# Proving the surjective property in Chinese Remainder Theorem

If $$\textrm{R}$$ is a commutative ring and $$\left\{\textrm{I}_i\right\}_{i=1}^n$$ are proper ideals of $$\textrm{R}$$ with $$\textrm{I}_i+\textrm{I}_j = \textrm{R}$$ for all $$1 \leq i \neq j \leq n$$, then the map $$\phi$$ from $$\textrm{R}$$ to $$\displaystyle\frac{\textrm{R}}{\textrm{I}_1}\oplus\displaystyle\frac{\textrm{R}}{\textrm{I}_2}\oplus\cdots\oplus \displaystyle\frac{\textrm{R}}{\textrm{I}_n}$$ given by

\begin{align*} \phi(r) = (r+\textrm{I}_1, r+\textrm{I}_2,\ldots,r+\textrm{I}_n) \end{align*} is an onto ring homomorphism with

\begin{align*} Ker(\phi) = \bigcap\limits^n_{i=1}\textrm{I}_i \end{align*}

I have proven that $$\phi$$ is a ring homomorphism with kernel as given above. But I have proven that the map $$\phi$$ is onto, only when the ring $$\textrm{R}$$ has unity. What about the case when $$\textrm{R}$$ does not have a unity ? Would the theorem still be valid ? I have checked that this would be true if $$n=2$$. But for any $$n\geq 3$$, the proof become cumbersome. Most book that I've come across mentions that the ring in the Chinese Remainder Theorem should be a commutative ring with unity. If the theorem doesn't hold for $$n \geq 3$$ and a commutative ring with no unity, is their any counter-example?