(x + I) + (y + I) = (x + y) + I, and (x +I) . (y + I) = xy + 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 ned to check that . is commutative, associative and distributive over +.
Now,
i) . is commutative: (a + I). (b + I) = ab + I = ba + I = (b + I). (a + 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 ∈R/I. then (a + I). ((b +I) + (c+ I)) = (a + I) [(b + c) + I}
= a(b + c) + I = (ab + ac) + I = (ab + I) + (ac +I).
= (a + I). (b + I) + (a + I). (c + 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 and I’.
Example 10: Let R = Z and I = nZ. What is R/I?
Solution: In Sec--- you have seen that nZ is an ideal of Z. From Unit 2 you know that
Z/nZ = {nZ, I + nZ, + nZ, …, (n – 1) + nZ}
= {0− ,1−,..., n −1}, the same as the set of equivalence classes modulo n.
So, R/I is the ring Zn.
Now let us look at an ideal of Zn, where n = 8.
Example 11: Let R = Z8 show that I = {0,4}
−
−
is an ideal of R. Construct the Cayley tables for + and . in R/I.
Solution: I =4R
−
, and hence is an ideal of R. From group theory you know that the number of elements in R/I = o(R/I) =
) (
) (
I o
R o =
2 8= 4.
You can see that these elements are 0 + I ={0,4}
−
− , 1 + 1 = {1,5}
−
−
, 2 + 1 = {2,6}
−
−
, 3 + 1 = {3,7}
−
−
.
The Cayley tables for + and . in R/I are + 0+− 1 1+− 1 2+− 1 3+− 1 0+− 1 0+− 1 1+− 1 2+− 1 3+− 1 1+− 1 1+− 1 2+− 1 3+− 1 0+− 1 2+− 1 2+− 1 3+− 1 0+− 1 1+− 1 3+− 1 3+− 1 0+− 1 1+− 1 2+− 1 Try this exercise now.
SELF ASSISMENT EXERCISE 18
Show that if R is a ring with identity, then R/I is a ring with identity fir any ideal 1 of R.
SELF ASSISMENT EXERCISE 19
If R is a ring with identity 1 and I is an ideal of containing 1, then what does R/I look like?
SELF ASSISMENT EXERCISE 20
Let N be the nil radical of R. Show that R/N has non-zero nilpotent elements.
4.0 CONCLUSION
You will realize the utility and importance of quotient rings after we discuss homomorphisms in the next unit and when we discuss polynomial rings (Block 4).
Now let us briefly summarize what we have done in this unit.
5.0 SUMMARY
In this unit we have discussed the following points, with the assumption that all rings are commutative.
1) The definition and examples of a subring.
2) The proof and use of the fact that a non-empty subset S of a ring R is subring of R iff x – y ∈S and xy ∈S ∀x, y ∈S.
3) The intersection of subring of a ring is a subring of the ring.
. 0+− 1 1+− 1 2+− 1 3+− 1 1
0−+ 0−+1 0−+1 0−+1 0−+1 1+− 1 0+− 1 1+− 1 2+− 1 3+− 1 2+− 1 0+− 1 2+− 1 0+− 1 2+− 1 3+− 1 0+− 1 3+− 1 2+− 1 1+− 1
4) The Cartesian product of subrings is a subring of the Cartesian product of the corresponding rings.
5) The definition and examples of an ideal.
6) The definition of an ideal generated by n elements.
7) The set of nilpotent elements in a ring is an ideal of the ring.
8) If I is an ideal of a ring R with identity and I ∈I, = R.
9) If I and J are ideal of a ring R, then I ∩J, I + J and IJ are also ideals of R.
10) The definition and examples of a quotient ring.
SOLUTIONS/ANSWERS
SELF ASSISMENT EXERCISE 1
∀x, y ∈R, x – y ∈R and xy ∈R. Thus, R is a subring of C. Similarly, you can check the other cases.
SELF ASSISMENT EXERCISE 2
Clearly, S is non-empty.
Also, for any x, y ∈S, x – y = x ∆ y (As pointed out in Example 2).
You can check that x ∆ y ∈S ∀ x, y ∈S.
Also, for any x, y ∈S, xy = x ∩y ∈S, as you can check.
Thus, S is a subring of ℘(X).
SELF ASSISMENT EXERCISE 3
Firstly, S ≠ø. Secondly, for any A =
b 0
0
a and C =
d 0
0
c in S,
A – C =
− 0 c
a
−d b
0 ∈S and AC =
bd 0
0
ac ∈S.
Thus, S is a subring of R.
The unit element of S =
1 0
0
1 = the unit element of R.
SELF ASSISMENT EXERCISE 4
Both {0} and R are non-empty and satisfy (a) and (b) of Theorem 1.
SELF ASSISMENT EXERCISE 5
Since A is a subring of B, A ≠ø and ∀x, y ∈A, x – y ∈A and xy ∈A.
here the addition and multiplication are those defined on B. but these are the same as those defined on C since B is a subring of C. Thus, A satisfies Theorem 1, and hence is a subring of C.
