Page 168 - DMTH403_ABSTRACT_ALGEBRA

Unit 16: Ideals

When n = 1, the ideal we get is called a principal ideal. Thus, if a  R, then Ra = < a > is a principal Notes
ideal of R. In the next unit you will be using principal ideals quite a lot.

Tasks 1. Let R be a ring with identity. Show that < 1 > = R.

2. Find the principal ideals of Z generated by 3 and 5 .
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Definition: An element a of a ring R is called nilpotent if there exists a positive integer n such
that a” = 0.

2 2
0
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For example, 3 and 6 are nilpotent elements of Z , since 3   and 6  36  0. Also, in any
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ring R, 0 is a nilpotent element.
Now consider the following example.
Example: Let R be a ring. Show that the set of nilpotent elements of R is an ideal of R.
This ideal is called the nil radical of R.
Solution: Let N = { a  R | a = 0 for some positive integer n }. Then 0 EN.
n
Also, if a, b  N, then a = 0 and b = 0 for some positive integers m and n.
m
n
m n

r





Now, (a b) m n   m n C a ( b) m n r
r
r 0

For each r = 0, 1, ....., m + n, either r  n or m + n – r  m, and hence, either a = 0 or b m+n-r = 0. Thus,
r
the term a b m+n-r = 0. S0 (a – b) m+n = 0.
r
Thus, a – b  N whenever a, b  N.
Finally, if a  N, a” = 0 for some positive integer n, and hence, for any
r  R, (ar) = a r = 0, i.e., ar  N.
n
n n
So, N is an ideal of R.
Let us see what the nil radicals of some familiar rings are. For the rings Z, Q, R or C, N = {0}, since
the power of any non-zero element of these rings is non-zero.
For Z , N = {0,2}.
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Theorem 1: Let R be a ring with identity 1. If I is an ideal of R and I E I, then I = R.
Proof: We know that I  R. We want to prove that R  I. Let r E R. Since 1 E I and I is an ideal of
R, r = r . l  I. So, R  I. Hence I = R.
Using this result we can immediately say that Z is not an ideal of Q. Does this also tell us whether
Q is an ideal of R or not’? Certainly Since 1  Q and Q  R, Q can’t be an ideal of R.

Now let us shift our attention to the algebra of ideals. In the previous section we proved that the
intersection of subrings is a subring. We will now show that the intersection of ideals is an ideal.
We will also show that the sum of ideals is an ideal and a suitably defined product of ideals is an
ideal.

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