Let $R$ be a commutative ring and let $I, J$ be ideals of $R$.
Our professor mentioned in algebra class that the $R$-module isomorphism $R/I \otimes_R R/J \cong R/(I + J)$ can be directly proved from the $R$-module isomorphism $R/I \otimes_R N \cong N/IN$ by setting $N = R/J$, where $N$ is an $R$-module.
I'm not sure how that works. If I simply set $N = R/J$, I get $(R/J)/I(R/J)$ on the RHS. Is there an easy way to see that $(R/J)/I(R/J)$ is isomorphic to $R/(I+J)$ as $R$-modules using the isomorphism theorems somehow?
Consider the natural homomorphism of $R$-modules $\phi\colon R/J\to R/(I+J)$ such that $\phi(r+J)=r+(I+J)$. Its kernel consists of cosets $r+J$ such that $r\in I+J$, so $r=x+y, x\in I, y\in J$, so $r+J=x+J$, $x\in I$. So $\ker\phi=(I+J)/J\ge I(R/J)$. But $x+J=x(1+J)$ (I assume that $R$ contains $1$). So $(I+J)/J\subseteq I(R/J)$. Hence $\ker\phi=I(R/J)$ and an isomorphism theorem applies.