Originally Posted by

**demode** Let $\displaystyle R$ be a commutative ring with identity, and let $\displaystyle I$ be any ideal of $\displaystyle R$. Define the radical __RADICAL__ of I be the set $\displaystyle N(I)=\{ r \in R: r^n \in I \}$ for some positive integer $\displaystyle n$.

Prove that $\displaystyle N(I)$ is an ideal of $\displaystyle R$.

__My Attempt:__

I think I should use the "ideal test"; that is to show that (i) $\displaystyle a-b \in N(I)$ whenever $\displaystyle a,b \in N(I)$, and (ii) show that $\displaystyle ar$ and $\displaystyle ra$ are in $\displaystyle N(I)$ whenever $\displaystyle a \in N(I)$ and $\displaystyle r \in R$.

(i) Let $\displaystyle a,b \in N(I)$, then that means $\displaystyle a,b \in R$ such that $\displaystyle a^n, a^m \in I$. Since $\displaystyle I$ is an ideal of $\displaystyle R$, $\displaystyle a^n-b^m \in I$, and I *think* this implies that $\displaystyle a-b \in N(I)$.

But first I think we must show that $\displaystyle a-b \in R$. How do we show this?

(ii) Since I is an ideal of R we know that for any $\displaystyle r \in R$ and $\displaystyle a \in N(I)$, $\displaystyle ar$ and $\displaystyle ra$ are both in $\displaystyle N(I)$. Is this correct?