Prove that if $M$ is flat over $R$ and $N$ is flat over $S$ then $M\otimes N$ is flat over $S$

Question

We have an $M$ right $R$-module and $N$ both left $R$-module and right $S$-module, how could I show that if $M$ is flat over $R$ and $N$ is flat over $S$ then $M\otimes N$ is flat over $S$?

Answer 1

This is really just from the definition.

Flat $M$ means that if $$\varphi:A\rightarrow B$$ is an injective $S$-module map, then the induced map $$1_N\otimes \varphi : N\otimes_S A \rightarrow N\otimes_S B$$ is injective. Now tensor again with $M$ to get $$1_M\otimes(1_N\otimes \varphi) : M\otimes_R (N\otimes_S A)\rightarrow M\otimes_R (N\otimes_S B),$$ and this is injective by flatness of $N$. But this isn't the map you want: the map you want to prove injectivity for is actually $$1_{M\otimes N}\otimes \varphi: (M\otimes_RN)\otimes_S A \rightarrow (M\otimes_RN)\otimes_S B.$$

But now you have to argue that these two maps are the same under the identification $M\otimes (N\otimes A)\simeq (M\otimes N)\otimes A$. Write down the natural square and check that it commutes, you'll see what I mean.