The Stacks Project


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.

  1. The localization $S^{-1}R$ is a flat $R$-algebra.
  2. 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.
  3. 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 $R$.
  4. 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 $R$.
  5. 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 $R$.
  6. 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 $A$.
  7. 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)

    There are no comments yet for this tag.

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

    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.




    In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

    This captcha seems more appropriate than the usual illegible gibberish, right?