For every finitely generated ideal $I\subset A$ the map $I\otimes_A M\rightarrow A\otimes_A M$ is injective, then $M$ is flat. In the book I'm reading, there haven't been given any proof for the following theorem, since it's "obvious". I'm trying to prove it myself, but seem to have a bit of troubles.

Let $A$ ring and $M$ an $A$-module. If for every finitely generated
  ideal $I\subset A$ the map $I\otimes_A M\rightarrow A\otimes_A M$ is
  injective, then $M$ is flat.

So I know, what I have to prove is, that given an exact sequence $\ldots\rightarrow N'\rightarrow N \rightarrow N''\rightarrow \ldots $, I want to show that $\ldots\rightarrow N'\otimes_A M\rightarrow N \otimes_A M \rightarrow N''\otimes_A M\rightarrow \ldots $ is exact. I know that you can split long exact sequences into short exact sequences, and that tensor product is right exact. So I guess it would be enough to show that given $N\rightarrow N'$ injective, then $N \otimes_A M\rightarrow  N'\otimes_A M$ should be injective?
I'd also say that submodules of $A$ are ideals of $A$, hence we have injective $N \otimes_A M\rightarrow A \otimes_A M\cong M$ and $N'\otimes_A M\rightarrow A\otimes_A M\cong M$. However this is where I'm stuck.
 A: Note that the hypothesis is that for every finitely generated ideal $I$ of $A$,
$\mathrm{Tor}_1^A(A/I,M) = 0$. Indeed, from the exact sequence
$0\to I\to A\to A/I\to 0$ we get the exact sequence
$$0\to \mathrm{Tor}_1^A(A/I,M)\to I\otimes_A M\to A\otimes_A M \to A/I \otimes_A M\to0$$ 
since $A$ is flat. Thus $\mathrm{Tor}_1^A(A/I,M)$ is the kernel of your map, and hence zero iff it is injective. Now I claim that 


*

*Since every ideal is the filtered colimit of its f.g. ideals, tensor products commute with these, and filtered colimits are exact, your hypothesis holds for every ideal.

*Every module is the filtered colimit of its cyclic submodules, which are all isomorphic to $A/I$ for some ideal $I$.

*Because $\otimes_A M$ commutes with filtered colimits, so does $\mathrm{Tor}_1^A(?,M)$, which means that for any module $M'$,
$$\mathrm{Tor}_1^A(M',M) = \mathrm{colim}_{I} \mathrm{Tor}_A^1(A/I,M) = 0$$

*Finally, since $\mathrm{Tor}_1^A(?,M)$ vanishes identically, this means that $M$ is flat. 


Certainly you would have to prove $1$ (which is not hard), $2$ follows from this, while $3$ follows by the LES of a SES, in a similar way as the first part of this answer.
A: Hint:
First, as any ideal of $A$ is the union of its finitely generated sub-ideals and the direct limit is an exact functor which commutes with tensor products, the same hypothesis is true foe any ideal.
Next, any finitely generated $A$-module is a quotient of a finitely generated free module. So you can try to prove that, if $K$ is a submodule of a finitely generated free module $L$ of rank $n$, then
$$K\otimes_A M\longrightarrow L\otimes_A M$$
is injective. This can be done  by induction on the rank $n$.
For the case $n=2$, consider a submodule $K$ of $A^2$. Denote $K_1$ the intersection of $K$ with the first factor  $A\times\{0\}$ of $A^2$, and $M_K2$ the canonical projection of $M$ onto the second factor  $\{0\}\times A$.
Then do some diagram hunting in the commutative diagram:
\begin{alignat}6
K_1&\otimes_A M&{}\longrightarrow{}& &K&\otimes_A M&{}\longrightarrow {} K_2&\otimes_A M\\
&\downarrow &&&&\downarrow &&\downarrow\\[-0.5ex]
L_1&\otimes_A M&{}\longrightarrow{}& &A^2&\otimes_A M&\longrightarrow  L_2&\otimes_A M
\end{alignat}
