This question was asked in my commutative algebra assignment and I need help in solving it.
Prove that if S is a multiplicatively closed subset of A, then $S^{-1} A$ is a flat A-module.
Attempt: I have given that the sequence $0\to J \to K \to L \to 0$ is an exact sequence of modules and I have to show that $0\to J \otimes S^{-1}A \to K \otimes S^{-1} A \to L \otimes S^{-1} A \to 0$ is an exact sequence of A-modules. I know definition of exact sequences and relevant theorems but I am not sure how exactly should I proceed to prove the definition of exact sequence. I took maps f' ,g' ,h' ,i' for $0\to J \otimes S^{-1} A$ ,..., $L \otimes S^{-1} A \to 0$ and maps f,g,h,i for $0\to J$ ,..., $L \to 0$ respectively.
Unfortunately, I am unable to think how should I proceed with such a proof? Can you please give some hints on how should I start the proof.
I shall be really thankful.
Hint. For any $A$-module, there is a canonical isomorphism of $S^{-1}A$ modules $M\otimes_A S^{-1}A\simeq S^{-1}M$, and your problem boils down to prove that , if $0\longrightarrow J\overset{f}{\longrightarrow}K\overset{g}{\longrightarrow}L\longrightarrow 0$ is an exact sequence of $A$-modules, then
$0\longrightarrow S^{-1}J\overset{S^{-1}f}{\longrightarrow}S^{-1}K\overset{S^{-1}g}{\longrightarrow}S^{-1}L\longrightarrow 0$ is exact.
Here, we have denoted by $S^{-1}u$ the map of $S^{-1}A$-modules induced canonically by the $A$-linear map $u:M\to N$.
Now, it is enough to prove that $\ker(S^{-1}u)=S^{-1}(\ker(u)), im(S^{-1}u)=S^{-1}(im(u))$ (why?), which is not difficult...