Ring homomorphisms preserve nilpotency

Let \varphi : R \rightarrow S be a homomorphism of rings. Prove that if x \in R is nilpotent, then \varphi(x) \in S is nilpotent.


Suppose x^n = 0. Then \varphi(x)^n = \varphi(x^n) = \varphi(0) = 0, so that \varphi(x) is nilpotent.

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