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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.

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    $\begingroup$ Why not use the map from the CRT, show the kernel is $I\cap J$, and conclude that the image is isomorphic to the quotient, using the isomorphism theorem? The “meat” of the CRT is not the map, it’s the fact that it is surjective when $I+J=R$. $\endgroup$
    – Arturo Magidin
    Commented Mar 14, 2021 at 19:39
  • $\begingroup$ What exactly do you mean by the map from CRT? You mean the natural isomorphism $\phi: R \to R/I \times R/J$, and if so what would this map resemble? $\endgroup$
    – John Miller
    Commented Mar 14, 2021 at 21:34
  • $\begingroup$ What do you, “resemble”? You send $r$ to $(r+I,r+J)$. That’s the map that the CRT tells you is an surjective when $I+J=R$. $\endgroup$
    – Arturo Magidin
    Commented Mar 14, 2021 at 21:45
  • $\begingroup$ Ah yeah sorry I didn't realise CRT applied to rings as well, thanks for the help I finished the proof. I will add it below. $\endgroup$
    – John Miller
    Commented Mar 14, 2021 at 21:48

1 Answer 1

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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 $$ R/(I\cap J) \cong S. $$

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