Let $R$ be a commutative ring. Using the definition that two ideals $I, J \subseteq R$ are relatively prime if $I + J = R$.
I want to show that for two relatively prime ideals $I, J \subseteq R$, it holds true that:
$I \cap J = I J$, and that the transformation: $f: R/I J \to R/I \oplus R/J$ is an isomorphism.
Thanks in advance. I discovered another thread (Intersection of ideals generated by two relatively prime elements) where the first statement was dealt with in case that R is a principal ideal domain and the two ideals therefore generated by only one element. I'm not sure how to prove it generally, with R just being any commutative ring and the ideals not necessarily being principal.
First, we need to show that your homomorphism is surjective. Since these two ideals are relatively prime, you can find $e_{1} \in I$ and $e_{2} \in J$ such that $e_{1}+e_{2}=1$ $f(e_{1})=(e_{1}+I,e_{2}+J)=(e_{1}+I,1-e_{2}+J)=(\overline{0},\overline{1})$. You can do the similar thing to show $f(e_{2})=(\overline{1},\overline{0})$. Can you deduce surjectivity then? The kernel is $I \cap J$. We need to show $I \cap J =IJ$. Notice that $IJ \subset I \cap J$ is clear. Now let $x \in I \cap J$. Write $x=x \times 1=x(e_{1}+e_{2})=xe_{1}+xe_{2}$. Could you see what to do next ?