Lemma 10.107.14 (08YS)—The Stacks project

Lemma 10.107.14. For a ring homomorphism $R \to S$ the following are equivalent

$R \to S$ is an epimorphism of rings,
for any $S$-modules $N_1, N_2$ we have $\mathop{\mathrm{Hom}}\nolimits _ S(N_1, N_2) = \mathop{\mathrm{Hom}}\nolimits _ R(N_1, N_2)$, and
the restriction functor $\text{Mod}_ S \to \text{Mod}_ R$ is fully faithful.

Proof. Observe that (2) and (3) are equivalent by definition.

Assume (1), let $N_1, N_2$ be $S$-modules, and let $\varphi : N_1 \to N_2$ be an $R$-linear map. For any $x \in N_1$ consider the map $S \otimes _ R S \to N_2$ defined by the rule $g \otimes g' \mapsto g\varphi (g'x)$. Since both maps $S \to S \otimes _ R S$ are isomorphisms (Lemma 10.107.1), we conclude that $g \varphi (g'x) = gg'\varphi (x) = \varphi (gg' x)$. Thus $\varphi $ is $S$-linear.

Assume (2). Let $N_1 = S \otimes _ R S$ viewed as an $S$-module via the left $S$-module action and let $N_2 = S \otimes _ R S$ with the right $S$-module action. Since $N_1 = N_2$ as $R$-modules, by (2) we see that the two $S$-module structures on $S \otimes _ R S$ coincide. This implies (1) by Lemma 10.107.1. $\square$

Comments (2)

Comment #11029 by Kiran Kedlaya on December 14, 2025 at 11:45

This lemma can be strengthened to say that $R \to S$ is an epimorphism if and only if $\mathrm{Mod}_S \to \mathrm{Mod}_R$ is fully faithful. The point is that the full faithfulness implies that the two different $S$ -module structures on $S \otimes_R S$ coincide, and now tag 04VN implies that $R \to S$ is epic.

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