Page 169 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Theorem 2: If I and J are ideals of a ring R, then
(a) IJ,

(b) I + J = { a + b | a  I and b  J }, and
(c) IJ = { x  R | x is a finite sum a b + ... + a b , where ai 1 and bi  J } are ideals of R.
1 1
m m
Proof: (a) From Theorem 2 you know that I  J is a subring of R. Now, if a  l  J, then a  I and
a  J. Therefore, ax  I and a  J for all x in R. So ax  I  J for all a  I  J and x  R. Thus, I  J
is an ideal of R.

(b) Firstly, 0 = 0 + 0  l + J  I + J = f.
Secondly, if x, y  1 + J, then x = a + b and y = a + b for some a , a ,  I and b , b  J.
2
1
1
2
1
2
2
1
So x – y = (a + b ) – (a + b ) = (a – a ) (b – b )  I + J.
1
2
1
2
1
2
1
2
Finally, let x  I + J and r  R. Then x = a + b for some a  I and b  J. Now
xr = (a + b)r = ar +br  I + J, as a  I implies ar  I and b  J implies br  J for all r  R.
Thus, I + J is an ideal of R.
(c) Firstly, IJ  , since I   and J  .
Next, let x, y  IJ. Then x = a b + ... + a b and
1 1 m m
y = a’ b’ + ... + a’ b’ for some a , ..., a, a’ ,.. ., a’,  I and b ,.., b , b’ ,...., , b’  J.
1
n
l
1
n
i
n
1
m
1
 x – y = (a b + ... + a b ) – (a’ b’ + ... + a’ b’ )
1 1
m m
1
n
1
n
= a b + ... + a b + (– a’ )b’ + ... + (– a’ )b’ n
1
1
1 1
m m
n
which is a finite sum of elements of the form ab with a  I and b  J.
So, x – y  IJ.
Finally, let x  IJ say x = a b + ... + a b with a,  1 and b, E J. Then, for any r E R
1 1
n n
xr = (a b + ... + a b )r = a (b r ) ... + a (b r),
n
n
1
n n
1 1
1
which is a finite sum of elements of the form ab with a  1 and b  J.
(Note that b  J  b r  J for all r in R.)
i
i
Thus, IJ is an ideal of R.
Over here, we would like to remark that if we define IJ = { ab | a E I, b  J }, then IJ need not even
be a subring, leave alone being an ideal. This is because if x, y E IJ, then with this definition of IJ
it is not necessary that x – y  IJ.
Let us now look at the relationship between the ideals obtained. Let us first look at the following
particular situation:
R = Z, I = 22 and J = 10Z. Then I  J = J, since J  I. Also, any element of I + J is ot the form x = 2n
+ 10m, where n, m  2. Thus, x = 2(n + 5m)  2Z. On the other hand, 2Z = f  I + J. Thus , I + J =
< 2 , 1 0 > = < 2 >.
Similarly, you can see that IJ = < 20 >.
Note that IJ  IJ  I  I + J.
In fact, these inclusions are true for any I and J. We show the relationship in figure 16.1.
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