Lemma 10.81.11. Let $R \to S$ be a faithfully flat ring map. Then $R \to S$ is universally injective as a map of $R$-modules. In particular $R \cap IS = I$ for any ideal $I \subset R$.
Proof. Let $N$ be an $R$-module. We have to show that $N \to N \otimes _ R S$ is injective. As $S$ is faithfully flat as an $R$-module, it suffices to prove this after tensoring with $S$. Hence it suffices to show that $N \otimes _ R S \to N \otimes _ R S \otimes _ R S$, $n \otimes s \mapsto n \otimes 1 \otimes s$ is injective. This is true because there is a retraction, namely, $n \otimes s \otimes s' \mapsto n \otimes ss'$. $\square$
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