$M$ and $N$ flat, then $M\otimes N$ flat

Question

I want to show that if $M$ and $N$ are flat $R$-modules, then $M\otimes_R N$ is flat.

By flat we mean that if $0\to A\to B$ is exact, then $0\to A\otimes_RM\to B\otimes_RM$ is exact.

I am assuming that I want to use the fact that tensor products are associative, i.e. $A\otimes_R (M\otimes_R N) \cong (A\otimes_R M)\otimes_R N$ and then somehow use that $M$ is flat, but I have no idea how to work with the maps that I need.

Answer 1

We start with $$ 0\to A\to B $$ exact. Since $M$ is flat, by definition we have that $$ 0\to A\otimes M\to B\otimes M $$ is exact. Then, since $N$ is flat, $$ 0\to(A\otimes M)\otimes N\to (B\otimes M)\otimes N $$ is exact. Now use the associativity, and you're done.