= Solution
{c}
The <Chinese remainder theorem> for two comaximal two-sided <ideals> $I,J$ of a ring states that the natural <ring homomorphism> induces
$$
\boxed{R/(I\cap J)\cong R/I\times R/J\qquad(I+J=R).}
$$
Indeed, the map $r\mapsto(r+I,r+J)$ has <kernel of a ring homomorphism> $I\cap J$. Choose $i\in I$, $j\in J$ with $i+j=1$. Given target representatives $a,b\in R$, the element $aj+bi$ is congruent to $a$ modulo $I$ and to $b$ modulo $J$, proving <surjectivity>. The <first isomorphism theorem for rings> gives the result. For a commutative ring, $I\cap J=IJ$: if $x\in I\cap J$, then $x=xi+xj\in IJ$, while the reverse inclusion follows from the ideal properties.
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