Question: Determine the nilradical $N := \sqrt{(0)}$ in $\Bbb Z/n\Bbb Z$ and in $(\Bbb Z/n\Bbb Z) /N$ .
My attempt: $\sqrt{(0)}=\{x+n\Bbb Z\in\Bbb Z/n\Bbb Z\ |\ \exists m\in\Bbb N\ x^m\in n\Bbb Z\}$ for the nilradical of $\Bbb Z/n\Bbb Z$
But the second case seems weird.
$$\sqrt{(0)}=\{(x+n\Bbb Z)+N\in (\Bbb Z/n\Bbb Z)/N \ |\ \exists k\in \Bbb N\ \text{s.t.}\ (x+n\Bbb Z)^k\in N\}\\=\{(x+n\Bbb Z)+N\in (\Bbb Z/n\Bbb Z)/N \ |\exists k\in \Bbb N\ \text{s.t.}\ \exists m\in\Bbb N\text{ s.t.} ((x+n\Bbb Z)^k)^m\in n\Bbb Z\}\\=\{(x+n\Bbb Z)+N\in (\Bbb Z/n\Bbb Z)/N \ |\exists l=k+m\in\Bbb N\text{ s.t.} (x+n\Bbb Z)^l\in n\Bbb Z\}\\=\sqrt{(0)}/N=\sqrt{(0)}$$
I don't know if this makes sense... I'm not sure if I correctly understand what $(\Bbb Z/n\Bbb Z)/N$ is.
For whatever ideal $I$ in $R\ ,\ $ $I=0+I\in R/I$ that's what quotienting is, hence my confusion
$\DeclareMathOperator\rad{rad}$ Let $\varrho:\Bbb Z\to\Bbb Z/n\Bbb Z$ be the canonical projection. Since pre-image commutes with radical, we have \begin{align} &\varrho^{-1}N=\sqrt{n\Bbb Z}=\operatorname{rad}(n)\Bbb Z,& &\operatorname{rad}(n):=\prod_{p\mid n}p \end{align} where the product runs over the prime divisors $p$ of $n$. Since $\varrho$ is a surjective ring homomorphism we have $$N=\operatorname{rad}(n)\Bbb Z/n\Bbb Z$$
Let $\psi:\Bbb Z/n\Bbb Z\to(\Bbb Z/n\Bbb Z)/N$ be the canonical projection and $N'$ be the nilradical of $(\Bbb Z/n\Bbb Z)/N$. Then $$\psi^{-1}N'=\sqrt N=N$$ hence $N'=\{0\}$.