# Understanding flat modules

I have problems working with the concept of flat modules. I am studying it with Atiyah-McDonald's , so I know the characterization of them as

Given $$A$$ a conmutative unity ring, an $$A$$-module $$N$$ being flat is equivalent to
(1) Any exact sequence $$0 \rightarrow M'' \rightarrow M \rightarrow M' \rightarrow 0$$ satisfies that its tensored sequence $$0 \rightarrow M'' \otimes N \rightarrow M \otimes N \rightarrow M' \otimes N \rightarrow 0$$ is exact.
(2) For every $$A$$-modules $$M, M'$$, if $$f: M' \rightarrow M$$ is injective then $$f \otimes 1 : M' \otimes N \rightarrow M \otimes N$$ is also injective.

For example, when I try to prove that given $$A$$-modules $$M$$ and $$K \subset M$$, if $$M$$ and $$M/K$$ are flat then $$K$$ is flat I am getting stuck. My first idea consists on using (1), taking an exact sequence $$0 \rightarrow P'' \rightarrow P \rightarrow P' \rightarrow 0$$, then $$0 \rightarrow P'' \otimes M \rightarrow P \otimes M \rightarrow P' \otimes M \rightarrow 0$$ and $$0 \rightarrow P'' \otimes M/K \rightarrow P \otimes M/K \rightarrow P' \otimes M/K \rightarrow 0$$ are exact and my intuition says that here should be a way to apply the Snake lemma, but I do not get it. Also, trying it with (2) it seems confusing, so any possible help would be appreciated.

EDIT

In order to prove the claim, I have considered the diagram

$$\begin{array}{c} & & 0 \rightarrow & P'' \otimes M & \xrightarrow{i \otimes Id_M} & P \otimes M & \xrightarrow{p \otimes Id_M} & P' \otimes M & \rightarrow 0 \\ & & & \downarrow_{Id_{P''} \otimes \pi} & & \downarrow_{Id_P \otimes \pi} & & \downarrow_{Id_{P'} \otimes \pi} & & \\ & & 0 \rightarrow & P'' \otimes M/K & \xrightarrow{i \otimes Id_{M/K}} & P \otimes M/K & \xrightarrow{p \otimes Id_{M/K}} & P' \otimes M/K & \rightarrow 0 \end{array}$$

where $$\pi$$ is the quotient morphism. Applying on it the Snake Lemma, the following sequence is exact: $$0 \rightarrow \ker Id_{P''} \otimes \pi \rightarrow \ker Id_{P} \otimes \pi \rightarrow \ker Id_{P'} \otimes \pi \rightarrow \operatorname{coker} Id_{P''} \otimes \pi \rightarrow \\ \rightarrow\operatorname{coker} Id_{P} \otimes \pi \rightarrow \operatorname{coker} Id_{P'} \otimes \pi \rightarrow 0$$

If I could prove that $$\ker Id_p \otimes \pi = P \otimes K$$ (then similarly for $$P', P''$$) and that $$\operatorname{coker} Id_P \otimes \pi = 0$$, then it would be finished, but I am struggling with it, I do not have ideas since tensorial product is new and confusing for me yet, so any possible help or advice without using Homological algebra or more sophisticated concepts would be very apreciated.

• One way is to first show that for a flat module $M$ and any module $N$, $\operatorname{Tor}^1(M,N)=0$ and then use snake lemma. Commented Apr 30 at 19:03
• Why not try $0 \rightarrow F_{\bullet}(K) \rightarrow F_{\bullet}(M)\rightarrow F_{\bullet}(M/K) \rightarrow 0$ and pass it by $- \otimes P$ ?
– Ubik
Commented May 22 at 10:48
• @RHspqr Sorry, I do not know what is $F_\bullet$ Commented May 22 at 11:05

Lemma 1 (Right exactness) Let $$A$$ be a ring and $$M$$ is an $$A$$-module. If $$P'' \longrightarrow P \longrightarrow P' \longrightarrow 0$$ is an exact sequence of modules, then $$M \otimes P'' \longrightarrow M \otimes P \longrightarrow M \otimes P' \longrightarrow 0$$ is also exact.

