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\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*}

This is [Theorem 2.1, Chase].

Proposition 10.89.6. Let $R$ be a ring. The following are equivalent

  1. $R$ is coherent,

  2. any product of flat $R$-modules is flat, and

  3. for every set $A$ the module $R^ A$ is flat.

Proof. Assume $R$ coherent, and let $Q_\alpha $, $\alpha \in A$ be a set of flat $R$-modules. We have to show that $I \otimes _ R \prod _\alpha Q_\alpha \to \prod Q_\alpha $ is injective for every finitely generated ideal $I$ of $R$, see Lemma 10.38.5. Since $R$ is coherent $I$ is an $R$-module of finite presentation. Hence $I \otimes _ R \prod _\alpha Q_\alpha = \prod I \otimes _ R Q_\alpha $ by Proposition 10.88.3. The desired injectivity follows as $I \otimes _ R Q_\alpha \to Q_\alpha $ is injective by flatness of $Q_\alpha $.

The implication (2) $\Rightarrow $ (3) is trivial.

Assume that the $R$-module $R^ A$ is flat for every set $A$. Let $I$ be a finitely generated ideal in $R$. Then $I \otimes _ R R^ A \to R^ A$ is injective by assumption. By Proposition 10.88.2 and the finiteness of $I$ the image is equal to $I^ A$. Hence $I \otimes _ R R^ A = I^ A$ for every set $A$ and we conclude that $I$ is finitely presented by Proposition 10.88.3. $\square$


Comments (2)

Comment #1448 by jp on

Reference: this proposition is a result from the thesis of Stephen Chase (1960). It was published as Theorem 2.1 in his "Direct Products of Modules," Trans. AMS 97, 1960, pp. 457–473.

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  • 2 comment(s) on Section 10.89: Coherent rings

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