The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.80.3. Let $M$ be an $R$-module. Then $M$ is flat if and only if the following condition holds: for every finitely presented $R$-module $P$, if $N \to M$ is a surjective $R$-module map, then the induced map $\mathop{\mathrm{Hom}}\nolimits _ R(P, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(P, M)$ is surjective.

Proof. First suppose $M$ is flat. We must show that if $P$ is finitely presented, then given a map $f: P \to M$, it factors through the map $N \to M$. By Lemma 10.80.2 the map $f$ factors through a map $F \to M$ where $F$ is free and finite. Since $F$ is free, this map factors through $N \to M$. Thus $f$ factors through $N \to M$.

Conversely, suppose the condition of the lemma holds. Let $f: P \to M$ be a map from a finitely presented module $P$. Choose a free module $N$ with a surjection $N \to M$ onto $M$. Then $f$ factors through $N \to M$, and since $P$ is finitely generated, $f$ factors through a free finite submodule of $N$. Thus $M$ satisfies the condition of Lemma 10.80.2, hence is flat. $\square$


Comments (0)


Post a comment

Your email address will not be published.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 toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

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 input field. As a reminder, this is tag 058F. Beware of the difference between the letter 'O' and the digit '0'.