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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?

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