• Mar 21st 2011, 06:38 PM
Seda1
If I is an Ideal for communitive ring R, prove Rad(I) is and Ideal of R and that I is an ideal of Rad(I)

For the first, I looked at for r ∈ R, a ∈ Rad(I). Therefore a^n ∈ I for some positive integer n.

(ra)^n = r^n*a^n ∈ I since r^n ∈ R and since I is an Ideal.

Therefore since (ra)^n ∈ I, ra ∈ Rad(I). So Rad(I) is an Ideal of R.

But for the second, I let a ∈ I, b ∈ Rad(I), therefore b^n ∈ I, but I don't know how to prove ba ∈ I.

• Mar 21st 2011, 09:46 PM
tonio
Quote:

Originally Posted by Seda1
If I is an Ideal for communitive ring R, prove Rad(I) is and Ideal of R and that I is an ideal of Rad(I)

For the first, I looked at for r ∈ R, a ∈ Rad(I). Therefore a^n ∈ I for some positive integer n.

(ra)^n = r^n*a^n ∈ I since r^n ∈ R and since I is an Ideal.

Therefore since (ra)^n ∈ I, ra ∈ Rad(I). So Rad(I) is an Ideal of R.

But for the second, I let a ∈ I, b ∈ Rad(I), therefore b^n ∈ I, but I don't know how to prove ba ∈ I.

Why would you have to prove $ba\in I$? This is always true since $I$ is an ideal...
All you need to prove is $I\subset Rad(I)$ (why?)