Page 181 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Proof: (a) Since (0) is an ideal of R , by Theorem 3(b) we know that f-1({o}) is an ideal of R . But
2
1
f ({o}) = Ker f.
-1
Thus, we have shown that Ker f is an ideal of R .
1
(b) Since R is a subring of R , f(R ) is a subring of R , by Theorem 2(a). Thus, Im f is a subring of
1
2
1
1
R .
2
This result is very useful for showing that certain sets are ideals. For example, from Theorem 5
you can immediately say that {0,3} is an ideal of Z . As we go along you will see more examples
6
of this use of Theorem 5.
Let us look a little more closely at the kernel of a homomorphism. In fact, let us prove a result
analogous to Theorem 4 of Unit 6.
Theorem 6: Let f : R – R be a homomorphism. Then f is injective iff Ker f = {0}
2
1
Proof: f is injective iff f is an injective group homomorphism from (R , +) into (R , + ). This is true
1
2
iff Ker f = {0}, by Theorem 4 of Unit 6. So, our result is proved.
So far we have seen that given a ring homomorphism f : R — S, we can obtain an ideal of R,
namely, Ker f. Now, given an ideal I of a ring R can we define a homomorphism f so that
Ker f = I?

The following theorem answers this question. Before going to the theorem recall the definition
of quotient rings.

Theorem 7: If I is an ideal of a ring R, then there exists a ring homomorphism f : R  R/I whose
kernel is I.

Proof: Let us define f : R  R/I by f(a) = a + I for all a  R. Let us see iff is a homomorphism. For
this take any a, b  R. Then

f(a + b) = (a + b) + I = (a + I) + (b + I) = f(a) + f(b), and
f(ab) = ab + I = (a + I) (b + I) = f(a) f(b).
Thus, f is a homomorphism.

Further, Kerf = {a  R | f(a) = 0 + I} = { a  R | a + I = I }
= {a  R | a  I} = I .
Thus, the theorem is proved.

Also note that the homomorphism f is onto.
We call the homomorphism defined in the proof above the canonical (or natural) homomorphism
from R onto R/I.
Now let us look at the behaviour of the composition of homomorphisms. We are sure you find
the following result quite unsurprising.
Theorem 8: Let R , R and R be rings and f : R — R , and g : R  R be ring homomorphisms.
2
1
2
3
1
2
3
Then their composition gof : R  R given by (gof (x) = g(f(x)) for all x  R is a ring
1
1
3
homomorphism.
The proof of this result is on the same lines as the proof of the corresponding result in
Unit 6.
174 LOVELY PROFESSIONAL UNIVERSITY