Page 179 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes

Tasks 1. Let A and B be two rings. Show that the projection map P : A × B  A : p(x, y)
= x is a homomorphism. What are Ker p and Im p?

2. Is f : Z + 2Z  Z + 2Z : f(a + 2b ) = a – 2b a homomorphism?
3. Show that the map  : C[0, 1]  R × R :  (f = (f(0), f(1)) is a homomorphism.

Having discussed many examples, let us obtain some basic results about ring homomorphisms.

17.2 Properties of Homomorphisms

Let us start by listing some properties that show how a homomorphism preserves the structure
of its domain. The following result is only a restatement of Theorem 1 of Unit 6.
Theorem 1: Let f : R + R be a homomorphism from a ring R into a ring R . Then
1
2
2
1
(a) f(0) = 0,
(b) f(– x) = – f(x)  x  R , and
1
(c) f (x – y) = f(x) – f(y)  x, y  R . 1
Proof: Since f is a group homomorphism from (R , + ) to (R , + ), we can apply Theorem 1 of
1
2
Unit 6 to get the result.
Theorem 2: Let f : R – R be a ring homomorphism. Then
1
2
(a) if S is a subring of R , f(S) is a subring of R ;
1
2
(b) if T is a subring of R , f (T) is a subring of R .
-1
2
1
Proof: We will prove (b) and leave the proof of (a) to you. Let us use Theorem 1 of Unit 16.
Firstly, since T   , f (T)  . Next, let a, b  f (T). Then f(a), f(b)  T
-1
-1
 f(a) – f(b)  T and f(a) f(b)  T
 f(a – b)  T and f(ab)  T
 a – b  f (T) and ab  f (T)
-1
–1
 f (T) is a subring.
-1
Now, it is natural to expect an analogue of Theorem 2 for ideals. But consider the inclusion i : Z
– R : i(x) = x. You know that 22 is an ideal of Z. But is i(2Z) (i.e., 22) an ideal of R? No. For example,
1 1 1
2  22,  R, but 2.   2Z. Thus, the homomorphic image of an ideal need not be an ideal.
4 4 2
But, all is not lost. We have the following result.
Theorem 3: Let f : R – R be a ring homomorphism.
1 2
(a) Iff is surjective and I is an ideal of R , then f (I) is an ideal of R,.
1
(b) If I is an ideal of R , then f (1) is an ideal of R and Ker f  f (J).
-1
-1
2
1
Proof: Here we will prove (a)
Firstly, since I is a subring of R , f(1) is a subring of R .
1
2

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