## Tag `00HT`

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

Lemma 10.38.19. Let $R$ be a ring. Let $S \subset R$ be a multiplicative subset.

- The localization $S^{-1}R$ is a flat $R$-algebra.
- If $M$ is an $S^{-1}R$-module, then $M$ is a flat $R$-module if and only if $M$ is a flat $S^{-1}R$-module.
- Suppose $M$ is an $R$-module. Then $M$ is a flat $R$-module if and only if $M_{\mathfrak p}$ is a flat $R_{\mathfrak p}$-module for all primes $\mathfrak p$ of $R$.
- Suppose $M$ is an $R$-module. Then $M$ is a flat $R$-module if and only if $M_{\mathfrak m}$ is a flat $R_{\mathfrak m}$-module for all maximal ideals $\mathfrak m$ of $R$.
- Suppose $R \to A$ is a ring map, $M$ is an $A$-module, and $g_1, \ldots, g_m \in A$ are elements generating the unit ideal of $A$. Then $M$ is flat over $R$ if and only if each localization $M_{g_i}$ is flat over $R$.
- Suppose $R \to A$ is a ring map, and $M$ is an $A$-module. Then $M$ is a flat $R$-module if and only if the localization $M_{\mathfrak q}$ is a flat $R_{\mathfrak p}$-module (with $\mathfrak p$ the prime of $R$ lying under $\mathfrak q$) for all primes $\mathfrak q$ of $A$.
- Suppose $R \to A$ is a ring map, and $M$ is an $A$-module. Then $M$ is a flat $R$-module if and only if the localization $M_{\mathfrak m}$ is a flat $R_{\mathfrak p}$-module (with $\mathfrak p = R \cap \mathfrak m$) for all maximal ideals $\mathfrak m$ of $A$.

Proof.Let us prove the last statement of the lemma. In the proof we will use repeatedly that localization is exact and commutes with tensor product, see Sections 10.9 and 10.11.Suppose $R \to A$ is a ring map, and $M$ is an $A$-module. Assume that $M_{\mathfrak m}$ is a flat $R_{\mathfrak p}$-module for all maximal ideals $\mathfrak m$ of $A$ (with $\mathfrak p = R \cap \mathfrak m$). Let $I \subset R$ be an ideal. We have to show the map $I \otimes_R M \to M$ is injective. We can think of this as a map of $A$-modules. By assumption the localization $(I \otimes_R M)_{\mathfrak m} \to M_{\mathfrak m}$ is injective because $(I \otimes_R M)_{\mathfrak m} = I_{\mathfrak p} \otimes_{R_{\mathfrak p}} M_{\mathfrak m}$. Hence the kernel of $I \otimes_R M \to M$ is zero by Lemma 10.23.1. Hence $M$ is flat over $R$.

Conversely, assume $M$ is flat over $R$. Pick a prime $\mathfrak q$ of $A$ lying over the prime $\mathfrak p$ of $R$. Suppose that $I \subset R_{\mathfrak p}$ is an ideal. We have to show that $I \otimes_{R_{\mathfrak p}} M_{\mathfrak q} \to M_{\mathfrak q}$ is injective. We can write $I = J_{\mathfrak p}$ for some ideal $J \subset R$. Then the map $I \otimes_{R_{\mathfrak p}} M_{\mathfrak q} \to M_{\mathfrak q}$ is just the localization (at $\mathfrak q$) of the map $J \otimes_R M \to M$ which is injective. Since localization is exact we see that $M_{\mathfrak q}$ is a flat $R_{\mathfrak p}$-module.

