# Atiyah Macdonald proposition 2.19

In Atiyah Macodnald, ones defines a $$A$$ module $$N$$ to be flat $$A$$ module if for all $$A$$ modules and homomorphisms $$M' \to ^f M \to ^g M''$$ that are exact, if you tensor with $$N$$ (and take the maps $$f \otimes 1$$, $$g \otimes 1$$) it remains exact.

A moment before defining, we prove a nice theorem that if $$M' \to ^f M \to ^g M'' \to 0$$ is exact (of $$A$$ modules,homomorphisms), and $$N$$ is any $$A$$ module, then the sequence tensored with $$N$$ (And the same maps mentioned above) remains exact.

Finally to my question: We then claim that the following are equivalent for an $$A$$ module $$N$$

1. $$N$$ is flat

2. If $$0 \to M' \to M \to M'' \to 0$$ is any exact sequences of $$A$$ modules, the tensored sequence (with the same maps we mentioned a couple times $$f \otimes 1$$..) is exact.

3. If $$f:M' \to M$$ is injective, then $$f \otimes 1:M' \otimes N \to M \otimes N$$ is injective.

1 $$\to$$ 2 $$\to$$ 3 is clear. $$3 \to 1$$ is also clear using the proposition. I have a proof for 3->1 that says that $$\frac {M}{Im_f(M')} \otimes N$$ is naturally isomorphic to $$\frac {M \otimes N}{Im_{f \otimes 1} (M \otimes N)}$$, so that the injectiveness passes from one to another and we get 1.

Atiyh takes 2 $$\to$$ 1 for granted- probably something easier I'm missing? I'm guessing he means given $$M' \to ^f M \to ^g M''$$, think of it instead as $$0 \to \frac {M'}{Ker} \to ^{\bar{f}} M \to ^g Im_{g}(M) \to 0$$ and then he knows this tensored with $$N$$ is exact, and prove in a similiar I showed $$3 \to 1$$ it means the original tensored with $$N$$ is exact, but it seems even more work.

Am I missing something silly?

• @Aaron Notice the theorem only applies when the sequence has length 4 and ends with a 0, it doesn't claim to hold for all exact sequences of length 3 (which is indeed false)
– Andy
Sep 27, 2018 at 3:02
• Sorry, I wasn't being careful enough with my reading (and was implicitly converting your exact sequences into short exact sequences). Let me look at the question more carefully. Sep 27, 2018 at 3:08

That said, your idea is correct, and the application later on is that you can view a long exact sequence $$\cdots \longrightarrow M_{i-1} \longrightarrow M_i \stackrel{f_i}{\longrightarrow} M_{i+1} \longrightarrow \cdots$$ as a collection of short exact sequences
$$0\longrightarrow M_{i-1}/\ker(f_{i-1}) \longrightarrow M_i \longrightarrow \operatorname{im}(f_i)\longrightarrow 0$$