Quotient ring First Fundamental theorem

1. Quotient Ring
Let I be an ideal of a ring R then the set R/I = { a + I : aϵ R } is called cosets of I in R is
A ring called Quotient ring where addition and multiplication are defined as
(a + I ) + (b + I) = (a + b) + I a , b ϵ R
(a + I ) + (b + I) = ab + I a , b ϵ R
Note
(i) If R is a commutative ring with unity then R/I is also a commutative ring with
unity
(ii) 1 + I is the multiplicative identity of R/I and 0 + I = I is the additive identity of R/I

2. Theorem
If I is an ideal of a ring R then R/I is a ring
Proof
First we show that R/I is an abelian group under addition
R/I = { a + I : a ϵ R }
Let a + I , b + I ϵ R/I
(a + I) + (b + I) = ( a + b) + I
a , b ϵ R => a + b ϵ R
=> ( a + b) + I ϵ R
(a + I) + (b + I) ϵ R/I
Clouser law holds in R/I under addition

3. Now
let ( a + I) , (b + I) , (c + I) ϵ R/I ; a , b , c ϵ R
( a + I) , (b + I) , (c + I) = ( a + I) , (b + I) + I
= [ a + (b + c )] + I
= [ ( a + b ) + c ] + I
= ( a + b) + I + (c + I)
= [ (a + I ) + (b + I )] + ( c + I)
Thus associative law holds in R/I under addition
Since 0 ϵ R thus 0 + I ϵ R/I
a + I ϵ R/I
(0 + I) + ( a + I) = ( 0 + a ) + I = a + I
and
(a + I) + ( 0 + I ) = ( a + I ) + I = a + I
a ϵ R ; - a ϵ R (R is ring)
Thus a + I ϵ R/I and ( - a) + I ϵ R/I

4. ( a + I) and ( - a ) + I are the additive inverse of each other ; since
(a + I ) + ((-a) + I) = (a + (-a)) + I
= 0 + I
Thus each element of R/I has its additive inverse in R/I
For ( a + I ) , ( b + I ) ϵ R/I
( a + I) + (b + I) = (a + b) + I
= (b + a ) + I
= (b + I) + (a + I)
Thus R/I is commutative under addition
Hence R/I is abelian under addition

5. now we show that R/I is semi group under multiplication
Let
(a + I) , (b + I) ϵ R/I a , b ϵ R
(a + I)(b + I ) = ab + I
since a , b ϵ R => ab ϵ R
Thus ab + I ϵ R/I
I . e (a + I ) (b + I) ϵ R/I
Clouser law holds in R/I under multiplication
For (a + I) , (b + I) , (c + I) ϵ R/I a,b,c ϵ R
(a + I) [ (b + I)(c + I)] = (a + I) (bc + I)
= a (bc) + I
= (ab) c + I
= (ab +I) (c + I)
Thus associative law holds in R/I under multiplication
Hence R/I is semi group under multiplication

6. Now we show that both left and right distributive laws holds in R/I
Let
(a + I) , (b + I) , (c + I) ϵ R/I for a,b,c ϵ R
and
(a + I) [ (b + I) + (c + I)] = (a + I) [ (b + c) + I ]
= a ( b + c ) + I
= (ab + I ) + (ac + I )
=(a + I ) ( b + I ) + ( a + I )( c + I )
Left distributive law holds in R/I
Also
[ (b + I) + (c + I)] (a + I) = [ (b + c ) + I ] (a + I)
= (b + c)a + I
= (ba + ca) + I
= (ba + I ) + (ca + I)
= (b + I)(a + I) + (c + I) (a + I)
Right distributive law holds in R/I
Hence R/I is a ring

7. Theorem
If I is an ideal of a ring R then the mapping ɸ : R  R/I defined by
ɸ (a) = a + I
Is a homomorphism
Proof
for a , b ϵ R  a + b ϵ R
ɸ ( a + b) = (a + b) + I
= (a + I) + (b + I)
= ɸ (a) + (b)
And
ɸ(ab) = ab + I
= (a + I) + (b + I)
= ɸ(a) ɸ(b)
Hence ɸ is a homomorphism

8. Theorem
Let I be an ideal of a ring R; then there always exists an epimorphism
(homomorphism + Onto) ɸ: R  R/I with ker ɸ = I
Proof
Define a mapping ɸ : R  R/I define by
ɸ(a) = a + I a ϵ R
for a ,b ϵ R  a + b ϵ R
ɸ (a + b) = (a + b) + I
= (a + I ) + ( b + I )
= ɸ(a) + ɸ (b)
And
ɸ(ab) = ab + I
= (a + I) + (b + I)
= ɸ(a) ɸ(b)
Hence ɸ is a homomorphism

9. Now we show that ɸ is onto
For each a + I ϵ R/I ; there exists an elements a ϵ R
such that ɸ (a) = a +I
Hence ɸ is an onto mapping
Thus ɸ is an epimorphism
Now we have to show that ker ɸ = I
let a ϵ ker ɸ  ɸ (a) = I I is additive identity of R/I
but ɸ(a) = a + I
a + I = I  a ϵ I
ker ɸ I ------- (1)
Now let b ϵ I
 ɸ (b) = b+ I
 b + I = I  ɸ(b) = I
 b ϵ ker ɸ
I ker ɸ -------(2)
From 1 and 2 ker ɸ = I



10. Theorem : 1st Fundamental theorem
Let I be an ideal of a ring R and Ѱ : R  R’ be an epimorphism with ker Ѱ = I
Then R/I ≈ R’
Proof
Define a mapping ɸ : R  R’ by
ɸ (a + I) = Ѱ (a) a ϵ R
First we show that ɸ is well defined for this
Let
a + I = b + I
 a – b ϵ I
 a – b ϵ ker Ѱ I = ker Ѱ
 Ѱ (a - b) = 0’ where 0’ ϵ R’
 Ѱ (a) – Ѱ(b) = 0’
 Ѱ(a) = Ѱ (b)
 ɸ (a + I) = ɸ (b+ I)
hence ɸ is well defined

11. To show that ɸ is homomorphism let
ɸ [(a + I ) + ( b + I ) ] = ɸ [ (a + b) + I ]
= Ѱ (a + b )
= Ѱ(a) + Ѱ (b)
= ɸ(a + I ) + ɸ ( b + I )
Also
ɸ[(a + I)(b + I)] = ɸ [ab + I]
= Ѱ (ab)
= Ѱ(a) + Ѱ(b)
= ɸ(a + I) ɸ(b + I)
Thus ɸ is a homomorphism
To show that ɸ is onto
Let r’ ϵ R’ be any element of R
Since Ѱ is onto (epimorphism) there exists an elements r ϵ R such that Ѱ (r) = r’
 ɸ (r + I) = r’
Thus there will be exists an element r + I ϵ R/I
Such that ɸ (r + I) = r’
ɸ is onto

12. to show that ɸ is one – one
Let
ɸ (a + I ) = ɸ (b + I)
Ѱ(a) = Ѱ (b)
Ѱ(a) – Ѱ (b) = 0’
Ѱ(a – b ) = 0’
a – b ϵ ker Ѱ
but ker Ѱ = I
Thus a - b ϵ I
Now a ϵ b + I ------(i)
but a ϵ a + I ----- (ii)
From i and ii
a + I = b + I
This shows that ɸ is one – one
ɸ is an isomorphism from R/I  R’
Hence
R/I ≈ R’