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).

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