 Solution

For pairwise comaximal ideals $I_1,\dots ,I_r$ of $R$, reduction gives the Chinese remainder theorem for ideals:Its kernel is ${\bigcap }_i{I}_i$. For two comaximal ideals, write $1=u+v$ with $u\in I$, $v\in J$. If $x\in I\cap J$, then $x=xu+xv\in IJ$, showing $I\cap J=IJ$. Induction gives the product formula: a product of ideals individually comaximal with $J$ is still comaximal with $J$, by multiplying identities modulo $J$.

Put $J_i={\prod }_{j\ne i}{I}_j$. Since $I_i+J_i=R$, choose $e_i\in J_i$ with $e_i\equiv 1(\mathrm{mod}{I}_i)$. The element ${\sum }_i{a}_i{e}_i$ has any prescribed residues $a_i(\mathrm{mod}{I}_i)$, proving surjectivity. Its residue class modulo the product is unique because we already identified the kernel. In a Dedekind domain, powers of distinct nonzero prime ideals are pairwise comaximal, giving the prime-power form. The theorem requires comaximality; it is not a statement about arbitrary lists of ideals.