The image reads:
If $T_n$ is exact, that is to say, tensoring with $N$ transforms all exact sequences into exact sequences, then $N$ is said to be a flat $A$-module.
Proposition 2.19: The following are equivalent for any $A$ module $N$:
(i) $N$ is flat
(ii) If $0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0$ is any exact sequence of $A$ modules, the tensored sequence $0 \rightarrow M' \otimes N \rightarrow M \otimes N \rightarrow M'' \otimes N \rightarrow 0$ is exact
As far as I understand, (i) and (ii) are identical: (i) says $N$ is flat, (ii) spills out the definition of flat. What is there to prove? The book says:
(i) is implied and implied by (ii) by taking a long exact sequence and splitting into short exact sequences
I don't understand what long exact sequence I need to consider, because to me, the proposition looks like a tautology.
Best Answer
Atiyah-Macdonald says "exact sequence" to mean "long exact sequence". The definition of flatness says that tensoring preserves long exact sequences. Characterisation (ii) tells us that it if a module preserves short exact sequences on tensoring, then it must preserve long exact sequences on tensoring.
So if we are given an LES, we can break it up into SESes, apply flatness-on-SES to prove flatness-on-LES.
I was confused, since I implicitly assume "exact sequence = SES" as I am following multiple books at the same time.