Interpreting an $S^{-1}R$-module as an $R$-module. Can one do it?
I'm trying to prove that $S^{-1}I$ is an injective $S^{-1}R$-module whenever $I$ is an injective $R$-module.
So I need to start with a situation where I have:
(i) $S^{-1}R$-modules $M,N$.
(ii) a homomorphism $j:M\to S^{-1}I$
(iii) an injective homomorphism $i:M\to N$.
From this I want to strip the situation down, (by embedding $I\to S^{-1}I$).
I almost have a solution but I think it comes down to being able to interpret $M$ as an $R$-module.  Can this be done or do I have to remove elements of $M$ to make it work?
 A: As Mike explains, $M$ may be seen as  an $R$-module $M_0$ and moreover $S^{-1}M_0=M$ as $S^{-1}R$-modules.
However it is not clear to me  how that solves your problem, because there is no reason why the morphism of $S^{-1}R$-modules $j:M\to S^{-1}I$ should map $M=M_0$ into $I$ and we cannot say that the diagram of $S^{-1}R$-modules comes from a diagram of $R$-modules.    
However if $R$ is noetherian and $I$ is finitely generated over $R$ then indeed $S^{-1}I$ is $S^{-1}R$-injective:     
Proof
We use the criterion that $I$ is injective iff $Ext^1_R(N,I)=0$ for all $R$-modules $N$.
So we should like to prove that $Ext^1_{S^{-1}R}(M,S^{-1}I)=0$ for all $S^{-1}R$-modules $M$.
But these modules are of the form $S^{-1}M_0$, as explained above.
Since 
$$  Ext^1_{S^{-1}R}(S^{-1}M_0,S^{-1}I)= S^{-1} Ext^1_R(M_0,I)                 $$
under our hypotheses and since $Ext^1_R(M_0,I)=0$ by injectivity of $I$ and application of the criterion, we conclude that $S^{-1}I$ is an injective $S^{-1}R$-module by  invoking the criterion again.
