Which one of the following ideals is radical?

Question

For a commutative ring $A$ and an ideal $I$, $N(I)=\{x\in A\mid x^n\in I \ \mbox{for some integer}\ n\}$. Then which of these satisfy $N(I)=I$:

1. $A=\mathbb{Z}, I=(2)$,
2. $A=\mathbb{Z}[x], I=(x^2+2)$,
3. $A=\mathbb{Z}_{27}, I=([18])$.

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Obviously, the first one is true, but I don't know how to deal with 2, 3.

Answer 1

1. $A = \mathbb Z$, $I = (2)$: Since $2$ is prime, then $(2)$ is a prime ideal which means $N(I) = I$ since the radical of a prime ideal is itself.
2. $A = \mathbb Z[x]$, $I = (x^2 + 2)$: Since $\mathbb Z$ is a UFD, then $\mathbb Z[x]$ is a UFD in which case irreducibles are prime. Notice that $x^2 + 2$ is irreducible by Eisenstein's criterion, so it's prime, which means $(x^2 + 2)$ is prime, hence $N(I) = I$ since the radical of a prime ideal is itself.
3. $A = \mathbb Z_{27}$, $I = (\overline{18})$: Notice that $\overline{18} = \overline {9}$ as $9 = 27 - 18$ which means $I = (\overline{18}) = (\overline{9})$. Notice that $\overline 3^2 = \overline 9$ and hence $N(I) = (\overline 3)$.