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February 6th, 2012, 02:58 AM  #1 
Newbie Joined: Feb 2012 Posts: 6 Thanks: 0  Radical of an Ideal
I'm having trouble understanding what it means to be a radical of an ideal. Does the rad(J), where J is an ideal, contain J itself? I've defined rad(J)={x in A: a^n is in J for some n in Z}. In particular, I can't seem to show for any ideal J, rad(J) is an ideal of A.

February 6th, 2012, 03:12 AM  #2 
Newbie Joined: Feb 2012 Posts: 6 Thanks: 0  Re: Radical of an Ideal
Sorry for not including in the previous post, but would I want to start off with something like this: Assume J is an ideal of A. Let a^n and b^m be elements in J. Then would I want to show (a+b)^(n+m) is an element in J, which would imply J is in rad(J)? 
February 15th, 2012, 04:04 AM  #3 
Newbie Joined: Feb 2012 Posts: 17 Thanks: 0  Re: Radical of an Ideal
I'm not sure where you're getting your definition of 'radical' if it's causing this much confusion. Atiyah & MacDonald define the radical, for an ideal in a ring : It should be obvious that ; just let The proof idea you wrote here is the way you're supposed to show that is closed under addition. And yes, you would like to show that. Once you have that, the fact that has the 'ideal' property (i.e. ) is not very difficult at all (look back at the definition of and use the fact that itself is an ideal) 

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