , Then S is said to
be a Sub ring of R
If S satisfied all the axioms of ring under the induced binary
operations .
i.e
S is called a Sub ring of R if
1. a-b ϵ s Ɐ a,b ϵ s
2. ab ϵ s Ɐ a,b ϵ s
e.g
set of even integers { 0,±2,±4,±6…….} is a Sub ring Ring of
integers
{0,±1,±2,±3………..}

7. Theorem.
Let S be a non empty Subset of a ring (R,+,.) then S is Sub ring
of R
if and only if
1. a-b ϵ s Ɐ a,b ϵ s
2. ab ϵ s Ɐ a,b ϵ s
Proof
Suppose that S is a Sub ring of R i.e S satisfies all the axioms of a
ring .
let a,b ϵ s
Since S is abelian group under addition , thus for b ϵ S ; - b ϵ S (
addition inverse)
a , - b ϵ S
a + (-b) ϵ S (closure property)
a – b ϵ S
Condition (i) is proved
Also S is semi group under Multiplication
thus for a , b ϵ S
ab ϵ S
Condition (ii) is proved

8. Conversely :
Suppose Condition (i) and (ii) holds in S we have to show that
S satisfies all the axioms of a ring
let a,b ϵ S
by condition (i) a - b ϵ S
This shows that S is subgroup of R under Addition
Also S sub set of R and commutative law holds in R under Addition
Thus S is an abelian subgroup of R under Addition
Also for a,b ϵ S => ab ϵ S by Condition (ii)
thus S is closed under Multiplication
Since S is sub set of R and associative law holds in R thus it is holds in S
Thus S is semi group under Multiplication
Again
Since S is sub set of R and distributive law holds in R thus it is holds in S
Thus S is semi group under Multiplication
From above we have proved that’s
(i) S is abelian subgroup of R under addition
(ii) S is semi group of under multiplication
(iii) Distributive laws holds in S

9. Def :
If (R , + , . ) is a ring then the set of elements of R which commute with
every element of P forms the centre of R
Centre of R = { x ϵ R : ax = xa Ɐ a ϵ R }
Theorem :
centre of R is always non – empty . Since at least it contains
the identity element which commute with every element of R
now suppose
X1 , X2 ϵ centre of R
I . E a X1 = X1 a and a X2 = X2 a Ɐ a ϵ R
Then

10. ( X1 - X2 ) = a X1 - X2 a Distributive law
= a X1 - a X2
= a (X1 - X2 ) Distributive law
X1 - X2 ϵ Center of R
Also
( X1 X2 ) a = X1 (X2a) Associative law
= X1 (aX2)
= (X1 a)X2 Associative law
= (aX1 )X2
= a(X1 X2 ) Associative law
X1 X2 ϵ Center of R
For
X1 X2 ϵ Center of R
and
X1 - X2 ϵ Center of R
Thus Centre of R is a sub ring of R