TITLE: General Form of $S^{-1}A$ - modules
QUESTION [7 upvotes]: I have been trying to show that if a ring $A$ is absolutely flat then so is the localisation $S^{-1}A$ by any multiplicative set. Now while trying to do this, I asked myself the following: Is there a description for the general form of an $S^{-1}A$ - module?
What I am trying to do to prove my original statement  is this: If $A$ is absolutely flat, consider the exact sequence 
$$0 \stackrel{f'}{\longrightarrow} M' \stackrel{f}{\longrightarrow} M$$
of $S^{-1}A$ modules. Then suppose we tensor this with some $S^{-1}A$ module $N$ and consider the map
$$f \otimes 1: M'\otimes_{S^{-1}A } N \longrightarrow M \otimes_{S^{-1}A} N.$$
Now we know that if $N$ is flat as an $S^{-1}A$ - module, then it is flat as an $A$ - module so that the map $f|_A \otimes 1 : M' \otimes_A N \longrightarrow M \otimes_A N$ is injective. We write $f|_A$ for $f$ viewed as an injective $A$ - module homomorphism, rather than an injective $S^{-1}A$ - module homomorphism. Since localisation preserves exactness, we have that the map 
$$S^{-1}(f|_A \otimes 1) : S^{-1}(M' \otimes_A N) \longrightarrow S^{-1}(M \otimes_A N)$$ is injective. If I can produce some kind of $S^{-1}A$ isomorphism between $M' \otimes_{S^{-1}A} N$ and $S^{-1}(M'\otimes_A N)$, I should be done. This is what I am trying to do, which is why I asked for a general description for what an $S^{-1}A$ - module looks like.
Thanks.

REPLY [3 votes]: A solution is given in $\S 7.7.2$ of my commutative algebra notes.
It goes like this: let $N$ be a $S^{-1} R$-module.  Then there is an $R$-module $M$ 
such that $N \cong_{S^{-1} R} S^{-1} M := S^{-1} R \otimes_R M$.  (This is essentially what Georges is arguing for in his answer.  In my notes, I think this fact appears as an exercise.)  By definition of absolutely flat, $M$ is a flat $R$-module, and thus by the local characterization of flatness (which appears earlier in the chapter) so is $S^{-1} M$.
Added: I think the intermediate result used is more important than the fact that it was used to prove, and it is also relevant to the OP's question "Is there a description for the general form of an $S^{-1} A$-module?"
The key notion here is the following: if $\varphi: R \rightarrow S$ is a ring homomorphism, then then an $S$-module $N$ is said to be extended from R (via $\varphi$, although this is usually suppressed) if there is an $R$-module $M$ such that $N \cong_S M \otimes_R S$. 
Off the top of my head, there are two general classes of homomorphisms $\varphi$ such that every $S$-module is extended from $R$:
$\bullet$ $\varphi: R \rightarrow S^{-1} R$ is a localization map (discussed above).
$\bullet$ $\varphi: R \rightarrow R/I$ is a quotient map.  (This is easy.)
However there are many important results about particular classes of $S$-modules being extended (or not!) from $R$.  For instance, suppose $R = k$ is a field, so $S$ is a $k$-algebra.  Then an $S$-module is extended from $k$ iff it is free.  In particular, when $S = k[t_1,\ldots,t_n]$, a famous conjecture of Serre can be stated as: every finitely generated projective $S$-module is extended from $k$.  It would be equivalent to say that every finitely generated projective $S$-module is free, but the above formulation is indeed the path taken in all proofs of Serre's Conjecture I have seen, so this is somehow the more insightful way to say it.