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Thread: 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 $\displaystyle I$ is radical if and only if $\displaystyle R/I$ has no nilpotent elements.

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

    An ideal I is called radical if $\displaystyle \forall \ x \in R$ such that $\displaystyle x^n \in I$ (for some $\displaystyle n>0$), then $\displaystyle x \in I$ 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; Mar 26th 2011 at 09:53 AM. Reason: clarity
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    For your questions, use:

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

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


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