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For a commutative ring $\mathbb{M}$ and ideal $\mathbb{A}$, let $N(A)$={x in M|there exists a non-negative integer $ n $ such that $x^{n}$ in $\mathbb{A}$}. Which of following is true for $N(A)=A$?

I. $M=\mathbb Z, A=(2)$

II. $M= \mathbb Z[x]$, $A=(x^{2}+2)$

III. $M= \mathbb Z/27 \mathbb Z, A=(18+27 \mathbb Z)$

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    @SteveD I was wondering how you know this is the actual GRE question... (I tend to believe your observation though, since the OP asked a lot of at least _GRE-type_ questions.)2013-08-29

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I: Yes. We always have $A \subset N(A)$. For the other direction, let $x \in N(A)$ that is, $x^n \in (2) = A$ for some $n$. Assume $x \notin A$. Then $x$ does not have a factor of $2^k$. But then $x^n$ does not have a factor of $2^j$. Which would be a contradiction hence $x \in A$.

Hope this helps. Now try to answer the other cases in a similar fashion.

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    Well I g$u$ess I can expand on my answer some time later today.2012-11-08