My Math Forum Radical of an Ideal

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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 $\alpha$ in a ring $R$: $r(\alpha)= \{ x \in R : x^n \in \alpha \mbox{ for some } n > 0 \}=$ It should be obvious that $\alpha \subset r(\alpha)$; just let $n= 1$ The proof idea you wrote here is the way you're supposed to show that $r(\alpha)$ is closed under addition. And yes, you would like to show that. Once you have that, the fact that $r(\alpha)$ has the 'ideal' property (i.e. $ab \in r(\alpha) \, \forall \, a \in \alpha, \b \in R$ ) is not very difficult at all (look back at the definition of $r(\alpha)$ and use the fact that $\alpha$ itself is an ideal)

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