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

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (3)

Comment #1403 by Fred Rohrer on

Comment #1404 by jojo on

Comment #1413 by Johan on

There are also: