Chinese remainder theorem for rings

Kernel. $r\in\ker\varphi$ iff $r\in I$ and $r\in J$, i.e. $r\in I\cap J$.

Surjectivity. Let $(a+I,b+J)\in R/I\times R/J$. Since $I+J=R$, there exist $i\in I$, $j\in J$ with $i+j=1$. Set $r=aj+bi$. Then $r\equiv a(1-i)+bi\equiv a\pmod I$ (since $aj\equiv a\pmod I$ because $i\in I$). Similarly, $r\equiv b\pmod J$. So $\varphi(r)=(a+I,b+J)$.

Conclusion. By the first isomorphism theorem, $R/(I\cap J)\cong R/I\times R/J$.

Application. $\mathbb{Z}/mn\mathbb{Z}\cong\mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/n\mathbb{Z}$ for $\gcd(m,n)=1$.