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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*}

A flat module is faithfully flat if and only if it has nonzero fibers.

Lemma 10.38.15. Let $M$ be a flat $R$-module. The following are equivalent:

  1. $M$ is faithfully flat,

  2. for all $\mathfrak p \in \mathop{\mathrm{Spec}}(R)$ the tensor product $M \otimes _ R \kappa (\mathfrak p)$ is nonzero, and

  3. for all maximal ideals $\mathfrak m$ of $R$ the tensor product $M \otimes _ R \kappa (\mathfrak m) = M/{\mathfrak m}M$ is nonzero.

Proof. Assume $M$ faithfully flat. Since $R \to \kappa ({\mathfrak p})$ is not zero we deduce that $M \to M \otimes _ R \kappa ({\mathfrak p})$ is not zero, see Lemma 10.38.14.

Conversely assume that $M$ is flat and that $M/{\mathfrak m}M$ is never zero. Suppose that $N_1 \to N_2 \to N_3$ is a complex and suppose that $N_1 \otimes _ R M \to N_2\otimes _ R M \to N_3\otimes _ R M$ is exact. Let $H$ be the cohomology of the complex, so $H = \mathop{\mathrm{Ker}}(N_2 \to N_3)/\mathop{\mathrm{Im}}(N_1 \to N_2)$. By flatness we see that $H \otimes _ R M = 0$. Take $x \in H$ and let $I = \{ f \in R \mid fx = 0 \} $ be its annihilator. Since $R/I \subset H$ we get $M/IM \subset H \otimes _ R M = 0$ by flatness of $M$. If $I \not= R$ we may choose a maximal ideal $I \subset \mathfrak m \subset R$. This immediately gives a contradiction. $\square$


Comments (1)

Comment #892 by Kestutis Cesnavicius on

Suggested slogan: A flat module is faithfully flat iff it has nonzero fibers

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  • 1 comment(s) on Section 10.38: Flat modules and flat ring maps

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