Lemma 10.82.11. 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, $n \otimes s \otimes s' \mapsto n \otimes ss'$. $\square$

Comment #1404 by jojo on

I would say keep the proof, it's easier to read this way than having to go somewhere else to check it ;)

Comment #1413 by on

OK, guys, I sent the first occurence of this lemma (113.4.9) to the obsolete chapter. Thanks! See here.

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