Page 184 - DMTH403_ABSTRACT_ALGEBRA

Unit 17: Ring Homomorphisms

Thus, f(n) = nf(1)  n  Z ..... (1) Notes

Now, since f is a non-trivial homomorphism, f(m) # 0 for some m  Z.
Then, f(m) = f(m . 1) = f(m) f(1).
Cancelling f(m) on both sides we get f(1) = 1.
Therefore, from (1) we see that

f(n) = n  n  Z, i.e., f = I .
z
This theorem has an important corollary.

Corollary: Let R be a ring isomorphic. to Z. If f and g are two isomorphisms from R onto Z, then
f = g.
Proof: The composition f.g-’ is an isomorphism from Z. onto itself. Therefore, by Theorem 10,
fog = Iz, i.e., f = g.
-1
We are now in a position to prove the following result.
Theorem 11: Let R be a ring and f and g be homomorphisms from R onto Z such that Ker f =
Ker g. Then f = g.
Proof: By Theorem 9 we have isomorphisms
 : R/Ker f  Z and  : R/Ker g  Z.
r
g
Since Ker f = Ker g,  and  are isomorphisms of the same ring onto Z. Thus, by the corollary
r
g
above,  =  .
r
g
Also, the canonical maps r : R  R/Ker f and  : R  R/Ker g are the same since Ker f = Ker g.
g
 f = r o  =  o  = g.
g
f
g
Let us halt our discussion of homomorphisms here and briefly recall what we have done in this
unit. Of course, we have not finished with these functions. We will be going back to them again
and again in the future units.
Self Assessment
1. If R + R be two rings and f : R  R be a ring ................. then we define imf = {f(x) | x  R }.
2
1
1
2
1
(a) isomorphisms (b) automorphism
(c) homomorphism (d) polynomial
2. If im f = R , f is called an ................. or onto homomorphism, then R is called the
2
2
homomorphic image of RZ.
(a) epimorphism (b) hemomorphism
(c) isomorphism (d) analogous
3. Two rings are isomorphic if and only if they are algebraically .................
(a) designed (b) identical
(c) onto (d) isomorphic
4. A homomorphism that is ................. is an isomorphism
(a) subjective (b) bijective

(c) onto (d) injective

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