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

**joestevens** · March 26th 2011, 09:22 AM

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.

**Tinyboss** · March 26th 2011, 09:52 AM

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.

**FernandoRevilla** · March 26th 2011, 09:56 AM

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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