Chinese Remainder Theorem for commutative rings with identity
Let $R$ be a commutative ring with identity. If $I, J$ are ideals of $R$ satisfying $I+J=R$, then there is an isomorphism of rings: $$R/(I\cap J) \cong R/I \times R/J.$$
I am interested in the converse of this. I saw the following two cases:
Converse V1
If we have positive integers $m, n$ with $(m,n)\neq 1$, then $$ \mathbb{Z}/mn\mathbb{Z} \not\cong \mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/n\mathbb{Z}.$$
This one is easy if we consider characteristics.
Converse V2
Let $F$ be a field. If we have polynomials $f, g \in F[x]$ with $(f)+(g)\neq F[x]$, then $$ F[x]/(f(x)g(x))\not\cong F[x]/(f(x))\times F[x]/(g(x)).$$
This one is answered here. The idea is counting number of ideals of both sides. It is easy to see that if we have factorization of ideals into prime ideals (Dedekind Domain), the same idea applies.
However, we have this example:
Example If $R=\prod_{i=1}^{\infty} \mathbb{Z}$ and $I=J=(0)$, then $I+J\neq R$ and $$ R/(I\cap J) \cong \prod_{i=1}^{\infty} \mathbb{Z} \cong \prod_{i=1}^{\infty} \mathbb{Z}\times \prod_{i=1}^{\infty} \mathbb{Z} \cong R/I\times R/J.$$
Thus, the converse of CRT does not hold in general for "commutative ring with identity". We have seen, however, the converse of CRT holds for "Dedekind Domain". My question is
Quetion Do we have a commutative ring with identity which is not a "Dedekind Domain" such that the converse of CRT holds?