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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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What have you tried? At least the first 2 are a matter of simply applying the definition. – Mariano Suárez-Alvarez Oct 22 '12 at 8:57
I mean, this is a little ridiculous. This user is just posting verbatim GRE math questions! Why do we continue to answer?? – user641 Oct 22 '12 at 18:19
@SteveD I'm a statistics major student, so I do not learn analysis, abstract algebra in class. I'm learning analysis and abstract algebra by myself in very short time and My professors also do not familiar with that topic. No one (offline)can help me. I really appreciate to your help : ) – John Hass Oct 23 '12 at 1:24
But learning is about trying things yourself. You are just posting exact GRE questions, with no motivation, nor anything you yourself have tried. At least show some effort! – user641 Oct 23 '12 at 11:44
@SteveD Thanks for your comment. It's helpful. – John Hass Oct 23 '12 at 11:46
up vote 2 down vote accepted

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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I don't see why the third one is not true by simply using the "similar fashion". – Goku Nov 7 '12 at 21:39
Well I guess I can expand on my answer some time later today. – Rudy the Reindeer Nov 8 '12 at 7:22

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