Equivalent definitions of flat modules

Question

According to my lecture notes, these three definitions of flat modules are equivalent:

a) For every exact sequence of $R$-modules $N’ \rightarrow N \rightarrow N’’$, the sequence $N’ \otimes_R M \rightarrow N \otimes_R M \rightarrow N’’ \otimes_R M $ is exact.

b) For every short exact sequence of $R$-modules $0 \rightarrow N’ \rightarrow N \rightarrow N’’ \rightarrow 0$, the sequence $0 \rightarrow N’ \otimes_R M \rightarrow N \otimes_R M \rightarrow N’’ \otimes_R M \rightarrow 0$ is exact.

c) For every exact sequence of $R$-modules $0 \rightarrow N’ \rightarrow N \rightarrow N’’$, the sequence $0 \rightarrow N’ \otimes_R M \rightarrow N \otimes_R M \rightarrow N’’ \otimes_R M $ is exact.

I can see that c) implies b) because of the right-exactness of the tensor product, but I can’t see why b) implies a).

I thought about “splitting” the exact sequence $N’ \rightarrow N \rightarrow N’’$ into short exact sequences, for instance $0 \rightarrow \ker(\alpha) \rightarrow N’ \rightarrow \mbox{im}(\alpha) \rightarrow 0$ and $0 \rightarrow \ker(\beta) \rightarrow N \rightarrow \mbox{im}(\beta) \rightarrow 0$ and trying to apply b) to these, then to patch them together to obtain a) using that $\ker(\beta)=\mbox{im}(\alpha)$, but to no avail (here $\alpha:N’ \rightarrow N$ and $\beta: N \rightarrow N’’$).

I realize this has been asked before on this site, but this was in 2012 and I don’t understand why the answer solves the problem (is there a reason why $\mbox{coker} (\alpha) \cong N’’$?).

Also, I don’t see why a) implies c).

Answer 1

b) $\implies$ a)

b) means that taking kernels commutes with tensoring by $M$ (and tensor products always commute with taking images). Write $\alpha_M : N' \otimes_R M \to N \otimes_R M$ and $\beta_M : N \otimes_R M \to N'' \otimes_R M$ for the induced maps after tensoring.

Hence, $$ \ker \alpha_M = M \otimes_R \ker \alpha = M \otimes_R \text{im } \beta = \text{im } \beta_M $$

a) $\implies$ c)

You can just divide the exact sequence into pieces. That is,

$$ 0 \to N' \otimes_R M \to N \otimes_R M \\ N' \otimes_R M \to N \otimes_R M \to N'' \otimes_R M \\ N \otimes_R M \to N'' \otimes_R M \to 0 $$

are all exact, so

$$ 0 \to N' \otimes_R M \to N \otimes_R M \to N'' \otimes_R M \to 0 $$ is exact.