# Equivalent definitions of flat modules

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).

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.

• Thank you! But for a) implies c), what we want to prove is that $0 \rightarrow N’ \otimes_R M \rightarrow N \otimes_R M$ is exact, right? We are already assuming that $N’ \otimes_R M \rightarrow N \otimes_R M \rightarrow N’’ \otimes_R M$ is… unless I’m not following somewhere. Commented Jan 5, 2023 at 21:35
• @dahemar I actually wrote the right thing but the wrong conclusion. Just updated it! Commented Jan 8, 2023 at 20:55