Theorem 1 (Chinese Remainder Theorem)   Let $R$ be a ring and $I_1,I_2, ..., I_n$ pairwise comaximal ideals such that $R =I_j +R^2 $ for all $j$ . The homomorphism:
is surjective and $ker f = I_1 \cap I_2\cap \cdots \cap I_n$ .

Proof. Clearly $f$ is a homomorphism with kernel $I_1\cap I_2\cap \cdots \cap I_n$ . It remains to show the surjectivity.
We have:
Moreover,
Continuing, we obtain that $R = I_1 + \bigcap_{j\ne 1}I_j$ . We show similarly that:\begin{equation*} R = I_2 + \bigcap_{j\ne 2}I_j = I_3 +\bigcap_{j\ne 3}I_j = \cdots = I_n + \bigcap_{j\ne n} I_j.\end{equation*}Given elements $a_1, a_2, ..., a_n$ , we can find $x_j\in I_j$ and $y_j\in \bigcap_{j\ne k} I_k$ such that $a_j = x_j + y_j$ .
Take $a:=\sum_{i=1}^n x_i = a_j\pmod{I_j}$ .
Hence\begin{equation*} f(a) = (a_1 +I_1, a_2+I_2, ..., a_n+I_n),\end{equation*}and we conclude that $f$ is surjective as required.