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 10.38.14. Let $R$ be a ring. Let $M$ be an $R$-module. The following are equivalent

  1. $M$ is faithfully flat, and

  2. $M$ is flat and for all $R$-module homomorphisms $\alpha : N \to N'$ we have $\alpha = 0$ if and only if $\alpha \otimes \text{id}_ M = 0$.

Proof. If $M$ is faithfully flat, then $0 \to \mathop{\mathrm{Ker}}(\alpha ) \to N \to N'$ is exact if and only if the same holds after tensoring with $M$. This proves (1) implies (2). For the other, assume (2). Let $N_1 \to N_2 \to N_3$ be a complex, and assume the complex $N_1 \otimes _ R M \to N_2 \otimes _ R M \to N_3\otimes _ R M$ is exact. Take $x \in \mathop{\mathrm{Ker}}(N_2 \to N_3)$, and consider the map $\alpha : R \to N_2/\mathop{\mathrm{Im}}(N_1)$, $r \mapsto rx + \mathop{\mathrm{Im}}(N_1)$. By the exactness of the complex $-\otimes _ R M$ we see that $\alpha \otimes \text{id}_ M$ is zero. By assumption we get that $\alpha $ is zero. Hence $x $ is in the image of $N_1 \to N_2$. $\square$


Comments (2)

Comment #2340 by Federico Scavia on

Typo in the first line of the proof, an is missing in the exact sequence.

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

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