Suppose we let R be a commutative ring with identity, and let I be any ideal of R. And we define the RADICAL of I to be the set N(I) = {[tex]r \in R[/tex]: [tex]r^n \in I[/tex] for some positive integer n}.(adsbygoogle = window.adsbygoogle || []).push({});

I need the proof that:

- If I contains a unit of R, then show that I = R.

- N(N(I))=N(I)

An integral domain is a commutative, unital ring that contains no zero-divisors. So I'm guesing R must be an ID. If R is a ring with identity 1_{R}, then [tex]a \in R[/tex] is a unit if ab = ba = 1_{R}for some b in R, and b is the inverse of a. Now if we suppose I contains a, how do we show that I=R?

For the second proof we can assume N(I) is an ideal of R.

Any help or suggestions are appreciated.

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# Commutative Ring

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