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 10.98.10 (Variant of the local criterion). Let $R \to S$ be a local homomorphism of Noetherian local rings. Let $I \not= R$ be an ideal in $R$. Let $M$ be a finite $S$-module. If $\text{Tor}_1^ R(M, R/I) = 0$ and $M/IM$ is flat over $R/I$, then $M$ is flat over $R$.

Proof. First proof: By Lemma 10.98.8 we see that $\text{Tor}_1^ R(\kappa , M)$ is zero where $\kappa $ is the residue field of $R$. Hence we see that $M$ is flat over $R$ by Lemma 10.98.7.

Second proof: Let $\mathfrak m$ be the maximal ideal of $R$. We will show that $\mathfrak m \otimes _ R M \to M$ is injective, and then apply Lemma 10.98.7. Suppose that $\sum f_ i \otimes x_ i \in \mathfrak m \otimes _ R M$ and that $\sum f_ i x_ i = 0$ in $M$. By the equational criterion for flatness Lemma 10.38.11 applied to $M/IM$ over $R/I$ we see there exist $\overline{a}_{ij} \in R/I$ and $\overline{y}_ j \in M/IM$ such that $x_ i \bmod IM = \sum _ j \overline{a}_{ij} \overline{y}_ j $ and $0 = \sum _ i (f_ i \bmod I) \overline{a}_{ij}$. Let $a_{ij} \in R$ be a lift of $\overline{a}_{ij}$ and similarly let $y_ j \in M$ be a lift of $\overline{y}_ j$. Then we see that

\begin{eqnarray*} \sum f_ i \otimes x_ i & = & \sum f_ i \otimes x_ i + \sum f_ ia_{ij} \otimes y_ j - \sum f_ i \otimes a_{ij} y_ j \\ & = & \sum f_ i \otimes (x_ i - \sum a_{ij} y_ j) + \sum (\sum f_ i a_{ij}) \otimes y_ j \end{eqnarray*}

Since $x_ i - \sum a_{ij} y_ j \in IM$ and $\sum f_ i a_{ij} \in I$ we see that there exists an element in $I \otimes _ R M$ which maps to our given element $\sum f_ i \otimes x_ i$ in $\mathfrak m \otimes _ R M$. But $I \otimes _ R M \to M$ is injective by assumption (see Remark 10.74.9) and we win. $\square$


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