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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?

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    You have a canonical map $\varphi$ from $R$ to $S^{-1}R$ that allows you to view an $S^{-1}R$-module as an $R$-module. The action of $r\in R$ is given by the action of $\varphi(r).$2012-02-07

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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.