If $R$ is a commutative ring with identity and $I$ and $J$ are ideals of $R$ such that $R/I \times R/J$ is isomorphic to $R/(I\cap J)$ as $R$-modules, then $I + J = R$.
I know this is the converse of CRT and there are many posts on this. However, I can't find a post that provides a complete answer to this.
This is the solution I found.
However, I find it to be very complicated. What does $\otimes_R$ mean? What does tensor product mean? I have not learn tensor product.
I prefer a solution without it and using only concepts from elementary ring theory course.
Is there simpler answer to this?
No need to use Zorn's Lemma. The proof is quite elementary.
Lemma 1. If $A,B,C$ are $R$-modules such that $A \oplus B$ is cyclic, then every bilinear map $\beta : A \times B \to C$ is trivial.
Proof. Let $(u,v)$ be a generator of $A \oplus B$. Then $u$ is a generator of $A$ and $v$ is a generator of $B$. It suffices to prove $\beta(u,v)=0$. Choose some $r \in R$ such that $(u,0)=r(u,v)$. This means $u=ru$ and $rv=0$. Hence, $$\beta(u,v) = \beta(ru,v)=\beta(u,rv)=\beta(u,0)=0. ~~~\checkmark$$
Lemma 2. For any two ideals $I,J \subseteq R$, there is a surjective bilinear map $$R/I \times R/J \to R/(I+J).$$
Proof. Just check that $(a \bmod I,b \bmod J) \mapsto a \cdot b \bmod I+J$ is well-defined, bilinear and surjective. $\checkmark$
Corollary. If $R/I \oplus R/J$ is cyclic, then $I+J=R$.
Proof. By Lemma 2, we have a surjective bilinear map $R/I \times R/J \to R/(I+J)$. By Lemma 1, it is trivial. Hence, by surjectivity, $R/(I+J)$ is trivial. This means $I+J=R$. $\checkmark$
The proof above is motivated by the following more usual proof, which uses universal objects (exterior powers and tensor products):
If $R/I \oplus R/J$ is cyclic, then
$0=\Lambda^2(R/I \oplus R/J)\\ ~~\cong \underbrace{\Lambda^2(R/I)}_{=0} \otimes \Lambda^0(R/J) \,\oplus\, \Lambda^1(R/I) \otimes \Lambda^1(R/J) \,\oplus\, \Lambda^0(R/I) \otimes \underbrace{\Lambda^2(R/J)}_{=0}\\ ~~ \cong R/I \otimes R/J\\ ~~ \cong R/(I+J). ~~~ \checkmark$