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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.2012-10-22
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    I mean, this is a little ridiculous. This user is just posting verbatim GRE math questions! Why do we continue to answer??2012-10-22
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    @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 : )2012-10-23
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    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!2012-10-23
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    @SteveD Thanks for your comment. It's helpful.2012-10-23
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    @navigetor23 I rolled back because of the change of notation not because of the title. I see no reason why we should go edit people's questions to use different notation. I disagree with it and I also think the edit should have been rejected. Especially if the answers are written using the notation used by the OP. If you want to teach a user what notation is being used _generally_ feel free to do so in a comment. Cheers2012-11-08
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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

1 Answers 1

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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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    I don't see why the third one is not true by simply using the "similar fashion".2012-11-07
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    Well I guess I can expand on my answer some time later today.2012-11-08