Notation: By two ideals $A,B$ in $R$ are comaximal we mean $A+B=R$.
Assume $R$ is a commutative ring with $1$, and
$\{A_i\}_{1\le i\le n}$ are pairwise-comaximal ideals in $R$.
The Chinese Remainder Theorem states $A_1$ is coprime with $\prod\limits_{2\le i\le n} A_i$.
I wonder if for all $m\in \{1,\cdots, n-1\}$,
$\prod\limits_{1\le i\le m} A_i$ and $\prod\limits_{m+1\le i\le n} A_i$ are comaximal?
The proof for the Chinese Remainder Theorem seems failing to be rewritten to handle such case. So I believe some counterexample exists, but I can not find any.
Best Answer
Maybe you can try the following lemma : We can Prove that if I is an ideal that is comaximal with the ideals $A_{j}$ for all $j \in \{1,....n\}$ then I and $\prod_{j} A_j$ are comaximal.
Now back to your question. Let $m \in \{1,....,n\}$.
For all $j \in \{1,...,m\}$ : $A_j$ is comaximal with $\prod_{m+1\le i\le n} A_i$. I think you can apply the lemma above with $I=\prod_{m+1\le i\le n} A_i$.