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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?

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  • $\begingroup$ How does the counterexample fit with this theorem? $\endgroup$ – user153312 Dec 8 '15 at 0:13
  • $\begingroup$ The theorem is for the natural injective homomorphism, but I am asking for just isomorphism. $\endgroup$ – Sungjin Kim Dec 15 '15 at 2:31
  • $\begingroup$ If you are still interested in this question , then I might mention that If $I,J$ are ideals of a commutative ring $R$ with unity and $f : R/{I \cap J} \to R/I \times R/J$ is the natural injective ring homomorphism given by $f(x + I \cap J)=(x+I ,x+J) , \forall x \in R$ , then $f$ is surjective if and only if $I+J=R$ . This is Proposition 1.10 of Chapter 1 of Atiyah , Macdonald's Commutative Algebra book . It is easy to prove , you should also try it yourself $\endgroup$ – user Aug 26 '17 at 14:48
  • $\begingroup$ It is the converse of CRT, meaning that the injective homomorphism is not necessarily assumed to be the natural one. $\endgroup$ – Sungjin Kim Aug 26 '17 at 23:17
  • $\begingroup$ yeah I understand that ... it's a good question $\endgroup$ – user Aug 27 '17 at 1:51

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