Page 182 - DMTH403_ABSTRACT_ALGEBRA

Unit 17: Ring Homomorphisms

17.3 The Isomorphism Theorems Notes

We discussed group isomorphisms and various results involving them. In this section we will
do the same thing for rings. So, let us start by defining a ring isomorphism.
Definition: Let R and R be two rings. A function f : R  R is called a ring isomorphism (or
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simply an isomorphism) if
(i) f is a ring homomorphism,
(ii) f is 1 – 1, and
(iii) f is onto.

Thus, a homomorphism that is bijective is an isomorphism.
An isomorphism of a ring R onto itself is called an automorphism of R.
Iff : R  R is an isomorphism, we say that R is isomorphic to R , and denote it by R  R .
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Remark: Two rings are isomorphic if and only if they are algebraically identical. That is,
isomorphic rings must have exactly the same algebraic properties. Thus, if R is a ring with
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identity then it cannot be isomorphic to a ring without identity. Similarly, if the only ideals of
R are {0} and itself, then any ring isomorphic to R must have this property too.
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And now, let us go back to Unit 6 for a moment. Over there we proved the Fundamental Theorem
of Homomorphism for groups, according to which the homomorphic image of a group G is
isomorphic to a quotient group of G, Now we will prove a similar result for rings, namely, the first
isomorphism theorem or the Fundamental Theorem of Homomorphism for rings.
Theorem 9 (The Fundamental Theorem of Homomorphism): Let f : R  S be a ring
homomorphism. Then R/Ker f  Im f. In particular, iff is surjective, then R/Ker f  S.
Proof: Firstly, note that K/Ker f is a well defined quotient ring since Ker f is an ideal of R. For
convenience, let us put Ker f = I. Let us define
 : R/I – S by (x t I) = f(x).

As in the case of Theorem 8 of Unit 6, we need to check that , is well defined, i.e., if
x + I = y + I then (x + I) = (y + I).
Now, x t I = y + Ix – y  I = Ker f  f(x – y) = 0  f(x) = f(y)
(x + I) = (y + I).

Thus,  is well defined.
Now let us see whether , is an isomorphism or not.
(i) , is a homomorphism : Let x, y  R. Then
((x + I) + (y + I)) = (x + y + I) = f(x + y) = f(x) + f(y)

= (x + I) + (y + I), and
((x + 1) (y + 1)) = (xy + 1) = f(xy) = f(x) f(y)
= (x + 1)(y + 1)
Thus,  is a ring homomorphism.

(ii) Im  = Im f : Since (x + I) = f(x)  Im f  x  R, Im y  Im f. Also, any element of Im f is
of the form f(x) = (x + I) for some x  R. Thus, Im f  Im . So, Im  = Im f.

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