Past exam of the algebra and analysis II for mathematics and philosophy course of the University of Oxford / 2013 / 3 / b / Solution

Define $\Phi :R\to R/I\times R/J$ by $\Phi (r)=(r+I,r+J)$. It is a ring homomorphism withSince $I+J=R$, choose $i\in I,j\in J$ such that $i+j=1$. For arbitrary $a,b\in R$, the element $r=aj+bi$ satisfies $r-a=i(b-a)\in I$ and $r-b=j(a-b)\in J$. Hence $\Phi (r)=(a+I,b+J)$, proving surjectivity. The first isomorphism theorem now givesThis is the two-ideal Chinese remainder theorem for ideals. The hypothesis is comaximality of the two ideals; no assertion that every pair of ideals has this property is needed.