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I would like to solve the following exercise (2.26) from Atiyah & MacDonald's "Introduction to Commutative Algebra":

If $M$ is an $A$-module (where $A$ is a commutative ring), then: $$M \text{ is flat} \iff \text{Tor}_1(M,A/\mathfrak{a}) = 0 \text{ for all finitely generated ideals } \mathfrak{a}\subset A$$

I managed to prove that: $$M \text{ flat } \iff \text{Tor}_1(M,A/\mathfrak{a}) = 0 \text{ for all ideals } \mathfrak{a}\subset A$$

Any solutions or suggestions on how to proceed would be much appreciated.

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up vote 5 down vote accepted

So you want to show that $\operatorname{Tor}_1(M,A/\mathfrak{a}) = 0$ for all finitely generated ideals $\mathfrak{a} \subset A$ implies that $\operatorname{Tor}_1(M,A/\mathfrak{a})$ for all ideals $\mathfrak{a} \subset A$. Let's suppose we have such an ideal $\mathfrak{a}$. The trick is to 'approximate' $\mathfrak{a}$ by finitely generated ideals by writing it as an inductive limit $\mathfrak{a} = \varinjlim \mathfrak{a}_n$ of finitely generated ideals (this can always be done). Then, assuming I haven't convinced myself of anything false, we can write $\operatorname{Tor}_1(M,A/\mathfrak{a}) = \operatorname{Tor}_1(M,A/\varinjlim \mathfrak{a}_n) = \operatorname{Tor}_1(M,\varinjlim A/\mathfrak{a}_n) = \varinjlim \operatorname{Tor}_1(M,A/\mathfrak{a}_n)$, thus reducing the question to the finitely generated case. Now you can use the assumption that each $\operatorname{Tor}_1(M,A/\mathfrak{a}_n)$ vanishes.

(disclaimer: I haven't thought hard about commutative algebra for a while, so apologies if I've overlooked anything important)

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Thank you for your help. – Dave Feb 7 '12 at 21:52
I have in fact tried this approach. The step I'm having trouble with is: $\text{Tor}_1(M,A/\varinjlim\mathfrak{a}_n)=\text{Tor}_1(M,\varinjlim A/\mathfrak{a}_n)$ (or $A/\varinjlim\mathfrak{a}_n\cong\varinjlim A/\mathfrak{a}_n$). It would be great if you could clarify this part. Everything else is clear. – Dave Feb 7 '12 at 22:06
I guess first I should make clear that the inductive limit $\varinjlim \mathfrak{a}_n$ is over the set of finitely generated ideals of $\mathfrak{a}$, partially ordered by inclusion. So to show that $\varinjlim A/\mathfrak{a}_n \cong A/\mathfrak{a}$, first note that we have a natural map $A/\mathfrak{a}_n \to A/\mathfrak{a}$ for each $n$, since $\mathfrak{a}_n \subset \mathfrak{a}$. (continued...) – Alex Amenta Feb 7 '12 at 23:07
If $\mathfrak{a}_n \subset \mathfrak{a}_m$, then these maps into $A/\mathfrak{a}$ commute with the natural map $A/\mathfrak{a}_n \to A/\mathfrak{a}_m$, yielding a map $\varinjlim A/\mathfrak{a}_n \to A/\mathfrak{a}$. Now you just need to show that it's an isomorphism! – Alex Amenta Feb 7 '12 at 23:07
+1 @AlexanderAmenta Welcome to Math.SE!!!!!!!!!!!!!!!!!!!!!!!!!!!!! – user38268 Feb 8 '12 at 5:59

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