Is it true that:
Let $R$ be a ring and let $I, J$ be ideals of $R$. Then for some subring of $S \subseteq R/I \times R/J$ we have
$$
R/(I \cap J) \cong S.
$$
This would be equivalent to the Chinese Remainder Theorem for non-coprime ideals. I've tried constructing a function $\phi: R/(I \cap J) \to R/I \times R/J$ and showing its injective.
Best Answer
Proof
We construct the natural homomorphism $ \varphi: R \to R/I \times R/J $ such that $$ \varphi(x) = (x+I,x+J). $$ We know that $$ \ker \varphi = \{x\in R: x\in I, x\in J\} = I \cap J. $$ Since the $\text{im } \varphi$ is a subset $S$ of $R/I \times R/J$ so by the first isomorphism theorem $$ R/(I\cap J) \cong S. $$