This proves (7) and (6). The other statements follow in a straightforward way from the last statement (proofs omitted). $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 8848–8876 (see updates for more information).

```
\begin{lemma}
\label{lemma-flat-localization}
Let $R$ be a ring. Let $S \subset R$ be a multiplicative subset.
\begin{enumerate}
\item The localization $S^{-1}R$ is a flat $R$-algebra.
\item If $M$ is an $S^{-1}R$-module, then $M$ is a flat $R$-module
if and only if $M$ is a flat $S^{-1}R$-module.
\item Suppose $M$ is an $R$-module. Then
$M$ is a flat $R$-module if and only if $M_{\mathfrak p}$ is a flat
$R_{\mathfrak p}$-module for all primes $\mathfrak p$ of $R$.
\item Suppose $M$ is an $R$-module. Then $M$ is a flat $R$-module if
and only if $M_{\mathfrak m}$ is a flat
$R_{\mathfrak m}$-module for all maximal ideals $\mathfrak m$ of $R$.
\item Suppose $R \to A$ is a ring map, $M$ is an $A$-module,
and $g_1, \ldots, g_m \in A$ are elements generating the unit
ideal of $A$. Then $M$ is flat over $R$ if and only if each localization
$M_{g_i}$ is flat over $R$.
\item Suppose $R \to A$ is a ring map, and $M$ is an $A$-module.
Then $M$ is a flat $R$-module if and only if the localization
$M_{\mathfrak q}$ is a flat $R_{\mathfrak p}$-module
(with $\mathfrak p$ the prime of $R$ lying under $\mathfrak q$)
for all primes $\mathfrak q$ of $A$.
\item Suppose $R \to A$ is a ring map, and $M$ is an $A$-module.
Then $M$ is a flat $R$-module if and only if the localization
$M_{\mathfrak m}$ is a flat $R_{\mathfrak p}$-module
(with $\mathfrak p = R \cap \mathfrak m$)
for all maximal ideals $\mathfrak m$ of $A$.
\end{enumerate}
\end{lemma}
\begin{proof}
Let us prove the last statement of the lemma.
In the proof we will use repeatedly that localization is exact
and commutes with tensor product, see Sections \ref{section-localization}
and \ref{section-tensor-product}.
\medskip\noindent
Suppose $R \to A$ is a ring map, and $M$ is an $A$-module.
Assume that $M_{\mathfrak m}$ is a flat $R_{\mathfrak p}$-module
for all maximal ideals $\mathfrak m$ of $A$ (with
$\mathfrak p = R \cap \mathfrak m$). Let $I \subset R$ be an ideal.
We have to show the map $I \otimes_R M \to M$ is injective.
We can think of this as a map of $A$-modules.
By assumption the localization
$(I \otimes_R M)_{\mathfrak m} \to M_{\mathfrak m}$ is injective
because
$(I \otimes_R M)_{\mathfrak m} =
I_{\mathfrak p} \otimes_{R_{\mathfrak p}} M_{\mathfrak m}$.
Hence the kernel of $I \otimes_R M \to M$ is zero by
Lemma \ref{lemma-characterize-zero-local}.
Hence $M$ is flat over $R$.
\medskip\noindent
Conversely, assume $M$ is flat over $R$. Pick a prime $\mathfrak q$
of $A$ lying over the prime $\mathfrak p$ of $R$. Suppose that
$I \subset R_{\mathfrak p}$ is an ideal. We have to show that
$I \otimes_{R_{\mathfrak p}} M_{\mathfrak q} \to M_{\mathfrak q}$
is injective. We can write $I = J_{\mathfrak p}$ for some
ideal $J \subset R$. Then the map
$I \otimes_{R_{\mathfrak p}} M_{\mathfrak q} \to M_{\mathfrak q}$
is just the localization (at $\mathfrak q$) of the map
$J \otimes_R M \to M$ which is injective. Since localization is exact
we see that $M_{\mathfrak q}$ is a flat $R_{\mathfrak p}$-module.
\medskip\noindent
This proves (7) and (6). The other statements follow in a straightforward
way from the last statement (proofs omitted).
\end{proof}
```

## Comments (0)

## Add a comment on tag `00HT`

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 lower-right corner).

All contributions are licensed under the GNU Free Documentation License.

There are no comments yet for this tag.

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