$R$ noetherian, $I$ injective $R$-module $\Rightarrow$ $S^{-1}I$ is injective over $S^{-1}R$

Question

I am trying to prove that if $R$ is a noetherian ring, $S$ a multiplicative part and $I$ an injective $R$-module, then $S^{-1}I$ is an injective $S^{-1}R$-module.

So far I thought: I reduce to check that I can lift a map $\varphi:S^{-1}\mathfrak{a} \rightarrow S^{-1}I$ to a map from the whole ring $S^{-1}R$ (here $\mathfrak{a}$ is an ideal of $R$). Since $R$ is noetherian, we have $\mathfrak{a}=(a_1, \ldots , a_n)$. The map $f$ gives us elements $f\left( \frac{a_i}{1}\right)=\frac{b_i}{s}$ with $b_i \in I$ (since the generators are finite, I chose a common denominator). Now, the rough idea is to define a map $\psi:(sa_1, \ldots , sa_n)\rightarrow I$ sending $sa_i$ to $b_i$, lift it to a map $\Psi$ from $R$ using injectivity of $I$ and then taking the map $S^{-1}\Psi$ to conclude. I guess I can go by induction on the minimal number of generators for $\mathfrak{a}$, so that the base of induction is fine, while I am left just with the inductive step. I am stuck in showing that I can choose $b_n$ so that the map $\psi$ is well defined...

Am I going in the right direction? Any hint or answer is welcome.

Answer 1

It is enough to show that the homomorphism $\operatorname{Hom}_{R_S}(R_S,I_S) \rightarrow \operatorname{Hom}_{R_S}(J_S,I_S)$ is surjective for every ideal $J$ of $R$.

We have an exact sequence

$0 \rightarrow J \rightarrow R \rightarrow R/J \rightarrow 0$.

Since $I$ is injective, the functor $\operatorname{Hom}_R(-,I)$ is exact. Hence we have an exact sequence

$0 \rightarrow \operatorname{Hom}_R(R/J,I) \rightarrow \operatorname{Hom}_R(R,I) \rightarrow \operatorname{Hom}_R(J,I) \rightarrow 0$.

Now, localization is exact, hence we have new exact sequence

$0 \rightarrow (\operatorname{Hom}_R(R/J,I))_S \rightarrow (\operatorname{Hom}_R(R,I))_S \rightarrow (\operatorname{Hom}_R(J,I))_S \rightarrow 0$.

Finally, using the fact that for any finitely generated $R$-module $M$ (where $R$ must be Noetherian) we have $(\operatorname{Hom}_R(M,I))_S = \operatorname{Hom}_{R_S}(M_S,I_S)$, we see that our last exact sequence is

$0 \rightarrow \operatorname{Hom}_{R_S}(R_S/J_S,I_S) \rightarrow \operatorname{Hom}_{R_S}(R_S,I_S) \rightarrow \operatorname{Hom}_{R_S}(J_S,I_S) \rightarrow 0$,

which proves what we wanted.