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?

  • $\begingroup$ @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) $\endgroup$
    – Andy
    Sep 27, 2018 at 3:02
  • 1
    $\begingroup$ 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. $\endgroup$
    – Aaron
    Sep 27, 2018 at 3:08

1 Answer 1


If you have multiple propositions that you want to show are equivalent to each other, as you do here, it is sufficient to prove cyclic implication. So you have 1->2->3->1. If you want to show 2->1, you've already shown it, because 2->3->1. While direct proofs may be possible, they are unnecessary.

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

$$0\longrightarrow M_{i-1}/\ker(f_{i-1}) \longrightarrow M_i \longrightarrow \operatorname{im}(f_i)\longrightarrow 0 $$

which hae been woven together. Then, any additive functor which preserves the exactness of short exact sequences will preserve the long exact sequence because the images of the short exact sequences can be woven back together to reconstruct the long exact sequence.

  • $\begingroup$ I know, it's important to me though to understand if I'm missing arguments that make life easier because then in the future they can help me $\endgroup$
    – Andy
    Sep 27, 2018 at 3:17

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