## 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.