Conjecture on a stronger form of Chinese Remainder Theorem (on ideals): $\prod_{i=1}^m A_i+\prod_{i=m+1}^n A_i\overset{?}{=}R$?

Question

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.

Answer 1

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 $I=\prod_{m+1\le i\le n} A_i$.

Answer 2

Let $R$ be a commutative ring and $A_1,\dots,A_n$ be pairwise coprime, i.e. $A_i+A_j = (1)$ if $i \neq j$. Then the two product ideals $\prod_{i=1}^{m} A_i$ and $\prod_{i=m+1}^{n} A_i$ are coprime for all $m =1,\dots,n$.

Let $x_{(i,j)}+y_{(i,j)}=1$ with $i \neq j$, $x_{(i,j)} \in A_i$ and $y_{(i,j)} \in A_j$. This is possible since $A_i$ and $A_j$ are coprime. Now consider the product $$ 1= \prod_{i=1,\dots,m}\prod_{j=m+1,\dots,n} (x_{(i,j)}+y_{(i,j)}). $$ Each summand is in $\prod_{i=1}^{m} A_i$ or in $\prod_{i=m+1}^{n} A_i$.