Lemma 47.3.3 (08YV)—The Stacks project

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Table of contents
Part 3: Topics in Scheme Theory
Chapter 47: Dualizing Complexes
Section 47.3: Injective modules
Lemma 47.3.3 (cite)

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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 10.107.14. $\square$

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