Generalization of Chinese Remainder Theorem

Question

Q: With ring $R$, if $I, J \subseteq R$ are ideals such that $I+J=R$, then the map $R/(I \cap J) \to R/I \times R/J$ given by $a + (I \cap J) \mapsto (a+I, a+J)$ is an isomorphism, broadly generalizing the Chinese Remainder Theorem.

Can someone help me get started on this one?

Answer 1

Consider the canonical maps $\pi_{I}:R\rightarrow R/I$ and $\pi_{I}:R\rightarrow R/J$ and form the morphism $\varphi(r)=(\pi_{I}(r),\pi_{J}(r))$.

As Censi LI explained in his answer, because $I+J=R$ you can find $i,j$ such as $i+j=1$ which then gives $\varphi(i)=(0,1)$ and $\varphi(j)=(1,0)$, so that $\varphi$ is surjective.

The kernel of $\varphi$ is precisely $I\cap J$ so $\varphi$ factors, by the first isomorphism theorem, into an bijective morphism $\tilde{\varphi}:R/(I\cap J) \rightarrow R/I \times R/J$ which is precisely the one in your question.

The crucial point is the one in Censi LI's answer but I felt that showing that your morphism can be though of as a factored morphism was important.

Answer 2

Hint: Denote the canonical map $R/(I \cap J) \to R/I \times R/J$ by $\varphi$, since $I+J=R$, we can find $i\in I$ and $j\in J$ such that $i+j=1_R$, then $\varphi(\bar i)=(0,1),\ \varphi(\bar j)=(1,0)$.