Page 176 - DMTH403_ABSTRACT_ALGEBRA

Unit 17: Ring Homomorphisms

Definition: Let (R , +, . ) and (R ,+ . . ) be two rings and f : R – R be a map. We say that f is a ring Notes
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homomorphisms if
f(a + b) = f(a) 4 – f(b), and
f(a . b) = f(a) . f(b) for all a, b in R .
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Note that the + and . occurring on the left hand sides of the equations in the definition above are
defined on R , while the + and . occurring on the right hand sides are defined on R .
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So, we can say that f : R – R is a homomorphism if
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(i) the image of a sum is the sum of the images, and
(ii) the image of a product is the product of the images.
Thus, the ring homomorphism f is also a group homomorphisms from (R ,+ ) into (R , +).
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Just as we did in Unit 6, before giving some examples of homomorphisms let us define the
kernel and image of a homomorphism. As is to be expected, these definitions are analogous to
the corresponding ones in Unit 6.
Definition: Let R and R be two rings and f : R – R be a ring homomorphism. Then we define
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(i) the image of f to be the set lm f = {f(x) | x  R },
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(ii) the kernel off to be the set Ker f = {x  R | f(x) = 0).
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Note that lm f  R and Ker f  R ,
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If Im f = R , f is called an epimorphism or an onto homomorphism, and then R is called the
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homomorphic image of R .
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Now let us look at some examples.
Example: Let R be a ring. Show that the identity map I is a ring homomorphism. What are
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Ker I and Im I ?
R
R
Solution: Let x, y  R. Then
I (x + y) = x + y = I (x) + I (y), and
R
R
R
I (xy) = xy = I {(X) I (y).
R
R
R
.
Thus, I (xy) = xy = I (x) I (y).
R
R
R
Thus, IR is a ring homomorphism.
Ker I = { x  R | I (x) = 0 }
R
R
=.{x  R | x = 0)
= {0}
I I = {(I (x) [ x  R ]}
m
R
R
={x | x  R ]
= R.
Thus, I , is a surjection, and hence an epimorphism.
R
Example: Let s  N. Show that the map f : Z – Z, given by f(m) = m for all m  Z is a
homomorphism. Obtain Ker f and Im f also.

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