Nil radical of an ideal on a commutative ring This is a problem of an exercise list:

Let $J$ be an ideal of a commutative ring A. Show that $N(N(J))=N(J)$,
  where $N(J)=\{a \in A; a^n \in J$ for some $n \in \mathbb{N}\}$.

What I did:
$N(J)=\{a \in A; a^n \in J$ for some $n \in \mathbb{N}\}$
$N(N(J))=\{a \in A; a^k \in N(J)$ for some $k \in \mathbb{N}\}$
If $a \in N(J)$, then $a^1 \in N(J)$ $\Rightarrow$ $a \in N(N(J)) \Rightarrow N(J) \subseteq N(N(J))$.
On the other hand, if $a \in N(N(J))$, then $a^k \in N(J) \Rightarrow (a^k)^n \in J \Rightarrow a^{kn} \in J \Rightarrow a \in N(J) \Rightarrow N(N(J)) \subseteq N(J)$, for some $k,n\in\mathbb{N}$.
The problem is: I didn't use the hypothesis that A is a commutative ring. What did I do wrong?
Thanks! 

You did everything correctly. And I think the definition of nilradical does make sense (as a set) even in the noncommutative sense (you should be careful which ideals you are considering, left, right or two-sided). I guess the problem will be that $N(J)$ will no longer be an ideal. Indeed, take a (non-commutative) ring generated by two letters $a,b$ and with relation $a^2=0$. Then take two-sided ideal $J=(a)$, generated by $a$. Then $a\in N(J)$, but $ab\notin N(J)$ since $(ab)^n=abababa...b$ is not zero for any $n$. So $N(J)$ is not an ideal.
