Let $I_{1}, I_{2}, \ldots, I_{n}$ be a finite collection of two-sided ideals of $R$ such that $I_{i}+I_{j}=R$ for all $i \neq j$. If $a_{1}, a_{2}, \ldots, a_{n}$ are any elements of $R$ prove that there exists $r \in R$ with $r \equiv a_{i} \bmod I_{i}$ for all $i .$ Deduce that $R /\left(\bigcap_{1}^{n} I_{i}\right) \cong \oplus \sum_{1}^{n} R / I_{i} .$ This is the Chinese Remainder Theorem.