# A step in the proof of Atiyah' s exercise 2.26

Let $N$ be an $A$-module. The exercise want to prove

$N$ is flat iff $\operatorname{Tor}_{1}(A/\alpha,N)=0$ for all finitely generated ideals $\alpha$ in $A$

From the hint ，I know $N$ is flat iff $\operatorname{Tor}_{1}(M,N)=0$ for all finitely generated $A$-modules $M$. If $M$ is finitely generated, let $x_{1},...,x_{n}$ be a set of generators of $M$, and let $M_{i}$ be the submodule of $M$ generated by $x_{1},...,x_{i}$.

Then the book says by considering the module $M_{i}/M_{i-1}$ and using exercise 2.25 (exercise 2.25 is: let $\;0\rightarrow N'\rightarrow N \rightarrow N''\rightarrow 0$ be a ses and $N''$ flat, then $N$ is flat iff and only if $N'$ is flat. ), deduce that $N$ is flat if $\operatorname{Tor}_{1}(M,N)=0$ for all $M$ generated by a single element.

My question is how to use the exercise 2.25 to deduce that condition. I can only get that if $\operatorname{Tor}_{1}(M,N)=0$ for all $M$ generated by a single element, then $M_{i-1}\otimes N\rightarrow M_{i}\otimes N$ is injective.

Hope this is not a silly question.

Use induction on the number of generators. From the short exact sequence $$0\to M_{i-1}\to M_{i}\to M_{i}/M_{i-1} \to 0$$ (note that $$M_{i}/M_{i-1}$$ is a cyclic module), you deduce the long exact sequence of Tors: \begin{align} \DeclareMathOperator{\Tor}{Tor} \dotsm\to\Tor^R_1(M_{i-1},N)\to \Tor^R_1(M_{i},N) &\to\underbrace{\Tor^R_1(M_{i}/M_{i-1},N)}_{ =\,0}\to\\[0.5ex] \to M_{i-1}\otimes_R N\to M_{i}\otimes_R N&\to M_{i}/M_{i-1}\otimes_R N \to 0 \end{align} and use the inductive hypothesis.
• Is the SES supposed to be $0 \rightarrow M_{i-1} \rightarrow M_i \rightarrow M_i / M_{i-1} \rightarrow 0$? Apr 21, 2018 at 21:21
• the second exact sequence is incorrect. Should it be $$\DeclareMathOperator{\Tor}{Tor} \Tor^R_1(M_{i-1},N)\to \Tor^R_1(M_{i},N) \to \Tor^R_1(M_{i}/M_{i-1},N) \to M_{i-1}\otimes_R N\to M_{i}\otimes_R N \to (M_{i}/M_{i-1})\otimes_R N$$ Jul 18, 2018 at 3:34