# Intersection and isomorphism of two relatively prime ideals

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.

• This statement is one version of the Chinese Remainder Theorem; knowing that phrase will help you search for proofs. Does your ring $R$ have a unit element $1$? If so, choosing $x\in I$ and $y\in J$ such that $x+y=1$ will help you go a long way. For example, if $z\in I\cup J$, then $z=1z=xz+yz\in IJ$. May 9 '15 at 20:14

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 ?
• Well I think I see how to continue. $x e_1 \in I J$ (because $e_1 \in I$ and $x$ in both $I$ and $J$, therefore we consider it as an element of $J$, so therefore their product is in $IJ$). And similar, $x e_2 \in I J$ aswell. Therefore their sum $x e_1 + x e_2$ must be in $I J$ too, by definition of $I J$. Is this correct? May 10 '15 at 10:20