Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 1E / i / Solution

Let $e^2=e$ with $e\ne 0,1$. Then $e(1-e)=0$, and the ideals $eR$ and $(1-e)R$ are nonzero rings with identities $e$ and $1-e$. DefineThis is a ring homomorphism. Its inverse is $(x,y)\mapsto x+y$, because $ex=x$, $(1-e)y=y$, and the cross terms vanish. Hence $R\cong eR\times (1-e)R$, proving (ii).

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