Lemma 47.3.3. Let $R \to S$ be an epimorphism of rings. Let $E$ be an $S$-module. If $E$ is injective as an $R$-module, then $E$ is an injective $S$-module.

**Proof.**
This is true because $\mathop{\mathrm{Hom}}\nolimits _ R(N, E) = \mathop{\mathrm{Hom}}\nolimits _ S(N, E)$ for any $S$-module $N$, see Algebra, Lemma 10.106.14.
$\square$

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