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Tag 01U4

Chapter 28: Morphisms of Schemes > Section 28.24: Flat morphisms

Lemma 28.24.2. Let $f : X \to S$ be a morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent sheaf of $\mathcal{O}_X$-modules. The following are equivalent

  1. The sheaf $\mathcal{F}$ is flat over $S$.
  2. For every affine opens $U \subset X$, $V \subset S$ with $f(U) \subset V$ the $\mathcal{O}_S(V)$-module $\mathcal{F}(U)$ is flat.
  3. There exists an open covering $S = \bigcup_{j \in J} V_j$ and open coverings $f^{-1}(V_j) = \bigcup_{i \in I_j} U_i$ such that each of the modules $\mathcal{F}|_{U_i}$ is flat over $V_j$, for all $j\in J, i\in I_j$.
  4. There exists an affine open covering $S = \bigcup_{j \in J} V_j$ and affine open coverings $f^{-1}(V_j) = \bigcup_{i \in I_j} U_i$ such that $\mathcal{F}(U_i)$ is a flat $\mathcal{O}_S(V_j)$-module, for all $j\in J, i\in I_j$.

Moreover, if $\mathcal{F}$ is flat over $S$ then for any open subschemes $U \subset X$, $V \subset S$ with $f(U) \subset V$ the restriction $\mathcal{F}|_U$ is flat over $V$.

Proof. Let $R \to A$ be a ring map. Let $M$ be an $A$-module. If $M$ is $R$-flat, then for all primes $\mathfrak q$ the module $M_{\mathfrak q}$ is flat over $R_{\mathfrak p}$ with $\mathfrak p$ the prime of $R$ lying under $\mathfrak q$. Conversely, if $M_{\mathfrak q}$ is flat over $R_{\mathfrak p}$ for all primes $\mathfrak q$ of $A$, then $M$ is flat over $R$. See Algebra, Lemma 10.38.19. This equivalence easily implies the statements of the lemma. $\square$

    The code snippet corresponding to this tag is a part of the file morphisms.tex and is located in lines 4217–4238 (see updates for more information).

    \begin{lemma}
    \label{lemma-flat-module-characterize}
    Let $f : X \to S$ be a morphism of schemes.
    Let $\mathcal{F}$ be a quasi-coherent sheaf of $\mathcal{O}_X$-modules.
    The following are equivalent
    \begin{enumerate}
    \item The sheaf $\mathcal{F}$ is flat over $S$.
    \item For every affine opens $U \subset X$, $V \subset S$
    with $f(U) \subset V$ the $\mathcal{O}_S(V)$-module $\mathcal{F}(U)$ is flat.
    \item There exists an open covering $S = \bigcup_{j \in J} V_j$
    and open coverings $f^{-1}(V_j) = \bigcup_{i \in I_j} U_i$ such
    that each of the modules $\mathcal{F}|_{U_i}$ is
    flat over $V_j$, for all $j\in J, i\in I_j$.
    \item There exists an affine open covering $S = \bigcup_{j \in J} V_j$
    and affine open coverings $f^{-1}(V_j) = \bigcup_{i \in I_j} U_i$ such
    that $\mathcal{F}(U_i)$ is a flat $\mathcal{O}_S(V_j)$-module, for all
    $j\in J, i\in I_j$.
    \end{enumerate}
    Moreover, if $\mathcal{F}$ is flat over $S$ then for
    any open subschemes $U \subset X$, $V \subset S$ with $f(U) \subset V$
    the restriction $\mathcal{F}|_U$ is flat over $V$.
    \end{lemma}
    
    \begin{proof}
    Let $R \to A$ be a ring map. Let $M$ be an $A$-module.
    If $M$ is $R$-flat, then for all primes
    $\mathfrak q$ the module $M_{\mathfrak q}$ is flat over $R_{\mathfrak p}$
    with $\mathfrak p$ the prime of $R$ lying under $\mathfrak q$. Conversely, if
    $M_{\mathfrak q}$ is flat over $R_{\mathfrak p}$ for all primes $\mathfrak q$
    of $A$, then $M$ is flat over $R$. See
    Algebra, Lemma \ref{algebra-lemma-flat-localization}.
    This equivalence easily implies the statements of the lemma.
    \end{proof}

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