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

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.30

Solution: 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 $x \in \mathfrak{N}(R)$.