= Solution
Define $\Phi:R\to R/I\times R/J$ by $\Phi(r)=(r+I,r+J)$. It is a <ring homomorphism> with
$$
\ker\Phi=I\cap J.
$$
Since $I+J=R$, 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 gives
$$
\boxed{R/(I\cap J)\cong R/I\times R/J.}
$$
This 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.
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