## Tag `00HC`

Chapter 10: Commutative Algebra > Section 10.38: Flat modules and flat ring maps

Lemma 10.38.4. A composition of (faithfully) flat ring maps is (faithfully) flat. If $R \to R'$ is (faithfully) flat, and $M'$ is a (faithfully) flat $R'$-module, then $M'$ is a (faithfully) flat $R$-module.

Proof.The first statement of the lemma is a particular case of the second, so it is clearly enough to prove the latter. Let $R \to R'$ be a flat ring map, and $M'$ a flat $R'$-module. We need to prove that $M'$ is a flat $R$-module. Let $N_1 \to N_2 \to N_3$ be an exact complex of $R$-modules. Then, the complex $R' \otimes_R N_1 \to R' \otimes_R N_2 \to R' \otimes_R N_3$ is exact (since $R'$ is flat as an $R$-module), and so the complex $M' \otimes_{R'} \left(R' \otimes_R N_1\right) \to M' \otimes_{R'} \left(R' \otimes_R N_2\right) \to M' \otimes_{R'} \left(R' \otimes_R N_3\right)$ is exact (since $M'$ is a flat $R'$-module). Since $M' \otimes_{R'} \left(R' \otimes_R N\right) \cong \left(M' \otimes_{R'} R'\right) \otimes_R N \cong M' \otimes_R N$ for any $R$-module $N$ functorially (by Lemmas 10.11.7 and 10.11.3), this complex is isomorphic to the complex $M' \otimes_R N_1 \to M' \otimes_R N_2 \to M' \otimes_R N_3$, which is therefore also exact. This shows that $M'$ is a flat $R$-module. Tracing this argument backwards, we can show that if $R \to R'$ is faithfully flat, and if $M'$ is faithfully flat as an $R'$-module, then $M'$ is faithfully flat as an $R$-module. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 8318–8325 (see updates for more information).

```
\begin{lemma}
\label{lemma-composition-flat}
A composition of (faithfully) flat ring maps is
(faithfully) flat.
If $R \to R'$ is (faithfully) flat, and $M'$ is a
(faithfully) flat $R'$-module, then $M'$ is a
(faithfully) flat $R$-module.
\end{lemma}
\begin{proof}
The first statement of the lemma is a particular case of the
second, so it is clearly enough to prove the latter. Let
$R \to R'$ be a flat ring map, and $M'$ a flat $R'$-module.
We need to prove that $M'$ is a flat $R$-module. Let
$N_1 \to N_2 \to N_3$ be an exact complex of $R$-modules.
Then, the complex $R' \otimes_R N_1 \to
R' \otimes_R N_2 \to R' \otimes_R N_3$ is exact (since $R'$
is flat as an $R$-module), and so the complex
$M' \otimes_{R'} \left(R' \otimes_R N_1\right)
\to M' \otimes_{R'} \left(R' \otimes_R N_2\right)
\to M' \otimes_{R'} \left(R' \otimes_R N_3\right)$ is
exact (since $M'$ is a flat $R'$-module). Since
$M' \otimes_{R'} \left(R' \otimes_R N\right)
\cong \left(M' \otimes_{R'} R'\right) \otimes_R N
\cong M' \otimes_R N$ for any $R$-module $N$ functorially
(by Lemmas \ref{lemma-tensor-with-bimodule} and
\ref{lemma-flip-tensor-product}), this complex is isomorphic
to the complex
$M' \otimes_R N_1 \to M' \otimes_R N_2 \to M' \otimes_R N_3$,
which is therefore also exact. This shows that $M'$ is a flat
$R$-module. Tracing this argument backwards, we can show
that if $R \to R'$ is faithfully flat, and if $M'$ is
faithfully flat as an $R'$-module, then $M'$ is faithfully
flat as an $R$-module.
\end{proof}
```

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