Let $\displaystyle R$ be a commutative ring and $\displaystyle N$ be the ideal of $\displaystyle R$. Show that $\displaystyle \sqrt{N} = \{a\in R | a^n \in N \text{ for some } n\in \mathbb{Z^+}\}$ is also the ideal of $\displaystyle R$.
Let $\displaystyle a, b \in \sqrt{N}$. Then, $\displaystyle a^i \in N$ and $\displaystyle b^j \in N$ for some $\displaystyle i, j \in \mathbb{Z^+}$.
Basically you need to argue the followings in order to show that $\displaystyle \sqrt{N}$ is indeed an ideal of R.
1. Since $\displaystyle 0 \in N$, it follows immediately that $\displaystyle 0 \in \sqrt{N}$.
Since $\displaystyle a^i \in N$, we see that $\displaystyle (-a)^i = (-1)^i(a)^i \in N$. Thus $\displaystyle -a \in \sqrt{N}$.
2. $\displaystyle a+b \in \sqrt{N}$.
Since $\displaystyle a^i \in N$ and $\displaystyle b^j \in N$ for some $\displaystyle i, j \in \mathbb{Z^+}$, we see that $\displaystyle (a+b)^{i+j} \in N$ (Use a binomial exapansion ).
3. $\displaystyle ab \in \sqrt{N}$.
Argue that $\displaystyle (ab)^{ij} = (a^i)^j(b^j)^i \in N$.
4. $\displaystyle ra \in \sqrt{N}$ or $\displaystyle ar \in \sqrt{N}$ for any $\displaystyle r \in R$.
Argue that $\displaystyle (ra)^{i} = (r^i)(a^i) \in N$ or $\displaystyle (ar)^{i} = (a^i)(r^i) \in N$. Since R is a commutative ring (its ideals are two-sided), you need to show either of them.