Here is the definition of an internal direct sum of modules:
An R-Module $M$ is the internal direct sum of submodules $M_1, M_2$ if:
$a)$ $M=M_1+M_2$
$b)$ $M_1 \cap M_2 = \{0\}$
I am trying to prove using this definition that $R/I\cong J_{1}/I \oplus J_{2}/I$ iff $R=J_1+J_2$ and $J_1 \cap J_2 = I$.
Best Answer
First, notice that by assumptions the direct sum $J_1/I \oplus J_2/I$ is well defined. Let $\phi: J_1/I \to J_2/I \to R/I$ defined by $\phi(j_1+I,j_2+I)=j_1+j_2+I$. Then it is easy to check that $\phi$ is well-defined and is surjective since $R=J_1+J_2$, so we are left to prove $\phi$ is injective.
Assume $\phi(j_1+I,j_2+I)=0+I$, this is equivalent $j_1+j_2 \in I$. But $I \subset J_1$ so, $j_1+j_2 \in J_1$, so $j_2 \in J_1$. As $j_2 \in J_2$, we get $j_2 \in J_1 \cap J_2=I$. Thus, $j_2+I=0+I$. A similar argument shows $j_1+I=0$, so $\phi$ is injective.