Page 171 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Thus, is well defined on R/I.
Now our aim is to prove the following result.

Theorem 3: Let R be a ring and I be an ideal in R. Then R/I is a ring with respect to addition and
multiplication defined by
(X + I) + (y + I) = (x + y) + I, and

( x + I ) . ( y + I ) = x y + I  x,y  R.
Proof: As we have noted earlier, (R/I, +) is an abelian group. So, to prove that R/I is a ring we
only need to check that . is commutative, associative and distributive over +.
Now,
(i) . is commutative : (a + I). (b + I) = ab + I = ba + I = (b + I), (a 4- I) for all a + I,b + I  R/I.

(ii) . is associative : ‘ ’ a, b, c  R
((a + I). (b + I)). (c + I) = (ab + I). (C + I)

= (ab)c + I
= a(bc) + I
= (a + I) . ((b + I) . (c + I))
(iii) Distributive law : Let a + I, b + I, c + I RA. Then
(a + I). ((b + I) + (c + 1)) = (a + I) [(b + 6) + I]

= a(b + c) + I
= (ab + ac) + I
= (ab + I) + (ac + I).

= (a + I). (b + I) + (a + I).(c -1 I)
Thus, R/I is a ring.
This ring is called the quotient ring of R by the ideal I.
Let us look at some examples. We start with the example that ‘gave rise to the terminology
‘R mod I’.

Example: Let R = Z and I = nZ. What is R/I?
Solution: You have seen that nZ is, an ideal of Z. From Unit 2 you know that
Z/nZ = { nZ., I + nZ, ..., (n – 1) + nZ }.

= {0,1,....,n 1}, the same as the set of equivalence classes modulo n.

So, R/I is the ring Z .
n
Now let us look at an ideal of Z , where n = 8.
n
Example: Let R = Z . Show that I = {0,4} is an ideal of R. Construct the Cayley tables for
8
+ and, in R/I.

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