## The quotient of a product is module isomorphic to the product of quotients

Let $R$ be a ring. Let $\{A_i\}_{i=1}^n$ be a finite family of left $R$-modules, and for each $A_i$ let $B_i \subseteq A_i$ be an $R$-submodule. Prove that $(\prod A_i)/(\prod B_i) \cong_R \prod A_i/B_i$.

In this previous exercise, we showed that the mapping $\Phi : \prod A_i/\prod B_i \cong_R \prod A_i/B_i$ given by $\Phi((a_i) + \prod B_i) = (a_i + B_i)$ is a well-defined group homomorphism. It suffices to show that $\Phi$ preserves scalar multiplication. To that end, let $r \in R$. Then $\Phi(r \cdot ((a_i) + \prod B_i)) = \Phi(r \cdot (a_i) + \prod B_i)$ $= \Phi((r \cdot a_i) + \prod B_i)$ $= (r \cdot a_i + B_i)$ $= r \cdot (a_i + B_i)$ $= r \cdot \Phi((a_i) + \prod B_i)$ as desired.