SELF ASSISMENT EXERCISE 6
There are several examples. We take {I}. in fact, any finite subset of Z, apart from {0}, will do.
SELF ASSISMENT EXERCISE 7
1 + i and 2
1are elements of the union.
But 1 + i - 2 1 =
2
1 + i ∉Z + iZ ∪Q is not a subring of C.
SELF ASSISMENT EXERCISE 8
2Z X R, 3Z X {0} are two among infinitely many examples.
SELF ASSISMENT EXERCISE 9
note that the two sets are 3 Z6
−
and 2 Z6
−
. From Example 4 you know that they are subrings of Z6. Now, by element wise multiplication you can check that rx ∈3 Z6
− ∀r ∈Z6 and x ∈3 Z6
−
. (for instance, 5−.3− = 15 = 3∈3Z6.)
You can similarly see that rx ∈2Z6 ∀r∈Z6 x∈2Z6. Thus, 3Z6and 2Z6are ideals of Z6.
SELF ASSISMENT EXERCISE 10
1a ≠ø, since 0 ∈1a
f, g∈1a⇒ (f – g) (a) = f(a) – g(a) = 0 ⇒ f - g∈1a.
f∈1a, g∈C[0, 1] ⇒(fg) (a) = f(a) g(a) \= 0.g(a) = 0⇒fg ∈1a.
∴1a is an ideal of C[0, 1].
SELF ASSISMENT EXERCISE 11
Ra is a subring of R (see Example 4).
Also for r∈R and xa ∈Ra, R(xa) = (rx)a ∈Ra.
∴Ra is an ideal of R.
SELF ASSISMENT EXERCISE 12
We know that <1> ⊆R. we need to show that R ⊆ <1>.
Now, for any r∈R, r = r.1 ∈<1>. Thus, R ⊆<1>.
∴R = <1>.
SELF ASSISMENT EXERCISE 13
3Z10={3x|x∈Z10}= {0,3,6,9,12,15,18,21,24,27} = {0,3,6,9,2,5,8,1,4,7}
= Z10.
5Z10 = {0,5}.
SELF ASSISMENT EXERCISE 14
Let the nil radical of Z8 be N. then 0∈N.
1∉N since 1n = 1 ≠ 0 for all n.
23 = 0 ⇒2∈N.
3n ≠0∀n. ∴3∉N.
Similarly, you can check that 4,6∈N and 5,7∉N.
∴N = {0,2,4,6}.
For any A ∈℘(X), An = A ∩A∩…. ∩A = A∀n.
Thus, An = ø iff A = ø. Thus, the nil radical of ℘(X) is { ø}.
SELF ASSISMENT EXERCISE 15
Firstly, 1 ≠ø since 0 ∈1.
Secondly, r, s ∈1 ⇒ ra = 0 = sa ⇒ (r – s)a = 0 ⇒ r – s ∈1.
Finally, r ∈1 and x ∈R ⇒ (rx)a = x(ra) = x0 = 0 ⇒ rx ∈1.
Thus, 1 is an ideal of R.
SELF ASSISMENT EXERCISE 16
a) For any a ∈I and b ∈J, ab ∈I and ab ∈J.
Thus, ab∈I ∩J. since I ∩J is an ideal, any finite sum of such elements will lso be in I∩J. Thus, IJ ⊆I ∩J.
Clearly, I∩J ⊆I and I∩J⊆J.
Also, I ⊆I + J, J ⊆ I + J is obvious.
b) Let A be an ideal of R containing I as well as J. Then certainly I + J ⊆ A. Thus, (b) is proved.
c) Let B be an ideal of R such that B ⊆ I and B ⊆J. Then certainly, B ⊆I∩J. Thus, (c) is proved.
d) We want to show that I ∩J ⊆ IJ.
Let x∈I and x ∈J.
Since I ⊆ ∈R = I +J, I = I + j, for some i∈I and j ∈J.
∴x = x,I = xi + xj = ix + xj ∈IJ Thus, I ∩J⊆IJ.
SELF ASSISMENT EXERCISE 17
1 + I is the identity of R/I.
SELF ASSISMENT EXERCISE 18
from Theorem 4, you know that I = R.
∴R/I = {0}.
SELF ASSISMENT EXERCISE 19
Let x + N ∈R/N be a nilpotent element.
Then (x + N)n = N for some positive integer n.
⇒ xn∈N for some positive integer n.
⇒ (nn)m = 0 for some positive integer m.
⇒xnm = 0 for some positive integer nm.
⇒ x ∈N
⇒ x + N = 0 + N, the zero element of R/N.
Thus, R/N has no non-zero nilpotent elements.
6.0 TUTOR MARKED ASSIGNMENT
1. Show that if R is a ring with identity, then R/I is a ring with identity for any ideal I of R
2. Let N be nil radical of R. Show that R/N has no non-zero nilpotent element
7.0 REFERENCES/FURTHER READINGS
Blacksell: Topics in Algebra
UNIT 3 RING HOMOMORPHISMS