# Atiyah Macdonald 2.19 is a tautology?

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.

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.

• "exact sequence" might also just refer to sequences $A \to B \to C$. Though I'm not too familiar with Atiyah-Macdonald... Commented Mar 23, 2021 at 20:23
• @Gae.S. I was asking the question, then understood what I was missing. math.se is meant to be a repository of questions and answers. It is acceptable to answer your own question. I thought I should, since this seems like an easy thing to be confused by! Commented Mar 23, 2021 at 20:27