Exercise 2.26 Atiyah-Macdonald, flatness I'm stuck on this exercise.
$A$ is a commutative ring with unit. $N$ is an $A$-module. Then $N$ is flat $\Longleftrightarrow $ $\text{Tor}_{1}(A/a, N ) = 0 $ for every finitely generated ideal $a$ of $A$.
What I know: every module is isomorphic to the direct limit of its finitely generated submodules, I think I should use this information.
 A: Using the tensor-hom adjunction, one can see that $N$ is flat iff $N^c$ is injective, where $N^c=\mathrm{Hom}_{\mathbb{Z}}(N,\mathbb{Q}/\mathbb{Z})$ is the module of characters. By Baer's criterion, this happens if and only if $\mathrm{Ext}_A^1(A/a,N^c)=0$ for every ideal $a \subseteq A$. Using the isomorphism $\mathrm{Tor}_1^A(A/a,N)^c \simeq \mathrm{Ext}_A^1(A/a,N^c)$ and the fact that $M^c=0$ iff $M=0$ one can see that $N$ is flat if and only if $\mathrm{Tor}_1^A(A/a,N)=0$ for every ideal $a \subseteq A$. What remains is to reduce this to the finitely generated case.
Using the long exact sequence
$$\dots \rightarrow \mathrm{Tor}_1^A(A/a,N) \rightarrow a \otimes_A N\rightarrow A \otimes_A N\rightarrow A/a \otimes_A N\rightarrow 0$$
one can see that $\mathrm{Tor}_1^A(A/a,N)=0$ is equivalent to the fact that the functor $-\otimes N$ preserves injectivity of the inclusion $i: a\subseteq A$.
So assume for contradiction that $\mathrm{Tor}_1^A(A/a,N)=0$ for every finitely generated ideal $a\subseteq A$ and that there is an ideal $b \subseteq A$ such that $\mathrm{Tor}_1^A(A/b,N)\neq 0$, i.e. the induced map $j\otimes_A N: b \rightarrow A \otimes_AN$ is not injective ($j$ denotes the inclusion $b \subseteq A$). That is, there is a nonzero element $\sum_{k=0}^n b_k \otimes n_k$ belonging to the kernel of $j\otimes_A N$. 
Consider a finitely generated ideal $a:=\sum_{k=0}^n b_kA$. Then we have the inclusions
$$a\stackrel{i}\hookrightarrow b \stackrel{j}\hookrightarrow A, \; \;a \stackrel{l}\hookrightarrow A, \;\;l=j\circ i $$
and hence $$a \otimes_A N \stackrel{i\otimes_A N}\rightarrow b\otimes_A N \stackrel{j\otimes_A N}\rightarrow A \otimes_A N, \;\; a \otimes_A N \stackrel{l\otimes_A N}\hookrightarrow A \otimes_A N, $$
and $l \otimes_A N= (j \otimes_A N) \circ( i \otimes_A N)$ (and the injectivity of $l \otimes_A N$ follows from $\mathrm{Tor}_1^A(A/a,N)=0$).
This is clearly a contradiction, since $\sum_{k=0}^n b_k \otimes n_k$ is a nonzero element  of $a \otimes_A N$ (it is mapped to a nonzero element by $ i \otimes_A N$), which is sent by the injective map $ l \otimes_A N$ to $0$.
