## 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

- The sheaf $\mathcal{F}$ is flat over $S$.
- 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.
- 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 $j\in J, i\in I_j$.
- 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 4227–4248 (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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