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Math Help - Prove that I is radical if and only if R/I has no nilpotent elements.

  1. #1
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    Prove that I is radical if and only if R/I has no nilpotent elements.

    Prove that I is radical if and only if R/I has no nilpotent elements.

    First: Should the problem state "Prove that I is radical if and only if R/I has no nonzero nilpotent elements."?

    An ideal I is called radical if \forall \ x \in R such that x^n \in I (for some n>0), then x \in I also.

    Note: Rings are defined to be associative and commutative with unity.
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  2. #2
    Senior Member Tinyboss's Avatar
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    r+I=0 in R/I if and only if r is an element of I. Think nilpotent in the first case and radical in the second.
    Last edited by Tinyboss; March 26th 2011 at 10:53 AM. Reason: clarity
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    For your questions, use:

    (x+I)^n=0+I \Leftrightarrow x^n+I =0+I\Leftrightarrow x^n-0\in I\Leftrightarrow x^n\in I

    So, x+I\in R/I is nilpotent iff there exists n>0 such that x^n\in I


    Edited: Sorry, I didn't see the previous post.
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