Assume the following result
Let $R$ be a ring with $1$, $I$ a right ideal of $R$, and $M$ a left $R$-module. Then $$\frac{R}{I} \otimes_R M \cong \frac{M}{IM}$$as abelian groups, where $IM$ is the submodule of $M$ generated by $\{rx \mid r \in I, x \in M\}$.
Great. I want to use this result to prove the corollary:
Let $R$ be a ring with $1$, $I$ a right ideal of $R$, $J$ a left ideal of $R$. Then $$\frac{R}{I}\otimes_R\frac{R}{J} \cong \frac{R}{I+J}$$as abelian groups.
Ok, since $J$ is a left ideal of $R$, we have that $R/J$ has a natural left $R$-module structure. The first result then says that $$\frac{R}{I}\otimes_R \frac{R}{J} \cong \frac{R/J}{I(R/J)},$$so it all boils down to checking that $$\frac{R/J}{I(R/J)} \cong \frac{R}{I+J}.$$Since $J \subseteq I+ J$, the identity of $R$ passes to a surjective map $$\frac{R}{J} \to \frac{R}{I+J},$$but I'm screwing up right at the end. How does one check that the kernel of this last map is $I(R/J)$?
Say $r+J\to 0$ then $r\in I+J$, so we could write that as $$r=a+b$$ with $a\in I$ and $b\in J$ now we have $$a+b+J=a+J$$ so $$r+J=a(1+J)$$ and so $$r+J\in I\frac{R}{J}.$$