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! 
 A: You didn't do anything wrong.
The definition given for the radical of an ideal does not actually yield an ideal for noncommutative rings, because nilpotent elements needn't be closed under multiplication. There are a couple of ways to fix this, at least for the nilradical (radical of the zero ideal), both achieved by replacing the usual definition with an equivalent definition that does yield an ideal in noncommutative rings. One is the lower nilradical is the intersection of all prime ideals, and the upper nilradical is the ideal generated by all nil ideals. (I just got this from Wikipedia, really.)
A: 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.
