Lemma 47.3.8. Let R be a Noetherian ring. Let S \subset R be a multiplicative subset. If E is an injective R-module, then S^{-1}E is an injective S^{-1}R-module.
Proof. Since R \to S^{-1}R is an epimorphism of rings, it suffices to show that S^{-1}E is injective as an R-module, see Lemma 47.3.3. To show this we use Injectives, Lemma 19.2.6. Thus let I \subset R be an ideal and let \varphi : I \to S^{-1} E be an R-module map. As I is a finitely presented R-module (because R is Noetherian) we can find an f \in S and an R-module map I \to E such that f\varphi is the composition I \to E \to S^{-1}E (Algebra, Lemma 10.10.2). Then we can extend I \to E to a homomorphism R \to E. Then the composition
R \to E \to S^{-1}E \xrightarrow {f^{-1}} S^{-1}E
is the desired extension of \varphi to R. \square
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