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For a commutative ring $R$ and ideal $A$, let $N(A)=\{x \in R\mid $ there exists a nonnegative integer $n$ such that $x^n \in A\}$. For which of the following $R$ and $A$ is it true that $N(A)=A$ ?

I. $R=\Bbb Z,\ A=(2)$

II. $R=\Bbb Z[x],\ A=(x^2+2)$

III. $R=\Bbb Z/27\Bbb Z,\ A=(18+27\Bbb Z)$

Potato
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user101204
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1 Answers1

1

I. True

II. True : Since $(x^2+2)$ is an irreducible so $f(x)^n = p(x)(x^2+2)\ (i.e., \ f(x)\in N(A))$ implies $f(x)=q(x)(x^2+2)\ (i.e., \ f(x)\in A)$ . The converse is clear.

III Not true : $A = \{ 18, 9, 0\}$, and $3\in N(A)$ since $3^2=9$.

HK Lee
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