If $R/I \times R/J$ is isomorphic to $R/(I\cap J)$ as $ R $-modules, then $I + J = R$. 
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?
 A: 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
$0=\Lambda^2(R/I \oplus R/J)\\ ~~\cong \underbrace{\Lambda^2(R/I)}_{=0} \otimes \Lambda^0(R/J) \,\oplus\, \Lambda^1(R/I) \otimes \Lambda^1(R/J) \,\oplus\, \Lambda^0(R/I) \otimes \underbrace{\Lambda^2(R/J)}_{=0}\\ ~~ \cong R/I \otimes R/J\\ ~~ \cong R/(I+J). ~~~ \checkmark$
