## The nilradical of the quotient of a ring by its nilradical is trivial

Prove that if $R$ is a commutative ring and $\mathfrak{N}(R)$ is its nilradical, then $\mathfrak{N}(R/\mathfrak{N}(R)) = 0$.

Suppose $x + \mathfrak{N}(R) \in \mathfrak{N}(R/\mathfrak{N}(R))$. Then for some positive natural number $n$, we have $(x + \mathfrak{N}(R))^n = x^n + \mathfrak{N}(R) = \mathfrak{N}(R)$. Thus $x^n \in \mathfrak{N}(R)$. Then for some positive natural number $m$, we have $(x^n)^m = x^{nm} = 0$. Hence $x \in \mathfrak{N}(R)$.