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

is the desired extension of $\varphi $ to $R$. $\square$

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