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Let $R$ be a ring and $I$ and $J$ be two sided ideals, $I+J=R$. Are $R/I\cap J$ and $I/I\cap J \bigoplus J/I\cap J$ isomorphic?

My guess is yes, because internal and external direct product of two rings are isomorphism. Is my observation right?

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  • $\begingroup$ This is trivial.you can prove this as inner direct sum.right? $\endgroup$ – Sky Oct 17 '17 at 5:54
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First note that $I/I\cap J, J/I\cap J$ are ideals in $R/I \cap J$ therefore $$\phi : I/I\cap J \bigoplus J/I\cap J \to R/I \cap J \\ \phi (x+ I \cap J, y+I \cap J)= x+y +I \cap J$$ is well defined, and it is easy to see it is a homeomorphism.

Since $R=I+J$ you get immediately that $\phi$ is onto.

For 1-1, let us check the kernel. If $\phi (x+ I \cap J, y+I \cap J)=0 +I \cap J$ then $x+y \in I \cap J$. Since $x \in I$ you get that $y=(x+y)-x \in I$ and hence $y \in I \cap J$. Same way $x \in I \cap J$.

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It is true, but I'm not clear on your observation.

Maybe in your observation you meant this: "I see there are two ideals of $R/I\cap J$, namely $I/I\cap J$ and $J/I\cap J$, that have trivial intersection, and add up to $R/I\cap J$, therefore their sum is direct." Not only does that say they are isomorphic, it says the two sets are equal.

You could also just apply elementary ring isomorphism theorems.

The Chinese remainder theorem it is isomorphic to $R/I\times R/J$, and the second isomorphism theorem shows that each of these is isomorphic to what you described.

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