Note that $$M\otimes-$$ is not exact generally. Therefore, you can not expect the tensor functor preserves kernels. But the right exactness is equivalent to preserving cokernels. Also note that free modules $$\Rightarrow$$ projecitve modules $$\Rightarrow$$ flat modules.

This problem is also related to "pure exactness". A sequence is pure exact if it is exact and is still exact after tensoring any module.

Lemma 2. An $$A$$-module $$B'$$ is flat iff every exact sequence $$0 \longrightarrow B'' \longrightarrow B \longrightarrow B' \longrightarrow 0$$ is pure exact.

Proof of Lemma 2 ($$\Rightarrow$$): Let $$N'$$ be any $$A$$-module. There exists a free module $$N$$ and a module $$N''$$ such that $$0 \longrightarrow N'' \longrightarrow N \longrightarrow N' \longrightarrow 0$$ is exact. Then Lemma 1 implies that we have exact rows and columns in the following diagram.

$$\begin{array}{c} N'' \otimes B'' \longrightarrow & N'' \otimes B \longrightarrow & N'' \otimes B' \longrightarrow 0 \\ \downarrow & \downarrow & \downarrow \\ N \otimes B'' \longrightarrow & N \otimes B \longrightarrow & N \otimes B' \longrightarrow 0 \\ \downarrow & \downarrow & \downarrow \\ N' \otimes B'' \longrightarrow & N' \otimes B \longrightarrow & N' \otimes B' \longrightarrow 0 \\ \downarrow & \downarrow & \downarrow \\ 0 & 0 & 0 \end{array}$$ Since $$N$$ is free, the second row is short exact (adding $$0 \longrightarrow$$ on the left). Since $$B'$$ is flat, the third column is short exact. Then a diagram chase shows that the third row is also short exact.

End proof of Lemma 2.

Your proof is almost finished. You have exact sequences $$0 \longrightarrow \ker Id_{P''} \otimes \pi \longrightarrow \ker Id_{P} \otimes \pi \longrightarrow \ker Id_{P'} \otimes \pi \longrightarrow 0$$ and Lemma 2 implies that $$0 \longrightarrow K \longrightarrow M \longrightarrow M/K \longrightarrow 0$$ is pure exact. Tensoring $$P''$$, $$0 \longrightarrow P'' \otimes K \longrightarrow P'' \otimes M \longrightarrow P'' \otimes M/K \longrightarrow 0$$ is exact. So $$\ker Id_{P''} \otimes \pi = P'' \otimes K$$ and etc. Thus, $$0 \longrightarrow P'' \otimes K \longrightarrow P \otimes K$$ is exact.

Supplement:

If $$N'$$ is an $$A$$-module, then we can construct $$N = \oplus_{x \in N'} A$$ as free A-modules. A basis of $$N$$ is $$\{ q_x \}_{x \in N'}$$. Define $$N \longrightarrow N'$$ to be $$q_x \mapsto x$$. Obviously, this is surjective. Therefore, it gives an exact sequence $$0 \longrightarrow \ker \longrightarrow N \longrightarrow N' \longrightarrow 0.$$

• The proof only includes one direction of Lemma 2 and it is enough to solve the problem. Commented May 22 at 12:04
• Thanks for your answer! I have two questions to understund it propperly. First, why is there a free module $N$ and a module $N''$ such that we have that exact sequence? Second, I do not see how do you apply lemma 2 on the short exact sequence to conclude, why does it imply that equality? Commented May 22 at 15:48
• @Superdivinidad, I supplement some knowledge. Commented May 23 at 0:43
• thanks for the supplement! That explains the first question but the second is not solved yet, in fact this is was I asking in my edit, why $\ker Id_{P''} \otimes \pi = P'' \otimes K$? How can we prove it? This is what I do not see Commented May 23 at 7:05
• @Superdivinidad, I have written in my answer "Tensoring P''". After that, we have an exact sequence $0 \longrightarrow P'' \otimes K \longrightarrow P'' \otimes M \longrightarrow P'' \otimes M/K \longrightarrow 0$. Commented May 23 at 7:16