So I was wondering if the product of two modules is flat then one of them being flat implies the other one is flat too. I think the answer is affirmative and I will try to present my reasoning here. I would be really glad if I got a confirmation that my reasoning is correct and this is indeed true or if someone pointed out at a mistake.
Suppose $M \otimes N$ and $M$ are flat. We'll take an injection $0 \to K \overset{f}{\to} P$ and show that $0 \to K \otimes N \to P \otimes N$ is an injection, thus proving that $N$ is flat.
Suppose there is a nonzero $\sum_i k_i \otimes n_i \in K \otimes N$, such that $\sum_i f(k_i) \otimes n_i = 0 \in P \otimes N$. Then after tensoring with $M$ on the right we get that for any $m \in M$ $$\left(\sum_ik_i\otimes n_i\right)\otimes m \in \ker \left(f \otimes \mathrm{id}_{M \otimes N}\right).$$
Thus, because $f$ is injective and $M \otimes N$ is flat, $\forall m \in M \: \left(\sum_ik_i\otimes n_i\right)\otimes m = 0$.
But I believe it contradicts our assumption that $\sum_ik_i\otimes n_i \neq 0$ and $M$ is flat. Indeed, cyclic module generated by $\sum_ik_i\otimes n_i$ injects into $K \otimes N$ thus, by flatness of $M$, $\left(\sum_ik_i\otimes n_i\right) \otimes M$ must inject into $K \otimes N \otimes M$, but we just saw that it the whole submodule becomes zero in $K \otimes N \otimes M$. We have arrived at a contradiction, thus proving our claim to be true.
P.S. If this is true, than this very easily shows that wherever $M$ is a flat $R$-module and $I = \mathrm{Ann}\,M$, then $R/I$ is flat. That is because $R/I \otimes M = M$ is flat and $M$ is flat.
Thank you in advance for any help!
Best Answer
As pointed out in the comments, the quick way to see this cannot be correct is to observe that the zero module is flat. In particular, if $M=0$ then both $M$ and $M\otimes N=0$ are flat but $N$ could be anything. The error in your proof is that while it is true that $C\otimes M$ injects into $K\otimes N\otimes N$ where $C$ is the cyclic module generated by $\sum k_i\otimes n_i$ and every element of $C\otimes M$ goes to $0$ under this map, this does not tell you that $\sum k_i\otimes n_i=0$. Indeed, even if $C$ is nonzero, $C\otimes M$ could be $0$ (which is what happens if $M=0$).
What your argument shows is that if $M\otimes N$ is flat and $M$ is faithfully flat then $N$ is flat. (Faithfully flat means that both $M$ is flat and that if $M\otimes C=0$ for any module $C$ then $C=0$.)