## Tag `05CK`

Chapter 10: Commutative Algebra > Section 10.81: Universally injective module maps

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

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 19080–19085 (see updates for more information).

```
\begin{lemma}
\label{lemma-faithfully-flat-universally-injective}
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 $I \subset R$.
\end{lemma}
\begin{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,
$n \otimes s \otimes s' \mapsto n \otimes ss'$.
\end{proof}
```

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