# The Stacks Project

## Tag 05CK

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 19075–19080 (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}

Comment #1403 by Fred Rohrer (site) on April 13, 2015 a 9:07 pm UTC

Replace the proof by "This is the same as Lemma 051A."

Comment #1404 by jojo on April 14, 2015 a 5:51 am UTC

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 Johan (site) on April 15, 2015 a 6:53 pm UTC

OK, guys, I sent the first occurence of this lemma (051A) to the obsolete chapter. Thanks! See here.

There are also 2 comments on Section 10.81: Commutative Algebra.

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