Let $A\rightarrow B$ be a ring homomorphism, $M$ and $N$ - modules over $A$. How to prove, that $$ (M \otimes_A N) \otimes _A B = (M \otimes_A B) \otimes_B (N\otimes_A B) $$ as $B$-modules in the most simplest way. Of course, it's possible to define 2 maps and prove, that they are correct and are inverse to each other. Is there more understandable proof? For example, using some universal property?
Hint: $(M \otimes_A B) \otimes_B (N \otimes_A B) = M \otimes_A (B \otimes_B (N\otimes_A B))$. By extension of scalars, $N \otimes_A B$ is a $B$-module. Now if $R$ is any ring and $D$ is an $R$-module, what is $R \otimes_R D$?