# The Stacks Project

## Tag 0584

Lemma 10.38.9. Let $R$ be a ring. Let $S \to S'$ be a faithfully flat map of $R$-algebras. Let $M$ be a module over $S$, and set $M' = S' \otimes_S M$. Then $M$ is flat over $R$ if and only if $M'$ is flat over $R$.

Proof. Let $N \to N'$ be an injection of $R$-modules. By the faithful flatness of $S \to S'$ we have $$\mathop{\rm Ker}(N \otimes_R M \to N' \otimes_R M) \otimes_S S' = \mathop{\rm Ker}(N \otimes_R M' \to N' \otimes_R M')$$ Hence the equivalence of the lemma follows from the second characterization of flatness in Lemma 10.38.5. $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 8599–8605 (see updates for more information).

\begin{lemma}
\label{lemma-flatness-descends-more-general}
Let $R$ be a ring.
Let $S \to S'$ be a faithfully flat map of $R$-algebras.
Let $M$ be a module over $S$, and set $M' = S' \otimes_S M$.
Then $M$ is flat over $R$ if and only if $M'$ is flat over $R$.
\end{lemma}

\begin{proof}
Let $N \to N'$ be an injection of $R$-modules. By the faithful flatness
of $S \to S'$ we have
$$\Ker(N \otimes_R M \to N' \otimes_R M) \otimes_S S' = \Ker(N \otimes_R M' \to N' \otimes_R M')$$
Hence the equivalence of the lemma follows from the second characterization
of flatness in
Lemma \ref{lemma-flat}.
\end{proof}

There are no comments yet for this tag.

There is also 1 comment on Section 10.38: Commutative Algebra.

## Add a comment on tag 0584

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 lower-right corner).