Page 178 - DMTH403_ABSTRACT_ALGEBRA

Unit 17: Ring Homomorphisms

Now, (f + g) = (f + g) (1/2) = f(1/2) + g(1/2) = (f) + (g), and Notes

1
1   
g
f(fg) = (fg) (1/2) = f        (f) (g).

2
 2   
Thus,  is a homomorphism.
   a 0 


Example: Consider the ring R =  0 b  a, b R  under matrix addition and multiplication.
     
Show that the map I : Z  13 : f(n) =   n 0   is a homomorphism.
 0 n 
Solution: Note that f(n) = nI, where I is the identity matrix of order 2. Now you can check that f(n
+ m) = f(n) + f(m) and f(nm) = f(n) f(m)  n, m  Z. Thus, f is a homomorphism.

Example: Consider the ring  (X) of Unit 14.
Let Y be a non-empty subset of X .
Define f :  (X)   (Y) by f(A) I= AY for all A in  ( X ) . Show that f is a homomorphism.
Does Y’  Ker f ? What is Im f ?
Solution: For any A and B in  (X),
f(A  B) = f((A\ B)  (B\ A))

= ((A\B)  (B\AY))
= ((A\B)Y)  ((B\A)Y)
= ((AY)\(BY))((BY)\(AY))
= (f(A)\ f(B)(f(B) \f(A))

= f(A) A f(B), and
f(AB) = (AB)Y
= (AB)(YY)
= (AY)(BY), sinceis associative and commutative.

I = f(A)f(B).
So, f is a ring homomorphism from  (X) into  (Y).
Now, the zero element of  (Y) is . Therefore,
Ker f = { A   (X) | AY =  ) .  Y  Ker f.
c
We will show that f is surjective.
Now, Im f = {AY | A   (x)]
Thus, Im f  (Y). To show that  (Y)  Im f, take any B   (Y).
Then B   (X) and f(B) — BY = B. Thus, B  Im f.

Therefore, Im f =  (Y).
Thus, f is an onto homomorphism.

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