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.