Given a morphism of commutative rings $A\to B$ such that $B$ is a flat $A$-module and given $M$, $N$ two $B$-modules flat as $A$-modules, is the tensor product $M\otimes_B N$ flat over $A$??
The tensor product $M\otimes_A N$ is flat over $A$, the proof is not hard:
Given an exact sequence $0\to X\to Y\to Z\to 0$ of $A$-modules since $M$ is a flat $A$-module the sequence $0\to X\otimes_A M\to Y\otimes_A M\to Z\otimes_A M\to 0$ is exact. Again $N$ is a flat $A$-module, and since the tensor commute we have the statement.
I've tried some similar arguments but without any success, and I can't find a counterexample.
There is a morphism $M\otimes_A N\to M\otimes_B N$ (it is shown in here), but I don't know how to use it.