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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?


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@blue Is there a definition for noncommutative rings? – Carol Lafetá Aug 13 '14 at 1:54
I think the best analogue for the nilradical in the noncommutative case is the lower nilradical, which is equal to the intersection of all prime ideals, and is equal to the set of all strongly nilpotent elements. – rschwieb Aug 13 '14 at 12:53
up vote 4 down vote accepted

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.)

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And also, considering $[\begin{smallmatrix}0&1\\ 0&0\end{smallmatrix}]+[\begin{smallmatrix}0&0\\ 1&0\end{smallmatrix}]$, they needn't be closed under addition, either. – rschwieb Aug 14 '14 at 11:56

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.

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You mean that $ab \notin N(J)$, right? Because, if $J$ is an ideal, and $a \in J$, then $ab \in J$ too, right? Or am I confused here? – Carol Lafetá Aug 13 '14 at 2:06
Yes, I mean that $ab\notin N(J)$. Sorry, it was just a typo. – Sasha Patotski Aug 13 '14 at 2:07

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